The NCERT Solutions Class 7 Maths Chapter 2: Arithmetic Expressions (Ganita Prakash) wraps up its essential content with exercise 2.5, which challenges students to simplify arithmetic expressions with multiple steps! This exercise ties together all the skills so far: BODMAS (order of operations), working with brackets (parentheses, braces and square brackets), and integer properties.
The primary purpose of Exercise 2.5 is to practice developing a systematic and accurate process to simplify long chains of operations down to a perfect, single numerical value! If students are successful in this, they will be confident in their ability to tackle any arithmetic calculations in future math courses.
Access the stepwise detailed solutions for NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.5 in a PDF format. Use the solutions to check your final answers and understand the exact order of steps needed for correct simplifications.
Brackets→ Orders (or Powers)→ Division→ Multiplication→ Addition→ Subtraction.
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?
Sol. (a) Supplies of mangoes by Rahim in the market each day =9 kg
Supplies of mangoes by Shyam in the market each day =11 kg
Total supplies of mangoes in the market on each day =(9+11)kg
Therefore, total supplies of mangoes in the market in all 7 days
=7×(9+11)kg
=7×20
=140 kg
(b) Binu's per month earning = ₹ 20,000
Binu's total monthly expenditures =
₹ 5,000 on rent + ₹ 5,000 on food +
₹ 2,000 on other expenses
=5,000+5,000+2,000
Therefore, Binu's monthly savings
= ₹ 20,000−₹(5,000+5,000+2,000)
= ₹ 20,000 - ₹ 12,000
= ₹ 8,000
Thus, Binu's total yearly savings
=12×8000=96000
Hence, Binu will save ₹ 96000 by the end of the year.
(c) Since the snail climbs 3 cm up the post in daytime and slips down by 2 cm at night.
So, the distance climbed by the snail of the post =3−2=1 cm in a day.
∴ The distance climbed in 7 days =7 cm
The height of the post is 10 cm .
The distance climbed on the 8th day before slipping =7+3=10 cm
So, the snail will take 8 days to reach the top of the post and get the delicious treat
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
Sol. Number of days in a week except Tuesday and Saturday =7−2
Since Melvin reads a two-page story every day except Tuesday and Saturday. Therefore, number of stories read in a week =1×(7−2)
So, number of stories read in 8 weeks
=8×1×(7−2)
=8×(7−2) or (7−2)×8 [Expression (b)]
or 7×8−2×8 [Expression (g)]
Only expressions (b) and (g) describe this scenario.
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
Sol. (a) 1−2+3−4+5−6+7−8+9−10
=(1+3+5+7+9)+(−2−4−6−8−10)=25+(−30)=−5
OR
===1−2+3−4+5−6+7−8+9−10(1−2)+(3−4)+(5−6)+(7−8)+(9−10)(−1)+(−1)+(−1)+(−1)+(−1)−5[
(b) 1−1+1−1+1−1+1−1+1−1
===(1−1)+(1−1)+(1−1)+(1−1)+(1−1)0+0+0+0+00
OR
1−1+1−1+1−1+1−1+1−1=(1+1+1+1+1)+(−1−1−1−1−1)=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)
Sol. (a) 49−7+8=49−7+8
( ∵ All the terms on both sides are the same)
(b) 83×42>83×40
∴83×42−18>83×40−18
(c) 17×8>17×6
⇒−17×8<−17×6
∴145−17×8<145−17×6
(d) 23×(48−35)=23×48−23×35 and 35<23×3523×48−35>23× (48-35)
(e) (16−11)×12=16×12−11×12=−11×12+16×12
∴(16−11)×12=−11×12+16×12
(f) (76−53)×88=76×88−53×88=−(53−76)×88
∴(76−53)×88>88×(53−76)
(g) 43+15=42+1+15=42+16
⇒25×(43+15)=25×(42+16)
∴25×(42+16)=25×(43+15)
(h) 35×(27−15)=35×(28−16)
∴36×(28−16)>35×(27−15)
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
Sol. (a) 83-37-12=83-37-12+(1-1)
=(83+1)−37−1−12
=84−38−12 (option (i))
=34
OR
83−37−12=−37+83−12
=46−12
=34 (option (iv))
Hence, (i) and (iv) are equal to the given expression 83-37-12.
(b) (iv) 37×44+93+76
Rearrange the terms, and we get 93+37×44+76, which is equal to the given expression.
Hence, (iv) is equal to the given expression 93+37×44+76.
6. Choose a number and create ten different expressions having that value.
Sol. Let us choose number 24 and create ten different expressions having that value
i. 12+12=24
ii. 30−6=24
iii. 8×3=24
iv. 248=24
v. 25−1=24
vi. 6×4=24
vii. 50−26=24
viii. 28−4=24
ix. 240/10 =24
x. 5×5−1=24
(Session 2026 - 27)