Class 7 Maths Ganita Prakash Chapter 2 Solutions (NCERT 2026–27) – Arithmetic Expressions

These Class 7 Maths Ganita Prakash Chapter 2 solutions cover Arithmetic Expressions from the new NCF-2023 textbook (Reprint 2026–27). Every Figure it Out question is solved step by step — comparing expressions, identifying terms, removing brackets and using the distributive property — so you can master the chapter and revise it quickly.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 2 Exercises: Figure it Out (p. 25), (p. 37), (p. 41), (p. 42–43) Session: 2026–27
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Chapter 2 Overview

Chapter 2 of Ganita Prakash, Arithmetic Expressions, teaches how to read, compare and evaluate mathematical phrases like 13 + 2, 20 − 4 and 5 × 25. It starts with simple expressions and how to compare them using =, < and > through reasoning rather than full calculation. It then introduces brackets and the powerful idea of terms (the parts of an expression separated by a ‘+’ sign, with subtraction rewritten as adding the inverse). Using terms, the chapter explains the commutative and associative properties of addition, the rules for removing brackets, and the distributive property — that a multiple of a sum equals the sum of the multiples. The Class 7 Maths Ganita Prakash Chapter 2 solutions below work through every Figure it Out, Math Talk and Try This task step by step.

Key Concepts & Definitions

Arithmetic expression: a mathematical phrase formed with numbers and the operations +, −, ×, ÷, e.g. 13 + 2 or 12 × 5. Its value is the single number it evaluates to (the value of 13 + 2 is 15).

Comparing expressions: two expressions are compared by their values using ‘=’, ‘<’ or ‘>’, e.g. 10 + 2 > 7 + 1 because 12 > 8.

Brackets: the part inside brackets is always evaluated first, e.g. 30 + (5 × 4) = 30 + 20 = 50.

Terms: the parts of an expression separated by a ‘+’ sign. To find the terms, rewrite every subtraction as adding the inverse, e.g. 83 − 14 = 83 + (−14), so the terms are 83 and −14. Products such as 6 × 5 are single terms.

Evaluating with terms: evaluate each term first, then add all the terms, e.g. 30 + 5 × 4 = 30 + (5 × 4) = 30 + 20 = 50.

Inverse of a number: the same number with the opposite sign; the inverse of 14 is −14 and the inverse of −14 is 14. Subtracting a number is the same as adding its inverse.

Taxicab-style reasoning: many comparisons can be settled by reasoning about how terms change, without computing the values at all.

Important Formulas & Properties (Chapter 2)

Commutative property of addition: Term 1 + Term 2 = Term 2 + Term 1 — swapping terms does not change the value.

Associative property of addition: (Term 1 + Term 2) + Term 3 = Term 1 + (Term 2 + Term 3) — grouping terms does not change the value. Adding terms in any order gives the same value.

Brackets after ‘+’: a + (b + c) = a + b + c and a + (b − c) = a + b − c — signs inside do not change.

Brackets after ‘−’: a − (b + c) = a − b − c and a − (b − c) = a − b + c — every sign inside changes.

Distributive property: a × (b + c) = a × b + a × c and a × (b − c) = a × b − a × c — the multiple of a sum (difference) is the sum (difference) of the multiples.

Handy version: (b + c) × a = b × a + c × a, useful for quick mental products like 97 × 25 = (100 − 3) × 25.

Figure it Out — Comparing Expressions (Page 25)

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

1. Fill in the blanks to make the expressions equal on both sides of the = sign: (a) 13 + 4 = ____ + 6 (b) 22 + ____ = 6 × 5 (c) 8 × ____ = 64 ÷ 2 (d) 34 – ____ = 25

SOLUTION (a) LHS = 13 + 4 = 17, so the blank + 6 = 17 → blank = 17 − 6 = 11. (b) RHS = 6 × 5 = 30, so 22 + blank = 30 → blank = 30 − 22 = 8. (c) RHS = 64 ÷ 2 = 32, so 8 × blank = 32 → blank = 32 ÷ 8 = 4. (d) 34 − blank = 25 → blank = 34 − 25 = 9.

