# Maths Ace Class 7 Solutions Chapter 8

## 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 :

1st case,

SI = PRT/100

or, R = SI × 100/PT

or, R = 30000 × 100/15000 × 4

or, R = 5%

2nd 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 :

Letprinciple = P

P = S.I × 100/RT

= 480 × 100/4 × 5

= 4200 Rs

A = 4200 + 840

= 5040 Rs

Next Chapter Solution :

Updated: June 17, 2023 — 5:39 am