Q1: What is the 20th even number in the sequence 2, 4, 6, 8, ...?
Ans: 40 Explanation: The nth even number is 2n. For n = 20, 2 × 20 = 40.
Q2: Meera has 5 boxes and odd number cards (1, 3, 5, …). Can she pick 5 cards that add to 50?
Ans:No Explanation: Adding 5 odd numbers always gives an odd sum (each odd number adds an unpaired unit). Since 50 is even, it’s impossible.
Q3: Sana adds 6 odd numbers. Is the sum even or odd?
Ans:Even Explanation: Each odd number has one unpaired unit. For 6 odd numbers, the 6 unpaired units pair up (6 ÷ 2 = 3 pairs), so the sum is even.
Q4: Raj has a 14 × 19 grid. Is the number of small squares even or odd?
Ans: Even Explanation: The number of squares is 14 × 19. Since 14 is even and 19 is odd, even × odd = even, so the total is even.
Q5: What is the 30th odd number in the sequence 1, 3, 5, …?
Ans:59 Explanation: The nth odd number is 2n – 1. For n = 30, 2 × 30 – 1 = 60 – 1 = 59.
Q6: In a 3 × 3 magic square with numbers 1 to 9, what is the sum of each row?
Ans:15 Explanation: The sum of numbers 1 to 9 is 45. In a magic square, each of the 3 rows sums to the same value, so 45 ÷ 3 = 15.
Q7: Tara writes the Virahãnka sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. What is the next number?
Ans:144 Explanation: Each number in the Virahãnka sequence is the sum of the two previous numbers. So, 55 + 89 = 144.
Q8: In the puzzle K + K + K = LK, where K and L are digits, what is K?
Ans: 5 Explanation: 3K = LK (a two-digit number with units digit K). So, 3K = 10L + K. (expansion of a two digit number = 10p + q) Subtract K: 2K = 10L. Thus, K = 5L. Since K and L are digits, L = 1, K = 5. Check: 5 + 5 + 5 = 15, and LK = 15.
Q9: A bulb is ON. Vijay toggles it 63 times. Is the bulb ON or OFF?
Ans: OFF Explanation: Each toggle switches the bulb (ON to OFF or OFF to ON). Starting ON, 63 toggles (odd number) means it ends OFF.
Q10: In the Virahãnka sequence, two numbers are 377 and 610. What is the next number?
Ans: 987 Explanation: In the Virahãnka sequence, the next number is the sum of the two previous ones. So, 377 + 610 = 987.
Q1:Define vertically opposite angles. Are they always equal? Ans: Vertically opposite angles are the angles opposite each other when two lines intersect. Yes, they are always equal.
Q2:Two angles form a linear pair. If one is 125°, find the other. Ans: The other angle = 180° − 125° = 55°
Q3:Name a pair of lines that never intersect. Ans: Parallel lines
Q4:A transversal cuts two lines and ∠1 = ∠5. What kind of angles are these, and what can you conclude about the lines? Ans: Corresponding angles; the lines are parallel.
Q5: If ∠x and ∠y are supplementary and ∠x = 2∠y, find the angles. Ans: ∠y = 60°, ∠x = 120°
Q6:∠3 and ∠6 are equal and lie between two lines on opposite sides of the transversal. What type are they? Ans: Alternate interior angles
Q7: Which angles lie between the two lines and on the same side of a transversal? Ans: Co-interior (or consecutive interior) angles
Q8: How many pairs of corresponding angles are formed when a transversal cuts two lines? Ans: 4 pairs
Q1: If Priya’s age is p = 25 years, find Rahul’s age using the expression r = p – 4.
Ans: r = 21 years. Explanation: The expression r = p – 4 means Rahul’s age is 4 years less than Priya’s age. Substitute p = 25: r = 25 – 4 = 21 Thus, Rahul is 21 years old.
Q2: How many matchsticks are needed to make 9 T’s, given the expression 3n, where n is the number of T’s?
