Page 122

Figure it Out

Q1. The sum of four consecutive numbers is 34. What are these numbers?
Ans: Let four consecutive numbers be x, (x + 1), (x + 2) and (x + 3).
x + (x + 1) + (x + 2) + (x + 3) = 34
x + x + 1 + x + 2 + x + 3 = 34
4x + 6 = 34
4x = 34 – 6
4x = 28
x = 28/7 = 7.
So, (x + 1 ) = 7 + 1 = 8
(x + 2) = 7 + 2 = 9
(x + 3) = 7 + 3 = 10
Therefore, the given four consecutive numbers are 7, 8, 9, and 10.

Q2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Ans: If p is the greatest of five consecutive numbers, then the other four numbers are (p – 1), (p – 2), (p – 3), and (p – 4).

Q3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
(i) The sum of two even numbers is a multiple of 3.
Ans: Let the two even numbers be 2a + 2b
Sum = 2a + 2b = 2(a + b)
For 2(a + b) to be a multiple of 3, (a + b) must be multiple of 3.
Example:
2 + 4 = 6 → divisible by 3
2 + 8 = 10 → not divisible by 3 
Conclusion: Sometimes true.

(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Ans: If a number is divisible by 18, then it is also divisible by 9 because 9 is a factor of 18.
18a ÷ 9 = 2a → divisible by 9.
But if a number is divisible by 9, it is not always divisible by 18.
9b ÷ 18 = b/2 → not divisible by 9.
Example: 9 is divisible by 9 but not divisible by 18.
27 is divisible by 9 but not 18.
Conclusion: Sometimes true.

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Ans: Let the two numbers be a and b.
Not divisible by 6 means they do not satisfy 6∣a or 6∣b.
But their sum can still be divisible by 6.
Example: 2 and 4 → both not divisible by 6.
But, 2 + 4 = 6 → divisible by 6.
Conclusion: Sometimes true.

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Ans: Let the multiple of 6 be 6a, the multiple of 9 be 9b.
Sum: 6a + 9b = 3(2a + 3b)→ clearly divisible by 3.
Example:
6 + 9 = 15 → divisible by 3.
12 + 18 = 30 → divisible by 3.
Conclusion: Always true.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Ans: Let multiple of 6 be 6a, multiple of 3 be 3b.
Sum: 6a + 3b = 3(2a + b).
For it to be divisible by 9, 2a + b must be divisible by 3.
Example:
6 (6 × 1) + 3 (3 × 1) = 9 →divisible by 9
6 + 6 = 12 → not divisible by 9
Conclusion: Sometimes true.

Q4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Ans: L.C.M of 3 and 4 = 12.
All such numbers are given by the expression = 12a + 2.
Examples:
(i) 12 × 1 + 2 = 12 + 2 = 14.
(ii) 12 × 2 + 2 = 24 + 2 = 26.
(iii) 12 × 3 + 2 = 36 + 2 = 38.

Q5. “I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”

Ans: Grouped in 3’s leaves 1.
Pairing (2’s) leaves 1.
Grouped by 5 leaves 1.
Grouped by 7 is perfect.
Number ≤ 100.
L.C.M of 2, 3, and 5 = 30.
In all those cases, when we group them, 1 pebble remains.
So, the actual number of pebbles must be = 30 + 1 = 31, but 31 is not divisible by 7.
The next multiple of 30 is 2 × 30 = 60.
So, 60 + 1 = 61, but this is also not divisible by 7.
Similarly, the next number is 90 + 1 = 91.
And 91 is divisible by 7.
Hence, the number of pebbles I hold = 91.

Q6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?
Ans: A number that leaves remainder of 2 when divided by 6 can be written as 6k + 2.
Three such numbers are: (6a + 2), (6b + 2), (6c + 2).
(6a + 2) + (6b + 2) + (6c + 2) = 6(a + b + c) + 6 = 6(a + b + c + 1).
This sum is divisible by 6.
So yes, Tathagat’s claim is always true.
Example: Take 20, 26, 32 → sum = 78 → divisible by 6.
Take 2, 8, 14 → sum = 24 → divisible by 6.

Page 123

Q7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661 (ii) 4779 – 661
Ans: (i) 4779 + 661
= Remainder 5 + Remainder 3
= Remainder 8
8 divided by 7 → remainder 1.
(ii) 4779 – 661
= Remainder 5 – Remainder 3
= Remainder 2

Q8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Ans: A number that leaves a remainder of 2 when divided by 3 is = 3x + 2
A number that leaves a remainder of 3 when divided by 4 is = 4x + 3
A number that leaves a remainder of 4 when divided by 5 is = 5x + 4
L.C.M of 3, 4, and 5 = 60
All the numbers are the same, so 4x + 3 = 3x + 2
4x – 3x = 2 – 3
x = -1
Each remainder is 1 less than the divisor.
Hence, the number is 1 less than the L.C.M = (60 – 1) = 59.
So, 59 is the smallest number that satisfies all the given conditions.

Page 126

Figure it Out

Q1. Find, without dividing, whether the following numbers are divisible by 9.
(i) 123
Ans: Digit sum of the number 123 = (1 + 2 + 3) = 6
Now, (6 ÷ 9) is not divisible by 9.
So, the whole number 123 is not divisible by 9.

(ii) 405
Ans: Digit sum of the number 405 = (4 + 0 + 5) = 9
Now, (9 ÷ 9) = 1,divisible by 9.
So, the whole number 405 is divisible by 9.

(iii) 8888
Ans: Digit sum of the number 8888 = (8 + 8 + 8 + 8) = 32
Now, (32 ÷ 9) is not divisible by 9.
So, the whole number 8888 is not divisible by 9.

(iv) 93547
Ans: Digit sum of the number 93547 = (9 + 3 + 5 + 4 + 7) = 9
Now, (28 ÷ 9) is not divisible by 9.
So, the whole number 93547 is not divisible by 9.

(v) 358095
Ans: Digit sum of the number 358095 = (3 + 5 + 8 + 0 + 9 + 5) = 30
Now, (30 ÷ 9) is not divisible by 9.
So, the whole number 358095 is not divisible by 9.

Q2. Find the smallest multiple of 9 with no odd digits.
Ans: If we multiply 9 by odd digits, we will get odd digits as a result.
So, we will multiply 9 by only even digits.

• 18 ( 1 is odd)
• 36 ( 3 is odd)
• 72 (7 is odd)
• 90 (9 is odd)
• 108 ( 1 is odd)
• 216 ( 1 is odd)
• 288(2 + 8 + 8 = 18 → divisible by 9, and digits 2,8,8 are even)

Q3. Find the multiple of 9 that is closest to the number 6000.
Ans: First Divide 6000 by 9 → (6000 ÷ 9) →Quotient = 666 and Remainder = 6
So, 5994 is 6 less than 6000.
And next closest number is 667×9 = 6003, 6003 is 3 greater than 6000.
Hence, the closest to the number 6000 that is multiple of 9 = 6003.

Q4. How many multiples of 9 are there between the numbers 4300 and 4400?
Ans: First divide 4300 by 9 = (4300 ÷ 9) = Quotient = 477 and Remainder = 7
The smallest number is 478×9 = 302 which divisible by 9. (Between 4300 – 4400)
Now, divide 4400 by 9 = (4400 ÷ 9) = Quotient = 488 and Remainder = 8
And the largest number is 488×9 = 4392 which divisible by 9. (Between 4300 – 4400)
Therefore the multiple of 9 between the number 4300 to 4400 is = (488 – 478) + 1 = 10 + 1 = 11.

Page 131

Figure it Out

Q1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Ans: Let be the 8-digit number is = x
So, digital root of x = 5
Now the number 10 more than that is = (x + 10)
∴ Digital root of (x + 10) is = 5 + 1 + 0 = 6
Therefore 6 will be the digital root of 10 more than that number.

Q2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Ans: Let the first number of the sequence be 4.
Digital roots:

• 4 → 4
• 15 → 1 + 5 = 6
• 26 → 2 + 6 = 8
• 37 → 3 + 7 = 10 → 1 + 0 = 1
• 48 → 4 + 8 = 12 → 1 + 2 = 3
• 59 → 5 + 9 = 14 → 1 + 4 = 5
• 70 → 7 + 0 = 7
• 81 → 8 + 1 = 9
• 92 → 9 + 2 = 11 → 1 + 1 = 2
• 103 → 1 + 0 + 3 = 4
• 114 → 1 + 1 + 4 = 6

We can see that the pattern repeats every 9 terms.

Q3. What will be the digital root of the number 9a + 36b + 13?
Ans: We find the digital root by taking the expression modulo 9:
9a = 0( mod 9)
36 b = 0( mod 9)
13 = 4( mod 9)
So, 9a + 36b + 13 = 4 mod 9
Hence, the digital root is 4.

Q4. Make conjectures by examining if there are any patterns or relations between
(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.
Ans: (i) There is no fixed relation between Parity and digital root. Parity and digital root are independent. A number can be even or odd regardless of its digital root.
Example: Even number 14 has a digital root, 1 + 4 = 5, which is odd.
Also, even number 24 has a digital root, 2 + 4 = 6, which is even.
(ii) The digital root of a number is the same as the remainder when the number is divided by 9, except when the remainder is 0 — in that case, the digital root is 9.
If the digital root is 3, 6, or 9, then the number is divisible by 3.

Page 132-134

Figure it Out

Q1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.
Ans: We know that the digital root of multiples of 9 is always 9.
So, the digit root of the number 31z5 is = 9
Hence the value of z = 0 or 9.
Proceedings:
Therefore, 3 + 1 + z + 5 = 9
Or, 9 + z = 9
Or, z = 0
Now, the expression 3 + 1 + z + 5 = 9 + z must be divisible by 9.
If z = 0, then 9 + z = 9 is divisible by 9.
And when z = 9, then 9 + z = 18 is divisible by 9.
So, the value of z = 0 or 9.
And the numbers are 3105 and 3195.
That’s why there are two answers to this problem.

Q2. “I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.
Ans: 1st number = 12k + 8
2nd number = 12k – 4
Sum = 12k + 8 + 12k – 4 = 24k + 4
According to Snehal, it is always a multiple of 8.
If we put k = 1, 24 × 1 + 4 = 24, which is a multiple of 8.
k = 2, 24 × 2 + 4 = 48, which is a multiple of 8.
k = 3, 24 × 3 + 4 = 76, which is not a multiple of 8.
So, her claim is “Sometimes True”.

Q3.  When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.
Ans: Let the two numbers be 3a and 3b (multiples of 3)
Sum = 3a + 3b = 3(a + b)
This is always divisible by 3.
But it’s divisible by 6 only if the value of (a + b) is even.
So, we can conclude that

• If a + b is even -> sum divisible by 6
• If a + b is even -> sum divisible by 6

Q4. Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9 “.
(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?
Ans: (i) We know that a number is divisible by 9 if the sum of its digits is divisible by 9.
So, if we reverse the digits of a number which a multiple of 9, then it will still be divisible by 9 as the sum of the digits remains the same.
Example: 117 is divisible by 9, as well as 711 and 171.
(ii) As long as the sum of digits remains the same, any shuffle will work.

Q5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
Ans: LCM of 2 and 9 is 18.
So if the number is divisible by 2 and 9, then it is divisible by 18.
A number is divisible by 2 if the last digit is even or 0.
A number is divisible by 9 if the sum of its digits is divisible by 9.
In 48a23b, the sum of the digits apart from a and b is 4 + 8 + 2 + 3 = 17
So if a = 1 and b = 0, the number becomes 481230, which is divisible by both 2 and 9; thus divisible by 18.
No more values can be obtained.
So the values of a and b are 1 and 0, respectively.

Q6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Ans: LCM of 4 and 11 is 44.
So if a number is divisible by 4 and 11, then it is divisible by 44.
If the last two digits of a number are divisible by 4, then the whole number is divisible by 4.
A number is divisible by 11 if the difference between the sum of its digits in odd places and the sum of its digits in even places is either 0 or divisible by 11.
So, (3+7+8)-(p+q) = 18-p-q must be divisible by 11 or 0.
If the values of p = 7 and q = 0, then the number becomes 37708, which is divisible by both 4 and 11; thus divisible by 44.
Also, values that are possible:
p = 5, q = 2;
p = 3, q = 4;
p = 1, q = 6;

Q7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Ans: The first set of numbers which satisfies the conditions is 14, 15, 16.
LCM of 2, 3, and 4 = 12
If we add 12 to the first number, then the sequence repeats.
So, the next such sets are:
14, 15, 16
26, 27, 28
38, 39, 40
50, 51, 52
And so on

Q8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Ans: Step:
Divided 45000 by 36.
We will get 1250.
Now, to get the multiple between 45000 and 47000, we have to multiply 36 by more than 1250.
So, the five multiples of 36 between 45,000 and 47,000 are:
36 × 1251 = 45036
36 × 1252 = 45072
36 × 1253 = 45108
36 × 1254 = 45144
36 × 1255 = 45180

Q9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Ans: Let be the 5 consecutive even numbers are = x, (x + 2), (x + 4), (x + 6), (x + 8)
The middle number is (x + 4)
Therefore, (x + 4) = 5p
Or, x = 5p – 4
So, the other four numbers are =
1^st number → 5p – 4
2^nd number → 5p – 2
4^th number → 5p + 2
5^th number → 5p + 4

Q10. Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Ans: LCM of 3 and 5 is 15.
So if a number is divisible by both 3 and 5, then the number is divisible by 15.
A number is divisible by 3 if the sum of its digits is divisible by 3.
If the last digit of a number is 0 or 5, then the number is divisible by 5.
For the reverse number to be divisible by 6, it has to be divisible by 2 and 3.
So the first digit should be even, and the sum of the digits should be divisible by 3.
Let’s try 234105..
Here, some of the digits are 15, the last digit is 5, and in reverse order, the last digit is even.
So it satisfies all the conditions.

Q11. Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.
Ans: Let’s check Dipak’s hypothesis.
11 × 2 = 22
Now, the multiples of 22, such as 44, 66, 88, are all multiples of 11.
11 × 3 = 33
Now, the multiples of 33, such as 66, 99, 132, are all multiples of 11.
So, Deepak’s claim is not correct.
All doubled multiples of 11 are still multiples of 11.

Q12. Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning. 
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
Ans: ‘Always True’
Explanation: Let be the two numbers are = 6a and 3b.
So the product of these = (6a × 3b) = 18ab
It saws that 18ab is also divisible by 9. [18 is a multiple of 9]
Example: If a = 3 and b = 2
(18 × 3 × 2) = 108, so 108 is a multipleof p.

(ii) The sum of three consecutive even numbers will be divisible by 6.
Ans: ‘Always True’
Explanation: Let be the first consecutive even number = x
So the other consecutive even numbers = (x + 2) and (2 + 4)
Therefore, sum of these number = x + x + 2 + x + 4 = 3x + 6 = 3(x + 2)
Example: If x = 6, then 3(6 + 2) = 24, divisible by 6.
When x = 10, then 3(10 + 2) = 36, divisible by 6.

(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
Ans: ‘Always True’
Explanation: If a number is divisible by 6 it must be divisible by 2 and 3.
Checking divisibility by 2: We check the last digits of the number, if it is even then the number must be divisible by 2.
And checking divisibility by 3: We check the sum of the digits of the number if it is divisible by 3, then the number is also divisible by 3.
Here we can see that the last digit of both the numbers ‘abcdef’ and ‘badcef’ is the same and all the digits are the same, only their positions have changed.
So, if abcdef is a multiple of 6, then badcef should be a multiple of 6.

(iv) 8 (7b-3)-4 (11b+1) is a multiple of 12.
Ans: ‘Never True’
Explanation: 8 × (7b – 3) – 4 × (11b + 1)
= 56b –24 – 44b – 4
= 12b – 28
We see that 12b is a multiple of 12 but 28 is not a multiple of 12.
So, we say that 12b – 28 is not divisible by 12.

Q13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Ans: 

• 3,6,9 → sum = 18 → Divisible by 3
• 1,2,3 → sum = 6 → Divisible by 3
• 1,2,4 → sum = 7 → Not Divisible by 3

The sum of 3 numbers is divisible by 3 if their total sum is divisible by 3
So if all numbers leave the same remainder mod 3 (like 1, 4, 7 ), or their remainders sum to 3 or 6, the result is divisible by 3.

Q14. Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?
Ans: Yes, the product of two consecutive integers is always a multiple of 2.
Product of two consecutive numbers: n(n + 1)
If n even → even × odd = even
If n odd → odd × even = even
No, it sometimes can be divisible by 6, if the last digits are even and the sum of the digits is divided by 3.
2 × 3 = 6 [Divisible by 6]
3 × 4 = 12 [Divisible by 6]
4 × 5 = 20 [Not divisible by 6]
The product of 4 consecutive integers is divisible by 24.
The product of 5 consecutive integers is divisible by 120.

Q15. Solve the cryptarithms — 
(i) EF × E = GGG 
(ii) WOW × 5 = MEOW
Ans: (i) EF × E = GGG
=10E + F × E = 100 G + 10G + G
= (10E + F) × E = 111G
If E = 1, then 10 + F = 111G
[It is not possible because for any value of F, LHS can’t be equal to RHS]
If E = 2, then (20 + F) × 2 = 111G
[It is also not possible because for any value of F, LHS can’t be equal to RHS]
For E = 3, then (30 + F) × 3 = 111G
=90 + 3F = 111G
If F = 7 and G = 1, then LHS = RHS.
∴ The values of E, F, and G are 3, 7, and 1, respectively.
(ii) WOW × 5 = MEOW
Using the same process as the previous one.
(100W + 10O + W) × 5 = MEOW
⇒ (101 W + 10 O) × 5 = MEOW
⇒ 505 W + 50 O = MEOW
Let’s try possible values of W and O such that the result is a 4-digit number.
If we set W = 5 and O = 7, we obtain a 4-digit number.
505 × 5 + 50 × 7 = 2875
On the right-hand side, if MEOW = 2875
W = 5, O = 7
1000M + 100E + 10O + W = 1000M + 100E + 70 + 5 = 1000M + 100E + 75
If we take M = 2 and E = 8, then it satisfies the LHS.
So, the values of M, E, O, and W are 2, 8, 7, and 5, respectively.

