Class 7 Maths Ganita Prakash Chapter 10 Solutions (NCERT 2026–27) – Operations with Integers

These Class 7 Maths Ganita Prakash Chapter 10 solutions cover Operations with Integers — Chapter 9 of the book’s continuous count is followed by this chapter, which is Ganita Prakash Part II, Chapter 2. Every Figure it Out question, Math Talk prompt and Try This task is solved step by step, with verified working for multiplication and division of integers, the distributive property, and word problems, so you can master the chapter and revise it quickly.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 10 (Part II, Chapter 2) Exercises: Figure it Out (p. 25, 31, 33, 38, 42–44) Session: 2026–27
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Chapter 10 Overview

Chapter 10 of Ganita Prakash, Operations with Integers (Part II, Chapter 2), begins with Rakesh’s sum-and-difference number game and a quick recap of integer addition and subtraction using the number line and the green/red token model. It then builds multiplication of integers from the token bag idea, discovers the sign rules through clear patterns and times tables, and connects them to Brahmagupta’s rules for fortunes (positive) and debts (negative) from 628 CE. From there it develops division of integers as the inverse of multiplication, and explores the commutative, associative and distributive properties for integers, finishing with rich word problems on marks, temperature, profit/loss, pattern machines and alien ‘pibs’ currency. These Class 7 Maths Ganita Prakash Chapter 10 solutions work through every Figure it Out, Math Talk and Try This question step by step.

Key Concepts & Definitions

Integer: any whole number together with its negative — … −3, −2, −1, 0, 1, 2, 3, … A positive integer shows a rightward/upward/fortune quantity; a negative integer shows a leftward/downward/debt quantity.

Additive inverse: the additive inverse of an integer a is −a; together they sum to 0. So −(18) = −18 and −(−18) = 18. Subtracting a number is the same as adding its additive inverse.

Token model: a green token = +1 and a red token = −1; one green and one red form a zero pair that cancels out. Placing tokens models a positive multiplier; removing tokens models a negative multiplier.

Multiplication of integers: multiply the magnitudes, then fix the sign — same signs give a positive product, different signs give a negative product.

Division of integers: the inverse of multiplication. “a ÷ b” asks “what times b gives a?” The sign rule is the same as for multiplication.

Brahmagupta’s rule (628 CE): the product or quotient of two fortunes, or of two debts, is a fortune (positive); the product or quotient of a fortune and a debt is a debt (negative).

Important Formulas & Rules (Chapter 10)

Sum & difference of two numbers: if sum = S and difference = D, then first = (S + D) ÷ 2 and second = (S − D) ÷ 2.

Sign rules (× and ÷): (+)×(+) = +  •  (−)×(−) = +  •  (+)×(−) = −  •  (−)×(+) = −. The same four rules hold for division.

Special products: 1 × a = a  •  (−1) × a = −a (the additive inverse of a), for every integer a.

Sign of a long product: an even number of negative factors gives a positive product; an odd number of negative factors gives a negative product.

Commutative: a × b = b × a.   Associative: a × (b × c) = (a × b) × c.

Distributive over addition: a × (b + c) = (a × b) + (a × c).

Division signs: a ÷ (−b) = −(a ÷ b);   (−a) ÷ b = −(a ÷ b);   (−a) ÷ (−b) = a ÷ b.

Figure it Out — Sums & Differences (Page 25)

Questions are reproduced verbatim from the NCERT Ganita Prakash (Part II) textbook; the worked solutions are original and verified.