Ans: 27 matchsticks. Explanation: The expression 3n indicates each T requires 3 matchsticks. For n = 9 T’s: 3 x 9 = 27 Therefore, 27 matchsticks are needed.
Q3: The cost of one mango is ₹40 and the cost of 1 kg of sugar is ₹50. Write and evaluate an expression to calculate the total cost for 6 mangoes and 4 kg of sugar.
Ans: ₹440 Explanation: Cost of 6 mangoes = 6 × Cost of 1 mango = 6 × 40 = ₹240 Cost of 4 kg sugar = 4 × Cost of 1 kg sugar = 4 × 50 = ₹200 Total Cost = Cost of mangoes + Cost of sugar = 240 + 200 = ₹440
Q4: Find the perimeter of a square with side length s = 8 cm using the expression 4s.
Ans: 32 cm Explanation: The perimeter of a square is given by 4s, where s is the side length. Substitute s = 8: 4 x 8 = 32 The perimeter is 32 cm.
Q5: Evaluate the arithmetic expression 30 – 8 x 3.
Ans: 6 Explanation: Follow the order of operations (BODMAS/PEMDAS): multiplication before subtraction. Calculate 8 x 3 = 24, then: 30 – 24 = 6 The value of the expression is 6.
Q6: Find the value of the expression 4k + 5 when k = 3.
Ans: 17 Explanation: Substitute k = 3 into the expression 4k + 5: 4 x 3 + 5 = 12 + 5 = 17 The value is 17.
Q7: Simplify the expression 7p + 2p + 5p to find the total money earned from selling pens.
Ans: 14p Explanation: Combine like terms by adding the coefficients of p: 7p + 2p + 5p = (7 + 2 + 5)p = 14p The simplified expression for the total money earned is 14p.
Q8: Calculate Anjali’s score in the first round of a quiz if x = 5 and y = 2, using the expression 6x – 4y.
Ans: 22 Explanation: Substitute x = 5 and y = 2 into the expression 6x – 4y: 6 x 5 – 4 x 2 = 30 – 8 = 22 Anjali’s score is 22.
Q9: Find the number of matchsticks needed for Step 20 in a matchstick pattern using the expression 3y + 2, where y is the step number.
Ans: 62 matchsticks Explanation: Substitute y = 20 into the expression 3y + 2: 3 x 20 + 2 = 60 + 2 = 62 Thus, 62 matchsticks are needed for Step 20.
Q10: Determine the total amount paid for renting 4 chairs and 2 tables using the simplified expression 30c + 70t, where c represents number of chairs and t represents number of tables.
Ans: ₹260 Explanation: The expression 30c + 70t gives the net cost for renting c chairs at ₹30 each and t tables at ₹70 each after returns. Substitute c = 4 and t = 2: 30 x 4 + 70 x 2 = 120 + 140 = 260 The total amount paid is ₹260.
Q3. A chocolate bar weighs 0.85 kg. What is the total weight of 6 such bars?
Solution:
0.85 × 6 = 5.10 kg
Q4. Convert the following:
(a) 6.2 cm to mm (b) 46 mm to meters
Solution:
(a) 6.2 × 10 = 62 mm (b) 46 ÷ 1000 = 0.046 meters
Q5. Estimate the sum of 39.847 and 11.205 by rounding off the numbers. Then find the exact sum. Compare both.
Solution:
Estimate:
39.847 ≈ 40
11.205 ≈ 11
Estimated sum = 40 + 11 = 51
Exact sum = 39.847 + 11.205 = 51.052 Comparison: The estimate (51) is very close to the actual sum (51.052), showing estimation is useful for quick checks.
Q6. Express 328 grams in kilograms.
Solution:
328 ÷ 1000 = 0.328 kg
Q7. Rahul walked 2.9 km in the morning and 4.7 km in the evening. What was the total distance he walked? Express your answer as a decimal and a mixed fraction.
Solution:
2.9 + 4.7 = 7.6 km
7.6 = 7 + 6/10 =
Q8.A roll of cloth is 14.4 meters long. If it is divided equally among 6 tailors, how much does each get?