Q16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
Ans: The correct answer is option (iv).
Here are the reasons for the answer.

• Every multiple of 32 is also a multiple of 8 and 4.
• Every multiple of 8 is also a multiple of 4.
• But not every multiple of 4 is a multiple of 8 or 32.

Page 94

Q1. Find all the other angles inside the following rectangles.

Ans:
(i) 

∠1 + ∠9 = 90° ……… (All corner angles of a rectangle are 90°)
∠1 + 30° = 90°
∠1 = 90° – 30°
∠1 = 60°
∠1 = ∠5 = 60°………. (Alternate interior angles)
∠9 = ∠4 = 30°………. (Alternate interior angles)
In △AOB, OA = OB, then the angles opposite them are equal
∴ ∠9 = ∠7 = 30°
∠7 = ∠3 = 30°………. (Alternate interior angles)
In △AOD, OA = OD, then the angles opposite them are equal
∴ ∠2 = ∠1 = 60°
∠2 = ∠6 = 60°………. (Alternate interior angles)
In △AOB,
∠9 + ∠7 + ∠AOB = 180° ………. (Sum of angles of a triangle)
30° + 30° + ∠AOB = 180°
60° + ∠AOB = 180°
∠AOB = 180° – 60°
∠AOB = 120°
∠AOB = ∠COD = 120° ………… (Vertically opposite angles)
∠AOB + ∠AOD = 180° …………. (Linear pair)
120° + ∠AOD = 180°
∠AOD = 180° – 120°
∠AOD = 60°
∠AOD = ∠BOC = 60° ………… (Vertically opposite angles)
Thus, ∠1 = ∠5 = ∠2 = ∠6 = ∠AOD = ∠BOC = 60°.
∠AOB = ∠COD = 120°.
∠9 = ∠4 = ∠7 = ∠3 = 30°.
(ii)

 ∠POS = ∠ROQ = 110° ………… (Vertically opposite angles)
∠POS + ∠POQ = 180° …………. (Linear Pair)
110° + ∠POQ = 180°
∠POQ = 180° – 110°
∠POQ = 70°
∠POQ = ∠SOR = 70° ………… (Vertically opposite angles)
In △POS, OP = OS, then the angles opposite them are equal.
∴ ∠1 = ∠2 = a
In △POS,
∠1 + ∠2 + ∠POS = 180° …………. (Sum of angles of a triangle)
a + a + 110° = 180°
2a = 180° – 110°
2a = 70°
a = 35°
∴ ∠1 = ∠2 = a = 35°
∠1 = ∠5 = 35° ………… (Alternate interior angles)
∠2 = ∠6 = 35° ………… (Alternate interior angles)
Since ABCD is a rectangle, ∠P = 90°
∠9 = ∠1 + ∠8
90° = 35° + ∠8
∠8 = 90° – 35°
∠8 = 55°
∠8 = ∠4 = 55° ………… (Alternate interior angles)
In △POQ, OP = OQ, then the angles opposite them are equal i.e. ∠7 = ∠8 = 55°
∠7 = ∠2 = 55° ………… (Alternate interior angles)
Thus, ∠POS = ∠ROQ = 110°.
∠POQ = ∠SOR = 70°.
∠1 = ∠2 = ∠5 = ∠6 = 35°.
∠8 = ∠4 = ∠7 = ∠2 = 55°

Q2. Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of 
(i) 30° 
(ii) 40° 
(iii) 90° 
(iv) 140°
Ans: 
(i) 30°

(ii) 40° 

(iii) 90° 

(iv) 140°

Q3. Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. 
What is the figure APML? Reason and/or experiment to figure this out.
Ans: .  

The figure is a square.
Therefore, the opposite sides are parallel. The opposite sides are equal in length and all angles are 90°.

Q4. We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact 90° using these?
Ans: We have to tie the two sticks together at one end so that they can rotate like a hinge.
Then we have to cut a thread of length 12 units (for example, 12 cm ) and tie its ends to the free ends of the two sticks.
After that, we have to mark the thread into segments of 3 units, 4 units, and 5 units.
We will form a triangle with the thread:

• Keep the 3-unit side along one stick.
• Keep the 4-unit side along the other stick.
• The 5-unit side will naturally form the hypotenuse.

By the Pythagoras theorem (32 + 42 = 52), the angle between the sticks will be exactly 90°.

Q5. We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal a rectangle?
Ans: No, this can’t be the definition of a rectangle. A quadrilateral with opposite sides parallel and equal is a parallelogram, but not all parallelograms are rectangles. A rectangle needs all angles to be right angles.

Page 102

Figure it Out

Q1. Find the remaining angles in the following quadrilaterals.

Ans: (i) Here PR ∥ EA, and PE ∥ RA
Therefore, PEAR is a parallelogram.
∠P = ∠A = 40° …………. (Opposite angles of a parallelogram are equal)
∠P + ∠R = 180° …………. (The sum of the adjacent angles of a parallelogram is 180°)
40° + ∠R = 180°
∠R = 180° – 40°
∠R = 140°.
∠R = ∠E = 140° …………. (Opposite angles of a parallelogram are equal)
(ii) Here PQ ∥ SR, and PS ∥ QR
∴ PQRS is a parallelogram.
∠P = ∠R = 110° …………. (Opposite angles of a parallelogram are equal)
∠P + ∠S = 180° …………. (The sum of the adjacent angles of a parallelogram is 180°)
110° + ∠S = 180°
∠S = 180° – 110°
∠S = 70°.
∠S = ∠Q = 70° …………. (Opposite angles of a parallelogram are equal)
(iii) Here, XWUV is a rhombus (all sides equal). 
In △VUX, UV = UX, then the angles opposite them are equal.
∴ ∠UXV = ∠UVX = 30°
∠UXV = ∠WXV = 30° …………. (The diagonals of a rhombus bisect its angles)
Also, ∠UVX = ∠WVX = 30° …………. (The diagonals of a rhombus bisect its angles)
∠E = 2 × ∠UVX = 2 × 30° = 60°
∠V = ∠X = 60° …………. (Opposite angles of a rhombus are equal)
∠V + ∠U = 180° …………. (The sum of adjacent angles of a rhombus is 180°)
60° + ∠U = 180°
∠U = 180° – 60°
∠U = 120°
∠U = ∠W = 120° …………. (Opposite angles of a rhombus are equal)
(iv) Here, AEIO is a rhombus (all sides equal). 
In △EAO, AE = AO, then the angles opposite them are equal.
∴ ∠AOE = ∠AEO = 20°
∠AEO = ∠IEO = 20° …………. (The diagonals of a rhombus bisect its angles)
Also, ∠AOE = ∠IOE = 20° …………. (The diagonals of a rhombus bisect its angles)
∠E = 2 × ∠AEO = 2 × 20° = 40°
∠E = ∠O = 40° …………. (Opposite angles of a rhombus are equal)
∠E + ∠A = 180° …………. (The sum of adjacent angles of a rhombus is 180°)
40° + ∠A = 180°
∠A = 180° – 40°
∠A = 140°
∠A = ∠I = 140° …………. (Opposite angles of a rhombus are equal)

Q2. Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.
Ans: 

Steps of construction:
(i) Draw a line segment AC of length 7 cm and mark its midpoint as O.
(ii) At point O, draw an angle of 140° with respect to diagonal AC.
(iii) From O, along the 140° line in both directions, mark OD = 2.5 cm OD and OB = 2.5 cm using a compass.
(iv) Join D to A and C.
Join B to A and C.
ABCD is the required parallelogram.

Q3. Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.
Ans: 

Steps of construction:
(i) Draw a line segment AC of length 5 cm.
(ii) Draw the perpendicular bisector of AC, intersecting it at O.
(iii) With O as centre and radius 2 cm, mark points B (below) and D (above) on the perpendicular bisector.
(iv) Join A–D, D–C, C–B, and B–A.
ABCD is the required rhombus.

Page 107 & 108

Figure it Out

Q1. Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.
Ans: 

Since all sides of an equilateral triangle are equal.
Thus, the lengths of all sides of the given quadrilateral are equal.
∴ PQ = QR = RS = SP = 4 cm.
Also, the measure of all angles of an equilateral triangle is 60°.
∠P = ∠R = 60°
∠S = ∠PSQ + ∠RSQ = 60° + 60° = 120°.
∠Q = ∠PQR + ∠RQS = 60° + 60° = 120°.

Q2. Construct a kite whose diagonals are of lengths 6 cm and 8 cm.
Ans: 

(i) Draw a line segment AC = 6 cm.
(ii) Construct the perpendicular bisector of AC; let it meet AC at O (so O is the midpoint).
(iii) With centre O and radius 3 cm draw an arc to cut the bisector above AC; label that point D. With centre O and radius 5 cm draw an arc to cut the bisector below AC; label that point B.
(iv) Join A ⁣− ⁣B,  B ⁣− ⁣C,  C ⁣− ⁣D,  D ⁣− ⁣A.
ABCD is the required kite.

Q3. Find the remaining angles in the following trapeziums —
Ans: 
Since AB ∥ DC, and AD is a tranversal, then
∠A + ∠D = 180° …………. (Sum of angles on the same side of the transversal)
135° + ∠D = 180°
∠D = 180° – 135°
∠D = 45°
Also, since AB ∥ DC, and BC is a tranversal, then
∠B + ∠C = 180° …………. (Sum of angles on the same side of the transversal)
105° + ∠C = 180°
∠C = 180° – 105°
∠C = 75°

Since PQ ∥ SR, and PS is a tranversal, then
∠P + ∠S = 180° …………. (Sum of angles on the same side of the transversal)
∠P + 100° = 180°
∠P = 180° – 100° = 80°.
∠S = ∠R = 100° …………… (Angles opposite to equal sides are equal)
Also, since PQ ∥ SR, and QR is a tranversal, then
∠Q + ∠R = 180° …………… (Sum of angles on the same side of the transversal)
∠Q + 100° = 180°
∠Q = 180° – 100° = 80°.

Q4. Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions 
(i) What is the quadrilateral that is both a kite and a parallelogram?
(ii) Can there be a quadrilateral that is both a kite and a rectangle?
(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?
Ans: (i) A rhombus is a quadrilateral that is both a kite and a parallelogram.
(ii) A square is a quadrilateral that is both a kite and a rectangle.
(iii) No, every kite is not a rhombus.
Correct relationship: Every rhombus is a kite, but not every kite is a rhombus.

Q5. If PAIR and RODS are two rectangles, find ∠IOD.
Ans: Since PAIR and RODS are two triangles.
∠RIO = 90° ……… (Corner angle of a rectangle)
In △RIO,
∠IRO + ∠IOR + ∠RIO = 180° …………. (Sum of angles of a triangle)
30° + ∠IOR + 90° = 180°
120° + ∠IOR = 180°
∠IOR = 180° – 120° = 60°.
∴ ∠IOD = 90° – ∠IOR = 90° – 60° = 30°.

Q6. Construct a square with a diagonal 6 cm without using a protractor.
Ans: 

Steps of construction:
(i) Draw a line segment AC = 6 cm and mark its midpoint as O.
(ii) With O as centre and radius greater than half of AC, draw arcs above and below AC from points A and C.
(iii) Join the arc intersections to get a line perpendicular to AC and passing through O.
(iv) Again, with O as centre and radius equal to 3 cm, mark points B and D on the perpendicular line.
(v) Connect A–B–C–D–A.
Hence, ABCD is the required square with a diagonal of 6 cm.

Q7. CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).

Ans: 

Q8. If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.
Ans: Reasoning:

• A rhombus is a quadrilateral with four equal sides.
• If a rhombus has one angle of 90°, then:
• Its opposite angle is also 90° (opposite angles of a rhombus are equal).
• Each adjacent angle must also be 90° (sum of adjacent angles in a parallelogram/rhombus is 180°).
• Thus, all four angles are 90°.
• Since the quadrilateral has all sides equal and all angles right angles, it is a square.

Construction and measurement:

Steps of construction:
(i) Draw a line segment PQ of length 5 cm.
(ii) At point PP, construct a perpendicular line to PQ.
(iii) On this perpendicular, mark point S such that PS = 5 cm.
(iv) With S as centre and radius 5 cm, draw an arc to the right of PS.
(v) With Q as centre and radius 5 cm, draw an arc above PQ to intersect the arc from step (4) at point R.
Join Q–R, R–S, and S–P to complete the square PQRS.
Verification by measurement:
All sides: PQ = QR = RS = SP = 5 cm
All angles: ∠P =∠Q =∠R =∠S = 90°.
Conclusion: The figure constructed is a square.

Q9. What type of quadrilateral is one in which the opposite sides are equal? Justify your answer.
Hint: Draw a diagonal and check for congruent triangles.
Ans: If a quadrilateral has opposite sides equal, then it is a parallelogram. 
Geometric reasoning using a diagonal:

Given: Quadrilateral ABCD with AB = CD and BC = DA.
Draw diagonal AC.
In △ABC and △CDA,
AB = CD (given)
BC = DA (given)
AC = AC (common side)
By SSS congruence, △ABC ≅ △CDA.
From congruence, corresponding angles are equal:
∠BAC =∠DCA and ∠ACB =∠CAD.
But these are alternate interior angles.
∴ AB ∥ DC and AD ∥ BC.
Hence, ABCD is a parallelogram.

Q10. Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.
Ans: Yes, the sum of the angles in a quadrilateral will always be 360°.
Construction: Mark four non-collinear points as A, B, C, and D, and join them to form a quadrilateral ABCD.

Geometric reasoning:
In quad. ABCD, join BD to divide it into two triangles.
Now, In △BAD,
∠DBA + ∠BAD + ∠ADB = 180° ……….(1)……. (Sum of angles of a triangle)
In △BCD,
∠BCD + ∠CDB + ∠DBC = 180° ……….(2)……. (Sum of angles of a triangle)
Adding (1) and (2), we get
∠DBA + ∠BAD + ∠ADB + ∠BCD + ∠CDB + ∠DBC = 180° + 180°
(∠DBA + ∠DBC) + (∠ADB + ∠CDB) + ∠BAD + ∠BCD = 360°
∠ABC + ∠ADC + ∠BAD + ∠BCD = 360°
Thus, the sum of the angles of the given quadrilateral is 360°.

Q11. State whether the following statements are true or false. Justify your answers.
(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.
Ans: False.

A quadrilateral whose diagonals are equal and bisect each other is a rectangle. A square is a special case of a rectangle where all sides are also equal.

(ii) A quadrilateral having three right angles must be a rectangle.
Ans: True.

Three right angles force the fourth to be right as well and a quadrilateral with four right angles is a rectangle.

(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.
Ans: True.

If the diagonals bisect each other, then the two triangles formed by a diagonal are congruent, which gives pairs of opposite sides parallel. Hence the figure is a parallelogram.

(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.
Ans: False.

Squares, kites, and some other quadrilaterals also have perpendicular diagonals. Therefore, having perpendicular diagonals does not necessarily mean the quadrilateral is a rhombus.

(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.
Ans: True.

If both pairs of opposite angles are equal, then each pair of adjacent angles are supplementary, which implies opposite sides are parallel. Hence the quadrilateral is a parallelogram.

(vi) A quadrilateral in which all the angles are equal is a rectangle.
Ans: True

If all four angles are equal, each angle must be 360°/4 = 90°. A quadrilateral with four right angles is a rectangle.

(vii) Isosceles trapeziums are parallelograms.
Ans: False.

An isosceles trapezium has exactly one pair of parallel sides and the non-parallel sides equal while a parallelogram must have two pairs of parallel sides. So an isosceles trapezium is not a parallelogram.

NCERT Solutions: Proportional Reasoning-1

Page 162

Example 1:
(i) Are the ratios 3 : 4 and 72 : 96 proportional?
(ii) What is the HCF of 72 and 96?
Ans: 
(i) HCF of 3 and 4 = 1
∴ Ratio 3/4 is in its simplest form.
HCF of 72 and 96 = 24 (∵ 72 = 24 × 3 and 96 = 24 × 4)

∴ Ratio 72/96 in its simplest form is 3/4.
∴ Ratios 3 : 4 and 72 : 96 are proportional, because both ratios in their simplest form are the same.

(iii) HCF of 72 and 96 is equal to 24.

Example 2: Kesang wanted to make lemonade for a celebration. She made 6 glasses of lemonade in a vessel and added 10 spoons of sugar to the drink. Her father expected more people to join the celebration. So he asked her to make 18 more glasses of lemonade. 
Ans: Kesang’s original mixture: 6 glasses of lemonade → 10 spoons of sugar
Now she needs 18 more glasses of lemonade.
To keep the same sweetness, the ratios must be proportional:
6 : 10 :: 18 : ?
First term changes from 6 to 18.
Factor = 18 ÷ 6 = 3
Multiply sugar by the same factor
10 × 3 = 30
So, she should add 30 spoons of sugar to make 18 more glasses of lemonade with the same sweetness.

