Page No. 81
Example 1: Shabnam is 3 years older than Aftab. When Aftab’s age 10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, what will Shabnam’s age be? _______
Ans: Shabnam’s age will be 18 + 3 = 21 years
Use this expression to find A’ftab’s age if Shabnam’s age is 20.
Solution:
According to the expression,
Aftab’s age = Shabnam’s Age – 3
∴ Aftab’s age = 20 – 3 = 17 years
Page No. 82
Q: Given the age of Shabnam, write an expression to find Aftab’s age.
Ans: We know that Aftab is 3 years younger than Shabnam. So, Aftab’s age will be 3 less than Shabnam’s. This can be described as
Aftab’s age = Shabnam’s age – 3.
If we again use the letter a to denote Aftab’s age and the letter s to denote Shabnam’s age, then the algebraic expression would be: a = s – 3, meaning 3 less than s.
Q: Use this expression to find Aftab’s age if Shabnam’s age is 20.
Ans: If Shabnam’s age = 20,
Using above expression,
Aftab’s age = Shabnam’s age – 3.
= 20 – 3 = 17
Page No. 83
Example 3: Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is ₹35 and the price of 1 kg jaggery is ₹60.
How much should she pay if she buys 10 coconuts and 5 kg jaggery?
Ans: Cost of 10 coconuts = 10 × ₹35
Cost of 5 kg jaggery = 5 × ₹60
Total cost = 10 × ₹35 + 5 × ₹60 = ₹350 + ₹300 = ₹650.
How much should she pay if she buys 8 coconuts and 9 kg jaggery?
Ans: Cost of 8 coconuts = 8 × ₹35
Cost of 9 kg jaggery = 9 × ₹60
Total cost = 8 × ₹35 + 9 × ₹60 = ₹280 + ₹540 = ₹820.
Q: Use this expression: c × 35 + j × 60, to find the total amount to be paid for 7 coconuts and 4 kg jaggery.
Here, ‘c’ represents the number of coconuts and ‘j’ represents the number of kgs of jaggery.
Ans: Substitute c = 7 and j = 4 into the expression:
c × 35 + j × 60 = 7 × 35 + 4 × 60Now calculate:7 × 35 = 2454 × 60 = 240Now add them:245 + 240 = 485₹485 is the total amount to be paid. Page No. 84 & 85 Figure it OutQ1: Write formulas for the perimeter of:(a) triangle with all sides equal. (b) a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all side lengths and angle measures are equal) (c) a regular hexagonAns:(a) triangle with all sides equal.Perimeter = 3 × s, where s is the length of each side.(b) a regular pentagonPerimeter = 5 × s, where s is the length of each side.(c) a regular hexagonPerimeter = 6 × s, where s is the length of each side.Q2: Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe.Ans: Combined length = 20 + k, where k is the length of the other pipe in meters.Q3: What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table:
Ans:
Q4: Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially?(a) 10 + 8 + y(b) (10 + 8) × y(c) 10 × 8 × y(d) 10 + 8 × y(e) 10 × y + 8Ans: (d) 10 + 8 × yExplanation: The total time to grind y kg of grain consists of two parts:The initial 10 seconds for the roller mill to start running. After that, it takes 8 seconds for each kg of grain, so for y kg, it takes 8 × y seconds.Thus, the total time taken is: 10 + 8 × ySo the correct answer is (d) 10 + 8 × y.Q5: Write algebraic expressions using letters of your choice.(a) 5 more than a numberAns: Let the number be represented by x. The expression for 5 more than the number is: x + 5(b) 4 less than a numberAns: Let the number be represented by x. The expression for 4 less than the number is: x – 4(c) 2 less than 13 times a numberAns: Let the number be represented by x. The expression for 2 less than 13 times the number is: 13x – 2(d) 13 less than 2 times a numberAns: Let the number be represented by x. The expression for 13 less than 2 times the number is: 2x – 13Q6: Describe situations corresponding to the following algebraic expressions:(a) 8 × x + 3 × yAns: A shop sells pens at 8 rupees each and notebooks at 3 rupees each. If x pens and y notebooks are bought, the total cost is 8x + 3y rupees.(b) 15 × j – 2 × kAns: A person earns 15 rupees per hour for j hours of work but pays 2 rupees per k units of electricity used. The net income is 15j – 2k rupees.Q7: In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.