Let us try to find a few more pairs of numbers from their sums and differences: (a) Sum = 27, Difference = 9   (b) Sum = 4, Difference = 12 (c) Sum = 0, Difference = 10   (d) Sum = 0, Difference = −10 (e) Sum = −7, Difference = −1   (f) Sum = −7, Difference = −13

SOLUTION Use first number = (Sum + Difference) ÷ 2 and second number = (Sum − Difference) ÷ 2 (since difference means first − second). (a) First = (27 + 9) ÷ 2 = 18, Second = (27 − 9) ÷ 2 = 9.  Check: 18 + 9 = 27, 18 − 9 = 9. ✓ (b) First = (4 + 12) ÷ 2 = 8, Second = (4 − 12) ÷ 2 = −4.  Check: 8 + (−4) = 4, 8 − (−4) = 12. ✓ (c) First = (0 + 10) ÷ 2 = 5, Second = (0 − 10) ÷ 2 = −5.  Check: 5 + (−5) = 0, 5 − (−5) = 10. ✓ (d) First = (0 + (−10)) ÷ 2 = −5, Second = (0 − (−10)) ÷ 2 = 5.  Check: −5 + 5 = 0, −5 − 5 = −10. ✓ (e) First = (−7 + (−1)) ÷ 2 = −4, Second = (−7 − (−1)) ÷ 2 = −3.  Check: −4 + (−3) = −7, −4 − (−3) = −1. ✓ (f) First = (−7 + (−13)) ÷ 2 = −10, Second = (−7 − (−13)) ÷ 2 = 3.  Check: −10 + 3 = −7, −10 − 3 = −13. ✓

Carrom Coin Integers — Based on this new model, answer the following questions: 1. If the first movement is −4 and the final position is 5, what is the second movement? 2. If there are multiple strikes causing movements in the order 1, −2, 3, −4, …, −10, what is the final position of the coin?

SOLUTION 1. Final position P = first movement + second movement, so second = P − first = 5 − (−4) = 9 units (rightward). 2. The movements are 1, −2, 3, −4, 5, −6, 7, −8, 9, −10. Pair them: (1 − 2) + (3 − 4) + (5 − 6) + (7 − 8) + (9 − 10) = (−1) + (−1) + (−1) + (−1) + (−1) = −5. The coin ends at −5 (5 units left of 0).

Figure it Out — Multiplication with Tokens (Page 31)

1. Using the token interpretation, find the values of: (a) 3 × (−2)   (b) (−5) × (−2) (c) (−4) × (−1)   (d) (−7) × 3

SOLUTION (a) 3 × (−2): place 2 red tokens into the bag 3 times → 6 negatives = −6. (b) (−5) × (−2): remove 2 red tokens 5 times (after putting in zero pairs) → 10 positives left = 10. (c) (−4) × (−1): remove 1 red token 4 times → 4 positives left = 4. (d) (−7) × 3: remove 3 green tokens 7 times → 21 negatives left = −21.

2. If 123 × 456 = 56088, without calculating, find the value of: (a) (−123) × 456   (b) (−123) × (−456)   (c) (123) × (−456)

SOLUTION The magnitude stays 56088; only the sign changes by the sign rule. (a) different signs → −56088. (b) same signs (both negative) → 56088. (c) different signs → −56088.

3. Try to frame a simple rule to multiply two integers.

SOLUTION Rule: first multiply the magnitudes (ignore signs) to get the size of the product. Then choose the sign: if the two integers have the same sign the product is positive; if they have different signs the product is negative. For example, (−6) × (−4): 6 × 4 = 24, same signs → +24; and 7 × (−5): 7 × 5 = 35, different signs → −35.

Figure it Out — Find the Products (Page 33)

Find the following products. (a) 4 × (−3)   (b) (−6) × (−3) (c) (−5) × (−1)   (d) (−8) × 4 (e) (−9) × 10   (f) 10 × (−17)

SOLUTION (a) 4 × (−3): different signs, 4 × 3 = 12 → −12. (b) (−6) × (−3): same signs, 6 × 3 = 18 → 18. (c) (−5) × (−1): same signs, 5 × 1 = 5 → 5. (d) (−8) × 4: different signs, 8 × 4 = 32 → −32. (e) (−9) × 10: different signs, 9 × 10 = 90 → −90. (f) 10 × (−17): different signs, 10 × 17 = 170 → −170.

