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**Maths Ace Class 7 Solutions Chapter 8 Comparing Quantities**

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Maths Ace Prime Class 7 Math Book, Chapter 8, Comparing Quantities. Here students can easily find step by step solutions of all the problems for Comparing Quantities, Exercise 8.1, 8.2, 8.3, 8.4, 8.5 and 8.6 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 8 solutions.

**Comparing Quantities Exercise 8.1 Solution :**

**Question no – (1) **

**Solution :**** **

**(a) 655 m to 1 km**

= 655/1 × 100

= 635/1000

= 131 : 200

**(b) 35 minutes to 2 hours**

= 35/2 × 60

= 7 : 14

**(c) 845 g to 1 kg**

= 845/1 × 100

= 169 : 200

**(d) 12 hours to 1.5 days**

= 12/1.5 × 24

= 1 : 3

**Question no – (2) **

**Solution :**

**(a) 5 : 8 or 7 : 9**

= 5/8 = 5 × 9/8 × 9

= 45/72

= 7/9 = 7 × 8/9 × 8

= 56/72

**∴** 7/9 > 5/8

**(b) 49 : 35 or 45 : 25**

= 49/35 = 7/5

= 45/25 = 9/5

**∴** 9/5 > 7/5

**Question no – (3)**

**Solution :**

= 7/9 = 98/126

= 7/9 = 7/9

**Question no – (4)**

**Solution :**

= No, because the product extremes & the product of means are not equal.

**Question no – (5)**

**Solution :**

= 25/m = 18/36

or, m = 25 × 36/18

= 50

**Question no – (6)**

**Solution :**

Actual distance = x km

distance = 4.5/100000

According to question,

1/200000 = (4.5/100000)/x

or, x = 4.5/100000 × 200000

or, x = 9 km

So, the length is **9 km**.

**Question no – (8)**

**Solution :**

1st case – bottles are

= 24 × 6

= 144

2nd case bottles are

= 105 × 30

= 3150

So, 2nd is a better deal.

**Comparing Quantities Exercise 8.2 Solution :**

**Question no – (1)**

**Solution :**

Vineet paid

= 12 × 108/4

= 3264

So, Vineet paid 3264 Rs.

**Question no – (2)**

**Solution :**

Distance cover = 9 × 220/5

= 396

So, the distance is 396 km.

**Question no – (3)**

**Solution :**

Total annual salary of 157 employees

= 7, 10582 × 12

= 8526984

= Total annual salary of 1 employees

= 8536984/152

= 54,312 Rs

So, 54,312 is the annual salary of 1 employee.

**Question no – (4)**

**Solution :**

**Rahul earns ₹ 21,000 in 7 months.**

= Rahul earns 1 month

= 21000/7

= 3000

**(a) How much does he earn in 12 months**

**∴** Rahul earns 12 month

= 3000 × 12/36000

**(b) In how many months will he earn ₹ 33,000**

**∴** Rahul earn

= 33000 × 12/36000

= 11 months

**Question no – (5)**

**Solution :**

Total new people

= 25 + 10

= 35

**∴** rice needed for 35 people

= 35 × 55/25

= 77 kg

Therefore, rice will be needed for a week 77 kilograms.