Ans:
Page No. 87 Mind the Mistake, Mend the MistakeSome simplifications are shown below where the letter-numbers are replaced by numbers and the value of the expression is obtained.1. Observe each of them and identify if there is a mistake. 2. If you think there is a mistake, try to explain what might have gone wrong. 
3. Then, correct it and give the value of the expression. Ans:
Page No. 88 & 89
Example 5: Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.Let us first find the money earned by the sale of pencils.Q: The money earned by selling pencils on Day 1 is 5c. Similarly, the money earned by selling pencils on Day 2 is _____, and Day 3 is ______.Ans: On Day 2, the money earned = 3c.On Day 3, the money earned = 10c.Q2: If c = ₹50, find the total amount earned by the scale of pencils which is 18c. Ans: 18c = 18 × 50 = ₹900Q3: Write the expression for the total money earned by selling erasers. Then, simplify the expression.Ans: Total money earned by selling erasers will be 4d + 6d + d = 11dThe expression for the total money earned by selling erasers during these three days is: 18c + 11d.Q4: Can the expression 18c + 11d be simplified further? Ans: No, the expression cannot be simplified further because it contains different letter-numbers. It is already in its simplest form.In this problem, we saw the expression 5c + 3c + 10c getting simplified to the expression 18c. Q5: Check that both expressions take the same value when c is replaced by different numbers.Ans: We can check by substituting different values for c and comparing both expressions.For example:If c = 10, then the expression 5c + 3c + 10c simplifies to 5(10) + 3(10) + 10(10) = 50 + 30 + 100 = 180, which is equivalent to 18c = 18(10) = 180.This confirms that both expressions take the same value when c is replaced by different numbers.Page No. 91, 92 & 93Q: Could we have written the initial expression: (40x + 75y) – (6x + 10y) as (40x + 75y) + (– 6x – 10y)?Ans: Yes, the expression (40x + 75y) + (– 6x – 10y) is equivalent to the original expression.We can simplify it as follows: (40x + 75y) + (–6x – 10y) = (40x + 75y) – (6x + 10y)Thus, both forms represent the same expression.Example 8: Charu has been through three rounds of a quiz. Her scores in the three rounds are 7p – 3q, 8p – 4q, and 6p – 2q. Here, p represents the score for a correct answer and q represents the penalty for an incorrect answer. What do each of the expressions mean?Ans: If the score for a correct answer is 4 (p = 4) and the penalty for a wrong answer is 1 (q = 1), find Charu’s score in the first round. Charu’s score is 7 × 4 – 3 × 1. We can evaluate this expression by writing it as a sum of terms. 7 × 4 – 3 × 1 = 7 × 4 + – 3 × 1 = 28 + – 3 = 25 What are her scores in the second and third rounds? What if there is no penalty? What will be the value of q in that situation? What is her final score after the three rounds?Ans: Her final score will be the sum of the three scores: (7p – 3q) + (8p – 4q) + (6p – 2q). Since the terms can be added in any order, we can remove the brackets and write 7p + – 3q + 8p + – 4q + 6p + –2q = 7p + 8p + 6p + – (3q) + – (4q) + – (2q) (by swapping and grouping) = (7 + 8 + 6)p + – (3 + 4 + 2)q = 21p + – 9q = 21p – 9q. Charu’s total score after three rounds is 21p – 9q. Her friend Krishita’s score after three rounds is 23p – 7q.Q: Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q.Ans: To find some possible scores for Krishita, we need to break down the expression 23p – 7q into three parts (scores for each round). Here are some possible combinations:Round 1: 10p, Round 2: 8p – 3q, Round 3: 5p – 4qThese add up as 10p + (8p – 3q) + (5p – 4q) = 23p – 7q.Round 1: 15p, Round 2: 5p – 2q, Round 3: 3p – 5qThese add up as 15p + (5p – 2q) + (3p – 5q) = 23p – 7q.Round 1: 18p, Round 2: 4p – q, Round 3: p – 6qThese add up as 18p + (4p – q) + (p – 6q) = 23p – 7q.Q3: Can we say who scored more? Can you explain why? How much more has Krishita scored than Charu? This can be found by finding the difference between the two scores. 23p – 7q – (21p – 9q)Ans: To find how much more Krishita has scored than Charu, we subtract Charu’s score from Krishita’s score:23p – 7q – (21p – 9q)Now, simplify the expression:= 23p – 7q – 21p + 9q= (23p – 21p) + (9q – 7q)= 2p + 2q = = 2(p + q)So, Krishita has scored 2(p + q) marks more than Charu.Q4: Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take.
