CMDLearn

Learn

Class 8 Maths: We Distribute, Yet Things Multiply, 8 practice questions with answers

CMDLearn ·

These are 8 practice questions on Chapter 6, We Distribute, Yet Things Multiply, of the new NCERT Class 8 Maths textbook, Ganita Prakash, Part 1. Each has a worked answer and the textbook page it comes from. They cover working backwards from how much a product changes, extending a field's length and width, spotting a mistake in a quick multiplication, undoing the ×11 and ×101 shortcuts, expanding a product in a shopping problem, a quick way to square numbers ending in 5, and finding two numbers from the increase in their product. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the data is taken from the chapter, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Prakash, Grade 8, Part 1, Chapter 6, printed pages 136 to 157):

Working backwards with products

Question 1. Two numbers are multiplied. If only the first number is increased by 1, the product grows by 47. If only the second number is increased by 1, the product grows by 38. What are the two numbers and their product?

Answer: the first number is 38, the second is 47, and the product is 1786. In the product ab, increasing a by 1 gives (a + 1)b = ab + b, so the product grows by b, the other number. So the second number is 47, and in the same way the first number is 38. Their product is 38 × 47 = 1786. (Increasing one number by 1, pages 137 and 138.)

Question 2. A farmer's rectangular field is 67 m long and 58 m wide. She buys land to extend both the length and the width by 3 m. By how much does the area grow?

Answer: by 384 m². By Identity 1, (67 + 3)(58 + 3) = 67 × 58 + 3 × 58 + 67 × 3 + 3 × 3. So the area grows by 3 × 58 + 3 × 67 + 9 = 174 + 201 + 9 = 384 m². The old area was 67 × 58 = 3886 m². (Identity 1, page 139.)

Question 3. Ravi works out 73 × 79 in his head: "76² is 5776, and the numbers are 3 less and 3 more than 76, so the answer is 5776 – 3 = 5773." Find his mistake and give the right answer.

Answer: he should subtract 3² = 9, not 3; the answer is 5767. By Identity 1C, (76 – 3)(76 + 3) = 76² – 3² = 5776 – 9 = 5767. Ravi subtracted b instead of b². (Identity 1C, page 148.)

Shortcuts in reverse

Question 4. A number multiplied by 11 gives 6457. Using the one-line method for ×11, find the number.

Answer: 587. In the one-line method the units digit of the product is the units digit of the number, so the number ends in 7. The next digit of the product, 5, is the last digit of 7 + (the tens digit of the number); for it to end in 5, the tens digit must be 8, and 7 + 8 gives a carry of 1. The next digit, 4, is the last digit of 8 + (the hundreds digit) + 1; for it to end in 4, the hundreds digit must be 5, again with a carry of 1. The front digit is then 5 + 1 = 6. So the number is 587. Check: 587 × 11 = 5870 + 587 = 6457. (Multiplying by 11, pages 143 and 144.)

Question 5. A three-digit number multiplied by 101 gives 89183. Find the number.

Answer: 883. Write the number as cba. Then cba × 101 = cba00 + cba, so, column by column from the right: the units digit of the product is a, the tens digit is b, the hundreds digit is c + a (with any carry going on), the thousands digit is b plus that carry, and the front is c. In 89183 the units digit is 3, so a = 3, and the tens digit is 8, so b = 8. The hundreds digit is 1, so c + 3 ends in 1 and c = 8, with a carry of 1. Check the rest: the thousands digit is 8 + 1 = 9 and the front is 8, as in 89183. So the number is 883. Check: 883 × 101 = 88300 + 883 = 89183. (Multiplying by 101, page 144.)

Expanding in real settings

Question 6. A shop sells pens at ₹(x + 3) each. One day it sells (2x + 1) pens. Write the day's takings as an expanded expression, and find the takings when x = 58.

Answer: 2x² + 7x + 3 rupees; ₹7137 when x = 58. By the distributive property, (x + 3)(2x + 1) = x × 2x + x × 1 + 3 × 2x + 3 × 1 = 2x² + x + 6x + 3 = 2x² + 7x + 3. When x = 58, the pens cost ₹61 each and 117 are sold, so the takings are 61 × 117 = ₹7137, and 2 × 58² + 7 × 58 + 3 gives the same. (Distributive property and Identity 1, pages 136 to 139.)

Question 7. Explain why any number ending in 5, written as 10a + 5, has the square 100 × a(a + 1) + 25. Use this to find 85².

Answer: 85² = 7225. By Identity 1A, (10a + 5)² = 100a² + 2 × 10a × 5 + 25 = 100a² + 100a + 25 = 100 × a(a + 1) + 25. For 85, a = 8, so 85² = 100 × 8 × 9 + 25 = 7200 + 25 = 7225. (Identity 1A, page 145.)

Question 8. Two numbers differ by 46. When both are increased by 1, their product grows by 129. Find the numbers.

Answer: 87 and 41. When both numbers in ab are increased by 1, the product grows by a + b + 1. So a + b + 1 = 129, which gives a + b = 128. With a – b = 46, adding gives 2a = 174, so a = 87 and b = 41. Check: 87 – 41 = 46 and 87 + 41 + 1 = 129. (Increasing both numbers by 1, page 138.)

Practise more, and check your marks

Where CMDLearn fits

CMDLearn is building an AI tutor for Classes 1 to 10 that explains a step and checks the child's working, but leaves the writing to the child. It is still in development, and no child uses it yet. If you would like to hear when there is something to try, you can join the waitlist.

Sources

All opened and read on 22 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 135.

The properties and identities are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.