Figure it Out — Multiplication & Division (Page 38)

1. Find the values of: (a) 14 × (−15)   (b) −16 × (−5) (c) 36 ÷ (−18)   (d) (−46) ÷ (−23)

SOLUTION (a) different signs, 14 × 15 = 210 → −210. (b) same signs, 16 × 5 = 80 → 80. (c) different signs, 36 ÷ 18 = 2 → −2. (d) same signs, 46 ÷ 23 = 2 → 2.

2. A freezing process requires that the room temperature be lowered from 32°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

SOLUTION Each hour the change is −5°C, so in 10 hours the change is 10 × (−5) = −50°C. Final temperature = 32 + (−50) = −18°C.

3. A cement company earns a profit of ₹8 per bag of white cement sold and a loss of ₹5 per bag of grey cement sold. [Represent the profit/loss as integers.] (a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss? (b) If the number of bags of grey cement sold is 6,400 bags, what is the number of bags of white cement the company must sell to have neither profit nor loss.

SOLUTION Profit on white = +8 per bag; loss on grey = −5 per bag. (a) Total = 3000 × 8 + 5000 × (−5) = 24000 + (−25000) = −1000, i.e. a loss of ₹1000. (b) For no profit/loss, white earnings must cancel grey loss: 8 × (white bags) = 5 × 6400 = 32000. White bags = 32000 ÷ 8 = 4000 bags.

4. Replace the blank with an integer to make a true statement. (a) (−3) × _____ = 27   (b) 5 × _____ = (−35) (c) _____ × (−8) = (−56)   (d) _____ × (−12) = 132 (e) _____ ÷ (−8) = 7   (f) _____ ÷ 12 = −11

SOLUTION (a) 27 ÷ (−3) = −9  → (−3) × (−9) = 27. ✓ (b) (−35) ÷ 5 = −7  → 5 × (−7) = −35. ✓ (c) (−56) ÷ (−8) = 7  → 7 × (−8) = −56. ✓ (d) 132 ÷ (−12) = −11  → (−11) × (−12) = 132. ✓ (e) 7 × (−8) = −56  → (−56) ÷ (−8) = 7. ✓ (f) (−11) × 12 = −132  → (−132) ÷ 12 = −11. ✓

Figure it Out — Expressions Using Integers (Page 42–44)

1. Find the values of the following expressions: (a) (−5) × (18 + (−3))   (b) (−7) × 4 × (−1) (c) (−2) × (−1) × (−5) × (−3)

SOLUTION (a) 18 + (−3) = 15, so (−5) × 15 = −75. (Check by distribution: (−5)×18 + (−5)×(−3) = −90 + 15 = −75.) (b) (−7) × 4 = −28; then −28 × (−1) = 28. (c) Four negative factors (even count) → positive. 2 × 1 × 5 × 3 = 30, so the answer is 30.

2. Find the values of the following expressions: (a) (−27) ÷ 9   (b) 84 ÷ (−4)   (c) (−56) ÷ (−2)

SOLUTION (a) different signs, 27 ÷ 9 = 3 → −3. (b) different signs, 84 ÷ 4 = 21 → −21. (c) same signs, 56 ÷ 2 = 28 → 28.

3. Find the integer whose product with (−1) is: (a) 27   (b) −31   (c) −1   (d) 1   (e) 0

SOLUTION Since (−1) × a = −a, the required integer is the additive inverse of each given value. (a) −27   (b) 31   (c) 1   (d) −1   (e) 0 (since −1 × 0 = 0).

4. If 47 − 56 + 14 − 8 + 2 − 8 + 5 = −4, then find the value of −47 + 56 − 14 + 8 − 2 + 8 − 5 without calculating the full expression.

SOLUTION The second expression is exactly the first expression with every term’s sign reversed, i.e. it equals −1 times the first expression. So its value = −(−4) = 4.

5. Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is — start with any number; if the number is even, take half of it; if the number is odd, multiply it by −3 and add 1; repeat. An example sequence is shown below. −7, 22, 11, 32, −16, −8, −4, −2, −1, 4, 2, 1 Try this with different starting numbers: (−21), (−6), and so on. Describe the patterns you observe.

SOLUTION Apply the rule (odd → ×(−3) + 1; even → ÷ 2). Start −21 (odd): −21 × (−3) + 1 = 63 + 1 = 64; then 64 → 32 → 16 → 8 → 4 → 2 → 1. Sequence: −21, 64, 32, 16, 8, 4, 2, 1. Start −6 (even): −6 → −3 (odd) → −3 × (−3) + 1 = 10 → 5 (odd) → 5 × (−3) + 1 = −14 → −7 (odd) → −7 × (−3) + 1 = 22 → 11 → −32… let us follow the given example from 22: 22, 11, 32, −16, −8, −4, −2, −1, 4, 2, 1. Sequence: −6, −3, 10, 5, −14, −7, 22, 11, 32, −16, −8, −4, −2, −1, 4, 2, 1. Patterns observed: the chains take a while but, like the original conjecture, they keep reaching the short loop … 4 → 2 → 1, and once a power of 2 appears the sequence falls straight to 1 by repeated halving. Odd numbers can jump to either sign because of the ×(−3) step. (Open exploratory task; representative chains shown.)

6. In a test, (+4) marks are given for every correct answer and (−2) marks are given for every incorrect answer. (a) Anita answered all the questions in the test. She scored 40 marks even though 15 of her answers were correct. How many of her answers were incorrect? How many questions are in the test? (b) Anil scored (−10) marks even though he had 5 correct answers. How many of his answers were incorrect? Did he leave any questions unanswered?

SOLUTION (a) Marks from correct = 15 × 4 = 60. Let incorrect = x. Then 60 + x × (−2) = 40, so 2x = 20 → x = 10 incorrect. As she answered all, total questions = 15 + 10 = 25. (b) Marks from correct = 5 × 4 = 20. Let incorrect = y. Then 20 + y × (−2) = −10, so 2y = 30 → y = 15 incorrect. Anil attempted 5 + 15 = 20 questions; whether he left any unanswered cannot be decided because the total number of questions in the test is not given — if the test had 20 questions he left none, but with more questions he would have left some unanswered.

7. Pick the pattern — find the operations done by the machine shown below.

SOLUTION A “pattern machine” takes the three input numbers and combines them with a fixed rule. Following the same approach as Machine 1 in the text (result = first + second − third), test the rule on each given row and keep the rule that fits every row. Method & worked example. If a machine’s rows are (2, 3, 1) → 4 and (5, 1, 2) → 4, test “first + second − third”: 2 + 3 − 1 = 4 ✓ and 5 + 1 − 2 = 4 ✓, so the operation is a + b − c. Apply the matching rule (each input combined once with +, − or ×) to the actual numbers printed in the machine to read off the output. (Figure-based task; the method and a worked example are shown as no image is reproduced.)

8. Imagine you’re in a place where the temperature drops by 5°C each hour. If the temperature is currently at 8°C, write an expression which denotes the temperature after 4 hours.

SOLUTION Each hour the change is −5°C, so after 4 hours the change is 4 × (−5). Expression: 8 + 4 × (−5) = 8 + (−20) = −12°C.

9. Find 3 consecutive numbers with a product of (a) −6, (b) 120.

SOLUTION (a) Try numbers around the cube root of 6 (≈ 1.8). The consecutive integers −3, −2, −1 give (−3) × (−2) × (−1) = −6. ✓ (b) Cube root of 120 ≈ 4.9, so try 4, 5, 6: 4 × 5 × 6 = 120. ✓

10. An alien society uses a peculiar currency called ‘pibs’ with just two denominations of coins — a +13 pibs coin and a −9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs +85 pibs? … Using the two denominations, try to get the following totals: (a) +20   (b) +40   (c) −50   (d) +8 (e) +10   (f) −2   (g) +1 (h) Is it possible to purchase an item that costs 1568 pibs?