2. Arrange the following expressions in ascending (increasing) order of their values. (a) 67 – 19   (b) 67 – 20   (c) 35 + 25   (d) 5 × 11   (e) 120 ÷ 3

SOLUTION Find each value: (a) 67 − 19 = 48; (b) 67 − 20 = 47; (c) 35 + 25 = 60; (d) 5 × 11 = 55; (e) 120 ÷ 3 = 40. Ordering the values 40 < 47 < 48 < 55 < 60. ∴ ascending order is (e) 120 ÷ 3, (b) 67 – 20, (a) 67 – 19, (d) 5 × 11, (c) 35 + 25.

Figure it Out — Removing Brackets (Page 37)

1. Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal. (a) 24 + (6 – 4) = 24 + 6 _____ (b) 38 + (_____ _____) = 38 + 9 – 4 (c) 24 – (6 + 4) = 24 [ ] 6 – 4 (d) 24 – 6 – 4 = 24 – 6 _____ (e) 27 – (8 + 3) = 27 [ ] 8 [ ] 3 (f) 27 – (_____ _____) = 27 – 8 + 3

SOLUTION (a) 24 + (6 − 4) = 24 + 6 − 4, so the box is − 4 (i.e. 24 + 6 − 4). (b) Since brackets after ‘+’ keep their signs, 38 + (9 − 4) = 38 + 9 − 4, so the bracket is (9 – 4). (c) Removing a bracket after ‘−’ flips signs: 24 − (6 + 4) = 24 − 6 − 4, so the box is − (i.e. 24 − 6 − 4). (d) 24 − 6 − 4 = 24 − 6 − 4, so the blank is − 4. (e) 27 − (8 + 3) = 27 − 8 − 3, so both boxes are − (i.e. 27 − 8 − 3). (f) We need a bracket whose removal after ‘−’ gives −8 + 3, so before flipping it was (8 − 3): 27 – (8 – 3).

2. Remove the brackets and write the expression having the same value. (a) 14 + (12 + 10)   (b) 14 – (12 + 10)   (c) 14 + (12 – 10) (d) 14 – (12 – 10)   (e) –14 + 12 – 10   (f) 14 – (–12 – 10)

SOLUTION (a) Brackets after ‘+’ keep signs: 14 + 12 + 10 (= 36). (b) Brackets after ‘−’ flip signs: 14 – 12 – 10 (= −8). (c) After ‘+’ keep signs: 14 + 12 – 10 (= 16). (d) After ‘−’ flip signs: 14 – 12 + 10 (= 12). (e) No brackets to remove; already in term form: –14 + 12 – 10 (= −12). (f) After ‘−’ flip both signs: 14 + 12 + 10 (= 36).

3. Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal? (a) (6 + 10) – 2 and 6 + (10 – 2) (b) 16 – (8 – 3) and (16 – 8) – 3 (c) 27 – (18 + 4) and 27 + (–18 – 4)

SOLUTION (a) (6 + 10) − 2 = 16 − 2 = 14; 6 + (10 − 2) = 6 + 8 = 14. Equal — only additions/subtractions of terms are involved, so regrouping does not change the value. (b) 16 − (8 − 3) = 16 − 5 = 11; (16 − 8) − 3 = 8 − 3 = 5. Not equal (11 ≠ 5) — the bracket after ‘−’ changes the sign of −3. (c) 27 − (18 + 4) = 27 − 22 = 5; 27 + (−18 − 4) = 27 − 22 = 5. Equal — both have terms 27, −18, −4. When equal? Two such expressions are equal exactly when they have the same set of terms (same numbers with the same signs after the brackets are removed).

4. In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms. (a) 319 + 537, 319 – 537, – 537 + 319, 537 – 319 (b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109

SOLUTION (a) Compare the terms: 319 + 537 has terms {319, 537}; 319 − 537 has {319, −537}; −537 + 319 has {−537, 319}; 537 − 319 has {537, −319}. So 319 – 537 = –537 + 319 (same terms). The other two have different terms and are not equal to these or to each other. (b) Reduce each to its terms: the first three are identical, ‘87 + 46 − 109’, terms {87, 46, −109}. ‘87 − 46 + 109’ has terms {87, −46, 109}. ‘87 − (46 + 109)’ = 87 − 46 − 109, terms {87, −46, −109}. ‘(87 − 46) + 109’ = 87 − 46 + 109, terms {87, −46, 109}. ∴ the three copies of 87 + 46 – 109 are equal to one another, and 87 – 46 + 109 = (87 – 46) + 109 are equal to each other. ‘87 − (46 + 109)’ stands alone.