**Comparing Quantities Exercise 8.3 Solution :**

**Question number – (1)**

**Solution :**

**(a) 7/10**

= 7/10 × 100

= 70%

**(b) 5/8**

= 5/8 × 100

= 125/2%

**(c) 13/25**

= 13/25 × 100

= 52%

**(d) 14/33**

= 14/33 × 100

= 42.42%

**Question number – (2)**

**Solution :**

**(a) 0.334**

= 334/1000 × 100

= 33.4%

**(b) 1.24**

= 124/100 × 100

= 124%

**(c) 0.67**

= 67/100 × 100

= 67%

**(d) 0.596**

= 596/1000 × 100

= 5.96%

**Question number – (3)**

**Solution :**

**(a) 3 : 4**

= 3/4 × 100

= 75%

**(b)** **4 : 25**

= 4/25 × 100

= 16%

**(c)** **9 : 16**

= 9/16 × 100

= 225/4%

**(d) 1 : 2**

= 1/2 × 100

= 50%

**Question number – (4)**

**Solution :**

**(a) 28%**

= 28/100

= 0.28

**(b) 36%**

= 36/100

= 0.36

**(c) 45%**

= 45/100

= 0.45

**(d) 22.5%**

= 225/1000

= 0.225

**Question number – (5)**

**Solution :**

**(a) 32%**

= 32/100

= 4/25

**(b) 41%**

= 41/100

**(c) 55%**

= 17/100

**(d) 14.5%**

= 25/100

= 1/4

**Question number – (6)**

**Solution :**

**(a) 16%**

= 16/100

**(b) 20%**

= 20/100

**(c) 55%**

= 55/100

**(d) 14.5%**

= 145/1000

**Question number – (7)**

**Solution :**

Marks percentage = 66 × 100/80

= 165/2

= 82.5%

**Question number – (8)**

**Solution :**

Percentage of day for trip

= 100 × 8/30

= 80/3

= 26.66%

**Question number – (9)**

**Solution :**

For bottle A = 2 : 5 = 2 × 8/5 × 8 = 16/40

For bottle B = 3 : 8 = 3 × 5/8 × 5 = 15/40

**∴** 16/40 > 15/40

= A > B

**∴** Bottle A is high concentration

**Comparing Quantities Exercise 8.4 Solution :**

**Question number – (1)**

**Solution :**

**(a) 5% of ₹ 500**

= 500 × 5/100

= 25

**(b) 15% of 10L**

= 15/100 × 10

= 3/2 L

**(c) 22% of 25 kg**

= 22/100 × 25

= 11/2 kg

= 5.5 kg

**(d) 21 3/5% of 300 km**

= 300/100 × 108/5

= 64.8 km

**Question number – (2)**

**Solution :**

**(a) **₹2 of ₹ 40?

= 2/40 × 100

= 5%

**(b) 50 cm of 2 m?**

= 50/200 × 100

= 25%

**(c) 84 g of 3.36 kg?**

= 84/3600 × 100

= 2.5%

**(d) 75 paise of ₹ 5?**

= 75/500 × 100

= 15%

**Question number – (3)**

**Solution :**

**(a) 30% of it is 16500**

= 16500 × 100/30

= 55000

**(b) 25% of it is 20**

= 20 × 100/25

= 80

**Question number – (4)**

**Solution :**

House rent = 1000 × 15/100 = 1500

food = 10000 × 12/100 = 1200

bill payment = 10000 × 10/100 = 1000

**∴** He saves,

= 10000 – (1500 + 1200 + 1000)