If the expressions 5u and 5 + u are equal, then they should take the same values for any given value of u. But we can see that they do not. So, these two expressions are not equal.Ans:
Q6: Are the expressions 10y – 3 and 10(y – 3) equal?10y – 3, short for 10 × y – 3, means 3 less than 10 times y, 10(y – 3), short for 10 × (y – 3), means 10 times (3 less than y). Are the expressions 10y – 3 and 10(y – 3) equal? 10y – 3, short for 10 × y – 3, means 3 less than 10 times y, 10(y – 3), short for 10 × (y – 3), means 10 times (3 less than y).Let us compare the values that these expressions take for different values of y. 
Ans:
Q7: After filling in the two diagrams, do you think the two expressions are equal?Ans: We see that the values of 10y – 3 and 10(y – 3) are not equal for different values of y. So, the expressions 10y – 3 and 10(y – 3) are not equal.Figure it OutQ1: Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing. 
Ans: Hence, we see that while we add the expressions in different ways but the sums are always the same.Page No. 94Q2: Simplify each of the following expressions:(a) Simplify the expression: p + p + p + p, p + p + p + q, p + q + p – qAns:p + p + p + p = 4pp + p + p + q = 3p + qp + q + p – q = 2pSo, the simplified expressions are 4p, 3p + q and 2p.(b) Simplify the expression: p – q + p – q, p + q – p + qAns:p – q + p – q = (p + p) – (q + q) = 2p – 2qp + q – p + q = (p – p) + (q + q) = 0 + 2q = 2qSo, the simplified expressions are 2p – 2q and 2q.(c) Simplify the expression: p + q – (p + q), p – q – p – qAns:p + q – (p + q) = (p + q) – (p + q) = 0p – q – p – q = (p – p) – (q + q) = 0 – 2q = –2qSo, the simplified expressions are 0 and –2q.(d) Simplify the expression: 2d – d – d – d, 2d – d – d – cAns:2d – d – d – d = 2d – (d + d + d) = 2d – 3d = –d2d – d – d – c = (2d – d – d) – c = (2d – 2d) – c = 0 – c = –cSo, the simplified expressions are –d and –c.(e) Simplify the expression: 2d – d – (d – c), 2d – (d – d) – cAns:2d – d – (d – c) = (2d – d) – (d – c) = d – d + c = 0 + c = c2d – (d – d) – c = 2d – (0) – c = 2d – cSo, the simplified expressions are c and 2d – c.(f) Simplify the expression: 2d – d – c – cAns:2d – d – c – c = (2d – d) – (c + c) = d – 2cSo, the simplified expression is d – 2c.Mind the Mistake, Mend the MistakeSome simplifications of algebraic expressions are done below. The expression on the right-hand side should be in its simplest form. • Observe each of them and see if there is a mistake. • If you think there is a mistake, try to explain what might have gone wrong. • Then, simplify it correctly.
Ans: Q3: Take a look at all the corrected simplest forms (i.e. brackets are removed, like terms are added, and terms with only numbers are also added). Is there any relation between the number of terms and the number of letter-numbers these expressions have?Ans: YesIf the expression contains a term having only a number,the number of terms = number of letter-numbers + 1If an expression has no term that has only numbers,then number of terms = number of letter-numbersPage No. 95, 96 & 97Q1: Find out the formula of this number machine.
The formula for the number machine above is “two times the first number minus the second number”. When written as an algebraic expression, the formula is 2a – b. The expression for the first set of inputs is 2 × 5 – 2 = 8. Check that the formula holds true for each set of inputs.
Ans: Yes, the formula holds for each set of inputs.As, 2 × 8 – 1 = 15; 2 × 9 – 11 = 7; 2 × 10 – 10 = 10; and 2 × 6 – 4 = 8Q2: Find the formulas of the number machines below and write the expression for each set of inputs.