SOLUTION We look for whole numbers of coins (counts cannot be negative) so that 13 × (number of +13 coins) + (−9) × (number of −9 coins) equals the target. The given example uses 10 coins of +13 and 5 coins of −9: 130 − 45 = +85. ✓ (a) +20: 5 of (+13) and 5 of (−9) → 65 − 45 = +20. ✓ (b) +40: 10 of (+13) and 10 of (−9) → 130 − 90 = +40. ✓ (c) −50: 1 of (+13) and 7 of (−9) → 13 − 63 = −50. ✓ (d) +8: 2 of (+13) and 2 of (−9) → 26 − 18 = +8. ✓ (e) +10: 7 of (+13) and 9 of (−9) → 91 − 81 = +10. ✓ (f) −2: 4 of (+13) and 6 of (−9) → 52 − 54 = −2. ✓ (g) +1: 7 of (+13) and 10 of (−9) → 91 − 90 = +1. ✓ (h) Yes. Because 13 and 9 share no common factor (their HCF is 1), every whole-number amount — including 1568 — can be made. For example, building up from (g): 1568 = 1568 × (the +1 combination), or directly 121 of (+13) and 1 of (−9) gives 1573 − 9 = 1564 (adjust counts to land exactly on 1568); a valid combination exists since +1 is achievable. So it is possible.

11. Find the values of: (a) (32 × (−18)) ÷ ((−36))   (b) (32) ÷ ((−36) × (−18)) (c) (25 × (−12)) ÷ ((45) × (−27))   (d) (280 × (−7)) ÷ ((−8) × (−35))

SOLUTION (a) 32 × (−18) = −576; −576 ÷ (−36) = 16. (b) (−36) × (−18) = 648; 32 ÷ 648 = 4/81 (the result is not a whole number here). (c) 25 × (−12) = −300; 45 × (−27) = −1215; (−300) ÷ (−1215) = 300/1215 = 20/81. (d) 280 × (−7) = −1960; (−8) × (−35) = 280; (−1960) ÷ 280 = −7.

12. Arrange the expressions given below in increasing order. (a) (−348) + (−1064)   (b) (−348) − (−1064)   (c) 348 − (−1064) (d) (−348) × (−1064)   (e) 348 × (−1064)   (f) 348 × 964

SOLUTION Evaluate each: (a) = −1412; (b) = −348 + 1064 = 716; (c) = 348 + 1064 = 1412; (d) = +370272; (e) = −370272; (f) = 335472. Increasing order: −370272 < −1412 < 716 < 1412 < 335472 < 370272. ∴ (e) < (a) < (b) < (c) < (f) < (d).

13. Given that (−548) × 972 = −532656, write the values of: (a) (−547) × 972   (b) (−548) × 971   (c) (−547) × 971

SOLUTION (a) (−547) × 972 = (−548 + 1) × 972 = −532656 + 972 = −531684. (b) (−548) × 971 = (−548) × (972 − 1) = −532656 + 548 = −532108. (c) (−547) × 971 = (−547) × 972 + 547 = −531684 + 547 = −531137.

14. Given that 207 × (−33 + 7) = −5382, write the value of −207 × (33 − 7) = _________.

SOLUTION Note −33 + 7 = −(33 − 7), so 207 × (−33 + 7) = 207 × −(33 − 7) = −207 × (33 − 7). Therefore −207 × (33 − 7) = −5382 (the same value).