Q2: Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.
(a) 322 + 215 ____ 324 + 213
Ans: Left side: 322 + 215 = 537
Right side: 324 + 213 = 537
322 + 215 = 324 + 213 (They are equal)
(b) 571 + 432 ____ 570 + 434
Ans: Left side: 571 + 432 = 1003 Right side: 570 + 434 = 1004
Q1: Rounding 4,78,942 to the nearest thousand gives ______.
Ans: 4,79,000
As 942 is greater than 500 so we round up
Q2: Convert 7,300,000 into the Indian number system.
Ans: 73,00,000 (Seventy-three lakh)
Q3: Write the following numbers in Indian Place Value Notation:
One crore one lakh one thousand ten
Ans: 1,01,01,010
Step-by-step (Indian Number System):
One crore = 1,00,00,000
One lakh = 1,00,000
One thousand = 1,000
Ten = 10
Now add them all, we get: 1,01,01,010
Q4: True or False: 1 crore = 100 lakhs
Ans: True
1 crore = 100 lakhs.
Q5: Write the numbers 6,345,210 in words using the International system.
Ans: Six million three hundred forty-five thousand two hundred ten
Q6: Round 3,19,612 to the nearest ten thousand.
Ans: 3,20,000
(Since the thousands digit is 9, we round up)
Q7: Using any digits from 0 to 9 without repeating, write the greatest 5-digit number that is a multiple of 5.
Ans: A number is divisible by 5 if it ends in 0 or 5. For maximum value, use the largest digits: 9, 8, 7, 6. Case 1: Ends in 5: 98765 (digits 9, 8, 7, 6, 5, distinct, divisible by 5). Case 2: Ends in 0: 98760 (digits 9, 8, 7, 6, 0, distinct, divisible by 5). Compare: 98765 > 98760 (5 > 0 in units). So, 98765 is the greatest 5-digit number that is a multiple of 5.
Q8: Calculate the product in a quick way
116 x 5
Ans:
116 x (10/2) as 10/2 = 5 58 x 10 580
Q9: Arrange the following in ascending order: 45,210; 54,120; 40,250; 49,500
Q3:Anil wants to find the parity of the 10th term of the Virahãnka sequence. What is the parity? A) 0 B) 1 C) 2 D) 3
Ans: The Virahãnka-Fibonacci sequence is: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … Parities (odd = 1, even = 0): 1, 0, 1, 1, 0, 1, 1, 0, 1, 1. The pattern repeats every 3 terms: odd, even, odd (1, 0, 1). For the 10th term: 10 ÷ 3 = 3 remainder 1, so it corresponds to the 1st position in the cycle (odd, 1). Compute the 10th term: 89 (9th: 55, 10th: 55 + 34 = 89). Check: 89 ÷ 2 = 44.5 (odd). Parity = 1. Answer: B) 1.
Section B: Fill in the Blanks
Q4:Solve the cryptarithm: X4 + Y = Z11. The value of X is ______.
Ans:
X4 + Y = Z11: X4 (X tens, 4 units), Y (units), Z11 (Z hundreds, 1 tens, 1 units). Solve:
Units: 4 + Y = 1 (or 11 with carry).
Tens: X + carry = 1 (or 11).
Hundreds: 0 + carry = Z. Units: 4 + Y = 11, so Y = 7. Carry 1. Tens: X + 1 = 1, so X = 0. No carry to hundreds, Z = 0. Check: 04 + 7 = 11 (Z11 = 011, Z = 0, Y = 7).
Q5: Uneek wants to find all 5-beat rhythms (sums of 1’s and 2’s). The number of ways to write 5 as a sum of 1’s and 2’s is ______.
Ans: 8 Sol: The number of ways to write n as a sum of 1s and 2s is the nth Virahãnka-Fibonacci number: 1, 2, 3, 5, 8, … For n = 5, the 5th term is 8. List: 1+1+1+1+1, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, 1+2+2, 2+1+2, 2+2+1 (8 ways).