Ans: The formula for the number machines in the first row is “sum of first number and second number minus two,” and the expression is a + b – 2.The expression for each set of inputs is:5 + 2 – 2 = 5, 8 + 1 – 2 = 7, 9 + 11 – 2 = 18, 10 + 10 – 2 = 18, and a + b – 2.The formula for the number machines in the second row is “product of first number and second number plus one,” and the expression is a × b + 1.The expression for each set of inputs is:4 × 1 + 1 = 5, 6 × 0 + 1 = 1, 3 × 2 + 1 = 7, 10 × 3 + 1 = 31, and a × b + 1 = ab + 1.
Q3: Now, make a formula on your own. Write a few number machines as examples using that formula. Challenge your classmates to figure it out!Ans: Do it Yourself.
Example: Somjit noticed a repeating pattern along the border of a saree.
Q: Use the table to find what design appears at positions 99, 122, and 148.Ans: Using the table and the remainder rule:Position 99: Remainder is 0 (99 ÷ 3 = 33 remainder 0). Design is C.Position 122: Remainder is 2 (122 ÷ 3 = 40 remainder 2). Design is B.Position 148: Remainder is 1 (148 ÷ 3 = 49 remainder 1). Design is A.Page No. 98
Let us extend the numbers in the calendar beyond 30, creating endless rows.Q1: Will the diagonal sums be equal in every 2×2 square in this endless grid? How can we be sure?Ans: Yes, the diagonal sums will be equal in every 2 × 2 square in this endless grid.We can be sure because we checked a 2 × 2 square with the top left number as ‘a’. The numbers in the square are:Top left: aTop right: a + 1Bottom left: a + 7Bottom right: a + 8The diagonal sums are:First diagonal (a + (a + 8)) = 2a + 8Second diagonal ((a + 1) + (a + 7)) = 2a + 8Both sums are the same (2a + 8), and this works for any value of ‘a’. So, the diagonal sums are always equal in any 2 × 2 square.Q2: Given that we know the top left number, how do we find the other numbers in this 2 × 2 square?Ans: If the top left number is ‘a’, we can find the other numbers in the 2×2 square like this:Top right number: a + 1 (1 more than a)Bottom left number: a + 7 (7 more than a)Bottom right number: a + 8 (8 more than a, or diagonal to a)So, the 2×2 square looks like:Top left: aTop right: a + 1Bottom left: a + 7Bottom right: a + 8Page No. 99Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’.Ans: Let a = 2, then 
Here, the diagonal sums are 2 + 10 = 12 and 9 + 3 = 12And 2a + 8 = 2 × 2 + 8 = 12Hence, the diagonal sum is equal to 2a + 8.Consider a set of numbers from the calendar (having endless rows) forming under the following shape:
Q1: Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe?Ans: First, let’s find the sum of the numbers in the given shape: 8, 14, 15, 16, 22.Sum = 8 + 14 + 15 + 16 + 22 = 75.The number in the centre is 15. Compare the sum with the centre: 75 ÷ 15 = 5. The sum is 5 times the centre number.Now, let’s repeat this for another set of numbers with the same shape from the calendar. Let’s take the set with centre 8 (numbers around it: 1, 7, 9, 15).
