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

Chapter-wise Worksheets for Class 8 Mathematics: Ganita Prakash Part 1 Chapter 06 We Distribute Yet Things Multiply

Explore structured practice materials through the CBSE Class 8 Mathematics We Distribute Yet Things Multiply Practice Paper Worksheet Set 01 with Answers. Tailored for Class 8 learners, utilizing these Mathematics worksheets ensures thorough preparation and strengthens problem-solving accuracy before final school evaluations.

Practice Class 8 Mathematics Worksheets: Ganita Prakash Part 1 Chapter 06 We Distribute Yet Things Multiply

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Practice Paper 18 for CBSE Class 8 Mathematics Ganita Prakash Chapter 6 We Distribute, Yet Things Multiply carries 40 marks across 20 questions (Sections A to E), each followed by a worked solution. The questions focus on how a product changes when one factor is raised or lowered, quick products such as 86 × 114 and 107 × 93, squares like 163² and 165², expanding (15x − 6)², and proving that the square of a multiple of 12 is divisible by 144.

Section A: Multiple Choice Questions (1 Mark Each)

Question 1: If the product of two numbers is ab, then the decrease in the product when the second number is decreased by 6 is
(a) 6a
(b) ab − 6a
(c) ab − 6b
(d) 6b
Show Answer & Explanation

Answer: (a) 6a

Explanation:
1. Let the numbers be a and b; the product is ab.
2. New second number = b − 6. New product = a(b − 6) = ab − 6a.
3. Decrease = ab − (ab − 6a) = 6a.

Teacher's Note:
1. ab − 6a is the new product, not the decrease; read what is asked.
2. Reducing one factor by 6 removes 6 copies of the other factor.

Question 2: Using algebraic identity, the value of 96 × 104 is
(a) 9984
(b) 9964
(c) 9944
(d) 9924
Show Answer & Explanation

Answer: (a) 9984

Explanation:
1. Write 96 = 100 − 4 and 104 = 100 + 4.
2. (a − b)(a + b) = a² − b² gives 100² − 4² = 10000 − 16 = 9984.

Teacher's Note:
1. Square of the base minus square of the distance from the base.
2. 4² = 16, so the answer is 16 less than 10000.

Question 3: Which identity is best suited to calculate 163² mentally?
(a) (a − b)²
(b) (a + b)²
(c) (a + b)(a − b)
(d) a² + b²
Show Answer & Explanation

Answer: (b) (a + b)²

Explanation:
1. Split 163 as 160 + 3 (a round number plus a small number).
2. Use (a + b)² = a² + 2ab + b².
3. 163² = 160² + 2 × 160 × 3 + 3² = 25600 + 960 + 9 = 26569.
4. So the best identity is (a + b)².

Teacher's Note:
1. (170 − 7)² with (a − b)² also works, but 160 + 3 keeps the numbers small.
2. a² + b² alone is never the square of a sum.

Question 4: When one number is increased by 7 and the other decreased by 8, the change in product is
(a) 7b − 8a
(b) 7b − 8a − 56
(c) ab − 56
(d) 8a − 7b
Show Answer & Explanation

Answer: (b) 7b − 8a − 56

Explanation:
1. New product = (a + 7)(b − 8) = ab − 8a + 7b − 56.
2. Change = new product − ab = 7b − 8a − 56.

Teacher's Note:
1. Four products appear when two binomials are multiplied.
2. The term 7 × (−8) = −56 comes from multiplying the two changes.

Question 5: The square of an odd number always leaves remainder ___ when divided by 8.
(a) 3
(b) 5
(c) 7
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. Odd number = 2n + 1, so its square = 4n² + 4n + 1 = 4n(n + 1) + 1.
2. n(n + 1) is even, so 4n(n + 1) is a multiple of 8.
3. Hence the remainder is 1 (e.g. 25 = 8 × 3 + 1, 121 = 8 × 15 + 1).

Teacher's Note:
1. Test a few odd squares: 9, 25, 49, 81 all leave remainder 1.
2. One of two consecutive numbers is always even.

Question 6: Which expansion is correct?
(a) (15x − 6)² = 225x² − 36
(b) (15x − 6)² = 225x² − 180x + 36
(c) (15x − 6)² = 30x² − 36
(d) (15x − 6)² = 225x² − 90x + 6
Show Answer & Explanation

Answer: (b) (15x − 6)² = 225x² − 180x + 36

Explanation:
1. (a − b)² = a² − 2ab + b² with a = 15x, b = 6.
2. (15x)² = 225x², 2 × 15x × 6 = 180x, 6² = 36.
3. So (15x − 6)² = 225x² − 180x + 36.

