CBSE Class 8 Mathematics We Distribute Yet Things Multiply Practice Paper Assignment Set 03 with Answers

CBSE Class 8 Maths Ganita Prakash Chapter 6 We Distribute, Yet Things Multiply Practice Paper 21 contains 20 questions worth 40 marks in Sections A to E, each with a detailed answer. The paper tests products such as 95 × 105, 82 × 78 and 102 × 98, squares of 42, 134 and 141, expanding (9x − 4)², (x + 9)² and (y − 8)², the change in a product when one factor rises and the other falls, and why squares of even numbers, odd numbers and multiples of 6 behave as they do.

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, add twice x times 9, then add 81.

Teacher's Note:
1. Middle term: 2 × x × 9 = 18x.
2. Check with x = 1: 10² = 100 and 1 + 18 + 81 = 100.

Question 12. Using identity:
(i) Write the identity for (a + b)(a − b).
(ii) Find 82 × 78.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 82 × 78 = (80 + 2)(80 − 2) = 6400 − 4 = 6396.
In simple words: 82 and 78 are 2 on either side of 80, so the product is 6400 − 4 = 6396.

Teacher's Note:
1. The base here is 80, not 100.
2. Units check: 2 × 8 = 16 ends in 6.

Question 13. Increase in product:
(i) Find product of 21 and 34.
(ii) Find increase when both are increased by 1.
Answer:
(i) 21 × 34 = 714.
(ii) 22 × 35 = 770, so the increase = 770 − 714 = 56.
Identity check: a + b + 1 = 21 + 34 + 1 = 56.
In simple words: Increasing both numbers by 1 adds 21 + 34 + 1 = 56 to the product.

Teacher's Note:
1. (a + 1)(b + 1) = ab + a + b + 1.
2. Two methods, same answer: a reliable check.

Question 14. Even number property:
(i) Write an even number algebraically.
(ii) Write its square.
Answer:
(i) 2n, where n is a whole number.
(ii) (2n)² = 4n², so the square of an even number is a multiple of 4.
In simple words: An even number is 2n and its square is 4n².

Teacher's Note:
1. Square the 2 as well: (2n)² = 4n², not 2n².
2. Example: 14² = 196 = 4 × 49.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Using identities:
(i) Write the identity for (a − b)².
(ii) Expand (y − 8)².
(iii) Find the value of 42².
Answer:
(i) (a − b)² = a² − 2ab + b².
(ii) (y − 8)² = y² − 2 × y × 8 + 8² = y² − 16y + 64.
(iii) 42² = (40 + 2)² = 1600 + 160 + 4 = 1764 (using (a + b)²).
In simple words: Square of a difference has a minus middle term; 42² = 1600 + 160 + 4 = 1764.

Teacher's Note:
1. 42 is just above 40, so the sum identity is easier than the difference identity here.
2. 8² = 64 is the last term in (ii).

Question 16. Using identity (a + b)(a − b):
(i) Write the identity.
(ii) Express 76 × 64.
(iii) Find the value.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 76 × 64 = (70 + 6)(70 − 6).
(iii) = 70² − 6² = 4900 − 36 = 4864.
In simple words: 76 and 64 are 6 on either side of 70, so the product is 4900 − 36 = 4864.

Teacher's Note:
1. The base is (76 + 64) ÷ 2 = 70.
2. Units check: 6 × 4 = 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) Since n(n + 1) is even, 4n(n + 1) is a multiple of 8. So the square of an odd number is 1 more than a multiple of 8 (e.g. 15² = 225 = 224 + 1).
In simple words: Every odd square is a multiple of 8 plus 1.

Teacher's Note:
1. Explain why n(n + 1) is even: one of two consecutive numbers is even.
2. This also shows an odd square leaves remainder 1 on division by 4.

Section D: Long Answer Question (5 Marks)

Question 18. Find the value of 134² using algebraic identity.
(i) Write the identity used.
(ii) Express 134 suitably.
(iii) Expand using identity.
(iv) Simplify to obtain the value.
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) 134 = 100 + 34.
(iii) 134² = 100² + 2 × 100 × 34 + 34² = 10000 + 6800 + 1156.
(iv) 134² = 17956.
Check with 130 + 4: 16900 + 1040 + 16 = 17956.

In simple words: 134 = 100 + 34, so 134² = 10000 + 6800 + 1156 = 17956.

Teacher's Note:
1. 34² = (30 + 4)² = 900 + 240 + 16 = 1156.
2. Any split works; pick the one you can do mentally.

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

Question 19. Case Study 1: City Library Courtyard Expansion
A city library has a square open courtyard used for reading activities and small events. The side length of the courtyard is 52 m. Due to increased participation, the library committee decides to increase each side by 8 m to provide more seating space. A mathematics teacher suggests applying algebraic identities instead of lengthy multiplication to calculate the new area and the extra space created. This real-life situation shows how identities help in quick estimation, planning, and accurate decision-making.
(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 = 52² = (50 + 2)² = 2500 + 200 + 4 = 2704 m².
(iii) New side = 52 + 8 = 60 m. New area = (52 + 8)² = 2704 + 2 × 52 × 8 + 64 = 2704 + 832 + 64 = 3600 m².
(iv) Increase = 3600 − 2704 = 896 m² (= 832 + 64).

In simple words: The courtyard grows from 2704 m² to 3600 m², giving 896 m² of extra seating space.

Teacher's Note:
1. New area is easy to check: 60² = 3600.
2. The extra area is 2ab + b² with a = 52, b = 8.

Question 20. Case Study 2: Speed Calculation in Entrance Exams
While preparing for an entrance examination, a student practices mental mathematics to improve speed. One problem asks to evaluate 102 × 98 without using calculators. The student rewrites the numbers around 100 and applies the difference of squares identity. This approach helps the student calculate accurately within seconds, saving valuable exam time and improving confidence. The example highlights how conceptual understanding of algebraic identities is crucial for competitive and real-life problem-solving.
(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) 102 × 98 = (100 + 2)(100 − 2).
(iii) = 100² − 2² = 10000 − 4 = 9996.
(iv) Directly: 102 × 98 = 102 × 100 − 102 × 2 = 10200 − 204 = 9996. Both give 9996, so the result is verified.
In simple words: 102 × 98 = 100² − 2² = 9996.

Teacher's Note:
1. Writing 98 as 100 − 2 also makes the direct check quick.
2. The identity turns a multiplication into one easy subtraction.