Page 163-166

Example 3: Nitin and Hari were constructing a compound wall around their house. Nitin was building the longer side, 60 ft in length, and Hari was building the shorter side, 40 ft in length. Nitin used 3 bags of cement but Hari used only 2 bags of cement. Nitin was worried that the wall Hari built would not be as strong as the wall he built because she used less cement. 
Is Nitin correct in his thinking?
Ans: 
For Nitin, the ratio of the length of the wall to the number of bags of cement = 60/3 =20/1
For Hari, the ratio of the length of the wall to the number of bags of cement = 40/2 =20/1
∴ Ratios in simplest form are equal.
∴ Ratios 60 : 3 and 40 : 2 are in proportion.
∴ Walls are equally strong.
∴ Nitin is not correct in his thinking.
Note: Both ratios show that 1 bag of cement is required for a wall of 20 ft.

Example 4: In my school, there are 5 teachers and 170 students. The ratio of teachers to students in my school is 5 : 170. Count the number of teachers and students in your school. What is the ratio of teachers to students in your school? Write it below.
Ans: In my school:
Number of teachers = 20
Number of students = 600
So the ratio of teachers to students is 20 : 600
Divide both by 20:
20 ÷ 20 = 1
600 ÷ 20 = 30
So the simplest form of the ratio is 1 : 30
Therefore, Teachers : Students = 1 : 30.

Example 5: Measure the width and height (to the nearest cm) of the blackboard in your classroom. What is the ratio of width to height of the blackboard?
Can you draw a rectangle in your notebook whose width and height are proportional to the ratio of the blackboard?
Compare the rectangle you have drawn to those drawn by your classmates. Do they all look the same?
Ans: 
Let the width and height of the blackboard in my classroom be 300 cm and 150 cm, respectively.
∴ Ratio of width to height = 300/150 cm = 2/1
Draw a line AB equal to 6 cm.
Let the height of the rectangle be x cm, so that the width and height of the rectangle are proportional to the ratio of the blackboard.
∴ 2 : 1 :: 6 : x
⇒ 2/1 = 6/x 
⇒ 2x = 6
⇒ x = 3
∴ The height of the rectangle is 3 cm.
Draw AD and BC perpendicular to AB and of length 3 cm. Join CD.
ABCD is the required rectangle.
Comparing the rectangles drawn by other classmates, I find that the ratios of width and height of all rectangles drawn by other classmates are proportional. Their rectangle are all similar, but they do not look the same.

Example 6: When Neelima was 3 years old, her mother was 10 times her age. What is the ratio of Neelima’s age to her mother’s age? What would be the ratio of their ages when Neelima is 12 years old? Would it remain the same? 
Ans:
Age of Neelima = 3 years
Mother’s age is 10 times the age of Neelima.
∴ Age of mother = 10 × 3 = 30 years
∴ Ratio of Neelima’s age to her mother’s age = 3/30 = 1/10 = 1 : 10
Now, Neelima is 12 years old.
Age of Neelima is 12 years, after 12 – 3 = 9 years.
After 9 years, age of mother = 30 + 9 = 39 years
∴ Ratio of Neelima’s age to her mother’s age = 

Remark: It must be noted that ratios 14 : 13 are not in proportion.
The rule is that when a fixed number k is added (or subtracted) from a ratio a : b, then the new ratio a + k : b + k may or may not be in proportion with the original ratio a : b.
In other words, a : b :: a + k : b + k may not be true.

Example 7: Fill in the missing numbers for the following ratios that are proportional to 14 : 21.
(i) ____ : 42
(ii) 6 : ____
(iii) 2 : ____
What factor should we multiply 14 by to get 6? Can it be an integer? Or should it be a fraction? (Page 164)
Ans:
The given ratio is 14 : 21.
(i) The ratio is ____ : 42.
Let the missing number be x.
14 : 21 :: x : 42
⇒ 14/21 = x/42
⇒ 2/3 = x/42
⇒ 3x = 2 × 42 = 84
⇒ x = 28
∴ The missing number is 28.

(ii) The ratio is 6 : ____
Let the missing number be x.
∴ 14 : 21 :: 6 : x
⇒ 14/21 = 6/x
⇒ 2/3 = 6/x
⇒ 2x = 3 × 6 = 18
⇒ x = 9
∴ The missing number is 9.

(iii) The ratio is 2 : ____
Let the missing number be x.
∴ 14 : 21 :: 2 : x
⇒ 14/21 = 2/x
⇒ 2/3 = 2/x
⇒ 2x = 6
⇒ x = 3
∴ The missing number is 3.
Second Part: Ratios under consideration are 14 : 21 and 6 : ____
Now, 6/14 =3/7 and 
∴ 14 should be multiplied by 3/7 to get 6.
∴ Required factor = 3/7
∴ Missing term in the ratio 6 : ____ is 

∴ The ratio 6 : ____ is 6 : 9.
The factor here is 3/7 and this is a fraction.

Example 8: Filter coffee is a beverage made by mixing coffee decoction with milk. Manjunath usually mixes 15 mL of coffee decoction with 35 mL of milk to make one cup of filter coffee in his coffee shop. In this case, we can say that the ratio of coffee decoction to milk is 15 : 35. If customers want ‘stronger’ filter coffee. Manjunath mixes 20 mL of the decoction with 30 mL of milk. The ratio here is 20 : 30. Why is this coffee stronger?
(i) And when they want ‘lighter’ filter coffee, he mixes 10 mL of coffee and 40 mL of milk, making the ratio 10 : 40. Why is this coffee lighter?

(ii) The following table shows the different ratios in which Manjunath mixes coffee decoction with milk. Write in the last column if the coffee is stronger or lighter than the regular coffee.
Ans:
In one cup of regular filter coffee: Coffee decoction = 15 mL
Milk = 35 mL
∴ Ratio of coffee decoction to milk = 15 : 35 = 3 : 7
Here, 3 + 7 = 10
∴ Coffee decoction in 10 mL filter coffee = 3 mL
∴ Coffee decoction in 100 mL filter coffee =

In one cup of stronger filter coffee:
Coffee decoction = 20 mL
Milk = 30 mL
∴ Ratio of coffee decoction to milk = 20 : 30 = 2 : 3
Here, 2 + 3 = 5.
∴ Coffee decoction in 5 mL filter coffee = 2 mL
∴ Coffee decoction in 100 mL filter coffee 

Since 40 mL > 30 mL, the latter coffee is stronger.

(i) In a cup of lighter filter coffee:
Coffee decoction = 10 mL
Milk = 40 mL
∴ Ratio of coffee decoction to milk = 10 : 40 = 1 : 4
Here, 1 + 4 = 5.
∴ Coffee decoction in 5 mL filter coffee = 1 mL
∴ Coffee decoction in 100 mL filter coffee 

Since 20 mL < 30 mL, the third type of filter coffee is lighter.

(ii) 

S.No.1. Here 300 + 600 = 900
∴ Coffee decoction in 900 mL filter coffee = 300
∴ Coffee decoction in 100 mL filter coffee = 

Since  this filter coffee is stronger.

S.No.2. Here 150 + 500 = 650
∴ Coffee decoction in 650 mL filter coffee = 150

∴ Coffee decoction in 100 mL filter coffee = 

Since this filter coffee is lighter.

S.No. 3. Here 200 + 400 = 600
∴ Coffee decoction in 600 mL filter coffee = 200

∴ Coffee decoction in 100 mL filter coffee = 

Since  this filter coffee is stronger.

S.No. 4. Here 24 + 56 = 80
∴ Coffee decoction in 80 mL filter coffee = 24 mL
∴ Coffee decoction in 100 mL filter coffee 

Since 30 = 30, this filter coffee is regular.

S.No. 5. Here 100 + 300 = 400
∴ Coffee decoction in 400 mL filter coffee = 100 mL
∴ Coffee decoction in 100 mL filter coffee = 

Since 25 < 30, this filter coffee is lighter.

Figure it Out

Q1. Circle the following statements of proportion that are true.
(i) 4 : 7 :: 12 : 21
(ii) 8 : 3 :: 24 : 6
(iii) 7 : 12 :: 12 : 7
(iv) 21 : 6 :: 35 : 10
(v) 12 : 18 :: 28 : 12
(vi) 24 : 8 :: 9 : 3
Ans:  
(i) 4 : 7 :: 12 : 21.
This is true if 4/7 = 12/21
or if 4/7 = 4/7, which is true.
∴ The given statement is true.
(ii) 8 : 3 :: 24 : 6.
This is true if 8/3 = 24/6
or if 8/3 = 4 , which is false.
∴ The given statement is not true.
(iii) 7 : 12 :: 12 : 7.
This is true if 7/12 = 12/7, which is false.
∴ The given statement is not true.
(iv) 21 : 6 :: 35 : 10.
This is true if 21/6 = 35/10
or if 7/2 = 7/2, which is true.
∴ The given statement is true.
(v) 12 : 18 :: 28 : 12.
This is true if 12/18 = 28/12
or if 2/3 = 7/3
or 2 = 7, which is false.
∴ The given statement is not true.
(vi) 24 : 8 :: 9 : 3.
This is true if 24/8 = 9/3
Or if 3 = 3, which is true.
∴ The given statement is true.

Q2. Give 3 ratios that are proportional to 4 : 9. 
______ : ______   ______ : ______ ______ : ______ 
Ans: 
The given ratio is 4 : 9.

∴ 4 : 9 :: 8 : 18, 4 : 9 :: 12 : 27, and 4 : 9 :: 16 : 36
∴ Ratios 8 : 18, 12 : 27 and 16 : 36 are proportional to the given ratio 4 : 9.

Q3. Fill in the missing numbers for these ratios that are proportional to 18 : 24. 
3 : ______, 12 : ______, 20 : ______, 27 : ______
Ans: 

Q4. Look at the following rectangles. Which rectangles are similar to each other? You can verify this by measuring the width and height using a scale and comparing their ratios.
Ans: 
Using a scale, we measure the width and height of given rectangles.
Here, the ratio ‘Width : Height’ for given rectangles A, B, C, D, and E are respectively 1 : 3, 3 : 2, 9 : 4, 7 : 2 and 3 : 1.
These ratios are all distinct. The ratios of A and E are 1 : 3 and 3 : 1 respectively.
∴ For A and E, one side is 3 times the other side.
∴ Only rectangles A and E are similar.

Q5. Look at the following rectangle. Can you draw a smaller rectangle and a bigger rectangle with the same width to height ratio in your notebooks? Compare your rectangles with your classmates’ drawings. Are all of them the same? If they are different from yours, can you think why? Are they wrong?
Ans: 
For the given rectangle;
Width = 32 mm and height = 18 mm
∴ Ratio is 32 : 18.
We shall draw smaller and bigger rectangles and similar to the given rectangle by considering different ‘factors of change’.
Let the factor of change be 1/2.
∴ New width =  = 16 mm
and New height =  = 9 mm
A new, similar rectangle is shown in the figure.

Let ‘factor of change’ be 2.
∴ New width = 2 × 32 = 64 mm and new height = 2 × 18 = 36 mm
A new, similar rectangle is shown in the figure. The rectangles drawn by other classmates are all different, but they are all similar to the given rectangle.

Q6. The following figure shows a small portion of a long brick wall with patterns made using coloured bricks. Each wall continues this pattern throughout the wall. What is the ratio of grey bricks to coloured bricks? Try to give the ratios in their simplest form.

Ans:
(a) We consider one set of patterns in the given wall.

Number of grey bricks in one set of pattern = 2 + 3 + 4 = 9
Number of coloured bricks in one set of pattern = 3 + 2 + 1 = 6
∴ Ratio of grey bricks to coloured bricks = 9 : 6
We have 9 : 6 = 3 : 2
∴ Ratio in the simplest form = 3 : 2

(b) We use one set of patterns on the given wall
Number of grey bricks in one set of pattern
= 3 + 2 + 2 + 2 + 2 + 2 + 3
= 16
Number of coloured bricks in one set of pattern = 1 + (1 + 1) + (1 + 1) + (1 + 1) + (1 + 1) + (1 + 1) + 1
= 1 + 2 + 2 + 2 + 2 + 2 + 1
= 12
∴ Ratio of grey bricks to coloured bricks = 16 : 12
We have 16 : 12 = 4 : 3
∴ Ratio in the simplest form = 4 : 3.

Page 167

Q7. Let us draw some human figures. Measure your friend’s body—the lengths of their head, torso, arms, and legs. Write the ratios as mentioned below—

Now, draw a figure with head, torso, arms, and legs with equivalent ratios as above.
Ans: My friend’s body measurements:
(i) Head = 22 cm
(ii) Torso (neck to hip) = 50 cm
(iii) Arms (shoulder to fingertip) = 60 cm
(iv) Legs (hip to foot) = 80 cm
1. Head : Torso = 22 : 50
Simplify by dividing both by 2 → 11 : 25.
2. Torso : Arms = 50 : 60
Simplify by dividing both by 10 → 5 : 6.
3. Torso : Legs = 50 : 80
Simplify by dividing both by 10 → 5 : 8.
So the ratios are:

• Head : Torso = 11 : 25
• Torso : Arms = 5 : 6
• Torso : Legs = 5 : 8

Example 8: For the mid-day meal in a school with 120 students, the cook usually makes 15 kg of rice. On a rainy day, only 80 students came to school. How many kilograms of rice should the cook make so that the food is not wasted?
The ratio of the number of students to the amount of rice needs to be proportional. So, 120 : 15 :: 80 : ?
What is the factor of change in the first term? 
Ans: 
For 120 students, the rice required is 15 kg.
Let x kg of rice be required for 80 students.
∴ Ratios 120 : 15 and 80 : x are in proportion.
∴ 120 : 15 :: 80 : x
⇒ 120/15 = 80/x
⇒ x = 10
∴ 10 kg of rice is required.
Also, factor of change in the first term = 80/120 = 2/3

Alternative Method:
Factor of change in second term = x/15
Since the quality of food is the same, we have 2/3 = x/15
⇒ 3x = 30
⇒ x = 10
∴ 10 kg of rice is required.

Example 9: (i) A car travels 90 km in 150 minutes. If it continues at the same speed, what distance will it cover in 4 hours?
If it continues at the same speed, the ratio of the time taken should be proportional to the ratio of the distance covered.
(ii) 150 : 90 :: 4 : x
Is this the right way to formulate the question?
(iii) How can you find the distance covered in 240 minutes?
Ans: 
(i) We have, 4 hours = 4 × 60 = 240 minutes
In 150 minutes, the distance covered = 90 km
Let x km be covered in 4 hours, i.e., in 240 minutes.
∴ The ratios 150 : 90 and 240 : x are in proportion.
∴ 150 : 90 :: 240 : x
(ii) Since units must be the same in comparing ratios, the given proportion 150 : 90 :: 4 : ? is meaningless.
We have 4 hours = 240 minutes
∴ The proportion 150 : 90 :: 240 : ? is correct.
(iii) We have 150 : 90 :: 240 : x
⇒ 150/90 = 240/x
⇒ 5/3 = 240/x
⇒ 5x = 3 × 240
⇒ x = 144
∴ Distance covered in 4 hours = 144 km.

Example 10: A small farmer in Himachal Pradesh sells each 200 g packet of tea for ₹ 200. A large estate in Meghalaya sells each 1 kg packet of tea for ₹ 800. Are the weight-to-price ratios in both places proportional? Which tea is more expensive? Why? (Page 169)
Ans: 
We have 1 kg = 1000 g
In Himachal Pradesh, 200 g of tea costs ₹ 200.
∴ Ratio of weight to price in Himachal Prdesh = 200 : 200 = 1 : 1
In Meghalaya, 1000 g tea costs ₹ 800.
∴ Ratio of weight to price in Meghalaya = 1000 : 800 = 5 : 4
The ratios 1 : 1 and 5 : 4 are not proportional, because 1/1 ≠ 5/4
Price of 200 g tea in Himachal Pradesh = ₹ 200
Price of 1000 g tea in Himachal Pradesh = 

Since 1000 > 800, tea is more expensive in Himachal Pradesh.

Page 170-171

Q1. The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?
Ans: 
We know that:
1 million = 10 lakh = 10,00,000
and 1 year = 365/7 weeks.
940 million kilometres, i.e., 940 × 10,00,000 kilometres, are travelled by the Earth in 1 year, i.e., in 365/7 weeks.
Let the Earth travel x kilometres in 1 week
∴ The ratios 940 × 10,00,000 : 365/7 and x : 1 are in proportion.

⇒ x = 1,80,27,397 (nearly)
∴ In 1 week, Earth travels nearly 1,80,27,397 kilometres around the Sun.

Q2. A mason is building a house in the shape shown in the diagram. He needs to construct both the outer walls and the inner wall that separates two rooms. To build a wall of 10-feet, he requires approximately 1450 bricks. How many bricks would he need to build the house? Assume all walls are of the same height and thickness.
Ans:

Number of bricks required for a 10 ft wall =1450
∴ Ratio of length of wall to number of bricks = 10 : 1450

Total length of walls = AI + CH + DE + FG + IG + AF + CD
= 12 + (9 + 12) + 9 + 12 + (9 + 15) + (9 + 15) + 6
= 108 ft
Let x bricks be required for a 108 ft long wall.
∴ Ratio of length of wall to number of bricks = 108 : x
These ratios are in proportion.
∴ 10 : 1450 :: 108 : x
⇒ 10/1450 = 108x
⇒ 1/145 = 108x
⇒ x = 145 × 108 = 15,660
∴ Number of required bricks = 15,660.

Q: Puneeth’s father went from Lucknow to Kanpur in 2 hours by riding his motorcycle at a speed of 50 km/h. If he drives at 75 km/h, how long will it take him to reach Kanpur? Can we form this problem as a proportion 50 : 2 :: 75 : ____
Would it take Puneeth’s father more time or less time to reach Kanpur? Think about it. 
Ans: 
Time taken at the speed of 50 km/h = 2 hours
∴ Distance = Speed × Time
= 50 × 2
= 100 km
At speed of 75 km/h, time taken = 100/75 = 4/3 hours
∴ Ratio of speed to time in both cases are 50 : 2 and 75 : 4/3
Here, 50/2 = 25 and 

∴ The ratios 50 : 2 and 75 : 4/3 are not in proportion.