= 10000 – 3700

= 6300

**Question number – (5)**

**Solution :**

= 3L = 3 × 1000 = 3000 mL

= 5 boys drank = 5 × 300

= 1500 mL

**∴** Rent water contains % (3000 – 1500/3000 × 100)

= 1500/3000 × 100

= 50%

**Question number – (6)**

**Solution :**

Total salary

= 16000 × 100/32

= 50,000

**Question number – (7)**

**Solution :**

Spend to repayment

= 15000/45000 × 100

= 33.33%

**∴** Remaining salary

= 45000 – 15000

= 30000

**∴** Spend for food

= 30000 × 24/100

= 7200

**Question number – (8)**

**Solution :**

A Towns population

= 2400 – 2000/2400 × 100

= 400/24

= 16.66%

= 2300 – 1800/2300 × 100

= 500/2300 × 100

= 21.73%

**∴** Town B increase% is greater than Town A.

**Question number – (9)**

**Solution :**

Dheeraj’s salary incorrect,

= 6500 – 5000/6500 × 100

= 1500/6500 × 100

= 23.07%

Manoj’s salary incorrect,

= 5200 – 4000/5200 × 100

= 1200/52

= 23.07%

**∴** Both salary incorrect are same

**Question number – (10)**

**Solution :**

Let, Total student a passed = x

**Pass student,**

= x × 91/100 × 91h/100

**Fail student,**

= x – 91x/100

= 100x – 91x/100

According to questions –

= 9x/100 = 18

or, x = 18 × 100/9

= 200

**Question number – (11)**

**Solution :**

**(a) 39% of 5100**

= 5100 × 39/100

= 1989

**(b) 21% of 2900**

= 2900 × 21/100

= 609

**(c) 31% of 1100**

= 1100 × 31/10

= 341

**(d) 49% of 470**

= 470 × 49/100

= 230.9

**Comparing Quantities Exercise 8.5 Solution :**

**Question number – (1)**

**Solution :**

**(a) CP = ₹ 40,000 and SP = ₹ 50,000**

= Profit = 50000 – 40000

= 10000

**(b) CP = ₹ 45,500 and SP = ₹ 39,990**

= 45500 – 39990

= 5510

**Question number – (2)**

**Solution :**

Total C.P = 560000 + 15000 = 575000

SP = 700000

Profit,= 700000 – 575000

= 125000%

Profit = 125000/575000 × 100

= 21.74%

**Question number – (3)**

**Solution :**

Number of rose = 240

**∴** No of dozen = 240/12 = 20 dozen

**∴** CP of 1 dozen (12) = 10

**∴ **CP of 20 dozen

= 10 × 20

= 200

S.P of 20 dozen

= 240 × 150

= 360

**∴** Profit

= 360 – 200

= 160

Profit%

= 160/200 × 100

= 80%

**Question number – (4)**

**Solution :**

From the question we get,

C.P of piano = 1, 12000

C.P of guitar = 23,000

**∴** C.P of total instrument

= 112000 + 23000

= 135000

Loss = 3%

**∴** S.P of Total instrument

= 97/100 × 135000

= 130,950

**Question number – (5)**

**Solution :**

S.P of camera

= 35000

Profit = 10%

**∴** C.P of camera

= 35000/10 × 100

= 31818.18

**Question number – (6)**

**Solution :**

Amit’s total expenditure of car

= 2000000 + 3000 + 6000

= 2020000

**∴** S.P of car = 300000

Profit,

= 30000 – 259000

= 41000

Profit %,

= 41000/259000 × 100

= 15.83

**Question number – (7)**

**Solution :**

Total expenditure of car

= 2000000 + 15000 + 5000

= 2020000

SP of car = 2500000

**∴** Profit

= 2500000 – 2020000

= 480000

Profit%

= 480000/2020000 × 1000

= 23.76%

**Comparing Quantities Exercise 8.6 Solution :**

**Question number – (1)**

**Solution :**

**(a) P = Rs 10,000, T = 2 years, R = 8% per annum**

= S.I = PRT/100

or, S.I = 10000 × 8 × 2/100

or, S.I = 1600

**∴** Total amount

= 1000- + 1600

= 11,600

**(b) P = Rs 25,000, T = 3 years, R = 7% per annum**

S.I = PRT/100

or, S.I = 25000 × 7 × 3/100

or, S.I = 5250

**∴** Total amount

= 25000 + 5250

= 30,250

**Question number – (2)**

**Solution :**

**(a) P = Rs 10,000, T = 2 years, R = 8% per annum**

= S.I = PRT/100

or, S.I = 10000 × 8 × 2/100

or, S.I = 1600

**∴** Total amount

= 1000- + 1600

= 11,600

**(b) P = Rs 25,000, T = 3 years, R = 7% per annum**

S.I = PRT/100

or, S.I = 25000 × 7 × 3/100

or, S.I = 5250

**∴** Total amount

= 25000 + 5250

= 30,250

**Question number – (3) **

**Solution :**

SI = PRT/100

or, SI = 64000 × 12 × 5/100

or, SI = 384000

**∴** Total amount

= (640000 + 384000)

= 1024000

**Question number – (4) **

**Solution :**

**1**^{st} case,

SI = PRT/100

or, R = SI × 100/PT

or, R = 30000 × 100/15000 × 4

or, R = 5%

**2**^{nd}** case, **

S.I = PRT/100

or, S.I = 15000 × 4 × 18/100

or, S.I = 10800

**Question number – (5)**

**Solution :**

Let, P = x A = 2x Rate = R

S.I = A – P

According to questions =

P = SI × 100/PT

or, R = x × 100/x × 5

= 20%

P = SI × 100/PT

or, R = x × 100/x × 5

= 20%

**Question number – (6) **

**Solution :**

**S.I** = A – P

= 4526 – 3650

= 876

Let, Year = T

Now according to questions,

S.I = PRT/100

or, T = SI × 100/PR

= 876 × 100/3650 × 6

= 24/6

= 4

or, T = 4 years

**Question number – (7)**

**Solution :**

Let**, **principle = P

P = S.I × 100/RT

= 480 × 100/4 × 5

= 4200 Rs

**∴** A = 4200 + 840

= 5040 Rs

**Next Chapter Solution : **

👉 Chapter 9 👈