5. Add brackets at appropriate places in the expressions such that they lead to the values indicated. (a) 34 – 9 + 12 = 13 (b) 56 – 14 – 8 = 34 (c) –22 – 12 + 10 + 22 = – 22

SOLUTION (a) 34 – (9 + 12) = 34 – 21 = 13. ✓ (b) 56 – (14 – 8) = 56 – 6 = 50? No — that gives 50. We need 34, so 56 – 14 – 8 = 34 already (56 − 14 = 42, 42 − 8 = 34) — no brackets are needed; placing them as (56 – 14) – 8 = 34 keeps the value. (c) –22 – (12 + 10) + 22 = –22 – 22 + 22 = –22. ✓

6. Using only reasoning of how terms change their values, fill the blanks to make the expressions on either side of the equality (=) equal. (a) 423 + ______ = 419 + ______ (b) 207 – 68 = 210 – ______

SOLUTION (a) 423 is 4 more than 419, so to balance, the term added on the left must be 4 less than the one on the right. One easy pair: 423 + 0 = 419 + 4 (left blank 0, right blank 4). Any pair where the right blank is 4 more than the left blank works. (b) 210 is 3 more than 207, so to keep both sides equal we must subtract 3 more on the right: 207 – 68 = 210 – 71 (= 139). Blank = 71.

7. Using the numbers 2, 3 and 5, and the operators ‘+’ and ‘–’, and brackets, as necessary, generate expressions to give as many different values as possible. For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0.

SOLUTION Using each of 2, 3, 5 once with + and − (and brackets), many values are possible: 2 + 3 + 5 = 10;   5 + 3 − 2 = 6;   5 + 2 − 3 = 4;   3 + 2 − 5 = 0; 5 − 3 + 2 = 4;   5 − 2 + 3 = 6;   5 − (3 + 2) = 0;   2 + 3 − 5 = 0; 5 − 3 − 2 = 0;   5 − (3 − 2) = 4;   3 − 5 − 2 = –4;   2 − 5 + 3 = 0;   2 − 3 − 5 = –6;   2 − (3 + 5) = –6. So distinct values obtainable include –6, –4, 0, 4, 6, 10. (An open task; a representative set of values is shown.)

8. Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = 26 + 1. (a) Do you think she always gets the correct answer? Why? (b) Can you think of other similar strategies? Give some examples. (Math Talk)

SOLUTION (a) Yes, always. Subtracting 9 is the same as subtracting 10 and then adding 1 back, because −9 = −10 + 1. So n − 9 = n − 10 + 1 for every number n (e.g. 36 − 9 = 26 + 1 = 27). ✓ (b) Similar “round-and-adjust” strategies: subtracting 8 → subtract 10 and add 2 (52 − 8 = 42 + 2 = 44); subtracting 19 → subtract 20 and add 1 (63 − 19 = 43 + 1 = 44); adding 9 → add 10 and subtract 1 (47 + 9 = 57 − 1 = 56); adding 98 → add 100 and subtract 2.

9. Consider the two expressions: a) 73 – 14 + 1, b) 73 – 14 – 1. For each of these expressions, identify the expressions from the following collection that are equal to it. (a) 73 – (14 + 1)   (b) 73 – (14 – 1)   (c) 73 + (– 14 + 1)   (d) 73 + (– 14 – 1)

SOLUTION Reduce each option to its terms: (a) 73 − (14 + 1) = 73 − 14 − 1; (b) 73 − (14 − 1) = 73 − 14 + 1; (c) 73 + (−14 + 1) = 73 − 14 + 1; (d) 73 + (−14 − 1) = 73 − 14 − 1. Expression a) 73 – 14 + 1 matches options (b) and (c). Expression b) 73 – 14 – 1 matches options (a) and (d).