∴ We cannot write 50 : 2 :: 75 : 4/3.

At a speed of 75 km/h, Puneeth’s father will take 4/3 hours to reach Kanpur, which is less than 2 hours.

Example 11: Prashanti and Bhuvan started a food cart business near their school. Prashanti invested ₹ 75,000 and Bhuvan invested ₹ 25,000. At the end of the first month, they gained a profit of ₹ 4,000. They decided that they would share the profit in the same ratio as their investment. What is each person’s share of the profit?
Ans: 
Investment of Prashanti = ₹ 75,000
Investment of Bhuvan = ₹ 25,000
∴ Ratio of investment = 75000 : 25000 = 3 : 1
Total profit = ₹ 4,000
∴ Share of Prashanti =

= ₹ 3,000

∴ Share of Bhuvan = 

= ₹ 1,000

Verification: ₹ 3,000 + ₹ 1,000 = ₹ 4,000, the total profit.

Example 12: A mixture of 40 kg contains sand and cement in the ratio of 3 : 1. How much cement should be added to the mixture to make the ratio of sand to cement 5 : 2? 
Ans: 
Ratio of sand and cement = 3 : 1
Weight of mixture = 40 kg
∴ Weight of sand in mixture = 

Weight of cement in the mixture = 

Let the weight of cement in the new mixture be x kg, so that the new ratio is 5 : 2.
∴ Ratios 30 : x and 5 : 2 are in proportion.
⇒ 30/x = 5/2
⇒ 5x = 60
⇒ x = 12
Weight of cement added = 12 kg – 10 kg = 2 kg

Page 175

Q1. Divide ₹4,500 into two parts in the ratio 2 : 3.
Ans: 
Given ratio = 2 : 3
Amount to be divided = ₹ 4,500
∴ First part = 

= 2 × 900
= ₹ 1,800
∴ Second part = 

= 3 × 900
= ₹ 2,700
∴ Two parts are ₹ 1,800 and ₹ 2,700.
Verification:
1,800 : 2,700 = 1,800/2,700 = 18/27 = 2/3 = 2 : 3 and 1,800 + 2,700 = 4,500.

Q2: In a science lab, acid and water are mixed in the ratio of 1 : 5 to make a solution. In a bottle that has 240 mL of the solution, how much acid and water does the solution contain?
Ans:
Ratio of acid and water = 1 : 5
Quantity of solution = 240 mL
∴ Quantity of acid = 

= 40 mL
∴ Quantity of water = 

= 200 mL

∴ Quantities of acid and water in the solution are 40 mL and 200 mL.
Verification:
40 : 200 = 40/200 = 1/5 = 1 : 5 and 40 + 200 = 240.

Q3. Blue and yellow paints are mixed in the ratio of 3 : 5 to produce green paint. To produce 40 mL of green paint, how much of these two colours are needed? To make the paint a lighter shade of green, I added 20 mL of yellow to the mixture. What is the new ratio of blue and yellow in the paint?
Ans: Ratio of Blue and yellow paints in mixture = 3 : 5
To produce 40 mL of green paint, quantity of Blue pain needed = 3/(3 + 5) × 40 mL = 3/8 × 40 mL = 15 mL
∴ Quantity of yellow pain needed = 5/8 × 40 mL = 25mL
Now, add 20 mL of yellow to the mixture.
Then, the ratio of blue and yellow in the new mixture =15 : (25 + 20) = 15 : 45 = 1 : 3

Q4. To make soft idlis, you need to mix rice and urad dal in the ratio of 2 : 1. If you need 6 cups of this mixture to make idlis tomorrow morning, how many cups of rice and urad dal will you need?
Ans: Required ratio of rice and urad = 2 : 1 and total quantity of mixture = 6 cups
∴ Amount of rice required = 2/(2 + 1) × 6 = 2/3 × 6 = 4 cups
∴ Amount of urad required = 1/(2 + 1) × 6 = 1/3 × 6 = 2 cups

Q5. I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio of 3 : 5. I added another bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture?
Ans: 
Let the capacity of one bucket be x L.
Ratio of red paint and yellow paint = 3 : 5
∴ Quantity of red paint in the bucket =

∴ Quantity of yellow paint in the bucket =

One bucket of yellow paint is added to the mixture.
∴ New quantity of red paint in the mixture = 3x/8
∴ New quantity of yellow paint in the mixture = 

∴ New ratio of red paint and yellow paint in the mixture = 3x/8:13x/8 = 3 : 13

Page 176-177

Q1. Anagh mixes 600 mL of orange juice with 900 mL of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in its simplest form.
Ans: 
Quantity of orange juice = 600 mL
Quantity of apple juice = 900 mL
∴ Ratio of orange juice to apple juice = 600 : 900
Ratio in the simplest form = 600 : 900 = 2 : 3

Q2. Last year, we hired 3 buses for the school trip. We had a total of 162 students and teachers who went on that trip and all the buses were full. This year we have 204 students. How many buses will we need? Will all the buses be full?
Ans: 
Number of buses for 162 students and teachers = 3
Since the buses were full, the capacity of 1 bus = 162/3 = 54
∴ Ratio of number of seats to the number of buses is 54 : 1.
We have
54 : 1 = 2(54) : 2(1) = 108 : 2
54 : 1 = 3(54) : 3(1) = 162 : 3
54 : 1 = 4(54) : 4(1) = 216 : 4
∴ Capacity of 4 buses = 216
∴ For 204 students, we shall need 4 buses.
Since 216 – 204 = 12, we have 12 vacant seats in the buses.

Q3. The area of Delhi is 1,484 sq. km and the area of Mumbai is 550 sq. km. The population of Delhi is approximately 30 million and that of Mumbai is 20 million people. Which city is more crowded? Why do you say so?
Ans: 
Area of Delhi = 1,484 sq.km
Population of Delhi = 30 million
Area of Mumbai = 550 sq. km
Population of Mumbai = 20 million
∴ Ratio of area to population for Delhi = 1484 : 30
∴ Ratio of area to population for Mumbai = 550 : 20
Factor of change of area = 550/1484 = 0.371 (nearly)
Factor of change of population = 20/30 = 0.667 (nearly)
Since 0.667 > 0.371, Mumbai is more crowded than Delhi.

Alternative Method:
Ratio of area to population for Delhi = 1484 : 30
Let the density of Delhi and Mumbai be the same, and there be x people in Mumbai.
∴ The ratios 1,484 : 30 and 550 : x are in proportion.
∴ 1,484/30 = 550/x
⇒ 1484x = 30 × 550 = 16,500
⇒ x = 16500/1484 = 11.118
There should be 11.118 million people in Mumbai. But the population of Mumbai is 20 million.
∴ Mumbai is more crowded than Delhi.

Q4. A crane of height 155 cm has its neck and the rest of its body in the ratio 4 : 6. For your height, if your neck and the rest of the body also had this ratio, how tall would your neck be?
Ans:
The ratio of the height of the neck and the height of the rest of the body of a crane is 4 : 6.
My height is 65 inches, i.e., 165 cm.
Let the ratio of the height of my neck and the height of the rest of my body also be 4 : 6.
∴ Height of my neckcm = 66 cm

Q5. Let us try an ancient problem from Lilavati. At that time weights were measured in a unit named palas and niskas was a unit of money. “If palas of saffron costs 3/7 niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”
Ans: 
Here, the unit of weight in palas and the unit of money are niskas.
Cost of  palas of saffron = 37 niskas
∴ Ratio of weight to price is 

or 5/2 : 3/7
or 35 : 6
Let x palas of saffron be bought for 9 niskas.
∴ Ratio of weight to price is x : 9.
These ratios are in proportion.
∴ 35 : 6 :: x : 9
⇒ 35/6 = x/9
⇒ 6x = 35 × 9
⇒ x = 52.5
∴ 52.5 palas of saffron can be bought for 9 niskas.

Q6. Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s age when the ratio of her age to her brother’s age is 1 : 2?
Ans: 
The ages of Harmain and her brother are 1 year and 5 years.
Let x years, the ratio of their ages be 1 : 2.
Age of Harmain after x years = (1 + x) years.
Age of her brother after x years = (5 + x) years.
∴ After x years, ratio of their ages = 1 + x : 5 + x
The ratios are in proportion.
∴ 1 : 2 :: 1 + x : 5 + x

⇒ 5 + x = 2(1 + x)
⇒ 5 + x = 2 + 2x
⇒ 2x – x = 5 – 2
⇒ x = 3
∴ After 3 years, the age of Harmain = 1 + 3 = 4 years.
Verification:
After 3 years, age of her brother = 5 + 3 = 8 years
Also, the ratio of their ages = 4 : 8 = 1 : 2.

Q7. The mass of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?
Ans: 

The ratio of masses of gold and water, when their volumes are the same, is 37 : 2.

Mass of 1 litre of water = 1 kg
Let the mass of 1 litre of gold = x kg
∴ With equal volumes, the ratio of masses of gold and water is x : 1.
These ratios are in proportion.
∴ 37 : 2 :: 1 : x
⇒ 37/2 = x/1
⇒ x = 372
Thus, the mass of 1 litre of gold is 37/2 kg.

Q8. It is good farming practice to apply 10 tonnes of cow manure for 1 acre of land. A farmer is planning to grow tomatoes in a plot of size 200 ft by 500 ft. How much manure should he buy?
Ans: 
We have 1 ton = 1,000 kg
∴ 10 tonnes = 10 × 1,000 = 10,000 kg
Also, 1 acre = 43,560 sq. ft.
∴ Ratio of cow manure to area of land in kg and sq. ft. = 10,000 : 43,560
Size of plot = 200 ft. by 500 ft.
∴ Area of plot = 200 × 500 = 1,00,000 sq. ft.
Let cow manure be x kg.
∴ Ratio of cow manure to area of plot = x : 1,00,000
These ratios are in proportion.
∴ 10,000 : 43,560 :: x : 1,00,000

⇒ 43,560x = 10,000 × 1,00,000 = 1,00,00,00,000

⇒ x = 22956.84
∴ Required cow manure = 22956.84 kg = 22.95684 tonnes.

Q9. A tap takes 15 seconds to fill a mug of water. The volume of the mug is 500 mL. How much time does the same tap take to fill a bucket of water if the bucket has a 10-litre capacity?
Ans: 
Time taken by the tap for 500 mL of water = 15 seconds
∴ Ratio of volume to time = 500 : 15
We know 1 litre = 1,000 mL
10 litre = 10 × 1,000 = 10,000 mL
Let the time taken to fill a bucket of 10,000 mL be x seconds.
∴ Ratio of volume to time = 10,000 : x
These ratios are proportional.
∴ 500 : 15 :: 10,000 : x

⇒ 500x = 1,50,000
⇒ x = 300
∴ Time to fill bucket = 300 seconds
= 300/60 minutes
= 5 minutes.

Q10. One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land?
Ans: 
We know that 1 acre = 43,560 square feet.
∴ Cost of 43,560 sq. ft. land = ₹ 15,00,000
∴ Ratio of area of land to cost = 43,560 : 15,00,000
Let the cost of 2,400 sq. ft. of land be ₹ x.
∴ Ratio of area of land to cost = 2,400 : x
These ratios are proportional.
∴ 43,560 : 15,00,000 :: 2,400 : x

⇒ x = 82,644.63
∴ Cost of land = ₹ 82,644.63.

Q11. A tractor can plough the same area of a field 4 times faster than a pair of oxen. A farmer wants to plough his 20-acre field. A pair of oxen takes 6 hours to plough an acre of land. How much time would it take if the farmer used a pair of oxen to plough the field? How much time would it take him if he decides to use a tractor instead?
Ans: 
Ratio of efficiency of a tractor to a pair of oxen = 4 : 1
Time taken by a pair of oxen to plough 1 acre of field = 6 hours
∴ Time taken by a tractor to plough 1 acre field = 6/4 = 1.5 hours
∴ Time taken by a pair of oxen to plough 20 20-acre field = 20 × 6 = 120 hours
∴ Time taken by a tractor to plough a 20-acre field = 20 × 1.5 = 30 hours

Q12. The ₹10 coin is an alloy of copper and nickel called ‘cupro-nickel’. Copper and nickel are mixed in a 3 : 1 ratio to get this alloy. The mass of the coin is 7.74 grams. If the cost of copper is ₹906 per kg and the cost of nickel is ₹1,341 per kg, what is the cost of these metals in a ₹10 coin?
Ans: 
Ratio of copper and nickel in ₹ 10 coin = 3 : 1
Mass of one ₹ 10 coin = 7.74 grams
∴ Mass of copper in one ₹ 10 coin = 

= 5.805 grams

Mass of nickel in one ₹ 10 coin = 

= 1.935 grams

Cost of 1 kg copper = ₹ 906

∴ Cost of 1000 grams of copper = ₹ 906

∴ Cost of 5.805 grams copper = 

Cost of 1 kg nickel = ₹ 1341
∴ Cost of 1000 grams of nickel = ₹ 1341

∴ Cost of 1.935 grams nickel = 

∴ In one ₹ 10 coin, the cost of copper and the cost of nickel are respectively ₹ 5.26 and ₹ 2.59.

Page 48

Questions (Implied from Reema’s Curiosity):

Q1: Since when have humans been counting?

Ans: Humans have been counting since at least the Stone Age (around 10,000 years ago) to track quantities of food, livestock, trade goods, ritual offerings, and to predict events like lunar phases or seasons.

Q2: What was their need for counting?

Ans: The need arose to quantify resources (e.g., food, livestock), manage trade, record ritual offerings, and track time for events like new moons or seasonal changes.

Q3: What were they counting?

Ans: They counted food items, animals in livestock, trade goods, ritual offerings, and days for calendrical purposes.

Q4: Since when have people been writing numbers in the modern form?

Ans: The modern form (Hindu numerals, 0–9) originated in India around 2000 years ago, with the earliest known use in the Bakhshali manuscript (c. 3rd century CE). Aryabhata (c. 499 CE) formalized their use, and they spread globally by the 17th century.

Q5: How would the Mesopotamians have written 20, 50, 100?

Ans: The Mesopotamian (Babylonian) system was a base-60 positional system using symbols for 1 (⟐) and 10 (⟐). Numbers were grouped into powers of 60, with a placeholder for zero in later periods. 
Assuming the simplified notation from Section 3.4:

• 20: 20 = 20 × 1 = ⟐⟐ (two 10s).
• 50: 50 = 5 × 10 = ⟐⟐⟐⟐⟐ (five 10s).
• 100: 100 = 1 × 60 + 40 × 1 = ⟐,⟐⟐⟐⟐ (one 60 and four 10s). 

Note: Commas separate place values for clarity, though spacing was inconsistent in practice.

Page 51

Q1. How do we ensure that all cows have returned safely aftergrazing?

Ancient humans used sticks to count their herd of cows through a very practical, physical method:

• For each cow in the herd, one stick was set aside.
• After the cows went out to graze, the herder would collect a stick for each cow that returned.
• If a stick remained with no cow to match it, that meant a cow was missing; if all sticks matched, then all cows had returned.

Q2. Do we have fewer cows than our neighbour?

Ancient humans could compare herds without words by using physical objects—typically sticks, stones, or tallies—to represent each cow. Here’s how they would determine if they had fewer cows than their neighbor:

• Each person would create a collection of sticks, one stick for each cow in their herd.
• They would then place their collections side by side.
• By pairing off the sticks one by one from each herd, they could see which collection finished first.
• If your pile of sticks was exhausted before your neighbor’s, this showed you had fewer cows.
• The difference in the length (or count) of the two piles directly told you how many fewer cows you had.

This method worked because it relied entirely on one-to-one correspondence, requiring no written numbers or words—just an exact physical representation.

Q3. If there are fewer, how many more cows would we need so that we have the same number of cows as our neighbour?

If you find that you have fewer cows than your neighbor after comparing your piles of sticks (one stick per cow):

• Take your pile of sticks and your neighbor’s pile of sticks.
• Pair them one-to-one as far as possible.
• The number of sticks left (unpaired) in your neighbor’s pile tells you exactly how many cows you would need to add to your herd to have the same number as your neighbor.
• In other words, you simply count the extra sticks in your neighbor’s pile after pairing, and that is the answer.

How will you use such sticks to answer the other two questions (Q2 and Q3)?

Ans: Same as above.

Page 53

Q: How many numbers can you represent in this way using the sounds of the letters of your language?

Ans: Using only the sounds of English letters, you can represent at most 26 distinct numbers, since there are 26 letters. Letters don’t naturally map to numbers, so without creating combinations or a naming system, this method is limited. For numbers beyond 26, you’d need to combine letter sounds or use a more complex system.

Q: Do you see a way of extending this method to represent bigger numbers as well? How?

Ans: To represent bigger numbers using letter sounds, we need a fixed, ordered system—called a number system. While using letter sounds is convenient, it’s limited as there are only 26 letters. To extend it for bigger numbers, we must combine letter sounds or symbols to create a longer, standard sequence—just like Roman or Hindu number systems do.

Page 54

Figure it Out

Q1. Suppose you are using the number system that uses sticks to represent numbers, as in Method 1. Without using either the number names or the numerals of the Hindu number system, give a method for adding, subtracting, multiplying and dividing two numbers or two collections of sticks.

Ans: Using sticks (Method 1) to perform arithmetic operations:

• Addition:
  Combine two collections of sticks into a single larger collection. The total number of sticks now represents the sum.
• Subtraction:
  Remove sticks from one collection to match the number in another. The remaining sticks represent the difference.
• Multiplication:
  Think of multiplication as repeated addition. For example, to multiply 3 by 4, create 4 groups each having 3 sticks, then combine all the groups into one pile and count sticks.
• Division:
  Divide a collection of sticks into equal smaller groups. The number of groups you can make represents the quotient, and leftover sticks (if any) are the remainder.

