CBSE Class 8 Mathematics We Distribute Yet Things Multiply Practice Paper Assignment Set 01 with Answers

CBSE Class 8 Maths Ganita Prakash Chapter 6 We Distribute, Yet Things Multiply Practice Paper 17 is a 40-mark test of 20 questions in Sections A to E, solved here step by step. It practises the three main identities, (a + b)², (a − b)² and (a + b)(a − b), for mental calculation of products like 96 × 104 and squares like 115², along with the change in a product when its factors change and why every odd square is one more than a multiple of 8.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the increase in product when 18 and 27 are both increased by 1.
Answer:
Original product = 18 × 27 = 486.
New product = 19 × 28 = 532.
Increase = 532 − 486 = 46.
Check with the identity: (a + 1)(b + 1) − ab = a + b + 1 = 18 + 27 + 1 = 46.
In simple words: When both numbers go up by 1, the product goes up by (sum of the numbers + 1) = 46.

Teacher's Note:
1. (a + 1)(b + 1) = ab + a + b + 1, so the increase is a + b + 1.
2. The identity gives the answer without any long multiplication.

Question 12. Expand (9a − 4)².
Answer:
Use (x − y)² = x² − 2xy + y² with x = 9a, y = 4.
(9a)² = 81a², 2 × 9a × 4 = 72a, 4² = 16.
(9a − 4)² = 81a² − 72a + 16.
In simple words: Square the first term, subtract twice the product, and add the square of the last term.

Teacher's Note:
1. (9a)² is 81a², not 9a².
2. Only the middle term carries the minus sign.

Question 13. Evaluate 107 × 93 using identity.
Answer:
107 = 100 + 7 and 93 = 100 − 7.
Using (a + b)(a − b) = a² − b²:
107 × 93 = 100² − 7² = 10000 − 49 = 9951.
In simple words: Both numbers are 7 away from 100, so the answer is 100² − 7².

Teacher's Note:
1. The two numbers must be equally far on either side of the base.
2. Check: 107 × 93 ends in 1 because 7 × 3 = 21.

Question 14. Write the identity used to find the difference of two squares.
Answer:
a² − b² = (a + b)(a − b).
Example: 53² − 47² = (53 + 47)(53 − 47) = 100 × 6 = 600.
In simple words: The difference of two squares equals (sum of the numbers) × (difference of the numbers).

Teacher's Note:
1. The same identity read backwards helps multiply numbers like 98 × 102.
2. It works for any numbers or expressions a and b.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Using algebraic identities:
(i) Write the identity for (a + b)².
(ii) Expand (x + 5)².
(iii) Find the value of 25² using identity.
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) (x + 5)² = x² + 2 × x × 5 + 5² = x² + 10x + 25.
(iii) 25² = (20 + 5)² = 20² + 2 × 20 × 5 + 5² = 400 + 200 + 25 = 625.
In simple words: Square of a sum = first² + 2 × first × second + second²; so 25² = 400 + 200 + 25 = 625.

Teacher's Note:
1. The middle term 2ab is the most commonly forgotten part.
2. Choose a split where one part is a multiple of 10.

Question 16. Using identity (a + b)(a − b):
(i) Write the identity.
(ii) Express 72 × 68 using identity.
(iii) Find its value.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 72 × 68 = (70 + 2)(70 − 2).
(iii) = 70² − 2² = 4900 − 4 = 4896.
In simple words: 72 and 68 are both 2 away from 70, so 72 × 68 = 70² − 2² = 4896.

Teacher's Note:
1. The base is the average of the two numbers: (72 + 68) ÷ 2 = 70.
2. The answer is just below 70², as expected.

Question 17. Square of an odd number:
(i) Write an odd number in algebraic form.
(ii) Find its square.
(iii) State the conclusion.
Answer:
(i) An odd number is 2n + 1, where n is a whole number.
(ii) (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1.
(iii) n and n + 1 are consecutive numbers, so one of them is even and n(n + 1) is even. Then 4n(n + 1) is a multiple of 8.
Conclusion: the square of every odd number is one more than a multiple of 8 (for example 49 = 48 + 1, 81 = 80 + 1).
In simple words: Every odd square, like 9, 25, 49 or 81, is 1 more than a multiple of 8.

Teacher's Note:
1. Taking out 4n as a common factor shows the n(n + 1) pattern.
2. Product of two consecutive numbers is always even.

Section D: Long Answer Question (5 Marks)

Question 18. Find the value of 115² using algebraic identity.
(i) Write the identity used.
(ii) Express 115 suitably.
(iii) Expand using identity.
(iv) Simplify and find the value.
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) 115 = 100 + 15.
(iii) 115² = (100 + 15)² = 100² + 2 × 100 × 15 + 15² = 10000 + 3000 + 225.
(iv) 115² = 13225.
The two thin strips are 100 × 15 = 1500 each, together 3000.
In simple words: Split 115 into 100 + 15; 10000 + 3000 + 225 = 13225.

Teacher's Note:
1. The picture shows (a + b)² as one big square, one small square and two equal strips.
2. Check the last digits: 115² must end in 25, and 13225 does.

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

Question 19. Case Study 1: School Sports Ground Planning
A school has a square sports ground with each side measuring 45 m. Due to the introduction of new sports facilities, the management decides to increase the length of each side by 5 m. Instead of using lengthy calculations, the sports teacher asks students to apply algebraic identities learned in Class VIII to calculate the new area and the additional space created. This activity helps students understand how algebraic identities are useful in real-life planning and estimation problems.
(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 = 45² = 2025 m².
(iii) New side = 45 + 5 = 50 m. New area = (45 + 5)² = 45² + 2 × 45 × 5 + 5² = 2025 + 450 + 25 = 2500 m².
(iv) Increase = 2500 − 2025 = 475 m² (= 2 × 45 × 5 + 5² = 450 + 25).
Added space: two strips of 45 m × 5 m (225 m² each) and a corner of 5 m × 5 m (25 m²).
In simple words: The ground grows from 2025 m² to 2500 m², so 475 m² of extra space is created.

Teacher's Note:
1. The extra area is 2ab + b², not just b².
2. Always write square metres (m²) for areas.

Question 20. Case Study 2: Mental Maths Challenge
During a classroom mental-maths challenge, students are asked to calculate 98 × 102 without calculators. One student quickly applies the difference of squares identity by rewriting the numbers around 100. The teacher uses this example to explain how algebraic identities improve speed, accuracy, and confidence in calculations, especially in exams and competitions.
(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) 98 × 102 = (100 − 2)(100 + 2).
(iii) = 100² − 2² = 10000 − 4 = 9996.
(iv) Direct multiplication: 98 × 102 = 98 × 100 + 98 × 2 = 9800 + 196 = 9996. Both methods give 9996, so the result is verified.
In simple words: 98 and 102 are 2 below and above 100, so 98 × 102 = 10000 − 4 = 9996.

Teacher's Note:
1. Verification by the distributive property is a good habit in exams.
2. Choose the base as the number exactly halfway between the two factors.