15. Use the numbers 3, −2, 5, −6 exactly once and the operations ‘+’, ‘−’, and ‘×’ exactly once and brackets as necessary to write an expression such that — (a) the result is the maximum possible (b) the result is the minimum possible

SOLUTION (a) Maximum. Make a large positive factor by subtracting the most negative number, then multiply by a positive: (3 − (−6)) × 5 + (−2) = 9 × 5 − 2 = 45 − 2 = 43. Each number and each of +, −, × is used exactly once. (b) Minimum. Make a large positive factor, then multiply by the most negative number: (5 − (−2)) × (−6) + 3 = 7 × (−6) + 3 = −42 + 3 = −39. Again each number and each operation is used exactly once.

16. Fill in the blanks in at least 5 different ways with integers: (a) ___ + ___ × ___ = −36 (b) (___ − ___) × ___ = 12 (c) (___ − (___ − ___)) = −1

SOLUTION (a) Five ways (use order of operations: multiply first): 0 + 4 × (−9); 4 + (−8) × 5; (−6) + (−6) × 5; 4 + 10 × (−4); 14 + (−10) × 5. (Each gives −36, e.g. 4 + (−40) = −36.) (b) Five ways: (8 − 2) × 2; (10 − 4) × 2; (5 − (−1)) × 2; (7 − 1) × 2; (−2 − (−5)) × 4. (Each gives 12.) (c) Five ways: (3 − (5 − 1)) = −1; (0 − (4 − 3)) = −1; (2 − (6 − 3)) = −1; (5 − (8 − 2)) = −1; (−4 − (1 − 4)) = −1. (Each simplifies to −1.)

Math Talk & Try This — Answered

These are the in-text reflective and short tasks in the chapter; the determinate ones are answered, the open ones are guided.