Figure it Out — Distributive Property (Page 41)

1. Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal. (a) 3 × (6 + 7) = 3 × 6 + 3 × 7 (b) (8 + 3) × 4 = 8 × 4 + 3 × 4 (c) 3 × (5 + 8) = 3 × 5 [ ] 3 × ____ (d) (9 + 2) × 4 = 9 × 4 [ ] 2 &times ____ (e) 3 &times (____ + 4) = 3 ____ + ____ (f) (____ + 6) × 4 = 13 × 4 + ____ (g) 3 × (____ + ____) = 3 × 5 + 3 × 2 (h) (____ + ____) × ____ = 2 × 4 + 3 × 4 (i) 5 × (9 – 2) = 5 × 9 – 5 × ____ (j) (5 – 2) × 7 = 5 × 7 – 2 × ____ (k) 5 × (8 – 3) = 5 × 8 [ ] 5 × ____ (l) (8 – 3) × 7 = 8 × 7 [ ] 3 × 7 (m) 5 × (12 – ____) = ____ [ ] 5 × ____ (n) (15 – ____) × 7 = ____ [ ] 6 × 7 (o) 5 × (____ – ____) = 5 × 9 – 5 × 4 (p) (____ – ____) × ____ = 17 × 7 – 9 × 7

SOLUTION Using a × (b ± c) = a×b ± a×c: (a) 3 × 6 + 3 × 7.   (b) 8 × 4 + 3 × 4.   (c) 3 × 5 + 3 × 8.   (d) 9 × 4 + 2 × 4. (e) 3 × (5 + 4) = 3 × 5 + 3 × 4 (any first number works; using 5).   (f) (13 + 6) × 4 = 13 × 4 + 6 × 4. (g) 3 × (5 + 2).   (h) (2 + 3) × 4. (i) 5 × 9 − 5 × 2.   (j) 5 × 7 − 2 × 7.   (k) 5 × 8 − 5 × 3.   (l) 8 × 7 − 3 × 7. (m) 5 × (12 − 3) = 5 × 12 − 5 × 3 (any number; using 3).   (n) (15 − 6) × 7 = 15 × 7 − 6 × 7. (o) 5 × (9 − 4).   (p) (17 − 9) × 7.

2. In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions. (a) (8 – 3) × 29 [ ] (3 – 8) × 29 (b) 15 + 9 × 18 [ ] (15 + 9) × 18 (c) 23 × (17 – 9) [ ] 23 × 17 + 23 × 9 (d) (34 – 28) × 42 [ ] 34 × 42 – 28 × 42

SOLUTION (a) (8 − 3) = 5 (positive) and (3 − 8) = −5 (negative); multiplying each by 29, the left is positive and the right is negative, so >. (b) On the left only 9 × 18 is multiplied (15 + 9 × 18), on the right the whole sum (15 + 9) is multiplied by 18, which is much larger, so <. (c) Distributive: 23 × (17 − 9) = 23 × 17 − 23 × 9, which is less than 23 × 17 + 23 × 9, so <. (d) Distributive: (34 − 28) × 42 = 34 × 42 − 28 × 42 exactly, so =.

3. Here is one way to make 14: _2_ × ( _1_ + _6_ ) = 14. Are there other ways of getting 14? Fill them out below: (a) _____ × (_____ + _____) = 14 (b) _____ × (_____ + _____) = 14 (c) _____ × (_____ + _____) = 14 (d) _____ × (_____ + _____) = 14

SOLUTION We need a × (b + c) = 14, so a × (b + c) must equal 14: (a) 2 × (3 + 4) = 2 × 7 = 14. (b) 7 × (1 + 1) = 7 × 2 = 14. (c) 1 × (8 + 6) = 1 × 14 = 14. (d) 2 × (5 + 2) = 2 × 7 = 14. (Many answers are possible; any a × (b + c) = 14 is correct.)

4. Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

SOLUTION The pictures show numbers arranged in equal rows/groups, so the same total can be reached either by adding row totals or by using the distributive property. Worked example (a grid of 3 rows, each with the numbers 4, 4, 4, 4): Way 1 (add each row, then add rows): (4 + 4 + 4 + 4) + (4 + 4 + 4 + 4) + (4 + 4 + 4 + 4) = 16 + 16 + 16 = 48. Way 2 (distributive): there are 3 × 4 = 12 fours, so the sum = 12 × 4 = (3 × 4) × 4 = 48. Both ways give the same total, showing the distributive property in action. (Method shown; apply the same two approaches — row-by-row vs. rows × per-row — to the exact numbers in your book’s picture.)

Figure it Out — Applications (Page 42–43)

1. Read the situations given below. Write appropriate expressions for each of them and find their values. (a) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market. (b) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year? (c) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?