Sum = 1 + 7 + 8 + 9 + 15 = 40.The centre is 8. Compare the sum with the centre: 40 ÷ 8 = 5.The sum is again 5 times the centre number.Observation: The sum of the numbers in this shape is always 5 times the number in the centre.Q2: Will this always happen? How do you show this?Ans: Yes, this will always happen.To show this, let’s take the centre number as ‘a’. The numbers around it in the shape will be:Top: a – 7 (7 less than a, since it’s the previous row)Left: a – 1 (1 less than a, since it’s the previous column)Right: a + 1 (1 more than a, since it’s the next column)Bottom: a + 7 (7 more than a, since it’s the next row)So the numbers are: (a – 7), (a – 1), a, (a + 1), (a + 7).Sum = (a – 7) + (a – 1) + a + (a + 1) + (a + 7)= a – 7 + a – 1 + a + a + 1 + a + 7= (a + a + a + a + a) + (–7 + 7) + (–1 + 1)= 5a + 0 + 0= 5aThe sum is 5a, which is 5 times the centre number ‘a’. This works for any centre number ‘a’, so the sum is always 5 times the centre number.Q: Find other shapes for which the sum of the numbers within the figure is always a multiple of one of the numbers.Ans: Do it Yourself.Page No. 100Matchstick Patterns
Q1: How many matchsticks will there be in Step 33, Step 84, and Step 108? Of course, we can draw and count, but is there a quicker way to find the answers using the pattern present here?Ans: Observed Pattern:Step 1: 3 matchsticks (1 triangle)Step 2: 5 matchsticks (2 triangles) → +2 from previousStep 3: 7 matchsticks (3 triangles) → +2 from previous…Each new step adds 2 matchsticks.General Formula:The number of matchsticks (M) for Step n follows an arithmetic sequence:M = 3 + 2(n−1)Simplified:M = 2n + 1(Where n = Step Number)Calculations:Step 33:M = 2 × 33 + 1 = 66 + 1 = 67Step 84:M = 2 × 84 + 1 = 168 + 1 = 169Step 108:M = 2 × 108 + 1 = 216 + 1 = 217Page No. 101 & 102
There is a different way to count, or see the pattern. Let us take a look at the picture again.Q1: What are these numbers in Step 3 and Step 4?Ans: In step 3, there are 3 matchsticks placed horizontally and 4 matchsticks placed diagonally.In step 4, there are 4 matchsticks placed horizontally and 5 matchsticks placed diagonallyQ2: How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1?Ans: Horizontal: Always 2 matchsticks.Diagonal: 2y – 1 (Step 1: 1, Step 2: 3, Step 3: 5, Step 4: 7, so 2y – 1).Total = 2 + (2y – 1) = 2y + 1. Yes, they add up to 2y + 1.Figure it OutFor the problems asking you to find suitable expression(s), first try to understand the relationship between the different quantities in the situation described. If required, assume some values for the unknowns and try to find the relationship.Q1: One plate of Jowar roti costs 30 and one plate of Pulao costs 20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day?(a) 30x + 20y(b) (30 + 20) × (x + y)(c) 20x + 30y(d) (30 + 20) × x + y(e) 30x – 20yAns: (a) 30x + 20yExplanation: Cost of one plate of Jowar roti = ₹ 30∴ Cost of x plate of Jowar roti = 30xCost of one plate of Pulao = ₹ 20∴ Cost of y plate of Pulao = 20ySo, the expression for the total amount earned that day = 30x + 20yQ2: Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day?(a) p + q + r(b) p + q + 2r(c) 2 × (p + q + r)(d) p + q + r + 2(e) p + q + r + 1(f) 2 × (p + q)Ans: (a) p + q + rExplanation: Number of customers who bought only champak = pNumber of customers who bought only marigold = qNumber of customers who bought both = rAs Pushpita gave away a tiny national flag to every customer.So, the number of flags she gives away that day = p + q + r.Q3: A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.(a) Write an expression describing how far away the snail is from its starting position.Ans: During the day, the snail climbs up ‘u’ cm.During the night snail slips down d cm.So, the net distance covered in one day is u-d.So, in 10 days and 10 nights, the net distance covered by the snail = 10(u-d).Hence, the expression describing how far away the snail is from it starting position is 10(u – d) cm.(b) What can we say about the snail’s movement if d > u?Ans: If d > u, the snail moves downward overall, as it slips more than it climbs each day-night cycle.Q4: Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?Ans: Week 1: 5 km/day × 7 days = 35 kmWeek 2: (5 + z) km/day × 7 days = 7(5 + z) = 35 + 7z kmWeek 3: (5 + 2z) km/day × 7 days = 7(5 + 2z) = 35 + 14z kmTotal = 35 + (35 + 7z) + (35 + 14z) = 105 + 21z kmQ5: In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.