Q2. One way of extending the number system in Method 2 is by using strings with more than one letter—for example, we could use ‘aa’ for 27. How can you extend this system to represent all the numbers? There are many ways of doing it!

Ans: Extending a letter-based number system like Method 2 (‘a’ to ‘z’ representing 1 to 26):

To represent numbers beyond 26, use combinations of letters similar to how letters form words. For example:

• ‘a’ = 1, ‘b’ = 2, …, ‘z’ = 26
• ‘aa’ = 27, ‘ab’ = 28, ‘ac’ = 29, and so on.

This works like a base-26 system, where each position represents powers of 26, similar to how digits work in the decimal system (base-10).

You can keep extending to three letters, four letters, etc., allowing representation of all natural numbers.

Q3. Try making your own number system.

Ans: Do it Yourself!

Hint: Try these steps:

• Choose simple symbols or objects (like stones, shells, hand signs, or colors).
• Assign each symbol a specific value, starting from 1.
• Decide on a method to combine symbols to form larger numbers — for example, repetition to indicate quantity (like tally marks), or place value methods (like grouping symbols in sets).
• Create rules for arithmetic operations based on how you combine or separate these symbols.

For instance, you could use colored beads where each color counts as a certain number, and putting beads together adds their values; different bead strings could represent larger numbers.

This approach both honours ancient counting methods and encourages creative thinking about the concept of numbers.

Page 56

Q: Can you see how their number names are formed? 

Ans: The Gumulgal number system forms number names by counting in twos, combining the words “urapon” (1) and “ukosar” (2):

• 1: urapon
• 2: ukosar
• 3: ukosar-urapon (2 + 1)
• 4: ukosar-ukosar (2 + 2)
• 5: ukosar-ukosar-urapon (2 + 2 + 1)
• 6: ukosar-ukosar-ukosar (2 + 2 + 2)

Q: Can you see how the names of the other numbers are formed?

Ans: The pattern uses “ukosar” for each group of 2 and “urapon” for an additional 1, building numbers additively based on twos and ones.

• 3 = 2 + 1, 
• 4 = 2 + 2, 
• 5 = 2 + 2 + 1, 
• 6 = 2 + 2 + 2.

Gumulgal called any number greater than 6 ras.

Page 57

Q: Quickly count the number of objects in each of the following boxes:

Ans: Let’s count:

Observation:

• You can instantly recognize smaller quantities (1–4).
• For larger groups (like grapes, flowers, sticks), you likely had to count or estimate.
• This supports the idea that human perception easily handles up to 4 items instantly, but beyond that, counting is needed.

Page 58

Q: What could be the difficulties with using a number system that counts only in groups of a single particular size? How would you represent a number like 1345 in a system that counts only by 5s?

1. Lack of Flexibility

• You can only count in fixed steps (e.g., 5, 10, 15…), so it’s hard to represent numbers that aren’t exact multiples of that size, 
• Example: 1,234 can’t be represented accurately. 

2. Representation Becomes Lengthy or Complicated

• To represent numbers like 1, 2, or 3, you’d have to invent special symbols or combinations, since they don’t fit in the group size.
• Example: Representing the number 3 might need a symbol like “~~~” (three dashes), or a new symbol altogether, adding unnecessary complexity for small values.

3. Complex Arithmetic Systems

• Adding, subtracting, or comparing numbers becomes harder because there’s no positional value system or digits.
• Example: In our normal number system, adding 243 and 159 is easy because we use place values.
  For example:
  243 = 2 hundreds + 4 tens + 3 ones
  159 = 1 hundred + 5 tens + 9 ones
  We can align the digits and add them column by column.
• But in a system where you only use symbols or count in fixed steps (like 5s), there’s no such place value.
  For example, if ‘A’ = 5, then:
  1. A + A = 10 — but there’s no clear way to write 10 unless you define a new symbol for it.
  2. AAAAA (5 times A) = 25, and AAAAAA = 30 — but to compare them, you have to count each symbol every time.

4. Inefficiency for Large Numbers:

• Counting big numbers would require repeating the base unit multiple times, which is slow and impractical.

Representing 1345 in a System That Counts Only by 5s:

1345 can be written as: 1345 = (5 × 269) + 0

So, you would represent it as “269 groups of 5” and “0 extra units.”

In a simple group-of-5 system, you’d need 269 marks or symbols, each standing for 5, which is very inefficient for large numbers.

Page 59

Figure it Out

Q1. Represent the following numbers in the Roman system.

(i) 1222

(ii) 2999

(iii) 302

(iv) 715

Ans:

Explanation:

(i) 1222

Break it down:
1000 + 200 + 20 + 2 = M + CC + XX + II
Answer: MCCXXII

(ii) 2999

Break it down:
2000 + 900 + 90 + 9 = MM + CM + XC + IX
Answer: MMCMXCIX

(iii) 302

Break it down:
300 + 2 = CCC + II
Answer: CCCII

(iv) 715

Break it down:
700 + 10 + 5 = DCC + X + V
Answer: DCCXV

Page 60

Q: Do it yourself now: LXXXVII + LXXVIII

Ans: Step 1: Write all symbols together:
L + L + X + X + X + X + X + V + V + I + I + I + I + I

Step 2: Group and simplify:

• I + I + I + I + I = V
• V + V = X
• X + X + X + X + X = L

Step 3: Now combine:
L + L = C, plus remaining X and V

Final Answer: CLXV

Q: How will you multiply two numbers given in Roman numerals, without converting them to Hindu numerals? Try to find the product of the following pairs of landmark numbers: V × L, L × D, V × D, VII × IX.

Ans: You cannot multiply directly in Roman numerals on an abacus. You must:

1. Convert Roman → Hindu-Arabic
2. Multiply using abacus
3. Convert result back → Roman numeral

This method ensures accurate and efficient multiplication.

Note:

• _X = 10,000, so _XX = 20,000, and _XXV = 25,000
• Roman numerals above 3,999 use overlines to indicate multiplication by 1,000

Q: DAREDEVIL CONTEST: Multiply CCXXXI and MDCCCLII

• CCXXXI = 200 + 20 + 11 = 231
• MDCCCLII = 1000 + 700 + 50 + 2 = 1752

Multiplication in Roman numerals is impractical without conversion.
So, convert to Hindu-Arabic numerals:

231 × 1752 = 404712

Converting 404712 to Roman Numerals:

404712 = CD (400,000) + IV (4) + DCC (700) + XII (12)

So, Roman numeral: _CDIVDCCXII
(where CD means 400,000)

Page 60-70

Figure it out

Q1. A group of indigenous people in a Pacific island use different sequences of number names to count different objects. Why do you think they do this? 

Ans: Different sequences help keep track of various categories of objects. This method is practical in daily life, like using one set of words for counting coconuts and another for people. It allows them to focus on context, improve accuracy, and avoid confusion.

Q2. Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s. Come up with ways of performing the different arithmetic operations (+, –, ×, ÷)for numbers occurring in this system, without using Hindu numerals. Use this to evaluate the following:

(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasarukasar-urapon)

(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasarukasar)

(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)

(iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)

Ans: First, define base terms:

• urapon = 1
• ukasar = 2

So:

• ukasar-urapon = 3
• ukasar-ukasar = 4
• ukasar-ukasar-urapon = 5
• ukasar-ukasar-ukasar = 6
• ukasar-ukasar-ukasar-urapon = 7
• ukasar-ukasar-ukasar-ukasar = 8
• ukasar-ukasar-ukasar-ukasar-urapon = 9
• etc.

(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon)

• First term: 4 ukasar + 1 urapon = 9
• Second term: 3 ukasar + 1 urapon = 7

Add using parts:

• 4 + 3 = 7 ukasar
• 1 + 1 = 2 urapon = 1 ukasar

Total: 7 ukasar + 1 ukasar = 8 ukasar

Answer: 8 ukasar or ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar-ukasar)

• First term = 9
• Second term = 6

Subtract by cancelling:

• Remove 3 ukasar from 4 ukasar → 1 ukasar remains
• 1 urapon stays

Answer: ukasar + urapon = ukasar-urapon

(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)

Multiply:

• First term = 4 ukasar + 1 urapon = 9
• Second term = 2 ukasar = 4

Use repeated addition:

• 9 × 4 = 36
• 36 = 18 ukasar

Answer: 18 ukasar
→ Write out 18 repetitions of “ukasar” if needed, or say “18 ukasar” in Gumulgal form.

(iv) (ukasar × 8) ÷ (ukasar-ukasar)

• Numerator: 8 × ukasar = 8 × 2 = 16
• Denominator: ukasar-ukasar = 4

Divide:

• 16 ÷ 4 = 4
  → 4 = ukasar-ukasar

Answer: ukasar-ukasar

Q3: Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.

Ans: Features of the Hindu number system that make it efficient:

• Uses place value (units, tens, hundreds…)
• Has only 10 symbols (0–9) to write any number
• Includes zero, making calculations easier
• Easy to use for large numbers and arithmetic operations

Q4: Using the ideas discussed in this section, try refining the number system you might have made earlier.

Ans: Do it Yourself!

Hint: Tips to refine your number system

• Add symbols for higher values to reduce repetition
• Introduce place value (just like Hindu numerals)
• Define clear rules for operations (+, –, ×, ÷)
• Possibly create shortcuts or grouping patterns (like grouping by 5s or 10s) to improve speed and consistency.

Page 62

Q1: Represent the following numbers in the Egyptian system: 10458, 1023, 2660, 784, 1111, 70707.

1. 10,458

Let’s break it down:

• 1 × 10,000 
• 4 × 100
• 5 × 10
• 8 × 1 

Representation:

2. 1023

Let’s break it down:

• 1 × 1,000
• 0 × 100
• 2 × 10 
• 3 × 1

Representation:

3. 2660

Let’s break it down:

• 2 × 1,000 
• 6 × 100 
• 6 × 10
• 0 × 1

Representation:

4. 784

Let’s break it down:

• 7 × 100 
• 8 × 10 
• 4 × 1

Representation:

5. 1111

Let’s break it down:

• 1 × 1,000
• 1 × 100
• 1 × 10
• 1 × 1 

Representation:

6. 70707

Let’s break it down:

• 7 × 10,000
• 0 × 1,000
• 7 × 100 
• 0 × 10
• 7 × 1

Representation:

Q2: What numbers do these numerals stand for?

Ans: Using the Egyptian Number System:

• 2 × 10² = 2 × 100 = 200
• 7 × 10 = 70
• 3 × 1 = 3

Total value: 200 + 70 + 3 = 273

Ans: Using the Egyptian Number System:

• 4 × 10³ = 4 × 1,000 = 5,000
• 3 × 10² = 3 × 100 = 300
• 1 × 10 = 10
• 2 × 1 = 2

Total = 4,000 + 300 + 10 + 2 = 4,312

Q: Express the number 143 in this new system.

Ans: Let us start grouping, starting with the size 5³ = 125, as this is the largest landmark number smaller than 143. We get—

143 = 125 + 5 + 5 + 5 + 1 + 1 + 1.

Using the standard symbols,

So the number 143 in the new system is:

Page 63Figure it Out

Q1. Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50, 137, 293, 651.

Ans: Representing in base-5 system

1. 15

• 25 is too big.
• Largest power: 5 (5¹), can use 3 times (5×3 = 15).
• 15 = 5+5+5 = 3×5
• Base-5 symbols: 

2. 50

• Largest power 5²=25, fits twice.
• 5×2=50; nothing left.
• Base-5 symbols:

3. 137

• Largest power 5³=125 fits once. 137−125 = 12
• Next, 5¹=5 fits twice. 12−10 = 2
• Next, 5⁰ = 1 fits twice.
• 137 = 125 + 5 + 5 + 1 + 1
• Base-5 symbols:

4. 293

• 5⁴ = 625 is too big.
• 5³ = 125 fits twice (125 × 2 = 250). 293 − 250 = 43
• 5² = 25 fits once. 43 − 25 = 18
• 5¹ = 5 fits three times. 18 − 15 = 3
• 5⁰ = 1 fits three times.
• 293 = 125 + 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1
• Base-5 symbols: 

5. 651

• 5⁴ = 625 fits once. 651 − 625 = 26
• 5² = 25 fits once. 26 − 25 = 1
• 5⁰ = 1 fits once.
• 651 = 625 + 25 + 1
• Base-5 symbols:

Q2. Is there a number that cannot be represented in our base-5 system above? Why or why not?

Ans: No, every whole number can be represented in base-5.
This is because base-5 is a positional numeral system, and like base-10, it can represent any non-negative integer using combinations of digits 0–4.

Q3. Compute the landmark numbers of a base-7 system. In general, what are the landmark numbers of a base-n system? The landmark numbers of a base-n number system are the powers of n starting from n⁰ = 1, n, n², n³, …

Ans:  Landmark numbers of base-7:

• 7⁰ = 1
• 7¹ = 7
• 7² = 49
• 7³ = 343
• 7⁴ = 2401

In general, the landmark numbers of a base-n system are:
1, n, n², n³, n⁴, …
That is, powers of n, starting from n⁰ = 1.

Page 65

Figure it Out

Q1. Add the following Egyptian numerals:

Ans: Try it yourself!

Q2. Add the following numerals that are in the base-5 system that we created:

Remember that in this system, 5 times a landmark number gives the next one!

Ans: Let’s convert this to numerals

First numeral (left side)

This is: 1 circle, 2 hexagons, 1 square, 2 triangles

Value:

• 1 × 125 = 125
• 2 × 25 = 50
• 1 × 5 = 5
• 2 × 1 = 2

Sum: 125 + 50 + 5 + 2 = 182

Second numeral (right side):

Value:

• 3 × 125 = 375
• 1 × 25 = 25
• 2 × 5 = 10
• 2 × 1 = 2

Sum: 375 + 25 + 10 + 2 = 412

Page 66

Q: How to multiply two numbers in Egyptian numerals?

Let us first consider the product of two landmark numbers.

1. What is any landmark number multiplied by (that is 10)? Find the following products—

(i) 10 × 10 = 100

100 =

(ii) 100 × 10= 1,000

1,000 =

(iii) 1,000 × 10 = 10,000

10,000 =

(iv) 10,000 × 10 = 100,000

100,000 = 

Each landmark number is a power of 10 and so multiplying it with10 increases the power by 1, which is the next landmark number.

2. What is any landmark number multiplied by (10²)? Find the following products—

(i) 10 × 100 = 1,000

1,000 =

(ii) 100 × 100 = 10,000

10,000 =

(iii) 1,000 × 100 = 100,000

100,000 =

(iv) 10,000 × 100 = 1,000,000

1,000,000 =

Each landmark number represents a power of 10, so multiplying it by 10² increases the power by 2, resulting in the landmark number that is two steps higher.

Page 67

Q: Find the following products—

Here are the computations for each part:

(i) 10 × 100,000 = 1,000,000

1,000,000 =

(ii) 100 × 1,000 = 100,000

100,000 = 

(iii) 1,000 × 1,000 = 1,000,000

1,000,000 = 

(iv) 10,000 × 1,000,000 = 10,000,000,000 = 10¹⁰

Thus, the product of any two landmark numbers is another landmark number!

Q: Does this property hold true in the base-5 system that we created? Does this hold for any number system with a base?

Ans: In any place-value system, each “landmark number” is just a power of the base:

• In base-10 → 10, 100, 1,000 are powers of 10
• In base-5 → 5, 25, 125 are powers of 5

When you multiply a landmark by the base, it just moves to the next bigger landmark.

Q: What can we conclude about the product of a number and (10), in the Egyptian system?

(i)

Ans: 

As these are numbers, the distributive law holds. So,

(ii)

Ans: We can expandas

Applying the distributive property 

Now find the following products—

Ans: Applying the distributive property 

Page 69-70

Figure it Out

Q1. Can there be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times? Why not?

Ans: No, you cannot have one symbol appear 10 or more times in a standard Egyptian numeral.

Reason:
Egyptian numerals use a form of additive system—each symbol represents a power of ten (1, 10, 100, 1,000, etc.). To write a number, you repeat the symbol as many times as needed (up to 9). But as soon as you reach 10 of a symbol, you replace it with a single symbol of the next higher value.

Example:

• Nine arches (10s) = 90
• Ten arches (10s) would be written as one spiral (100), not 10 arches.

Q2. Create your own number system of base 4, and represent numbers from 1 to 16.

Ans: Try it yourself!

Hint: Let’s assign easy-to-draw symbols to each digit (in place of usual 0, 1, 2, 3):

• 0: • (dot)
• 1: | (vertical line)
• 2: = (double line)
• 3: ∆ (triangle)

Base-4 has places: 4¹, 4⁰, etc.

Q3. Give a simple rule to multiply a given number by 5 in the base-5 system that we created.

Ans: Simple Rule: To multiply a number by 5 in base-5, add a zero to the right of the number (just as multiplying by 10 in decimal adds a zero).

Example:

• In base-5, 213₅ × 5 = 2130₅.
• In symbols (using previous base-5 system):Suppose ■ stands for the base-5 digit “1”, then this rule means you just add a new place with a 0 (the lowest symbol or blank).

Why? 
Because each shift to the left increases the place value by one power of 5, so the new number is five times as large.

Page 73Figure it Out

Q1. Represent the following numbers in the Mesopotamian system using—

(i) 63

(ii) 132 

(iii) 200 

(iv) 60

(v) 3605

Ans: 

(i) 63  

(ii) 132 

(iii) 200 

(iv) 60

(v) 3605

Q: Look at the representation of 60. What will be the representation for 3,600?

Ans:The representation for 3,600 is a single mark or symbol in the “2” diamond.

