CBSE Class 8 Mathematics We Distribute Yet Things Multiply Practice Paper Assignment Set 02 with Answers

This CBSE Class 8 Maths Ganita Prakash Chapter 6 We Distribute, Yet Things Multiply Practice Paper 19 has 20 questions for 40 marks in Sections A to E, with a clear worked answer for each. It covers expanding (13x − 4)² and (y − 13)², computing 87 × 113, 88 × 112 and 103 × 97 with the difference of squares, squaring 57, 158 and 167, the increase in area of a growing square field, and reasoning about squares of odd numbers and multiples of 10.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Using identity:
(i) Write the identity for (a + b)².
(ii) Expand (x + 12)².
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) (x + 12)² = x² + 2 × x × 12 + 12² = x² + 24x + 144.
In simple words: Square the first term, add twice the product, add the square of 12.

Teacher's Note:
1. 12² = 144.
2. All three terms are positive for a sum.

Question 12. Using identity:
(i) Write the identity for (a + b)(a − b).
(ii) Find 88 × 112.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 88 × 112 = (100 − 12)(100 + 12) = 10000 − 144 = 9856.
In simple words: 88 and 112 are 12 away from 100, so the product is 10000 − 144.

Teacher's Note:
1. Average of 88 and 112 is 100.
2. Subtract the square of the gap, not the gap.

Question 13. Increase in product:
(i) Find the product of 26 and 41.
(ii) Find the increase when both are increased by 1.
Answer:
(i) 26 × 41 = 1066.
(ii) 27 × 42 = 1134, so the increase = 1134 − 1066 = 68.
By the identity: (a + 1)(b + 1) − ab = a + b + 1 = 26 + 41 + 1 = 68.
In simple words: Raising both numbers by 1 adds (26 + 41 + 1) = 68 to the product.

Teacher's Note:
1. The identity result matches the direct result.
2. This pattern always holds when both numbers increase by 1.

Question 14. Multiple of 10:
(i) Write a multiple of 10 algebraically.
(ii) Write its square.
Answer:
(i) 10n, where n is a whole number.
(ii) (10n)² = 100n².
In simple words: A multiple of 10 is 10n, and its square is 100n².

Teacher's Note:
1. Square the 10 too: 10² = 100.
2. This is why squares of multiples of 10 end in 00.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Using identities:
(i) Write the identity for (a − b)².
(ii) Expand (y − 13)².
(iii) Find the value of 57².
Answer:
(i) (a − b)² = a² − 2ab + b².
(ii) (y − 13)² = y² − 2 × y × 13 + 13² = y² − 26y + 169.
(iii) 57² = (50 + 7)² = 2500 + 700 + 49 = 3249. (Or (60 − 3)² = 3600 − 360 + 9 = 3249.)
In simple words: 57² can be found as 50² + 2 × 50 × 7 + 7² or as 60² − 2 × 60 × 3 + 3²; both give 3249.

Teacher's Note:
1. Either identity works; choose the one with easier numbers.
2. 13² = 169 is a useful square to know.

Question 16. Using identity (a + b)(a − b):
(i) Write the identity.
(ii) Express 84 × 76.
(iii) Find the value.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 84 × 76 = (80 + 4)(80 − 4).
(iii) = 80² − 4² = 6400 − 16 = 6384.
In simple words: 84 and 76 are 4 on either side of 80, so 84 × 76 = 6400 − 16 = 6384.

Teacher's Note:
1. The base is (84 + 76) ÷ 2 = 80.
2. Check the units digit: 4 × 6 = 24 ends in 4.

Question 17. Odd number reasoning:
(i) Write an odd number algebraically.
(ii) Find its square.
(iii) State the conclusion.
Answer:
(i) 2n + 1, where n is a whole number.
(ii) (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1.
(iii) n(n + 1) is a product of consecutive numbers, hence even, so 4n(n + 1) is a multiple of 8. The square of an odd number is 1 more than a multiple of 8.
In simple words: Every odd square leaves remainder 1 when divided by 8.

Teacher's Note:
1. Factorising 4n² + 4n as 4n(n + 1) reveals the hidden even factor.
2. Example: 13² = 169 = 168 + 1 = 8 × 21 + 1.

Section D: Long Answer Question (5 Marks)

Question 18. Find the value of 158² using algebraic identity.
(i) Write the identity used.
(ii) Express 158 suitably.
(iii) Expand using identity.
(iv) Simplify to obtain the value.
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) 158 = 100 + 58.
(iii) 158² = 100² + 2 × 100 × 58 + 58² = 10000 + 11600 + 3364.
(iv) 158² = 24964.
Check with (a − b)²: (160 − 2)² = 25600 − 640 + 4 = 24964.

In simple words: 158 = 100 + 58 gives 10000 + 11600 + 3364 = 24964.

Teacher's Note:
1. 58² = 3364 itself can be found as (60 − 2)² = 3600 − 240 + 4.
2. Two methods giving the same result is a strong check.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: National Stadium Square Layout
A national sports authority plans a square training field for athletes. Initially, each side of the field measures 62 m. To add safety zones and practice lanes, planners decide to increase each side by 9 m. During planning discussions, a mathematics consultant recommends using algebraic identities instead of long multiplication to calculate the new area and the additional space created. This approach allows planners to make quick, accurate decisions and clearly shows how mathematical identities are applied in large-scale infrastructure projects.
(i) Write the identity used.
(ii) Find original area.
(iii) Find new area.
(iv) Find increase in area.
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) Original area = 62² = (60 + 2)² = 3600 + 240 + 4 = 3844 m².
(iii) New side = 62 + 9 = 71 m. New area = (62 + 9)² = 3844 + 2 × 62 × 9 + 81 = 3844 + 1116 + 81 = 5041 m².
(iv) Increase = 5041 − 3844 = 1197 m² (= 1116 + 81).

In simple words: The field grows from 3844 m² to 5041 m², adding 1197 m².

Teacher's Note:
1. The added region is two 62 m × 9 m strips plus a 9 m × 9 m corner.
2. Check: 71² = (70 + 1)² = 4900 + 140 + 1 = 5041.

Question 20. Case Study 2: Ultra-Fast Calculation in Olympiad
During a mathematics olympiad, participants must solve calculations mentally under strict time limits. One problem asks students to evaluate 103 × 97. A well-trained student rewrites the numbers around 100 and applies the difference of squares identity. This method allows the student to arrive at the answer almost instantly, demonstrating how strong conceptual understanding of algebraic identities provides a decisive advantage in competitive mathematics.
(i) Identify the identity used.
(ii) Write the expression.
(iii) Find the value.
(iv) Verify the result.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 103 × 97 = (100 + 3)(100 − 3).
(iii) = 100² − 3² = 10000 − 9 = 9991.
(iv) Directly: 103 × 97 = 103 × 100 − 103 × 3 = 10300 − 309 = 9991. Both methods give 9991, so the result is verified.
In simple words: 103 and 97 are 3 on each side of 100, so 103 × 97 = 10000 − 9 = 9991.

Teacher's Note:
1. The verification uses the distributive property: 103 × (100 − 3).
2. Mental methods save time but always check reasonableness.