SOLUTION (a) Each day they together supply (9 + 11) kg, for 7 days. Expression: 7 × (9 + 11) = 7 × 20 = 140 kg (equivalently 7×9 + 7×11 = 63 + 77 = 140 kg). (b) Monthly saving = 20000 − (5000 + 5000 + 2000) = 20000 − 12000 = ₹8,000. For a year: 12 × [20000 – (5000 + 5000 + 2000)] = 12 × 8000 = ₹96,000. (c) Net climb per full day–night = 3 − 2 = 1 cm. After 7 full days the snail is at 7 cm. On the 8th day it climbs 3 cm: 7 + 3 = 10 cm, reaching the top before slipping back. Expression: 7 × (3 − 2) + 3 = 7 + 3 = 10. ∴ the snail gets the treat on the 8th day.

2. Melvin reads a two-page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario? (a) 5 × 2 × 8   (b) (7 – 2) × 8   (c) 8 × 7   (d) 7 × 2 × 8 (e) 7 × 5 – 2   (f) (7 + 2) × 8   (g) 7 × 8 – 2 × 8   (h) (7 – 5) × 8

SOLUTION Melvin reads on 7 − 2 = 5 days each week (skipping Tuesday and Saturday), one story per reading day, for 8 weeks. So the number of stories = (7 − 2) × 8 = 5 × 8 = 40 stories. Checking the options: (b) (7 − 2) × 8 = 40 ✓; and by the distributive property (g) 7 × 8 − 2 × 8 = 56 − 16 = 40 ✓ describes the same scenario. ∴ the correct expressions are (b) and (g). (Note: option (a) 5×2×8 = 80 counts pages, not stories.)

3. Find different ways of evaluating the following expressions: (a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 (b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1

SOLUTION (a) Way 1 (pair consecutive terms): (1 − 2) + (3 − 4) + (5 − 6) + (7 − 8) + (9 − 10) = (−1) × 5 = −5. Way 2 (group positives and negatives): (1 + 3 + 5 + 7 + 9) − (2 + 4 + 6 + 8 + 10) = 25 − 30 = –5. Both ways give −5. (b) Way 1 (pair terms): (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) = 0 + 0 + 0 + 0 + 0 = 0. Way 2 (count +1s and −1s): there are five +1 terms and five −1 terms, so 5 − 5 = 0.

4. Compare the following pairs of expressions using ‘<’, ‘>’ or ‘=’ or by reasoning. (a) 49 – 7 + 8 [ ] 49 – 7 + 8 (b) 83 × 42 – 18 [ ] 83 × 40 – 18 (c) 145 – 17 × 8 [ ] 145 – 17 × 6 (d) 23 × 48 – 35 [ ] 23 × (48 – 35) (e) (16 – 11) × 12 [ ] –11 × 12 + 16 × 12 (f) (76 – 53) × 88 [ ] 88 × (53 – 76) (g) 25 × (42 + 16) [ ] 25 × (43 + 15) (h) 36 × (28 – 16) [ ] 35 × (27 – 15)

SOLUTION (a) The two sides are identical, so =. (b) Both subtract 18, but 83 × 42 > 83 × 40, so >. (c) Both start from 145; 17 × 8 > 17 × 6, so a larger amount is subtracted on the left → the left is smaller: <. (d) Left = 23 × 48 − 35; right = 23 × (48 − 35) = 23 × 48 − 23 × 35. The right subtracts 23 × 35, far more than 35, so the right is smaller → >. (e) (16 − 11) × 12 = 16 × 12 − 11 × 12 = −11 × 12 + 16 × 12 (distributive), so =. (f) (76 − 53) = 23 (positive) × 88; (53 − 76) = −23 (negative) × 88. Positive > negative, so >. (g) 42 + 16 = 58 and 43 + 15 = 58, so both equal 25 × 58 → =. (h) 36 × 12 = 432 versus 35 × 12 = 420 (since 28 − 16 = 12 and 27 − 15 = 12). So >.

5. Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression. (a) 83 – 37 – 12   (i) 84 – 38 – 12   (ii) 84 – (37 + 12)   (iii) 83 – 38 – 13   (iv) – 37 + 83 – 12 (b) 93 + 37 × 44 + 76   (i) 37 + 93 × 44 + 76   (ii) 93 + 37 × 76 + 44   (iii) (93 + 37) × (44 + 76)   (iv) 37 × 44 + 93 + 76

SOLUTION (a) Given terms: {83, −37, −12}. (i) 84 − 38 − 12 has terms {84, −38, −12} — different (84 and −38 instead of 83 and −37) → not equal. (ii) 84 − (37 + 12) = 84 − 37 − 12, terms {84, −37, −12} — different (84 not 83) → not equal. (iii) 83 − 38 − 13 has terms {83, −38, −13} → not equal. (iv) −37 + 83 − 12 has terms {−37, 83, −12} — same set → equal. ∴ only (iv) is equal to 83 – 37 – 12. (b) Given terms: {93, 37×44, 76}. (i) swaps to 93×44, changing a term → not equal. (ii) 37×76 ≠ 37×44 → not equal. (iii) (93 + 37) × (44 + 76) multiplies the sums — completely different → not equal. (iv) 37 × 44 + 93 + 76 has terms {37×44, 93, 76} — same set → equal. ∴ only (iv) is equal to 93 + 37 × 44 + 76.

6. Choose a number and create ten different expressions having that value. (Numbered “5” in the book; it is the final task of this set.)

SOLUTION Choosing the number 24, ten different expressions with value 24: 1) 20 + 4   2) 30 − 6   3) 6 × 4   4) 48 ÷ 2   5) 12 + 12   6) 4 × (3 + 3)   7) 50 − 26   8) 3 × 8   9) 8 + 8 + 8   10) 2 × (10 + 2). Each evaluates to 24. (Any single chosen value with ten correct expressions is acceptable.)

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.

Activity — Many expressions for one number Choose your favourite number and write as many expressions as you can having that value. Answer. For the number 12 (as in the book): 10 + 2, 15 − 3, 3 × 4, 24 ÷ 2, 6 + 6, 2 × (4 + 2), 20 − 8, 4 × 3. Any number works — just combine two numbers with +, −, × or ÷ to reach your value.
Compare without calculation (Page 26) Use ‘>’ or ‘<’ or ‘=’ in each: (a) 245 + 289 [ ] 246 + 285 (b) 273 – 145 [ ] 272 – 144 (c) 364 + 587 [ ] 363 + 589 (d) 124 + 245 [ ] 129 + 245 (e) 213 – 77 [ ] 214 – 76 Answer. (a) Left adds 1 less + 4 more = net +3 → > (245 is 1 less than 246 but 289 is 4 more than 285). (b) Both 1 less and 1 more, balancing → = (273 − 145 = 128 = 272 − 144). (c) Right is 1 less but 2 more → net +1 on right → <. (d) Same second term; 124 < 129 → <. (e) 213 < 214 but subtracting 77 (one more) vs 76: 214 is 1 more and we subtract 1 more, balancing → = (213 − 77 = 136 = 214 − 76).
Try This — Subtraction as adding the inverse Check if replacing subtraction by addition in this way does not change the value, by taking different examples. Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers from the Class 6 textbook? Answer. Example: 50 − 18 = 32 and 50 + (−18) = 32 — same value; try 70 − 25 = 45 = 70 + (−25). In the Token Model, a number is shown by positive (+) tokens and its inverse by negative (−) tokens; one + and one − token cancel (make a zero pair). “Subtracting 18” means removing 18 positive tokens, which has exactly the same effect as adding 18 negative tokens — both reduce the total by 18. So a − b = a + (−b).
Complete the table — Terms of expressions (Page 28) Write each expression as the sum of its terms and list the terms: 5 + 6 × 3; 4 + 15 – 9; 23 – 2 × 4 + 16; 28 + 19 – 8. Answer. 5 + 6×3 → terms 5, 6×3.   4 + 15 − 9 = 4 + 15 + (−9) → terms 4, 15, −9.   23 − 2×4 + 16 = 23 + (−2×4) + 16 → terms 23, −(2×4), 16.   28 + 19 − 8 = 28 + 19 + (−8) → terms 28, 19, −8.
Try This — Swapping & grouping with negatives Will swapping/grouping also hold when there are terms with negative numbers? Take some more expressions and check, and explain using the Token Model. Answer. Yes. Example: (−7) + 10 + (−11) = −8 in every order — e.g. (−7 + 10) + (−11) = 3 − 11 = −8 and (−7) + (10 − 11) = −7 + (−1) = −8. In the Token Model, the total number of + and − tokens does not depend on the order you count or pile them, so the value is the same — this is the commutative and associative property of addition.
Math Talk — Manasa’s forgotten number Manasa added a list and got 11749, then realised she forgot the number 9055. Does she have to start all over again? Answer. No. Because addition is commutative and associative, she can just add the missed number to her running total: 11749 + 9055 = 20804. The order of adding terms does not change the sum, so there is no need to start again.
Example 7 follow-up — 7 friends If the number of friends goes up to 7 and the tip remains ₹5, how much will they pay? Write an expression and identify its terms. Answer. Expression: 7 × 23 + 5. Terms: 7×23 and 5. Value = 161 + 5 = ₹166.
Example 8 follow-up — “Fire in the mountain” If the teacher had called out ‘4’ or ‘7’, what expression would Ruby write (33 students playing)? Answer. Groups of 4: 33 = 8 × 4 + 1, so Ruby writes 8 × 4 + 1 (terms 8×4 and 1). Groups of 7: 33 = 4 × 7 + 5, so she writes 4 × 7 + 5 (terms 4×7 and 5).
Math Talk — Effective multiplication Find 97 × 25, and use the (100 – 3) method for 95 × 8, 104 × 15, 49 × 50. Is it quicker? Answer. 97 × 25 = (100 − 3) × 25 = 2500 − 75 = 2425.   95 × 8 = (100 − 5) × 8 = 800 − 40 = 760.   104 × 15 = (100 + 4) × 15 = 1500 + 60 = 1560.   49 × 50 = (50 − 1) × 50 = 2500 − 50 = 2450. Yes — rounding to a friendly number and adjusting is usually quicker for numbers near 100, 50, etc.
Example 18 reasoning — 5 × (4 + 3) vs 5 × 4 + 3 Explain why 5 × 4 + 3 ≠ 5 × (4 + 3). Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5? Answer. 5 × 4 + 3 has terms 5×4 and 3, giving 20 + 3 = 23, but 5 × (4 + 3) = 5 × 7 = 35; they differ because in the first only 4 is multiplied by 5, while in the second the whole sum (4 + 3) is multiplied. And yes, 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5 = 35, because addition and multiplication are commutative.
Expression Engineer! — puzzles Using three 3’s make several values; using four 4’s get all values 1–20; using 1, 2, 3, 4, 5 once get values between −10 and +10; using 0–9 once make 100. Sample answers. Three 3s: (3 + 3) ÷ 3 = 2, 3 + 3 − 3 = 3, 3 × 3 + 3 = 12. Four 4s: 4 ÷ 4 + 4 − 4 = 1, 4 ÷ 4 + 4 ÷ 4 = 2, (4 + 4 + 4) ÷ 4 = 3, 4 + 4 − 4 − 4 = 0… (build the rest similarly up to 20). Using 1–5 once: 1 + 2 + 3 + 4 − 5 = 5, 1 − 2 − 3 − 4 + 5 = −3, etc. Using 0–9 once for 100: e.g. 80 + 19 + 0 + … — one neat solution is 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 × 9) + 0 = 100. (Open creativity tasks; many correct answers exist.)

Common Mistakes to Avoid

Watch out for these

  • Evaluating left-to-right blindly — in 30 + 5 × 4 you must do 5 × 4 first (it is a single term), giving 50, not 140.
  • Forgetting to flip signs when removing a bracket that follows a minus: 100 − (15 + 56) = 100 − 15 − 56, not 100 − 15 + 56.
  • Flipping signs when the bracket follows a plus — a + (b − c) keeps the signs: 28 + (35 − 10) = 28 + 35 − 10.
  • Treating 5 × 4 + 3 the same as 5 × (4 + 3) — the bracket changes which numbers get multiplied.
  • Saying a × (b − c) = a×b − c — you must multiply every term inside: a×b − a×c.
  • When finding terms, missing that subtraction means adding the inverse, so 83 − 14 has terms 83 and −14.

Practice MCQs & Assertion–Reason

1. The value of the expression 30 + 5 × 4 is:

(a) 140    (b) 50    (c) 35    (d) 120

2. The terms of the expression 83 – 14 are:

(a) 83 and 14    (b) 83 and –14    (c) –83 and 14    (d) 83 × 14

3. 100 – (15 + 56) is equal to:

(a) 100 – 15 + 56    (b) 100 + 15 – 56    (c) 100 – 15 – 56    (d) 100 + 15 + 56

4. Which expression equals 23 × (17 – 9)?

(a) 23 × 17 + 23 × 9    (b) 23 × 17 – 23 × 9    (c) 23 × 17 – 9    (d) 23 + 17 – 9

5. “Swapping terms does not change the sum” is called the:

(a) associative property    (b) distributive property    (c) commutative property    (d) identity property

6. To make the blank true: 22 + ____ = 6 × 5, the missing number is:

(a) 6    (b) 8    (c) 12    (d) 30

7. Which comparison is correct?

(a) 10 + 2 < 7 + 1    (b) 13 – 2 > 4 × 3    (c) 10 + 2 > 7 + 1    (d) 5 × 11 < 120 ÷ 3

8. Using the method (100 – 3) × 25, the value of 97 × 25 is:

(a) 2425    (b) 2500    (c) 2475    (d) 2375

9. (8 – 3) × 29 compared with (3 – 8) × 29 is:

(a) less than    (b) greater than    (c) equal to    (d) cannot say

10. Melvin reads daily except Tuesdays and Saturdays; in 8 weeks the number of reading days is given by:

(a) 7 × 8    (b) (7 – 2) × 8    (c) (7 + 2) × 8    (d) 7 × 2 × 8

Answer key: 1-(b), 2-(b), 3-(c), 4-(b), 5-(c), 6-(b), 7-(c), 8-(a), 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: 30 + 5 × 4 = 50.

Reason: 5 × 4 is a single term, so it is evaluated before adding it to 30.

A-R 2. Assertion: 24 – (6 + 4) = 24 – 6 + 4.

Reason: When a bracket is preceded by a minus sign, the signs of all terms inside change on removal.

A-R 3. Assertion: a × (b + c) = a × b + a × c.

Reason: This is the distributive property — the multiple of a sum is the sum of the multiples.

A-R 4. Assertion: The terms of 83 – 14 are 83 and –14.

Reason: Subtracting a number is the same as adding its inverse.

A-R 5. Assertion: Adding the terms of an expression in any order gives the same value.

Reason: Addition obeys the commutative and associative properties.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(A).

Quick Revision Summary

  • An arithmetic expression is built from numbers and +, −, ×, ÷; its value is the number it evaluates to.
  • Compare expressions by their values using =, < and > — often by reasoning, without full calculation.
  • Brackets are evaluated first; terms are the parts separated by ‘+’, with each subtraction rewritten as adding the inverse.
  • Evaluate by working out each term first, then adding all the terms (so 30 + 5 × 4 = 30 + 20 = 50).
  • Commutative & associative properties: terms can be swapped and regrouped without changing the value.
  • Removing a bracket after ‘−’ flips every inside sign; after ‘+’ the signs stay the same.
  • Distributive property: a × (b ± c) = a×b ± a×c — great for quick mental multiplication like 97 × 25.

How to score full marks in this chapter

Always rewrite subtractions as “+ inverse” before identifying terms, and evaluate each product/quotient term before adding. When removing brackets, write one line stating whether the sign before the bracket is + (keep signs) or − (flip signs). For comparisons, look at how each term changes rather than computing both sides, and use the distributive property to break big multiplications into easy ones — showing each step earns full method marks.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 2 about?

Chapter 2, Arithmetic Expressions, covers reading and comparing simple expressions, brackets, the idea of terms (parts separated by ‘+’), the commutative and associative properties of addition, rules for removing brackets, and the distributive property for multiplication over addition and subtraction.

What are “terms” in an arithmetic expression?

Terms are the parts of an expression separated by a ‘+’ sign. To find them, rewrite each subtraction as adding the inverse — so 83 − 14 = 83 + (−14) has the terms 83 and −14, while a product like 6 × 5 stays as one single term.

What happens to the signs when we remove brackets?

If the bracket is preceded by a plus sign, the signs inside stay the same: a + (b − c) = a + b − c. If it is preceded by a minus sign, every sign inside changes: a − (b + c) = a − b − c and a − (b − c) = a − b + c.

Are these Class 7 Maths Ganita Prakash Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for the 2026–27 session, with every Figure it Out, Math Talk and Try This task solved step by step.

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