Ans: Top left: (w + 2) → (-5) → (w – 3) → (×3) → 3w – 9.Bottom left: (w + 2) → (-8) → (w – 6) → (-4) → (w – 10).Bottom right: (w + 2) → (+3) → (w + 5) → (×4) → (3w – 6). (which is not possible, there is an error in this part so cannot give the answer).Page No. 103Q6: A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations.(a) If t = 4, what is the time taken to travel from Yahapur to Vahapur?Ans: The train from Yahapur to Vahapur stops at 3 stations, and stops for 2 minutes at every station.Time taken in travelling = 4tAt t = 4, time taken in travelling = 4 × 4 = 16 minutesTime taken during stoppages = 3 × 2 = 6 minutesSo, the time taken to travel from Yahapur to Vahapur = 16 + 6 = 22 minutes(b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur?Ans: Let the time taken to travel from one station to another station = tSo, time taken to travel from Yahanpur to Vahapur = 4tAs there are three stoppages between these two stations, and the train stops for 2 minutes at each stoppage,therefore total time taken during stoppages = 2 × 3 = 6 minutesSo, the algebraic expression for total time taken is 4t + 6.Q7: Simplify the following expressions:(a) 3a + 9b – 6 + 8a – 4b – 7a + 16Ans: (3a + 8a – 7a) + (9b – 4b) + (-6 + 16) = 4a + 5b + 10(b) 3(3a – 3b) – 8a – 4b – 16Ans: 9a – 9b – 8a – 4b – 16 = (9a – 8a) + (-9b – 4b) – 16 = a – 13b – 16(c) 2(2x – 3) + 8x + 12Ans: 4x – 6 + 8x + 12 = (4x + 8x) + (-6 + 12) = 12x + 6(d) 8x – (2x – 3) + 12Ans: 8x – 2x + 3 + 12 = (8x – 2x) + (3 + 12) = 6x + 15(e) 8h – (5 + 7h) + 9Ans: 8h – 5 – 7h + 9 = (8h – 7h) + (-5 + 9) = h + 4(f) 23 + 4(6m – 3n) – 8n – 3m – 18Ans: 23 + 24m – 12n – 8n – 3m – 18 = (24m – 3m) + (-12n – 8n) + (23 – 18) = 21m – 20n + 5Q8: Add the expressions given below:(a) 4d – 7c + 9 and 8c – 11 + 9dAns: (4d + 9d) + (-7c + 8c) + (9 – 11) = 13d + c – 2(b) -6f + 19 – 8s and -23 + 13f + 12sAns: (-6f + 13f) + (-8s + 12s) + (19 – 23) = 7f + 4s – 4(c) 8d – 14c + 9 and 16c – (11 + 9d)Ans: 8d – 14c + 9 + 16c – 11 – 9d = (8d – 9d) + (-14c + 16c) + (9 – 11) = -d + 2c – 2(d) 6f – 20 + 8s and 23 – 13f – 12sAns: (6f – 13f) + (8s – 12s) + (-20 + 23) = -7f – 4s + 3(e) 13m – 12n and 12n – 13mAns: (13m – 13m) + (-12n + 12n) = 0(f) -26m + 24n and 26m – 24nAns: (-26m + 26m) + (24n – 24n) = 0Q9: Subtract the expressions given below:(a) 9a – 6b + 14 from 6a + 9b – 18Ans: (6a + 9b – 18) – (9a – 6b + 14) = (6a – 9a) + (9b – (-6b)) + (-18 – 14) = -3a + 15b – 32(b) -15x + 13 – 9y from 7y – 10 + 3xAns: (7y – 10 + 3x) – (-15x + 13 – 9y) = (3x – (-15x)) + (7y – (-9y)) + (-10 – 13) = 18x + 16y – 23(c) 17g + 9 – 7h from 11 – 10g + 3hAns: (11 – 10g + 3h) – (17g + 9 – 7h) = (-10g – 17g) + (3h – (-7h)) + (11 – 9) = -27g + 10h + 2(d) 9a – 6b + 14 from 6a – (9b + 18)Ans: (6a – 9b – 18) – (9a – 6b + 14) = (6a – 9a) + (-9b – (-6b)) + (-18 – 14) = -3a – 3b – 32(e) 10x + 2 + 10y from -3y + 8 – 3xAns: (-3y + 8 – 3x) – (10x + 2 + 10y) = (-3x – 10x) + (-3y – 10y) + (8 – 2) = -13x – 13y + 6(f) 8g + 4h – 10 from 7h – 8g + 20Ans: (7h – 8g + 20) – (8g + 4h – 10) = (-8g – 8g) + (7h – 4h) + (20 – (-10)) = -16g + 3h + 30Q10: Describe situations corresponding to the following algebraic expressions:(a) 8x + 3yAns: A shop sells pens at 8 rupees each and notebooks at 3 rupees each. If x pens and y notebooks are bought, the total cost is 8x + 3y rupees.(b) 15x – 2xAns: A person earns 15 rupees per hour for x hours but spends 2 rupees per hour for x hours on transport. The net income is 15x – 2x = 13x rupees.Q11: Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?