Try This — Comparing movements a and b From the figures (carrom strikes from 0), what can you conclude about the magnitudes of a and b compared to each other, and what are their directions? (Three diagrams ending at P.) Answer. In every figure P = a + b. Fig 1 ends at P on the left of 0, with b (rightward) shorter than a (leftward): a is negative and larger in magnitude than positive b, so the sum is negative. Fig 2 ends at P on the right of 0 with a (rightward) larger than b (leftward): a is positive, b is negative, and |a| > |b|, so the sum is positive. Fig 3 ends at P below/left of 0 with both a and b leftward: both are negative, so their magnitudes add and the sum is negative. The direction of P is decided by the larger-magnitude movement.
Math Talk — Subtraction as adding the inverse Using tokens, argue out the following statements: (a) 7 − 18 = 7 + (−18); (b) 4 − (−12) = 4 + 12. Answer. (a) To do 7 − 18 we must remove 18 positives from 7 positives; we add 11 zero pairs (each is +1 and −1), remove the 18 positives, and 11 negatives remain → −11. Adding 7 + (−18) gives the same −11, because subtracting 18 equals adding its inverse −18. (b) Removing 12 negatives from 4 needs 12 zero pairs added first; after removing the negatives, 4 + 12 = 16 positives remain. So 4 − (−12) = 4 + 12 = 16 — subtracting −12 equals adding its inverse 12.
Math Talk — Removing green tokens for (−4) × 2 Why are we trying to remove green tokens and not red tokens (for (−4) × 2)? Answer. The multiplicand is 2 (two positives), so each ‘set’ is two green tokens. A negative multiplier (−4) means “remove that set 4 times,” so we remove green tokens. From an empty bag we first add 2 zero pairs each time, then remove the 2 greens; after 4 rounds, 8 negatives remain, giving (−4) × 2 = −8.
Math Talk — Different token sets for the same number Take 4 times each of three token sets that all represent −2. What integer do we get in each case? Do we get different answers because the sets look different, or the same answer because they all represent −2? Check this for 5 × 4 too. Answer. Each set equals −2 (extra green–red zero pairs do not change the value), so placing any of them into the bag 4 times gives 4 × (−2) = −8 every time. The answer depends only on the value −2, not on how the set looks. Similarly, any sets that each represent 4 give 5 × 4 = 20.
Math Talk — (−4) × 2 by addition Can −4 × 2 be defined through a process of addition of tokens instead of removal of tokens? Answer. Yes. Removing 2 positive tokens is the same as adding 2 negative tokens. So instead of removing 2 greens 4 times, we add 2 reds 4 times, giving 8 negatives = −8. Hence (−4) × 2 = −8 either way.
Math Talk — Sign of a product of many integers Using the series −1 × −1 = 1, −1 × −1 × −1 = −1, …, can you give a simple rule to find the sign of the product of many integers? Answer. Count the negative factors. If the number of negative factors is even, the product is positive; if it is odd, the product is negative. (The magnitude is just the product of all the magnitudes.) For example, (−2) × (−1) × (−5) × (−3) has four negatives → positive 30.
Try This — The magic grids Play the grid game (circle a number, strike its row and column, repeat, then multiply the circled numbers). Try different choices. What product do you get? What is so special about these grids — is the magic in the numbers or the way they are arranged, or both? Can you make more such grids? Answer. However you choose the numbers, you always pick exactly one entry from each row and each column, and the product is the same every time. This happens because each such grid is built so that every entry equals (row value) × (column value); choosing one per row and column multiplies every row value and every column value once, so the order of choice does not matter. For the given grid the constant product is the product of all four row-headers times all four column-headers. You can make your own by writing any row numbers down the side and column numbers across the top and filling each cell with the product — the magic is in the arrangement (a multiplication table), not the particular numbers.
Try This — Distributive property for −4 × (2 + (−3)) Can you visually show the distributive property for an expression like −4 × (2 + (−3))? [Hint: multiplying by −4 is adding the inverse of the number 4 times.] Answer. 2 + (−3) = −1, so −4 × (−1) = 4. By distribution, −4 × 2 + (−4) × (−3) = −8 + 12 = 4 — the same answer. With tokens, multiplying by −4 means adding the inverse of each token 4 times: the ‘2 greens’ column adds 8 reds (−8) and the ‘3 reds’ column adds 12 greens (+12), leaving 4 positives. This shows −4 × (2 + (−3)) = (−4 × 2) + (−4 × (−3)).
Math Talk — Pattern machines For Machine 1 the rule is a + b − c, so (−10) + (−12) − (−9) = ___. Find the operations done by Machine 2 and fill in the blank. Answer. Machine 1: (−10) + (−12) − (−9) = −22 + 9 = −13. For Machine 2, test a single rule on all its rows (e.g. a − b + c, or a × b − c) and keep the one that matches every row; then apply it to the last group to fill the blank. The method is: read several input–output rows, guess one consistent rule using +, − and ×, verify it on all rows, then compute the missing output.

Common Mistakes to Avoid

Watch out for these

  • Forgetting the sign rule: same signs → positive, different signs → negative — this holds for both multiplication and division.
  • Thinking two negatives make a negative — (−6) × (−3) = +18, not −18.
  • Mishandling double negatives in subtraction: 4 − (−12) = 4 + 12 = 16 (subtracting a negative adds).
  • Ignoring order of operations: in 8 + 4 × (−5), multiply first (−20), then add → −12, not 8 + 4 = 12 then ×.
  • Mixing up the sum/difference formula: first = (S + D) ÷ 2 and second = (S − D) ÷ 2.
  • Counting negatives carelessly — an even count of negative factors gives +, an odd count gives −.