Q6: Ishan has number cards with values 1, 3, 5, 7, and 9. She wants to select 3 cards that sum to 21. The number of ways to select 3 cards summing to 21 is _______.
Ans:
Ishan selects 3 cards from 1, 3, 5, 7, 9 to sum to 21. List combinations:
Possible sums of 3 odd numbers (all cards are odd): Check combinations systematically.
Try: 9 + 7 + 5 = 21 (works).
Other combinations:
9 + 7 + 3 = 19 (too low).
9 + 7 + 1 = 17.
9 + 5 + 3 = 17.
9 + 5 + 1 = 15. 7 + 5 + 3 = 15.
7 + 5 + 1 = 13. 5 + 3 + 1 = 9. Only 9 + 7 + 5 = 21 is valid. Number of ways to select these 3 cards: 1 combination. Verify: No other triplets yield 21, as odd sums must be odd, and 21 is odd, but only this set works.
Section C: Word Problems
Q7: Priya and Rohan, two siblings born one year apart, celebrate their birthdays. Priya claims the sum of their ages is 25. Is this possible?
Ans: Since they are born one year apart, their ages are consecutive: n and n+1. We need n + (n+1) = 25. Solve: 2n + 1 = 25, so 2n = 24, n = 12. Ages: 12 and 13. Sum: 12 + 13 = 25, which is possible. Verify: Try other sums (e.g., 24: 2n + 1 = 24, 2n = 23, n = 11.5, not integer). Only 25 works for integer ages.
Q8:Write the next 3 numbers in the sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.
Ans: The Virahãnka-Fibonacci sequence has each term as the sum of the two previous terms: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … Calculate:
11th: 55 + 89 = 144.
12th: 89 + 144 = 233.
13th: 144 + 233 = 377. The next 3 numbers are 144, 233, 377. The second is 233. Verify: 55 + 89 = 144, 89 + 144 = 233, 144 + 233 = 377.
Q9: Two consecutive numbers in the Virahãnka sequence are 987 and 1597. What are the next 2 numbers in the sequence? Find the second of the next 2 numbers.
Ans: In the Virahãnka-Fibonacci sequence (Fn = Fn-1 + Fn-2):
Given 987 (Fn-1), 1597 (Fn), compute next terms:
Fn+1 = 987 + 1597 = 2584.
Fn+2 = 1597 + 2584 = 4181. The next 2 numbers are 2584, 4181. The second is 4181. Verify: 610 + 987 = 1597, 987 + 1597 = 2584, 1597 + 2584 = 4181.
Q10: A light bulb is OFF. A student toggles its switch 50 times. How many times is the bulb ON after 50 toggles? [Count ON state, 1 for ON, 0 for OFF.]
Ans: 25 Sol: Each toggle switches the bulb’s state (ON to OFF, OFF to ON). Starting ON: 1 toggle → OFF, 2 toggles → ON, etc. The state after n toggles is ON if n is even, OFF if n is odd. Here, n = 50 (odd, 50 ÷ 2 = 25 remainder 0). The light bulb is ON 25 times after 50 toggles.
Q11: Using the generalized form, find a magic square if the center number is 13.
Ans: 39 Sol: Odd numbers: 1, 3, 5, 7, … The nth odd number is 2n – 1 : n = 1 → 1, n = 2 → 3, n = 3 → 5. For the 20th odd number, n = 20: 2 × 20 – 1 = 40 – 1 = 39.
Q12: Vanshika wants to climb a 9-step staircase, taking either 1 or 2 steps at a time. In how many different ways can she reach the top?
Ans: 55 Sol: This is a classic Fibonacci-type problem. Let F(n) be the number of ways to climb n steps using 1 or 2 steps at a time. Then:F(n)=F(n−1)+F(n−2)
Because:
From step (n−1), she can take 1 step
From step (n−2), she can take 2 steps
Base Cases:
F(1) = 1 (only one way: 1)
F(2) = 2 (1+1 or 2)
Compute up to F(9):
There are 55 different ways for Vanshika to climb a 9-step staircase taking 1 or 2 steps at a time.