Page 76

Q: Represent the following numbers using the Mayan system:

(i) 77

(ii) 100 

(iii) 361 

(iv) 721

Ans: Try it Yourself!

Page 78

Q: Where does the Hindu/Indian number system figure in the evolution of ideas of number representation? What are its landmark numbers? And does it use a place value system?

Ans: The Hindu/Indian number system plays a crucial role in the evolution of number representation. It introduced two major innovations:

1. The concept of zero as a digit, and
2. A place value system based on base-10.

These ideas greatly simplified writing and calculating large numbers and influenced the development of modern numerals used worldwide today (often called Hindu-Arabic numerals).

Landmark numbers in this system are powers of 10, such as:

• 1 (10⁰)
• 10 (10¹)
• 100 (10²)
• 1,000 (10³), and so on.

Yes, the Hindu number system uses a place value system, meaning the position of a digit determines its value based on powers of 10. For example, in the number 345, the digit 3 represents 300 (3 × 100), 4 represents 40 (4 × 10), and 5 represents 5 (5 × 1).

This system laid the foundation for modern arithmetic and digital computation.

​Page 80Figure it Out

Q1. Why do you think the Chinese alternated between the Zong and Heng symbols? If only the Zong symbols were to be used, how would 41 be represented? Could this numeral be interpreted in any other way if there is no significant space between two successive positions?

Ans: Using Zong and Heng symbols:

• The Chinese number system used Zong (vertical) and Heng (horizontal) symbols to show place value clearly.
• They alternated the direction of the symbols at each place (units, tens, hundreds, etc.) to avoid confusion when reading the number.
• This made it easier to know which digit belonged to which place even when spaces were small or missing.In Chinese system: 41 = 4 tens and 1 unit.

Using only Zong symbols, it would be written as:
(Zong for 4) followed by (Zong for 1) → looks like: IIII I
Without alternating the symbol direction or keeping proper spacing, IIIII could be misread as 5 (i.e., 1 five) instead of 41.

The lack of direction or spacing removes the clue that one part is “tens” and the other is “units”.

Q2. Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the digits. Compare this system with that of the Gumulgal’s.

Ans: To form a base-2 place value system using ‘ukasar’ and ‘urapon’, we assign:
• ‘ukasar’ = 0
• ‘urapon’ = 1

This system works just like the binary number system, where each position from right to left represents increasing powers of 2. For example:
• The first place is 2⁰ = 1
• The next is 2¹ = 2
• Then 2² = 4 and so on.

So, we can represent numbers like this

A group of indigenous people in Australia called the Gumulgal had the following words for their numbers.

In the Gumulgal System, we can name every number, but in the binary system, only the numbers shown in the table have names.

Q3. Where in your daily lives, and in which professions, do the Hindu numerals, and 0, play an important role? How might our lives have been different if our number system and 0 hadn’t been invented or conceived of?

Ans: Hindu numerals Usage 

• ​Hindu numerals and the digit zero are used in everyday life, such as telling time, counting and managing money, reading prices, doing mathematics in school, writing phone numbers, etc. 
• Professions like banking, teaching, engineering, and science rely heavily on this number system.
• Zero is especially important because it helps in representing place value and making large numbers easy to write and understand

Without zero and the Hindu numeral system:

1. Basic calculations would be very difficult.
2. Trading and writing dates would be challenging
3. Technology like computers and calculators wouldn’t exist
4. Progress in all fields would be much slower

Q4. The ancient Indians likely used base 10 for the Hindu number system because humans have 10 fingers, and so we can use our fingers to count. But what if we had only 8 fingers? How would we be writing numbers then? What would the Hindu numerals look like if we were using base 8 instead? Base 5? Try writing the base-10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write it in base-2?

Ans: If we had only 8 fingers, we might have used base-8. Similarly, with 5 fingers, we could have used base-5.

Let’s convert the base-10 number 25 into different bases:

1. Base-8:
25 ÷ 8 = 3 remainder 1
3 ÷ 8 = 0 remainder 3
So, 25 in base-8 = 31

2. Base-5:
25 ÷ 5 = 5 remainder 0
5 ÷ 5 = 1 remainder 0
1 ÷ 5 = 0 remainder 1
So, 25 in base-5 = 100

3. Base-2:
25 ÷ 2 = 12 remainder 1
12 ÷ 2 = 6 remainder 0
6 ÷ 2 = 3 remainder 0
3 ÷ 2 = 1 remainder 1
1 ÷ 2 = 0 remainder 1
So, 25 in base-2 = 11001

In base-8 or base-5 systems, the Hindu numerals would have included only the digits needed (0 to 7 for base-8, and 0 to 4 for base-5).

Page 22

Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number v. 

(i) 10v (ii) 10 + v (iii) 2 × 10 × v (iv) 2¹⁰ (v) 2¹⁰v (vi) 10² v

Answer: 

The correct expression for the thickness of a sheet of paper after it is folded 10 times is (v) 2¹⁰v.

When a sheet of paper is folded, its thickness doubles with each fold. This is a form of exponential growth, not linear growth.

The process can be broken down as follows:

• Initial thickness: v
• After 1 fold: The paper has 2 layers, so the thickness is 2 × v, or 2¹v.
• After 2 folds: The paper is folded again, doubling the layers to 4. The thickness becomes 4 × v, or 2²v.
• After 3 folds: The thickness doubles again to 8 times the original, or 2³v.

Following this pattern, the thickness after ‘n’ folds is given by the formula:
Total Thickness = 2ⁿ × v

For 10 folds, you substitute n = 10 into the formula:
Total Thickness = 2¹⁰v

The other options are incorrect because they represent linear relationships, whereas the folding process is exponential. For instance, 10v would imply the thickness only increases by the original amount with each fold, rather than doubling the total current thickness

What is (– 1)⁵ ? Is it positive or negative? What about (– 1)⁵⁶? 

Answer:

• The expression (–1)⁵ equals –1, which is a negative number. When a negative number is raised to an odd exponent, the result is always negative.
  The calculation is: (–1) × (–1) × (–1) × (–1) × (–1) = –1.
• What about (–1)⁵⁶?
  The expression (–1)⁵⁶ equals +1, which is a positive number. When a negative number is raised to an even exponent, the result is always positive. This happens because the negative signs are multiplied an even number of times, causing them to cancel each other out in pairs.

Is (– 2)⁴ = 16? Verify.

Answer:

Yes, the statement (–2)⁴ = 16 is correct. To verify this, you multiply –2 by itself four times:

(–2) × (–2) × (–2) × (–2) = (4) × (–2) × (–2) = (–8) × (–2) = 16.

As with the previous example, raising the negative base (–2) to an even power (4) results in a positive number.

Figure it Out 

1. Express the following in exponential form:
(i) 6 × 6 × 6 × 6                                (ii) y × y 
(iii) b × b × b × b                             (iv) 5 × 5 × 7 × 7 × 7 
(v) 2 × 2 × a × a                              (vi) a × a × a × c × c × c × c × d

Answers:
(i) 6⁴
(ii) y² 
(iii) b⁴
(iv) 5² × 7³
(v) 2² × a²
(vi) a³ × c⁴ × d

2. Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648      (ii) 405           (iii) 540                (iv) 3600

Answers:
(i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³ × 3⁴
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5 
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5 
(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5²

3. Write the numerical value of each of the following:
(i) 2 × 10³       (ii) 7² × 2³          (iii) 3 × 4⁴         (iv) (– 3)² × (– 5)²
(v) 3² × 10⁴     (vi) (– 2)⁵ × (– 10)⁶

Answers:
(i) 2 × 10³ = 2 × 1000 = 2000 
(ii) 7² × 2³ = 49 × 8 = 392 
(iii) 3 × 4⁴ = 3 × 256 = 768 
(iv) (– 3)² × (– 5)² = 9 × 25 = 225 
(v) 3² × 10⁴ = 9 × 10000 = 90000 
(vi) (– 2)⁵ × (– 10)⁶ = – 32 × 1000000 = – 32000000

Page 24

Q. Use this observation to compute the following. (i) 2⁹ (ii) 5⁷ (iii) 4⁶

^Answer:

• (i) 2⁹
  Using the rule, this can be expressed as a product of powers. For example, since 9 = 4 + 5, we can write:
  2⁹ = 2⁴⁺⁵ = 2⁴ × 2⁵
  The final value is calculated by multiplying 2 by itself 9 times:
  2⁹ = 512
• (ii) 5⁷
  This can be expressed using the same logic. For example, since 7 = 3 + 4, we have:
  5⁷ = 5³⁺⁴ = 5³ × 5⁴
  The final value is calculated by multiplying 5 by itself 7 times:
  5⁷ = 78,125
• (iii) 4⁶
  This expression can be broken down as well. For example, since 6 = 3 + 3, we have:
  4⁶ = 4³⁺³ = 4³ × 4³
  The final value is calculated by multiplying 4 by itself 6 times:
  4⁶ = 4,096

Q. Write the following expressions as a power of a power in at least two different ways:
(i) 8⁶ (ii) 7¹⁵ (iii) 9¹⁴ (iv) 5⁸

Answers:
(i) 8⁶ = (8²)³ = (8³)² 
(ii) 7¹⁵ = (7³)⁵ = (7⁵)³
(iii) 9¹⁴ = (9²)⁷ = (9⁷)²
(iv) 5⁸ = (5²)⁴ = (5⁴)²

Page 25

Q. In the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles every day in this pond. After 30 days, the pond is completely covered with lotuses. On which day was the pond half full?

Answer: 

• The number of lotuses doubles daily.
• On day 30, the pond is fully covered.
• Since the lotuses double every day, the day before (day 29), the pond must have been half full.
• This is because doubling the lotuses from day 29 to day 30 makes the pond fully covered.
  The pond was half full on day 29.

Q. Write the number of lotuses (in exponential form) when the pond was — 

(i) fully covered (ii) half covered

Answer: 

Let’s assume we start with 1 lotus on day 1.

The number of lotuses doubles each day, so:

• On day 1: 1 lotus
• On day 2: 1 × 2 = 2 lotuses
• On day 3: 2 × 2 = 4 lotuses
• On day 4: 4 × 2 = 8 lotuses
• And so on.

This pattern shows the number of lotuses on day “n” is 2^(n-1).

For day 30 (fully covered):

• Number of lotuses = 2^(30-1) = 2²⁹.

For day 29 (half covered):

• Number of lotuses = 2^(29-1) = 2²⁸.
• (i) Fully covered (day 30): 2²⁹ lotuses
• (ii) Half covered (day 29): 2²⁸ lotuses

Q. There is another pond in which the number of lotuses triples every day. When both the ponds had no flowers, Damayanti placed a lotus in the doubling pond. After 4 days, she took all the lotuses from there and put them in the tripling pond. How many lotuses will be in the tripling pond after 4 more days?

Answer: 
In the first pond (Doubling Pond), the number of lotuses double every day, so for the first 4 days it doubles every day. 
So, after the first 4 days, the number of lotuses is 1 × 2 × 2 × 2 × 2 = 2⁴. 
In the second pond (Tripling Pond), the number of lotuses triple every day, so for the next four days, they triple every day.
So, after the next 4 days, the number of lotuses is 2⁴ × 3 × 3 × 3 × 3 = 2⁴ × 3⁴

^{Q. What if Damayanti had changed the order in which she placed the f lowers in the lakes? How many lotuses would be there?}

Answer: 

• Suppose she placed 1 lotus in the tripling pond first, for 4 days:1×3⁴
• Then moved it to the doubling pond for 4 days:3⁴×2⁴= (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2)

By regrouping it, this can be expressed as:

(3 x 2) x (3 x 2) x (3 x 2) x (3 x 2) = (3 x 2)⁴ = 6⁴

Q. Use this observation to compute the value of 2⁵ × 5⁵.

Answer:
2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵ = 100000

Q. Simplify  and write it in exponential form.

Answer: 

Look at the Expression:
This means:

Group the Terms:
You can pair each 10 in the numerator with a 5 in the denominator:

=105×105×105×105

Simplify Each Pair:

=2×2×2×2=2⁴

Page 27

Q. What is 2¹⁰⁰ ÷ 2²⁵ in powers of 2?

Answer: 2¹⁰⁰ ÷ 2²⁵ = 2^{(100 – 25)} = 2⁷⁵

Page 29

Q. We had required a and b to be counting numbers. Can a and b be any integers? Will the generalised forms still hold true?

Answer:
The general forms you identified, known as the laws of exponents, were initially observed for counting numbers (positive integers), but they do indeed hold true when the exponents a and b are any integers (positive, negative, or zero).

Let’s verify the two main rules you’re asking about with integer exponents.

1. Product of Powers Rule: nᵃ × nᵇ = nᵃ⁺ᵇ

nᵃ × nᵇ = nᵃ⁺ᵇ

This rule states that when you multiply powers with the same base, you add the exponents. Let’s test it with a negative exponent.

• Example: Consider 3⁵ × 3⁻².
• Using the definition of a negative exponent:
  3⁻² is the same as 1 / 3². So the expression is 3⁵ × (1/3²) = 3⁵ / 3².
  This means (3×3×3×3×3) / (3×3) = 3³ = 27.
• Using the generalized rule:
  We add the exponents: 3⁵⁺⁽⁻²⁾ = 3³ = 27.

As you can see, both methods yield the same result. The rule works perfectly with integers.

2. Power of a Power Rule: (nᵃ)ᵇ = nᵃᵇ

(nᵃ)ᵇ = nᵃᵇ

This rule states that to raise a power to another power, you multiply the exponents. Let’s test this with a negative exponent as well.

• Example: Consider (4²)⁻³.
• Using the definition of a negative exponent:
  The expression means 1 / (4²)³.
  This is 1 / (4² × 4² × 4²) = 1 / 4²⁺²⁺² = 1 / 4⁶.
• Using the generalized rule:
  We multiply the exponents: 4²ˣ⁽⁻³⁾ = 4⁻⁶.
  Since 4⁻⁶ is the same as 1 / 4⁶, the results match.

These rules hold true because of the mathematical definitions for zero and negative exponents:

• Zero Exponent: n⁰ = 1
• Negative Exponent: n⁻ᵃ = 1 / nᵃ

Q. Write equivalent forms of the following.
(i) 2^{– 4}    (ii) 10^{– 5}         (iii) (– 7)^–2          (iv) (– 5)^{– 3}        (v) 10^{– 100}

Answers:
(i) 2^{– 4} = 1/2⁴
(ii) 10^{– 5} = 1/10⁵
(iii) (– 7)^–2 = 1/(– 7)²
(iv) (– 5)^{– 3} = 1/(– 5)³
(v) 10^{– 100} = 1/10¹⁰⁰

Q. Simplify and write the answers in exponential form.
(i) 2^{– 4} × 2⁷    (ii) 3² × 3^{– 5} × 3⁶         (iii) p³ × p^–10         (iv) 2⁴ × (– 4) ^{– 2}         (v) 8^p × 8^q

Answers:
(i) 2^{– 4} × 2⁷ = 2^{(–4 + 7)} = 2³
(ii) 3² × 3^{– 5} × 3⁶ = 3^{(2 – 5 + 6)} = 3³
(iii) p³ × p^–10 = p^{(3 – 10)} = p^–7
(iv) 2⁴ × (– 4) ^{– 2} = 2⁴ × 1/(–4)² = 2⁴ × 1/16 = 16 × 1/16 = 1 = 2⁰ (or 4⁰) 
(v) 8^p × 8^q = 8^{(p + q)}

Page 30

Q. Can we say that 16384 (4⁷) is 16 (4²) times larger than 1,024 (4⁵)? 

Answer: Yes, since 4⁷ ÷ 4⁵ = 4^(7-5) = 4². 

Q. How many times larger than 4^–2 is 4²

Answer: 4² ÷ 4^-2 = 4 ^(2-(-2)) = 4⁽²⁺²⁾ = 4⁴

So, 4² is 4⁴ larger than 4^–2

Q. Use the power line for 7 to answer the following questions.

Answer: 

• 2,401 × 49 = ?
  • 2,401 is 7⁴ and 49 is 7².
  • 7⁴ × 7² = 7⁴⁺² = 7⁶
  • From the power line, 7⁶ is 117,649.
• 49³ = ?
  • 49 is 7².
  • (7²)³ = 7²ˣ³ = 7⁶
  • From the power line, 7⁶ is 117,649.
• 343 × 2,401 = ?
  • 343 is 7³ and 2,401 is 7⁴.
  • 7³ × 7⁴ = 7³⁺⁴ = 7⁷
  • From the power line, 7⁷ is 823,543.
• 16,807 / 49 = ?
  • 16,807 is 7⁵ and 49 is 7².
  • 7⁵ / 7² = 7⁵⁻² = 7³
  • From the power line, 7³ is 343.
• 7 / 343 = ?
  • 7 is 7¹ and 343 is 7³.
  • 7¹ / 7³ = 7¹⁻³ = 7⁻²
  • From the power line, 7⁻² is 1/49.
• 16,807 / 8,23,543 = ?
  • 16,807 is 7⁵ and 8,23,543 is 7⁷.
  • 7⁵ / 7⁷ = 7⁵⁻⁷ = 7⁻²
  • From the power line, 7⁻² is 1/49.
• 1,17,649 × (1 / 343) = ?
  • 1,17,649 is 7⁶ and 1/343 is 7⁻³.
  • 7⁶ × 7⁻³ = 7⁶⁻³ = 7³
  • From the power line, 7³ is 343.
• (1 / 343) × (1 / 343) = ?
  • 1/343 is 7⁻³.
  • 7⁻³ × 7⁻³ = 7⁻³⁻³ = 7⁻⁶
  • Since 7⁶ is 117,649, then 7⁻⁶ is 1/117,649.

