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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 :
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 :
Let, principle = P
P = S.I × 100/RT
= 480 × 100/4 × 5
= 4200 Rs
∴ A = 4200 + 840
= 5040 Rs
Next Chapter Solution :
👉 Chapter 9 👈