Q3:Anil wants to find the parity of the 10th term of the Virahãnka sequence. What is the parity? A) 0 B) 1 C) 2 D) 3
Ans: The Virahãnka-Fibonacci sequence is: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … Parities (odd = 1, even = 0): 1, 0, 1, 1, 0, 1, 1, 0, 1, 1. The pattern repeats every 3 terms: odd, even, odd (1, 0, 1). For the 10th term: 10 ÷ 3 = 3 remainder 1, so it corresponds to the 1st position in the cycle (odd, 1). Compute the 10th term: 89 (9th: 55, 10th: 55 + 34 = 89). Check: 89 ÷ 2 = 44.5 (odd). Parity = 1. Answer: B) 1.
Section B: Fill in the Blanks
Q4:Solve the cryptarithm: X4 + Y = Z11. The value of X is ______.
Ans:
X4 + Y = Z11: X4 (X tens, 4 units), Y (units), Z11 (Z hundreds, 1 tens, 1 units). Solve:
Units: 4 + Y = 1 (or 11 with carry).
Tens: X + carry = 1 (or 11).
Hundreds: 0 + carry = Z. Units: 4 + Y = 11, so Y = 7. Carry 1. Tens: X + 1 = 1, so X = 0. No carry to hundreds, Z = 0. Check: 04 + 7 = 11 (Z11 = 011, Z = 0, Y = 7).
Q5: Uneek wants to find all 5-beat rhythms (sums of 1’s and 2’s). The number of ways to write 5 as a sum of 1’s and 2’s is ______.
Ans: 8 Sol: The number of ways to write n as a sum of 1s and 2s is the nth Virahãnka-Fibonacci number: 1, 2, 3, 5, 8, … For n = 5, the 5th term is 8. List: 1+1+1+1+1, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, 1+2+2, 2+1+2, 2+2+1 (8 ways).
Q6: Ishan has number cards with values 1, 3, 5, 7, and 9. She wants to select 3 cards that sum to 21. The number of ways to select 3 cards summing to 21 is _______.
Ans:
Ishan selects 3 cards from 1, 3, 5, 7, 9 to sum to 21. List combinations:
Possible sums of 3 odd numbers (all cards are odd): Check combinations systematically.
Try: 9 + 7 + 5 = 21 (works).
Other combinations:
9 + 7 + 3 = 19 (too low).
9 + 7 + 1 = 17.
9 + 5 + 3 = 17.
9 + 5 + 1 = 15. 7 + 5 + 3 = 15.
7 + 5 + 1 = 13. 5 + 3 + 1 = 9. Only 9 + 7 + 5 = 21 is valid. Number of ways to select these 3 cards: 1 combination. Verify: No other triplets yield 21, as odd sums must be odd, and 21 is odd, but only this set works.
Section C: Word Problems
Q7: Priya and Rohan, two siblings born one year apart, celebrate their birthdays. Priya claims the sum of their ages is 25. Is this possible?
Ans: Since they are born one year apart, their ages are consecutive: n and n+1. We need n + (n+1) = 25. Solve: 2n + 1 = 25, so 2n = 24, n = 12. Ages: 12 and 13. Sum: 12 + 13 = 25, which is possible. Verify: Try other sums (e.g., 24: 2n + 1 = 24, 2n = 23, n = 11.5, not integer). Only 25 works for integer ages.
Q8:Write the next 3 numbers in the sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.
Ans: The Virahãnka-Fibonacci sequence has each term as the sum of the two previous terms: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … Calculate:
11th: 55 + 89 = 144.
12th: 89 + 144 = 233.
13th: 144 + 233 = 377. The next 3 numbers are 144, 233, 377. The second is 233. Verify: 55 + 89 = 144, 89 + 144 = 233, 144 + 233 = 377.
Q9: Two consecutive numbers in the Virahãnka sequence are 987 and 1597. What are the next 2 numbers in the sequence? Find the second of the next 2 numbers.