Powers of 10

Q. Write these numbers in the same way: (i) 172, (ii) 5642, (iii) 6374.

Answer: 

(i) 172
This number can be broken down by place value: 1 hundred, 7 tens, and 2 ones.
172 = (1 × 10²) + (7 × 10¹) + (2 × 10⁰)

(ii) 5642
This number is composed of 5 thousands, 6 hundreds, 4 tens, and 2 ones.
5642 = (5 × 10³) + (6 × 10²) + (4 × 10¹) + (2 × 10⁰)

(iii) 6374
This number is composed of 6 thousands, 3 hundreds, 7 tens, and 4 ones.
6374 = (6 × 10³) + (3 × 10²) + (7 × 10¹) + (4 × 10⁰)

Page 31 

Scientific NotationQ. Write the large-number facts we read just before in this form (scientific notation).
(i) The Sun is located 30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy. 
(ii) The number of stars in our galaxy is 1,00,00,00,00,000. 
(iii) The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg.

Answers:
(i) 3 × 10²² m (ii) 1 × 10¹¹ stars (iii) 5.976 × 10²⁴ kg

Page 32

Q. Can you say which of the three distances is the smallest?

Answer:

• The distance between the Sun and Saturn is 14,33,50,00,00,000 m = 1.4335 × 10¹² m.

• The distance between Saturn and Uranus is 14,39,00,00,00,000 m = 1.439 × 10¹² m.

• The distance between the Sun and Earth is 1,49,60,00,00,000 m = 1.496 × 10¹¹ m.

Compare and see which one has the least power of 10.

•Sun to Saturn: 1.4335 × 10¹² m -> 12

•Saturn to Uranus: 1.439 × 10¹² m -> 12

•Sun to Earth: 1.496 × 10¹¹ m -> 11

So, the distance between Sun and Earth is the smallest.

Q. Express the following numbers in standard form.
(i) 59,853      (ii) 65,950        (iii) 34,30,000         (iv) 70,04,00,00,000

Answers:
(i) 59,853 = 5.9853 × 10⁴
(ii) 65,950 = 6.595 × 10⁴
(iii) 34,30,000 = 3.43 × 10⁶
(iv) 70,04,00,00,000 = 7.004 × 10¹⁰

Page 38

Q. Calculate and write the answer using scientific notation:
(i) How many ants are there for every human in the world? 
(ii) If a flock of starlings contains 10,000 birds, how many flocks could there be in the world?

Answer:
(i) Global human population as of 2025 is 8.2 arab/8.2 billion (8.2 × 10⁹). 
Estimated population of ants globally is 20 padma/20 quadrillion (2 × 10¹⁶). 
Number of ants per human = (2 × 10¹⁶) / (8.2 × 10⁹) = (2 / 8.2) × 10(16^-9) ≈ 0.2439 × 10⁷ = 2.439 × 10⁶ ants per human. 
(ii) The estimated global population of starlings is around 1.3 arab/1.3 billion (1.3 × 10⁹). 
If a flock contains 10,000 birds (10⁴ birds). 
Number of flocks = (1.3 × 10⁹) / 10⁴ = 1.3 × 10^(9-4) = 1.3 × 10⁵ flocks.

Page 39

(iii) If each tree had about 10⁴ leaves, find the total number of leaves on all the trees in the world.

Answer:
The estimated number of trees (2023) globally stands at 30 kharab/3 trillion (3 × 10¹²). Total number of leaves = (3 × 10¹² trees) × (10⁴ leaves/tree) = 3 × 10⁽¹²⁺⁴⁾
= 3 × 10¹⁶ leaves.

(iv) If you stacked sheets of paper on top of each other, how many would you need to reach the Moon?

Answer:
Distance to the Moon is approximately 3,84,400 km = 3.844 × 10⁸ m. 
Thickness of one sheet of paper is 0.001 cm = 1 × 10^-5 m. 
Number of sheets = (3.844 × 108 m) / (1 × 10-5 m/sheet) = 3.844 × 10^{(8 – (-5))} = 3.844 

Page 40

Q. Think of some events or phenomena whose time is of the order of 
(i) 10⁵ seconds and (ii) 10⁶ seconds. Write them in scientific notation.

Answers:
(i) 10⁵ seconds ≈ 1.16 days. Example: A short trip, like a weekend getaway. 
(ii) 10⁶ seconds ≈ 11.57 days. Example: A two-week vacation.

Page 41

A fossil of Kelenken Guillermoi, a type of terror bird, is dated to 15 million years ago ( ≈_______________ seconds).

Answer:
15 million years = 15 × 10⁶ years. 
1 year ≈ 3.1536 × 10⁷ seconds. 
15 × 10⁶ years × 3.1536 × 10⁷ seconds/year ≈ 47.304 × 10¹³ seconds 
= 4.7304 × 10¹⁴ seconds.

Page 42

Plants on land started 47 crore/470 million years ago ( ≈ _______________ seconds).

Answer:
470 million years = 470 × 10⁶ years = 4.7 × 10⁸ years. 
1 year ≈ 3.1536 × 10⁷ seconds. 4.7 × 10⁸ years × 3.1536 × 10⁷ seconds/year 
≈ 14.822 × 10¹⁵ seconds = 1.4822 × 10¹⁶ seconds.

Q. Calculate and write the answer using scientific notation:
(i) If one star is counted every second, how long would it take to count all the stars in the universe? Answer in terms of the number of seconds using scientific notation. 
(ii) If one could drink a glass of water (200 ml) every 10 seconds, how long would it take to finish the entire volume of water on Earth?

Answers:
(i) The estimated number of stars in the observable universe is 2 × 10²³. Time to count = 2 × 10²³ seconds. 
(ii) Estimated number of drops of water on Earth is 2 × 10²⁵ drops (assuming 16 drops per millilitre). 
Volume of water on Earth = (2 × 10²⁵ drops) / (16 drops/ml) = 0.125 × 10²⁵ ml = 1.25 × 10²⁴ ml. 
Volume of one glass = 200 ml. 
Number of glasses = (1.25 × 10²⁴ ml) / (200 ml/glass) = 0.00625 × 10²⁴ glasses = 6.25 × 1021 glasses. 
Time to finish = (6.25 × 10²¹ glasses) × (10 seconds/glass) = 6.25 × 10²² seconds.

Figure it Out (Page 44)

Q1. Find out the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]

Answer:
2²²⁴ ÷ 4³² = 2²²⁴ ÷ (2²)³² = 2²²⁴ ÷ 2⁶⁴ = 2^(224-64) = 2¹⁶⁰. 
To find the units digit of 2¹⁶⁰, observe the pattern of units digits of powers of 2: 
2¹ = 2 
2² = 4 
2³ = 8 
2⁴ = 16 (units digit is 6) 
2⁵ = 32 (units digit is 2) 
The pattern of units digits is 2, 4, 8, 6, and it repeats every 4 powers. 
Divide the exponent 160 by 4: 160 ÷ 4 = 40 with a remainder of 0. 
A remainder of 0 means the units digit is the same as the 4th power in the cycle, which is 6. So, the units digit in the value of 2²²⁴ ÷ 4³² is 6.

Q2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?

Answer:
Initial bottles = 5 Bottles added per day = 5 (since a new container with 5 bottles is brought in) 
Total bottles after 40 days = Initial bottles + (Bottles added per day × Number of days) Total bottles = 5 + (5 × 40) = 5 + 200 = 205 bottles.

Q3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 64³ (ii) 192⁸ (iii) 32^–5

Answers:

(i) 64³

First, note that the base 64 can be written as 2⁶, 4³, or 8². Using the power of a power rule (nᵃ)ᵇ = nᵃᵇ, we find that 64³ = (2⁶)³ = 2¹⁸. We can now express 2¹⁸ in different ways.

• Way 1: By splitting the exponent into a sum (18 = 10 + 8):
  2¹⁰ × 2⁸
• Way 2: By changing the base to 4 (since 2² = 4):
  (2²)⁹ = 4⁹
• Way 3: By changing the base to 8 (since 2³ = 8):
  (2³)⁶ = 8⁶

(ii) 192⁸

First, find the prime factors of 192, which are 2⁶ × 3. Therefore, 192⁸ = (2⁶ × 3)⁸. Using the exponent rules, we can write this in several ways.

• Way 1: By distributing the exponent to each factor inside the parenthesis:
  (2⁶)⁸ × 3⁸ = 2⁴⁸ × 3⁸
• Way 2: By changing the base of the first term:
  (2²)²⁴ × 3⁸ = 4²⁴ × 3⁸
• Way 3: By grouping common exponents after splitting a power:
  2⁴⁰ × 2⁸ × 3⁸ = 2⁴⁰ × (2 × 3)⁸ = 2⁴⁰ × 6⁸

(iii) 32⁻⁵

First, recognize that 32 = 2⁵. Using the power of a power rule, 32⁻⁵ = (2⁵)⁻⁵ = 2⁻²⁵. This can be expressed in different forms.

• Way 1: By splitting the negative exponent into a sum (-25 = -10 + -15):
  2⁻¹⁰ × 2⁻¹⁵
• Way 2: By splitting the exponent into a sum of a negative and a positive integer (-25 = -30 + 5):
  2⁻³⁰ × 2⁵
• Way 3: By rearranging the exponents using the power of a power rule:
  (2⁻⁵)⁵ = (1/32)⁵

Q4. Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) Cube numbers are also square numbers. 
(ii) Fourth powers are also square numbers.
(iii) The fifth power of a number is divisible by the cube of that number. 
(iv) The product of two cube numbers is a cube number. 
(v) q46 is both a 4th power and a 6th power (q is a prime number).

Answers: 
(i) Cube numbers are also square numbers.

Only Sometimes True

Reasoning: A number n³ is a cube number. For it to also be a square number, it must be possible to write it as m² for some integer m.

• When it is true: Consider the cube number 64. 64 = 4³, and it is also a square number because 64 = 8². This works because the base number (4) is itself a perfect square (2²). If we take any base that is a square number, say k², its cube will be (k²)³ = k⁶. This can be rewritten as (k³)², which is a perfect square.
• When it is false: Consider the cube number 8. 8 = 2³, but 8 is not a perfect square. Similarly, 27 = 3³ is not a perfect square. This happens whenever the base is not a perfect square.

Since the statement is true for some numbers (like 1, 64, 729) but false for others (like 8, 27, 125), it is only sometimes true.

(ii) Fourth powers are also square numbers.

Always True

Reasoning: A fourth power of a number n is n⁴. Using the laws of exponents, we can rewrite this expression:
n⁴ = n²⁺² = n² × n² = (n²)²
Since n² is an integer (if n is an integer), (n²)² is the square of an integer. Therefore, any fourth power is also a perfect square.

• Example: 3⁴ = 81, which is 9² (and 9 = 3²).
• Example: 5⁴ = 625, which is 25² (and 25 = 5²).

(iii) The fifth power of a number is divisible by the cube of that number.

Always True

Reasoning: Let the number be n. Its fifth power is n⁵ and its cube is n³. To check for divisibility, we divide n⁵ by n³. Using the quotient rule for exponents (nᵃ / nᵇ = nᵃ⁻ᵇ):
n⁵ / n³ = n⁵⁻³ = n²
For any non-zero integer n, the result n² is also an integer. Since the division results in an integer, n⁵ is always divisible by n³.

(iv) The product of two cube numbers is a cube number.

Always True

Reasoning: Let the two cube numbers be a³ and b³. Their product is a³ × b³. Using the laws of exponents, we can combine this into a single term:
a³ × b³ = (a × b)³
Since a and b are integers, their product (a × b) is also an integer. Therefore, (a × b)³ is, by definition, a cube number.

Example: 8 × 27 = 2³ × 3³ = (2 × 3)³ = 6³ = 216.

(v) q⁴⁶ is both a 4th power and a 6th power (q is a prime number).

Never True

Reasoning: For an expression xᵃ to be a perfect bth power, the exponent a must be a multiple of b.

Is it a 4th power? For q⁴⁶ to be a 4th power, 46 must be divisible by 4. Since 46 ÷ 4 = 11.5, which is not an integer, q⁴⁶ is not a 4th power.

Is it a 6th power? For q⁴⁶ to be a 6th power, 46 must be divisible by 6. Since 46 ÷ 6 ≈ 7.67, which is not an integer, q⁴⁶ is not a 6th power.

Since q⁴⁶ is neither a 4th power nor a 6th power, it can’t possibly be both. Therefore, the statement is never true.

5. Simplify and write these in the exponential form.

Answer: 

• (i) 10⁻² × 10⁻⁵
  When multiplying powers with the same base, you add the exponents (nᵃ × nᵇ = nᵃ⁺ᵇ).
  10⁻² × 10⁻⁵ = 10⁻²⁺⁽⁻⁵⁾ = 10⁻⁷
• (ii) 5⁷ ÷ 5⁴
  When dividing powers with the same base, you subtract the exponents (nᵃ ÷ nᵇ = nᵃ⁻ᵇ).
  5⁷ ÷ 5⁴ = 5⁷⁻⁴ = 5³
• (iii) 9⁻⁷ ÷ 9⁴
  Using the same division rule as above:
  9⁻⁷ ÷ 9⁴ = 9⁻⁷⁻⁴ = 9⁻¹¹
• (iv) (13⁻²)⁻³
  To raise a power to another power, you multiply the exponents ((nᵃ)ᵇ = nᵃᵇ).
  (13⁻²)⁻³ = 13⁽⁻²⁾ˣ⁽⁻³⁾ = 13⁶
• (v) (m⁵n¹²)/(mn)⁹
  Assuming the expression is a fraction, first distribute the exponent in the denominator, and then apply the division rule for each base.
  1. Distribute the exponent: (m⁵n¹²)/(m⁹n⁹)
  2. Subtract exponents for each base: m⁵⁻⁹ n¹²⁻⁹
  3. Simplify: m⁻⁴n³

6. If 12² = 144 what is (i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²

Answer: 

• (i) (1.2)²
  This can be written as (12 × 10⁻¹)² = 12² × (10⁻¹)² = 144 × 10⁻² = 1.44
• (ii) (0.12)²
  This is (12 × 10⁻²)² = 12² × (10⁻²)² = 144 × 10⁻⁴ = 0.0144
• (iii) (0.012)²
  This is (12 × 10⁻³)² = 12² × (10⁻³)² = 144 × 10⁻⁶ = 0.000144
• (iv) 120²
  This can be written as (12 × 10¹)² = 12² × (10¹)² = 144 × 10² = 14,400

Figure it Out (Page 45)

7. Circle the numbers that are the same—

Answer: 

Explanation: 

Here is the simplification of each expression:

1. 2⁴ × 3⁶
   This expression is already in its simplest prime factor form.
2. 6⁴ × 3²
   First, express 6 as its prime factors (2 × 3).
   = (2 × 3)⁴ × 3²
   = (2⁴ × 3⁴) × 3²
   = 2⁴ × 3⁽⁴⁺²⁾
   = 2⁴ × 3⁶
3. 6¹⁰
   Express 6 as its prime factors (2 × 3).
   = (2 × 3)¹⁰
   = 2¹⁰ × 3¹⁰
4. 18² × 6²
   Express 18 (2 × 3²) and 6 (2 × 3) as their prime factors.
   = (2 × 3²)² × (2 × 3)²
   = (2² × 3⁴) × (2² × 3²)
   = 2⁽²⁺²⁾ × 3⁽⁴⁺²⁾
   = 2⁴ × 3⁶
5. 6²⁴
   Express 6 as its prime factors (2 × 3).
   = (2 × 3)²⁴
   = 2²⁴ × 3²⁴

Q. 8. Identify the greater number in each of the following— (i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰
Answer:
(i) To compare 4³ and 3⁴, we calculate their values.
4³ = 4 × 4 × 4 = 64.
3⁴ = 3 × 3 × 3 × 3 = 81.
Since 81 is greater than 64, 3⁴ is the greater number.

(ii) To compare 2⁸ and 8², we calculate their values.
2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256.
8² = 8 × 8 = 64.
Since 256 is greater than 64, 2⁸ is the greater number.

(iii) To compare 100² and 2¹⁰⁰, we can evaluate or estimate their values.
100² = 100 × 100 = 10,000.
2¹⁰⁰ can be written as (2¹⁰)¹⁰. Since 2¹⁰ = 1024, 2¹⁰⁰ = (1024)¹⁰.
Clearly, (1024)¹⁰ is a vastly larger number than 10,000. Therefore, 2¹⁰⁰ is the greater number.

Q. 9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?
Answer:
The code should consist of 10 digits.
The dairy needs to generate unique codes for 8.5 billion (8.5 × 10⁹) packets. Using the digits 0–9 provides 10 options for each position in the code. A code with ‘n’ digits can generate 10ⁿ unique combinations. We need to find the smallest integer ‘n’ where 10ⁿ is greater than or equal to 8.5 × 10⁹.

• If n = 9, 10⁹ = 1 billion, which is not enough codes.
• If n = 10, 10¹⁰ = 10 billion, which is more than 8.5 billion and can therefore provide a unique code for each packet.

Q. 10. 64 is a square number (8²) and a cube number (4³). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
Answer:
Yes, there are other numbers that are both perfect squares and perfect cubes. Such numbers can be described in general as perfect sixth powers.
A number that is a square has prime factors with even exponents, and a number that is a cube has prime factors with exponents that are multiples of three. For a number to be both, its prime factors’ exponents must be multiples of both 2 and 3, which means they must be multiples of 6.
Therefore, any number of the form n⁶, where ‘n’ is an integer, will be both a square and a cube.