Teacher's Note:
1. Square both the number and the variable: (15x)² = 225x².
2. The last term is +36, never −36.

Question 7: Using identities, the value of 86 × 114 is
(a) 9804
(b) 9844
(c) 9884
(d) 9764
Show Answer & Explanation

Answer: (a) 9804

Explanation:
1. 86 = 100 − 14 and 114 = 100 + 14.
2. 86 × 114 = 100² − 14² = 10000 − 196 = 9804.

Teacher's Note:
1. The base 100 is midway between 86 and 114.
2. Know squares up to 20 by heart: 14² = 196.

Question 8: The algebraic expression for decrease in area when the side of a square changes from m to m − 10 is
(a) 20m − 100
(b) m² − 100
(c) 20m + 100
(d) 10m − 100
Show Answer & Explanation

Answer: (a) 20m − 100

Explanation:
1. Old area = m². New area = (m − 10)² = m² − 20m + 100.
2. Decrease = m² − (m² − 20m + 100) = 20m − 100.

Teacher's Note:
1. Put the subtracted expression in brackets so every sign changes.
2. Check with m = 15: 225 − 25 = 200 and 20 × 15 − 100 = 200.

Question 9:
Assertion (A): The square of any multiple of 12 is divisible by 144.
Reason (R): Any multiple of 12 can be written as 12n.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A multiple of 12 is 12n; its square is (12n)² = 144n², which is divisible by 144. A is true.
2. R is true, and writing the number as 12n is exactly what proves A.
3. Answer: (a).

Teacher's Note:
1. (12n)² = 12² × n² = 144n².
2. Example: 36² = 1296 = 144 × 9.

Question 10:
Assertion (A): (a + b)² − (a − b)² is always divisible by 4 (for whole numbers a and b).
Reason (R): The expression simplifies to 4ab.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. (a + b)² − (a − b)² = (a² + 2ab + b²) − (a² − 2ab + b²) = 4ab. So R is true.
2. 4ab is 4 times a whole number, so A is true, and it is true because of R.
3. Answer: (a).

Teacher's Note:
1. Divisibility claims about letters need the letters to be whole numbers.
2. Example: a = 5, b = 2 gives 49 − 9 = 40 = 4 × 10.

Section B: Short Answer Questions (2 Marks Each)

Question 11. Using identity:
(i) Write the identity for (a − b)².
(ii) Expand (x − 14)².
Answer:
(i) (a − b)² = a² − 2ab + b².
(ii) (x − 14)² = x² − 2 × x × 14 + 14² = x² − 28x + 196.
In simple words: Square of a difference: first² − 2 × first × second + second².

Teacher's Note:
1. 14² = 196 is worth remembering.
2. The middle term is negative, the last term positive.

 

Question 12. Using identity:
(i) Write the identity for (a + b)(a − b).
(ii) Find 92 × 108.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 92 × 108 = (100 − 8)(100 + 8) = 100² − 8² = 10000 − 64 = 9936.
In simple words: 92 and 108 are 8 away from 100, so the product is 10000 − 64 = 9936.

Teacher's Note:
1. Find the base by averaging: (92 + 108) ÷ 2 = 100.
2. The product is always less than the base squared.

 

Question 13. Change in product:
(i) Find the product of 31 and 44.
(ii) Find the increase when both are increased by 1.
Answer:
(i) 31 × 44 = 1364.
(ii) New product = 32 × 45 = 1440. Increase = 1440 − 1364 = 76.
Shortcut: (a + 1)(b + 1) − ab = a + b + 1 = 31 + 44 + 1 = 76.
In simple words: Adding 1 to both numbers raises the product by their sum plus 1, which is 76.

Teacher's Note:
1. The shortcut avoids a second multiplication.
2. Use the direct method to check the shortcut.

 

Question 14. Multiple of 12:
(i) Write a multiple of 12 algebraically.
(ii) Write its square.
Answer:
(i) 12n, where n is a whole number.
(ii) (12n)² = 144n².
In simple words: A multiple of 12 is 12n, and its square is 144n².

Teacher's Note:
1. Square the coefficient as well as the letter.
2. So every such square is a multiple of 144.

 

Section C: Short Answer Questions (3 Marks Each)

Question 15. Using identities:
(i) Write the identity for (a + b)².
(ii) Expand (z + 14)².
(iii) Find the value of 64².
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) (z + 14)² = z² + 2 × z × 14 + 14² = z² + 28z + 196.
(iii) 64² = (60 + 4)² = 60² + 2 × 60 × 4 + 4² = 3600 + 480 + 16 = 4096.
In simple words: Split 64 into 60 + 4 and use the square-of-a-sum identity to get 4096.