Ans: In the Virahãnka-Fibonacci sequence (Fn = Fn-1 + Fn-2):
Given 987 (Fn-1), 1597 (Fn), compute next terms:
Fn+1 = 987 + 1597 = 2584.
Fn+2 = 1597 + 2584 = 4181. The next 2 numbers are 2584, 4181. The second is 4181. Verify: 610 + 987 = 1597, 987 + 1597 = 2584, 1597 + 2584 = 4181.
Q10: A light bulb is OFF. A student toggles its switch 50 times. How many times is the bulb ON after 50 toggles? [Count ON state, 1 for ON, 0 for OFF.]
Ans: 25 Sol: Each toggle switches the bulb’s state (ON to OFF, OFF to ON). Starting ON: 1 toggle → OFF, 2 toggles → ON, etc. The state after n toggles is ON if n is even, OFF if n is odd. Here, n = 50 (odd, 50 ÷ 2 = 25 remainder 0). The light bulb is ON 25 times after 50 toggles.
Q11: Using the generalized form, find a magic square if the center number is 13.
Ans: 39 Sol: Odd numbers: 1, 3, 5, 7, … The nth odd number is 2n – 1 : n = 1 → 1, n = 2 → 3, n = 3 → 5. For the 20th odd number, n = 20: 2 × 20 – 1 = 40 – 1 = 39.
Q12: Vanshika wants to climb a 9-step staircase, taking either 1 or 2 steps at a time. In how many different ways can she reach the top?
Ans: 55 Sol: This is a classic Fibonacci-type problem. Let F(n) be the number of ways to climb n steps using 1 or 2 steps at a time. Then:F(n)=F(n−1)+F(n−2)
Because:
From step (n−1), she can take 1 step
From step (n−2), she can take 2 steps
Base Cases:
F(1) = 1 (only one way: 1)
F(2) = 2 (1+1 or 2)
Compute up to F(9):
There are 55 different ways for Vanshika to climb a 9-step staircase taking 1 or 2 steps at a time.
Q1. Classify the following pairs of lines as Parallel (P), Perpendicular (⊥), or Intersecting (I): a) Opposite edges of a book b) Hands of a clock at 9:00 c) Letter “T” d) Railway tracks e) Corners of a window frame
Sol:a) Opposite edges of a book The top and bottom (or left and right) edges of a closed book run side by side and never meet, even if extended. ⇒ Parallel (P)
b) Hands of a clock at 9:00 At 9:00, the minute hand points at 12 and the hour hand points at 9. These hands form a right angle. ⇒ Perpendicular (⊥)
c) Letter “T” The vertical and horizontal lines in the letter T meet at a right angle. ⇒ Perpendicular (⊥)
d) Railway tracks Railway tracks are always the same distance apart and never cross each other. ⇒ Parallel (P)
e) Corners of a window frame The edges of a window meet at the corners to form right angles. ⇒ Perpendicular (⊥)
Section B: Numerical Based Questions
Q2.Two lines intersect to form four angles. If one angle is 70°, find the measures of the other three angles.
Sol: Vertically opposite angle = 70° (equal).
Linear pair angles: 180°−70°=110° (two angles of 110° each).
⇒ 70°, 110°, 110°.
Q3.In the figure below, line l∥m and transversal t cuts them. If ∠2=65°, find ∠6.
Sol: Since l∥m, corresponding angles are equal. Thus, ∠6=∠2=65°.
Section C: Theory Based Questions
Q4.Define: a) Parallel lines b) Perpendicular lines
Sol: a) Parallel lines are lines in the same plane that never meet, no matter how far they are extended. b) Perpendicular lines intersect at a right angle (90°).
Q5. Identify the type of angles formed when two lines intersect: a) Angles opposite each other b) Adjacent angles forming a straight line
Sol: a) Vertically opposite angles (they are equal). b) Linear pair (they add up to 180°).
Q6. Two railway tracks are said to be parallel. A boy standing on a bridge drops a straight stick that cuts across both tracks. What is the name of the stick in geometric terms? Also, name any two pairs of angles formed and say if they are equal.