Ans: Step 1 (0 fold): We get 0 + 2 = 2 piecesStep 2 (1 fold): We get 1 + 2 = 3 piecesStep 3 (2 folds): We get 2 + 2 = 4 piecesIn the same way, if the rope is folded 10 times and cut, we get 10 + 2 = 12 pieces.In the same way, when the rope is folded r times and cut, we get r + 2 pieces.Page No. 104 & 105
Q12: Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?Ans: For a chain of squares, each square after the first shares a side.1 square: 4 matchsticks2 squares: 4 + 3 = 7 matchsticks3 squares: 7 + 3 = 10 matchsticksPattern: 4 + 3(w – 1) = 3w + 1For 10 squares: 3 × 10 + 1 = 31 matchsticksFor w squares: 3w + 1 matchsticksQ13: Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour.
Ans: The sequence of red light: 1, 5, 9, …..In general, 4n – 3 positionsThe sequence of green light: 3, 7, 11,…..In general; 4n – 1 positionsThe sequence of yellow light: 2, 4, 6, …..In general, 2n positionsSince 90 and 190 are even numbers, it will be 2n positions.Now, 343 ÷ 4 = 85 quotient + 3 remainder.So, it matches a 4n-1 position.So, colour at positions 90, 190, and 343 are yellow, yellow, and green, respectively.Q14: Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?
Ans: Pattern: Step 1: 5 squares, Step 2: 9 squares, Step 3: 13 squaresThe formula for the nth term of an arithmetic sequence is:an = a1+(n−1)dWhere:an = number of squares at step na1=5a1=5 (first term)d=4 (common difference)Substitute these values:an=5+(n−1)×4an=5+4n−4an=4n+1Step 4:a4= 4×4+1=16+1=17Step 10:a10= 4×10+1=40+1=41Step 50:a50= 4×50+1=200+1=201Since 1 square has 4 vertices, the number of vertices (4n + 1) squares have 4(4n + 1) = 16n + 4.
Q15: Numbers are written in a particular sequence in this endless 4-column grid.(a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4).Ans: Expression to generate all the numbers in a given column (1, 2, 3, 4)Let r be the row number.Column 1: 1, 5, 9, 13,…… which starts at 1 and adds 4 each row.So, number in the rth row of column 1 = 4 × (r – 1) + 1Column 2: 4 × (r – 1) + 2Column 3: 4 × (r – 1) + 3Column 4: 4 × (r – 1) + 4If c is the column number, then the general formula to generate all numbers is 4 × (r – 1) + c.(b) In which row and column will the following numbers appear:(i) 124Ans: 124 ÷ 4 = 31 remainder 0 → Column 4, Row 31(ii) 147Ans: 147 ÷ 4 = 36 remainder 3 → Column 3, Row 37(iii) 201Ans: 201 ÷ 4 = 50 remainder 1 → Column 1, Row 51(c) What number appears in row r and column c?Ans: Number = 4(r – 1) + c(d) Observe the positions of multiples of 3. Do you see any pattern in it? List other patterns that you see.Ans: Every third number is a multiple of 3.We can observe that even numbers always appear in column 2 and column 4.Odd numbers always appear in column 1 and column 3.Every row has 2 odd and 2 even numbers.The sum of each row increases by 16.(e.g., Row 1: 1 + 2 + 3 + 4 = 10, Row 2: 5 + 6 + 7 + 8 = 26, Row 3: 9 + 10 + 11 + 12 = 42)
One and one-tenth and four-hundredths”:
“One and fourteen-hundredths”:
“One Hundred and Fourteen-hundredths”:
