NCERT Solutions for Class 7 Maths Chapter 2: Arithmetic Expressions – Ganita Prakash

Chapter 2 of the Class 7 NCERT Ganita Prakash textbook (Part 1) is titled “Arithmetic Expressions”. It covers forming and evaluating expressions, comparing expressions without full evaluation, identifying “terms” in an expression, the effect of brackets, and useful shortcuts (like the distributive law) for mental arithmetic. These Class 7 Mathematics Chapter 2 solutions are also useful as quick revision notes before exams.

Below are original, independently verified solutions to the chapter’s key questions.

2.1 Simple Expressions

Q. Fill in the blanks: (a) 13+4 = ___+6 (b) 22+___ = 6×5 (c) 8×___ = 64÷2 (d) 34−___ = 25
Solution: (a) 13+4=17, so 17=___+6 ⇒ 11. (b) 6×5=30, so 22+___=30 ⇒ 8. (c) 64÷2=32, so 8×___=32 ⇒ 4. (d) 34−___=25 ⇒ 9.

Q. Arrange in ascending order: 67−19, 67−20, 35+25, 5×11, 120÷3
Solution: Values: 67−19=48, 67−20=47, 35+25=60, 5×11=55, 120÷3=40. Ascending order: 40 < 47 < 48 < 55 < 60, i.e. 120÷3 < 67−20 < 67−19 < 5×11 < 35+25.

Terms in Expressions & Swapping/Grouping

Q. Show that subtracting a number is the same as adding its inverse (negative), using an example.
Solution: Take 56 and 17: 56−17 = 39. Also, 56+(−17) = 39. Same result — subtracting 17 is equivalent to adding −17.

Q. Does changing the order of terms change the value of an expression? Test with 4, 15, and −9.
Solution: 4+15+(−9) = 19−9 = 10. Reordered: (−9)+15+4 = 6+4 = 10. Same value — terms can be added in any order (this also holds for negative terms).

Q (Manasa’s sum). Manasa added a long list of numbers and got 11,749, but had forgotten to include one number, 9,055. Does she need to start over?
Solution: No — since addition can be done in any order, she just needs to add the missing number to her existing total: 11,749 + 9,055 = 20,804.

More Expressions and Their Terms

Q. If a dosa costs ₹23 each and 7 friends each order one, plus a ₹5 tip, write an expression and find the total.
Solution: Expression: 7 × 23 + 5. Terms: 7×23 and 5. Value = 161+5 = ₹166.

Q. Write ₹432 as a sum of terms in two different ways using place value.
Solution: (i) 432 = 4×100 + 1×20 + 1×10 + 2×1 = 400+20+10+2 = 432. (ii) 432 = 8×50 + 1×10 + 4×5 + 2×1 = 400+10+20+2 = 432. Both are valid decompositions using different “denominations.”

Figure It Out — Evaluating Expressions

Q. Find the value of: (a) 28−7+8 (b) 39−2×6+11 (c) 40−10+10+10 (d) 48−10×2+16+2 (e) 6×3−4×8×5
Solution: (a) 28−7+8 = 21+8 = 29. (b) 39−(2×6)+11 = 39−12+11 = 27+11 = 38. (c) 40−10+10+10 = 30+10+10 = 50. (d) 48−(10×2)+16+2 = 48−20+16+2 = 28+16+2 = 46. (e) (6×3)−(4×8×5) = 18−160 = −142.

Q (Story problems). Write a real-life story for each expression and find its value: (a) 89+21−10 (b) 5×12−6 (c) 4×9+2×6
Solution: (a) Riya has 89 stones, Siya gives her 21 more, then she gives away 10. Value: 89+21−10 = 110−10 = 100. (b) Radha buys 5 pens at ₹12 each with a ₹6 discount. Value: 5×12−6 = 60−6 = 54. (c) Sumit buys 4 pencils at ₹9 each and 2 erasers at ₹6 each. Value: 4×9+2×6 = 36+12 = 48.

Q (Elsa & Anna). Elsa and Anna each start with 100 gold coins. Elsa doubles hers; Anna keeps half of hers. Write an expression for their combined total.
Solution: Expression: 2×100 + 100÷2 = 200+50 = 250 coins.

Q (Metro tickets). Adult tickets cost ₹40, child tickets ₹20. Find the total cost for (i) 4 adults + 3 children (ii) two separate groups of 3 adults each.
Solution: (i) 4×40 + 3×20 = 160+60 = ₹220. (ii) 2×(3×40) = 2×120 = ₹240.

Removing Brackets

Q. Remove the brackets and confirm the value stays the same: (a) 14+(12+10) (b) 14−(12+10) (c) 14+(12−10) (d) 14−(12−10)
Solution: (a) 14+12+10 = 36. (b) 14−12−10 = −8. (c) 14+12−10 = 16. (d) 14−12+10 = 12. (Note: when removing a bracket preceded by a minus sign, every sign inside the bracket flips.)

Q. Compare (without necessarily evaluating both sides the same way): (a) (6+10)−2 vs. 6+(10−2) (b) 16−(8−3) vs. (16−8)−3
Solution: (a) (6+10)−2 = 16−2 = 14; 6+(10−2) = 6+8 = 14 — equal. (b) 16−(8−3) = 16−5 = 11; (16−8)−3 = 8−3 = 5 — not equal (subtraction is not associative the same way addition is).

Q. Insert brackets to make each statement true: (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 = 42−8 = 34. ✓ (c) −22−(12+10)+22 = −22−22+22 = −22. ✓

Q (Using 2, 3, 5). Using the numbers 2, 3, 5 with +, −, and brackets, find as many different values as possible.
Solution: (2+3)−5 = 0. 5+(3−2) = 6. −5+(2−3) = −6. (−2+3)−5 = −4. 2−3+5 = 4. 3−(5−2) = 0. This shows how the same three numbers can produce many different values depending on operation order and bracket placement.

Distributive Law & Mental Math Shortcuts

Q. Use the distributive law to quickly find: (a) 95×8 (b) 104×15 (c) 49×50
Solution: (a) 95×8 = (100−5)×8 = 800−40 = 760. (b) 104×15 = (100+4)×15 = 1500+60 = 1,560. (c) 49×50 = (50−1)×50 = 2500−50 = 2,450.

Q. Compare using reasoning (without full evaluation): (a) (8−3)×29 vs. (3−8)×29 (b) 15+9×18 vs. (15+9)×18
Solution: (a) (8−3)×29 = 5×29 = 145; (3−8)×29 = −5×29 = −145 — the first is greater. (b) 15+9×18 = 15+162 = 177; (15+9)×18 = 24×18 = 432 — the second is greater (grouping before multiplying scales the whole sum, not just one term).

Q (Word problems using the distributive law). (a) Rahim supplies 9 kg of mangoes/day and Shyam supplies 11 kg/day; the market is open all 7 days. Find the total weekly supply. (b) Binu earns ₹20,000/month, spending ₹5,000 rent + ₹5,000 food + ₹2,000 other. Find monthly and yearly savings.
Solution: (a) 7×(9+11) = 7×20 = 140 kg. (b) Expenses = 5000+5000+2000 = ₹12,000; monthly savings = 20,000−12,000 = ₹8,000; yearly savings = 12×8,000 = ₹96,000.

Q (Snail up the well). A snail climbs 3 cm every day but slips back 2 cm every night. A post is 10 cm high. How many days until it reaches the top?
Solution: Net progress per full day-night cycle = 3−2 = 1 cm. After 7 full cycles: 7 cm climbed. On day 8, the snail climbs 3 more cm, reaching 7+3=10 cm — exactly the top — before it can slip back that night. So it takes 8 days (not 10, since the final day’s climb happens before any further slip).

Q. Evaluate two different ways: 1−2+3−4+5−6+7−8+9−10
Solution: Method 1 (group positives and negatives): (1+3+5+7+9) + (−2−4−6−8−10) = 25 + (−30) = −5. Method 2 (pair consecutive terms): (1−2)+(3−4)+(5−6)+(7−8)+(9−10) = −1−1−1−1−1 = −5. Both methods agree.

Why This Chapter Matters (for Boards)

Comfort with terms, brackets, and the distributive law here directly sets up algebraic manipulation in later chapters (like expressions with letter-numbers) and mental-math efficiency for board exams.

More on This Chapter

See also: Class 7 Maths Part 1 NCERT Book and the Class 7 Maths Formulas Handbook.

Related pages: Extra Questions for Class 7 Maths Chapter 2 | Revision Notes for Class 7 Maths Chapter 2

Frequently Asked Questions

Q: What’s a “term” in an arithmetic expression?
A: A term is a part of an expression separated from the rest by a + or − sign (so a multiplication or division chunk like 7×23 counts as a single term). Terms can be added in any order without changing the expression’s value.

Q: What happens to the signs inside a bracket when you remove a bracket preceded by a minus sign?
A: Every sign inside the bracket flips (+ becomes −, and − becomes +). For example, 14−(12−10) becomes 14−12+10.

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.