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

Here is CBSE Class 8 Mathematics Ganita Prakash Chapter 6 We Distribute, Yet Things Multiply Practice Paper 20, a 20-question, 40-mark paper split into Sections A to E, with step-by-step solutions. Students practise products like 89 × 111, 84 × 116 and 101 × 99, squares such as 49², 147² and 156², expansions of (11x − 2)² and (p + 11)², the extra area of an enlarged square plaza, and divisibility proofs for squares of odd numbers and multiples of 8.

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 − 9)².
Answer:
(i) (a − b)² = a² − 2ab + b².
(ii) (x − 9)² = x² − 2 × x × 9 + 9² = x² − 18x + 81.
In simple words: Square x, subtract twice x times 9, and add 81.

Teacher's Note:
1. The last term is +81, positive.
2. Check with x = 10: 1² = 1 and 100 − 180 + 81 = 1.

Question 12. Using identity:
(i) Write the identity for (a + b)(a − b).
(ii) Find 84 × 116.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 84 × 116 = (100 − 16)(100 + 16) = 10000 − 256 = 9744.
In simple words: 84 and 116 are 16 on either side of 100, so the product is 10000 − 16² = 9744.

Teacher's Note:
1. 16² = 256.
2. The base is the average: (84 + 116) ÷ 2 = 100.

Question 13. Increase in product:
(i) Find the product of 23 and 38.
(ii) Find the increase when both numbers are increased by 1.
Answer:
(i) 23 × 38 = 874.
(ii) 24 × 39 = 936, so the increase = 936 − 874 = 62.
Identity check: (a + 1)(b + 1) − ab = a + b + 1 = 23 + 38 + 1 = 62.
In simple words: Adding 1 to both numbers adds their sum plus 1, i.e. 62, to the product.

Teacher's Note:
1. The increase does not depend on the product itself.
2. A quick identity check catches multiplication slips.

Question 14. Odd number property:
(i) Write an odd number algebraically.
(ii) Write its square.
Answer:
(i) 2n + 1, where n is a whole number.
(ii) (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1.
In simple words: An odd number is 2n + 1 and its square is 4n(n + 1) + 1.

Teacher's Note:
1. The factorised form shows the square is 1 more than a multiple of 8.
2. n = 3 gives 7² = 49 = 4 × 12 + 1.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Using identities:
(i) Write the identity for (a + b)².
(ii) Expand (p + 11)².
(iii) Find the value of 49².
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) (p + 11)² = p² + 2 × p × 11 + 11² = p² + 22p + 121.
(iii) 49 is just below 50, so use (a − b)²: 49² = (50 − 1)² = 2500 − 100 + 1 = 2401.
In simple words: 49² is easiest as (50 − 1)² = 2500 − 100 + 1 = 2401.

Teacher's Note:
1. For numbers just below a round number, the (a − b)² identity is quicker.
2. (40 + 9)² = 1600 + 720 + 81 = 2401 gives the same answer.

Question 16. Using identity (a + b)(a − b):
(i) Write the identity.
(ii) Express 78 × 62.
(iii) Find the value.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 78 × 62 = (70 + 8)(70 − 8).
(iii) = 70² − 8² = 4900 − 64 = 4836.
In simple words: 78 and 62 are 8 on either side of 70, so the product is 4900 − 64 = 4836.

Teacher's Note:
1. (78 + 62) ÷ 2 = 70 is the base.
2. Units check: 8 × 2 = 16 ends in 6.

Question 17. Multiple of 8 reasoning:
(i) Write a multiple of 8 algebraically.
(ii) Find its square.
(iii) State the conclusion.
Answer:
(i) 8n, where n is a whole number.
(ii) (8n)² = 64n².
(iii) 64n² is 64 times a whole number, so the square of every multiple of 8 is divisible by 64 (e.g. 40² = 1600 = 64 × 25).
In simple words: Squaring 8n gives 64n², so it is always divisible by 64.

Teacher's Note:
1. The proof works for every n at once, unlike checking examples.
2. Remember to square 8 as well as n.

Section D: Long Answer Question (5 Marks)

Question 18. Find the value of 147² using algebraic identity.
(i) Write the identity used.
(ii) Express 147 suitably.
(iii) Expand using identity.
(iv) Simplify to obtain the value.
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) 147 = 100 + 47.
(iii) 147² = 100² + 2 × 100 × 47 + 47² = 10000 + 9400 + 2209.
(iv) 147² = 21609.
Check with (a − b)²: (150 − 3)² = 22500 − 900 + 9 = 21609.

In simple words: 147 = 100 + 47, so 147² = 10000 + 9400 + 2209 = 21609.

Teacher's Note:
1. 47² = (50 − 3)² = 2500 − 300 + 9 = 2209.
2. The 150 − 3 split is even quicker for mental work.

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

Question 19. Case Study 1: Smart City Open Square Design
As part of a smart city initiative, planners design a square open plaza for cultural events and public gatherings. Initially, each side of the plaza measures 58 m. To accommodate larger crowds and facilities, planners decide to increase each side by 7 m. A group of students suggests using algebraic identities instead of lengthy multiplication to calculate the new area and additional space created. This real-life application helps planners make quick and accurate estimates during planning discussions and demonstrates the usefulness of algebra in urban development.
(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 = 58² = (60 − 2)² = 3600 − 240 + 4 = 3364 m².
(iii) New side = 58 + 7 = 65 m. New area = (58 + 7)² = 3364 + 2 × 58 × 7 + 49 = 3364 + 812 + 49 = 4225 m².
(iv) Increase = 4225 − 3364 = 861 m² (= 812 + 49).

In simple words: The plaza grows from 3364 m² to 4225 m², adding 861 m² of space.

Teacher's Note:
1. The increase can also be found as 65² − 58² = (65 + 58)(65 − 58) = 123 × 7 = 861.
2. Show units (m²) in the final answers.

Question 20. Case Study 2: Speed Maths in National-Level Exam
During a national-level competitive exam, students are required to solve numerical problems rapidly without calculators. One question asks for the value of 101 × 99. A well-prepared student rewrites the numbers around 100 and applies the difference of squares identity to obtain the answer instantly. This approach not only saves time but also reduces calculation errors, giving the student a clear advantage in the examination. The example highlights the importance of conceptual clarity and strategic thinking in 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) 101 × 99 = (100 + 1)(100 − 1).
(iii) = 100² − 1² = 10000 − 1 = 9999.
(iv) Directly: 101 × 99 = 101 × 100 − 101 = 10100 − 101 = 9999. Both give 9999, so the result is verified.
In simple words: 101 × 99 = 100² − 1² = 9999.

Teacher's Note:
1. When the gap is 1, the product is just 1 less than the base squared.
2. The direct check uses 99 = 100 − 1.