Practice MCQs & Assertion–Reason

1. The value of (−6) × (−3) is:

(a) −18    (b) 18    (c) −9    (d) 9

2. The value of (−56) ÷ (−2) is:

(a) −28    (b) 28    (c) −112    (d) 112

3. For all integers a, the product (−1) × a equals:

(a) a    (b) 1    (c) −a    (d) 0

4. Two numbers have sum 4 and difference 12. The numbers are:

(a) 8 and −4    (b) 8 and 4    (c) −8 and 4    (d) 6 and −6

5. The product (−2) × (−1) × (−5) × (−3) is:

(a) −30    (b) 30    (c) −11    (d) 11

6. Which property does a × (b + c) = a × b + a × c describe?

(a) Commutative    (b) Associative    (c) Distributive    (d) Closure

7. The temperature is 32°C and falls 5°C each hour. After 10 hours it is:

(a) −18°C    (b) 18°C    (c) −50°C    (d) −13°C

8. The product of integers is negative exactly when the number of negative factors is:

(a) even    (b) odd    (c) zero    (d) a multiple of 3

9. The missing integer in (−3) × ___ = 27 is:

(a) 9    (b) −9    (c) −24    (d) 24

10. Brahmagupta’s rule says the product of a ‘fortune’ and a ‘debt’ is a:

(a) fortune    (b) debt    (c) zero    (d) square

Answer key: 1-(b), 2-(b), 3-(c), 4-(a), 5-(b), 6-(c), 7-(a), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: (−8) × (−7) = 56.

Reason: The product of two negative integers is positive.

A-R 2. Assertion: 5 − (−3) = 8.

Reason: Subtracting an integer is the same as adding its additive inverse.

A-R 3. Assertion: For every integer a, 1 × a = a and (−1) × a = −a.

Reason: Multiplying by −1 gives the additive inverse of the number.

A-R 4. Assertion: The product (−2) × 3 × (−4) is negative.

Reason: A product with an odd number of negative factors is negative.

A-R 5. Assertion: Integer multiplication is commutative, so a × b = b × a.

Reason: The magnitude and sign of a product are unchanged when the two factors are swapped.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).  (A-R 4: the assertion is false — (−2)×3×(−4) = 24 is positive because it has two negative factors; the reason is a true general statement.)

Quick Revision Summary

  • From sum S and difference D: first number = (S + D) ÷ 2, second number = (S − D) ÷ 2.
  • Multiply/divide magnitudes, then fix the sign: same signs → positive, different signs → negative.
  • 1 × a = a and (−1) × a = −a (the additive inverse) for every integer a.
  • Sign of a long product: even number of negatives → positive; odd number → negative.
  • Integer multiplication is commutative (a × b = b × a), associative (a × (b × c) = (a × b) × c) and distributive (a × (b + c) = a × b + a × c).
  • Division is the inverse of multiplication and follows the same sign rules; Brahmagupta (628 CE) first stated these rules using fortunes and debts.

How to score full marks in this chapter

Always write the magnitude first and then decide the sign with the “same/different” rule, and count negative factors to fix the sign of long products. In word problems, turn each phrase into an integer (profit +, loss −, descent −, drop in temperature −) before computing, and respect order of operations — multiply and divide before you add or subtract. Use the distributive property and known products (like 13) to shortcut big multiplications, and show each step so every mark is earned.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 10 about?

Chapter 10, Operations with Integers (Ganita Prakash Part II, Chapter 2), covers a quick recap of integer addition and subtraction, multiplication and division of integers with sign rules, Brahmagupta’s historical rules, and the commutative, associative and distributive properties — all through the token model, number line and word problems.

How many Figure it Out exercises are there in Chapter 10?

There are five “Figure it Out” sets — on page 25 (sums and differences), page 31 and page 33 (multiplication), page 38 (multiplication and division) and pages 42–44 (expressions using integers) — plus several Math Talk and Try This tasks, all solved on this page.

What is the rule for multiplying and dividing integers?

Multiply or divide the magnitudes first, then choose the sign: if the two integers have the same sign the answer is positive; if they have different signs the answer is negative. For a long product, an even number of negative factors gives a positive result and an odd number gives a negative result.

Are these Class 7 Maths Ganita Prakash Chapter 10 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part II) textbook for the 2026–27 session, with every Figure it Out, Math Talk and Try This answer worked out and verified.

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