• 1 (since 1⁶ = 1, which is 1² and 1³)
• 64 (since 2⁶ = 64, which is 8² and 4³)
• 729 (since 3⁶ = 729, which is 27² and 9³)

Q. 11. A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?
Answer:
There are 60,466,176 possible codes.
This is calculated based on the following assumptions:

• The code has a fixed length of 5 characters.
• “Alphanumeric” includes the 10 digits (0–9) and the 26 uppercase letters of the English alphabet (A–Z), as shown in the examples. This gives a total of 36 possible characters for each position.
• Each position in the code is independent.

The total number of combinations is found by raising the number of character choices to the power of the code length:
Total codes = 36 × 36 × 36 × 36 × 36 = 36⁵ = 60,466,176.

Q. 12. The worldwide population of sheep (2024) is about 10⁹, and that of goats is also about the same. What is the total population of sheep and goats? (i) 2⁰⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹
Answer:
The correct expressions for the total population are (v) 2 × 10⁹ and (vi) 10⁹ + 10⁹.
The calculation is:
Total Population = (Sheep Population) + (Goat Population)
Total Population = 10⁹ + 10⁹
This sum can be simplified as 2 × (10⁹). Both expressions represent the same value, which is 2 billion.

Q. 13. Calculate and write the answer in scientific notation:
Answer:

(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
Total clothing = (8 × 10⁹ people) × 30 pieces/person = 240 × 10⁹ = 2.4 × 10¹¹ pieces of clothing.

(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
Total honeybees = (100 × 10⁶ colonies) × (50,000 bees/colony) = (1 × 10⁸) × (5 × 10⁴) = 5 × 10¹² honeybees.

(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
Total bacteria = (38 × 10¹² cells/person) × (8 × 10⁹ people) = (3.8 × 10¹³) × (8 × 10⁹) = 30.4 × 10²² = 3.04 × 10²³ bacterial cells.

(iv) Total time spent eating in a lifetime in seconds.
(Note: This assumes an average lifespan of 75 years and 1.5 hours spent eating per day.)
Total seconds = (75 years) × (365 days/year) × (1.5 hours/day) × (3600 seconds/hour) ≈ 148,000,000 seconds = 1.48 × 10⁸ seconds.

Q. 14. What was the date 1 arab/1 billion seconds ago?
Answer:
Assuming the current date is August 11, 2025, the date 1 billion (10⁹) seconds ago was December 4, 1993.
Here is the calculation:

• 1 billion seconds is equal to approximately 11,574 days (1,000,000,000 ÷ 86,400 seconds/day).
• Going back 11,574 days from August 11, 2025, lands on December 4, 1993. This calculation accounts for the exact number of days in each month and includes all leap years in the period (1996, 2000, 2004, 2008, 2012, 2016, 2020, and 2024).

Patterns and Properties of Perfect Squares

Page 4

Q. Find the squares of the first 30 natural numbers and fill in the table below.
Answer:Q. What patterns do you notice? Share your observations and make conjectures.

Answer:

The squares of the first 30 natural numbers are:

• 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100
• 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225, 16² = 256, 17² = 289, 18² = 324, 19² = 361, 20² = 400
• 21² = 441, 22² = 484, 23² = 529, 24² = 576, 25² = 625, 26² = 676, 27² = 729, 28² = 784, 29² = 841, 30² = 900
• Observation: The units digits of these square numbers are only 0, 1, 4, 5, 6, or 9. None of them end in 2, 3, 7, or 8.

Q. If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?

Answer: No. For example, the number 26 ends in 6, but it is not a perfect square. Just looking at the units digit is not enough to confirm if a number is a square, but it can tell us if a number is not a square.

Q. Write 5 numbers such that you can determine by looking at their units digit that they are not squares.

Answer: Any number ending in 2, 3, 7, or 8 is not a perfect square. Five examples are: 12, 33, 47, 58, and 102.

Page 5

Q. Which of the following numbers have the digit 6 in the units place?
(i) 38² (ii) 34² (iii) 46² (iv) 56² (v) 74² (vi) 82²

Answer:

To find the units digit (that is, the last digit) of the square of a number, we only need to look at the units digit of the original number, because squaring affects the last digit in a predictable way.

Here is a table showing what happens when we square numbers ending in each digit from 0 to 9:

From this table, we observe that:

A number’s square ends in 6 if the number itself ends in 4 or 6.
Checking Each Option:

1. 38²: The number ends in 8 → 8² ends in 4 → does not end in 6
2. 34²: The number ends in 4 → 4² ends in 6 → ends in 6
3. 46²: The number ends in 6 → 6² ends in 6 → ends in 6
4. 56²: The number ends in 6 → 6² ends in 6 → ends in 6
5. 74²: The number ends in 4 → 4² ends in 6 → ends in 6
6. 82²: The number ends in 2 → 2² ends in 4 → does not end in 6

Q. If a number contains 3 zeros at the end, how many zeros will its square have at the end?

Answer: Its square will have 6 zeros at the end. The number of zeros at the end of a square is always double the number of zeros at the end of the original number.

Q. What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?

Answer: The number of zeros at the end of a square is always double the number of zeros in the original number. This will always happen. Yes, we can say that perfect squares can only have an even number of zeros at the end.

Q. What can you say about the parity of a number and its square?

Answer: The square of an even number is always even. The square of an odd number is always odd.

Page 7

Q. Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?

Answer: Between n² and (n+1)², there are 2n non-perfect square numbers. For example, between 2² (4) and 3² (9), there are 2² = 4 numbers (5, 6, 7, 8). Between 3² (9) and 4² (16), there are 2³ = 6 numbers (10, 11, 12, 13, 14, 15).

Q. How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000?

Answer:

The largest square less than 1000 is 31² = 961.

Q. Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.

Answer:The sum of two consecutive triangular numbers is a perfect square.

• 1 + 3 = 4 = 2²
• 3 + 6 = 9 = 3²
• 6 + 10 = 16 = 4²
• The next term would be 

10 + 15 = 25 = 5²

Page 10: Figure it Out

1. Which of the following numbers are not perfect squares?
(i) 2032 (ii) 2048 (iii) 1027 (iv) 1089

Answer:

• (i) 2032 ends in 2. Not a perfect square.
• (ii) 2048 ends in 8. Not a perfect square.
• (iii) 1027 ends in 7. Not a perfect square.
• (iv) 1089 might be. lets find out:
  We know, 30²=900, so, 1089 is very close to 900, lets list the squares of the next numbers.
  31² = 961
  32² =  1024
  33² = 1089. So, 1089 is a perfect square.
• So, (i), (ii), and (iii) are not perfect squares.

2. Which one among 64², 108², 292², 36² has last digit 4?

Answer: A square has the last digit 4 if the original number’s last digit is 2 or 8.
Therefore, 108² and 292² will have the last digit 4.

3. Given 125² = 15625, what is the value of 126²?

(i)   15625 + 126  (ii) 15625 + 26²   (iii) 15625 + 253  (iv) 15625 + 251  (v) 15625 + 51²

Answer: From the question we know that, 125² = 15625. That means 15625 is the sum of 125 consecutive natural numbers. To find 126², we need to find the 126^th odd number and add it with 125^th odd number that is 15625.

126^th odd number = (2 x 126) -1 = 251.

Therefore, 126² = 15625 + 251 = 15876.

4. Find the length of the side of a square whose area is 441 m².

Answer: The length is √441. We know 20²=400 and the number ends in 1, so the root must end in 1 or 9. Trying 21, we get 21² = 441. The length is 21 m.

5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.

Answer: First, find the LCM of 4, 9, and 10.

• 4 = 2², 9 = 3², 10 = 2 × 5.
• LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180.
• The prime factorization of 180 is 2 × 2 × 3 × 3 × 5. To make it a perfect square, all prime factors must be in pairs. The factor 5 is not paired.
• We must multiply by 5: 180 × 5 = 900.
• The smallest square number is 900.

6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.

Answer:

Prime factorization of 9408: 

9408 = 2 × 4704 

= 2 × 2 × 2352  

= 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7.

In pairs: (2×2) × (2×2) × (2×2) × (7×7) × 3. 

The factor 3 is unpaired.

The smallest number to multiply by is 3.

The new number is 9408 × 3 = 28224.

The square root is 2 × 2 × 2 × 7 × 3 = 168.

7. How many numbers lie between the squares of the following numbers?
​(i) 16 and 17 (ii) 99 and 100

Answer: There are 2n numbers between n² and (n+1)².

• (i) Between 16² and 17²: 2 × 16 = 32 numbers.
• (ii) Between 99² and 100²: 2 × 99 = 198 numbers.

8. In the following pattern, fill in the missing numbers:

Answer: The pattern is a² + b² + (ab)² = (ab+1)².

9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.

Answer: There are 1025 tiny squares

1025 = 5 × 5 × 41

Page 12

Q. Complete the table below.

Answer:

Q. What patterns do you notice in the table above?

Answer: The last digit of a cube can be any digit from 0 to 9.

• Cube of a number ending in 1 ends in 1.
• Cube of a number ending in 2 ends in 8.
• Cube of a number ending in 3 ends in 7.
• Cube of a number ending in 4 ends in 4.
• Cube of a number ending in 5 ends in 5.
• Cube of a number ending in 6 ends in 6.
• Cube of a number ending in 7 ends in 3.
• Cube of a number ending in 8 ends in 2.
• Cube of a number ending in 9 ends in 9.
• Cube of a number ending in 0 ends in 0.

Q. We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are the possible last digits of cubes?

Answer: All digits from 0 to 9 are possible last digits for cubes.

Page 13

Q. Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?

Answer:

• 1-digit cubes: 1³, 2³ (1, 8) -> 2 cubes.
• 2-digit cubes: 3³, 4³ (27, 64) -> 2 cubes.
• 3-digit cubes: 5³, 6³, 7³, 8³, 9³ (125, 216, 343, 512, 729) -> 5 cubes.
• ​Observation: The number of cubes in a given range of digits is not as regular as squares.

Q. Can a cube end with exactly two zeroes (00)? Explain.

Answer: No. For a number to end in zero, it must be a multiple of 10. The cube of a multiple of 10 (like 10, 20, 30) will have a number of zeros that is a multiple of 3. For example, 10³ = 1000 (3 zeros), 20³ = 8000 (3 zeros). It is impossible for a perfect cube to end in exactly two zeros.

Q. The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each of these can be expressed as the sum of two positive cubes.

Answer:

• 4104: 4104 = 2³ + 16³ = 8 + 4096. Also, 4104 = 9³ + 15³ = 729 + 3375.
• 13832: 13832 = 2³ + 24³ = 8 + 13824. Also, 13832 = 18³ + 20³ = 5832 + 8000.

Page 14

Q. Can you tell what this sum is without doing the calculation?
91+93 +95 + 97 + 99 + 101 + 103 + 105 + 107 + 109.

Answer: This is a series of 10 consecutive odd numbers. The sum of n consecutive odd numbers starting from the correct term gives n³. The sum shown is 10³ that is, 1000.

Page 15

Q. Find the cube roots of these numbers: 
(i) ³√64              (ii) ³√512                 (iii) ³√729

Answer:

(i) ³√64

Step 1: Prime factorisation

64 = 2 × 2 × 2 × 2 × 2 × 2

Step 2: Group the factors into triplets

(2 × 2 × 2) × (2 × 2 × 2) 
cube root = 2 × 2 = 4

(ii) ³√512

Step 1: Prime factorisation

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Step 2: Group the factors into triplets

(2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) 

cube root = 2 × 2 × 2 = 8

(iii) ³√729

Step 1: Prime factorisation

729 = 3 × 3 × 3 × 3 × 3 × 3

Step 2: Group the factors into triplets

(3 × 3 × 3) × (3 × 3 × 3) 
cube root = 3 × 3 = 9

Page 16: Figure it Out

1. Find the cube roots of 27000 and 10648.

Answer:

(i) ³√27000

Step 1: Prime factorisation

27000 = 27 × 1000

  = (3 × 3 × 3) × (2 × 2 × 2 × 5 × 5 × 5)

Step 2: Group the prime factors into triplets

(3 × 3 × 3) × (2 × 2 × 2) × (5 × 5 × 5) = (3³) × (2³) × (5³)

cube root = 3 × 2 × 5 = 30

(ii) ³√10648

Step 1: Prime factorisation

10648 = 2 × 2 × 2 × 11 × 11 × 11 = 2³ × 11³

Step 2: Group the prime factors into triplets

(2 × 2 × 2) × (11 × 11 × 11)

cube root = 2 × 11 = 22

2. What number will you multiply by 1323 to make it a cube number?

Answer: Prime factorization of 1323 = 3 × 441 = 3 × 21² = 3 × (3×7)² = 3 × 3² × 7² = 3³ × 7².

• The factor 3 is a triplet, but 7 is only a pair. We need one more 7 to make it a triplet.
• You must multiply by 7.

3. State true or false. Explain your reasoning.
(i) The cube of any odd number is even.

False. 

When you cube an odd number, you multiply it by itself three times:
For example:

• 3³ = 3 × 3 × 3 = 27 (odd)
• 5³ = 5 × 5 × 5 = 125 (odd)

Rule:

• Odd × Odd = Odd
• So, Odd × Odd × Odd = Odd

Thus, the cube of an odd number is also odd, not even.

(ii) There is no perfect cube that ends with 8.
False. 2³ = 8, and any number ending in 2 will have a cube ending in 8 (e.g., 12³ = 1728).

(iii) The cube of a 2-digit number may be a 3-digit number.
False. The smallest 2-digit number is 10, and 10³ = 1000 (a 4-digit number). 
All other 2-digit cubes will be larger. 

(iv) The cube of a 2-digit number may have seven or more digits.
False. The largest 2-digit number is 99. 99³ = 970299 (a 6-digit number). So a 2-digit number cannot have 7 or more digits

(v) Cube numbers have an odd number of factors.
False. This is true for square numbers. For a number to have an odd number of factors, it must be a perfect square. Some cube numbers are also perfect squares (e.g., 64 = 8² = 4³), and these will have an odd number of factors. But most cubes (like 8, 27) are not perfect squares and have an even number of factors.

4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

Answer:

• 1331: Ends in 1, so root ends in 1. We know 10³=1000 and 20³=8000. 1331 lies between 1000 and 8000, and between 10 and 20 only 11 is a number that ends with 1. So, the root is 11.
• 4913: Ends in 3, so root ends in 7. We know 10³=1000 and 20³=8000. 4913 lies between 1000 and 8000, and between 10 and 20 only 17 is a number that ends with 1. So, the root is 17.
• 12167: Ends in 7, so root ends in 3. We know, 20³=8000 and 30³=27000. 12167 lies between 8000 and 27000. So, cube root of 12167 lies between 20 and 30. Only 23 is a number that ends with 3. So, the root is 23.
• 32768: Ends in 8, so root ends in 2. We know, 30³=27000 and 40³=64000. 32768 lies between these numbers, so cube root of 32768 lies between 30 and 40. 32 is the only number that ends with 2. So, the root is 32.

Page 17

5. Which of the following is the greatest? Explain your reasoning.
(i) 67³ – 66³               (ii) 43³ – 42³              (iii) 67² – 66²              (iv) 43² – 42²

Answer:

• We know n² – (n-1)² = 2n – 1.
• We know n³ – (n-1)³ = 3n² – 3n + 1.
• (iii) 67² – 66² = 2(67) – 1 = 133.
• (iv) 43² – 42² = 2(43) – 1 = 85.
• (i) 67³ – 66³ = 3(67)² – 3(67) + 1 = 3(4489) – 201 + 1 = 13467 – 200 = 13267.
• (ii) 43³ – 42³ = 3(43)² – 3(43) + 1 = 3(1849) – 129 + 1 = 5547 – 128 = 5419.
• Comparing the results, (i) 67³ – 66³ is the greatest.

Page 18: It’s Puzzle Time!

Look at the following numbers: 3    6    10    15    1 

They are arranged such that each pair of adjacent numbers adds up to a square.

Q. Try arranging the numbers 1 to 17 (without repetition) in a row in a similar way — the sum of every adjacent pair of numbers should be a square. Can you arrange them in more than one way? If not, can you explain why?

Answer:

Step 1: What sums are allowed?

Perfect squares less than or equal to 34 (because 17 + 16 = 33) are:

4,9,16,254, 9, 16, 25

So, any two neighbours in the arrangement must add up to 4, 9, 16, or 25.

Step 2: Try to find which numbers go together

List pairs of numbers from 1 to 17 whose sum is one of those perfect squares:

Sum = 4:

• 1 + 3

Sum = 9:

• 1 + 8
• 2 + 7
• 3 + 6
• 4 + 5

Sum = 16:

• 1 + 15
• 2 + 14
• 3 + 13
• 4 + 12
• 5 + 11
• 6 + 10
• 7 + 9
• 8 + 8 → not allowed (same number twice)

Sum = 25:

• 8 + 17
• 9 + 16
• 10 + 15
• 11 + 14
• 12 + 13

This tells us which numbers can be next to each other.

Step 3: Use logic and trial to arrange

Now, we try to connect these numbers step by step. After careful trial and checking, this arrangement works:

16,9,7,2,14,11,5,4,12,13,3,6,10,15,1,8,17

Let’s check that each pair adds up to a perfect square:

• 16 + 9 = 25
• 9 + 7 = 16
• 7 + 2 = 9
• 2 + 14 = 16
• 14 + 11 = 25
• 11 + 5 = 16
• 5 + 4 = 9
• 4 + 12 = 16
• 12 + 13 = 25
• 13 + 3 = 16
• 3 + 6 = 9
• 6 + 10 = 16
• 10 + 15 = 25
• 15 + 1 = 16
• 1 + 8 = 9
• 8 + 17 = 25

All pairs are correct, and every number from 1 to 17 is used exactly once.

Final Answer:

Yes, it is possible to arrange the numbers from 1 to 17 in this way. One such arrangement is:

16,9,7,2,14,11,5,4,12,13,3,6,10,15,1,8,17

No, we cannot arrange them in more than one way.