Teacher's Note:
1. Choose the split with a multiple of 10.
2. 64 = 2⁶, so 64² = 2¹² = 4096 is a nice check.

 

Question 16. Using identity (a + b)(a − b):
(i) Write the identity.
(ii) Express 86 × 74.
(iii) Find the value.
Answer:
(i) (a + b)(a − b) = a² − b².
(ii) 86 × 74 = (80 + 6)(80 − 6).
(iii) = 80² − 6² = 6400 − 36 = 6364.
In simple words: 86 and 74 are 6 on either side of 80, so the product is 6400 − 36 = 6364.

Teacher's Note:
1. Average of 86 and 74 is 80.
2. The base need not be 100; any convenient number works.

 

Question 17. Divisibility reasoning:
(i) Write a multiple of 12 algebraically.
(ii) Find its square.
(iii) State the conclusion.
Answer:
(i) 12n, where n is a whole number.
(ii) (12n)² = 12² × n² = 144n².
(iii) Since n² is a whole number, 144n² is a multiple of 144. Conclusion: the square of every multiple of 12 is divisible by 144 (e.g. 24² = 576 = 144 × 4).
In simple words: Squaring 12n gives 144n², so such squares are always divisible by 144.

Teacher's Note:
1. A general proof uses a letter, not one example.
2. Similarly, the square of a multiple of k is a multiple of k².

 

Section D: Long Answer Question (5 Marks)

Question 18. Find the value of 165² using algebraic identity.
(i) Write the identity used.
(ii) Express 165 suitably.
(iii) Expand using identity.
(iv) Simplify to obtain the value.
Answer:
(i) (a + b)² = a² + 2ab + b².
(ii) 165 = 100 + 65 (160 + 5 also works).
(iii) 165² = 100² + 2 × 100 × 65 + 65² = 10000 + 13000 + 4225.
(iv) 165² = 27225.
Check with 160 + 5: 25600 + 1600 + 25 = 27225.
100² = 10000 6500 6500 65² = 4225 100 65 100 65
In simple words: 165 = 100 + 65, so 165² = 10000 + 13000 + 4225 = 27225.

Teacher's Note:
1. A number ending in 5 has a square ending in 25.
2. Using two different splits is an easy way to check.

 

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

Question 19. Case Study 1: Square Courtyard in Heritage School
A heritage school plans to renovate its old square courtyard used for morning assemblies and cultural activities. Originally, each side of the courtyard measures 54 m. To allow better movement of students and installation of safety railings, the school management decides to increase the length of each side by 11 m. While estimating the additional area required for tiling and budgeting, a teacher suggests using algebraic identities instead of lengthy multiplication. This helps students understand how mathematics simplifies real-life planning tasks and promotes efficient problem-solving during large construction 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 = 54² = (50 + 4)² = 2500 + 400 + 16 = 2916 m².
(iii) New side = 54 + 11 = 65 m. New area = (54 + 11)² = 2916 + 2 × 54 × 11 + 11² = 2916 + 1188 + 121 = 4225 m².
(iv) Increase = 4225 − 2916 = 1309 m² (= 1188 + 121).
54² = 2916 594 594 121 54 m 11 m 54 m 11 m
In simple words: The courtyard grows from 2916 m² to 4225 m², an extra 1309 m² to be tiled.

Teacher's Note:
1. Extra area = 2ab + b² = 2 × 54 × 11 + 11².
2. Write units (m²) in every area answer.

 

Question 20. Case Study 2: Mental Maths in Scholarship Test
In a state-level scholarship test, students are expected to perform calculations quickly without calculators. One question asks candidates to evaluate 107 × 93. A confident student rewrites the numbers around 100 and applies the difference of squares identity to reach the answer rapidly. This strategy saves valuable time and reduces chances of error, giving the student an advantage over others. The scenario highlights the importance of mastering algebraic identities for speed, accuracy, and confidence in competitive examinations.
(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) 107 × 93 = (100 + 7)(100 − 7).
(iii) = 100² − 7² = 10000 − 49 = 9951.
(iv) Directly: 107 × 93 = 107 × 90 + 107 × 3 = 9630 + 321 = 9951. Both give 9951, so the result is verified.
In simple words: 107 and 93 are 7 on each side of 100, so the product is 10000 − 49 = 9951.

Teacher's Note:
1. Verification by ordinary multiplication confirms the identity method.
2. The trick works best when the numbers are close to a round number.

 

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CBSE Class 8 Mathematics Worksheets for Ganita Prakash Part 1 Chapter 06 We Distribute Yet Things Multiply

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