Maths Ace Class 8 Solutions Chapter 9

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Maths Ace Class 8 Solutions Chapter 9 Comparing Quantities

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 9, Comparing Quantities. Here students can easily find step by step solutions of all the problems for Comparing Quantities, Exercise 9.1, 9.2, 9.3, 9.4 and 9.5 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 9 solutions.

Comparing Quantities Exercise 9.1 Solution :

Question no – (1) 

Solution :

(a) As per the question,

Company profit increases from Rs 50,000 to Rs 80,000.

The profit of a company increases from Rs 50,000 to Rs 80,000.

∴ Increase profit,

= (80000 – 50000)

= 30000 Rs

∴ Percentage increase,

= (30000/50000 × 100)%

= 60%

Thus, percentage increase of the company will be 60%

(b) According to the given question,

Poaching of lions decreased from Rs 60,000 to Rs 40,000.

∴ Decreased

= (60000 – 40000)

= 20000

∴ Percentage decrease

= (20000/60000 × 100) %

= 33.33%

Hence, the percentage decrease in Poaching of lions will be 33.33%

(c) The price of a mobile phone increased from Rs 4500 to Rs 5000.

∴ Price increase,

= (5000 – 4500) Rs

= 500

∴ Percentage increase,

= (500/4500 × 100)

= 11.11 %

Therefore, the percentage increase will be 11.11%

(d) Mohan’s salary increased from Rs 25,000 to Rs 30,000 this year.

∴ Salary increase,

= (30000 – 25000)

= 5000

∴ Percentage increase,

= (5000/25000 × 100)%

= 20%

Therefore, the percentage increase will be 20%

Question no – (2) 

Solution : 

According to the given question,

Sales increased from Rs 60,000 per day in 2017 to Rs 82,500 per day in 2018.

∴ Sale increase,

= (82500 – 60000)

= 22500

∴ Percentage increase,

= (22500/60000 × 100)

= 37.5 %

Therefore, the percentage increase in sales will be 37.5%

Question no – (4) 

Solution : 

As per the question we know,

The price of silk was Rs 4000 per metre.

Due to lack of supply, the price went up by 6%.

Now, the new price of silk,

= (4000 + 4000 × 6/100) Rs

= 4000 + 240

= 4240 Rs

Therefore, the new price of silk per metre will be Rs. 4240

Question no – (5) 

Solution : 

According to the given question,

The population of a town was = 6,50,000.

Due to an epidemic, the population of the town went down by = 12%.

∴ Now, the current population of the town,

= 650000 – (650000 × 12/100)

= (650000 – 78000)

= 572000

Therefore, the current population of the town will be 572000.

Comparing Quantities Exercise 9.2 Solution :

Question no – (2) 

Solution :

As per the given question we know,

Sunita sold her old toys for Rs 475 and incurred a loss of 5%.

Selling Price = 475

Loss = 5%

Let, Cost Price = 100 Rs

95 …………. 475

1 …………. 475/95

∴ 100 …………. 475/95 × 100

= 500 Rs Cost Price

Now, the selling price to make profit 8%,

= 500 + (500 × 8/100) Rs

= 500 + 40

= 540 Rs

Therefore, she should sell them to Rs. 540 make a profit of 8%.

Question no – (3) 

Solution : 

According to the question we know,

Cost Price of saree = 1050;

Loss = 150

∴ Selling Price,

= (1050 – 150)

= 900

Now, the loss percentage,

= (150/1050 × 100)

= 14.28%.

Therefore, Rashmi’s loss percentage was 14.28%.

Question no – (4) 

Solution : 

According to the given question,

Raman bought eggs for Rs 6 each

Sold them making a profit of Rs 1.5 per egg.

Now, Selling Price of egg,

= (6 + 1.5) Rs

= 7.5 Rs

Therefore, Raman should sell the eggs at 7.5 Rs.

Question no – (5) 

Solution : 

According to the given question we know,

20 oranges price = 120 Rs

200 oranges price,

= 120/20 × 200

= 1200

∴ Fresh orange

= (200 – 25)

= 175

∴ Now, the selling price of oranges,

= (175 × 5)

= 875

∴ The loss,

= (1200 – 875)

= 325 Rs

∴ The loss percentage,

= (325/1200 × 100)%

= 27.08

Therefore, the loss percentage of fruit seller will be 27.08%

Question no – (6) 

Solution : 

As per the question we know,

Neha sold her old bicycle suffering a loss of 6%.

If she had sold her bicycle for Rs 90 more,

She would have made a profit of 3%.

Now, Let, CP of bicycle = 100

Selling Price, = 94

If he sell = (100 + 3) = 103

∴ (103 – 94) % = 90 Rs

∴ 9 % = 90 Rs

= 100 % = (90/9 × 100)

= 1000 Rs

Therefore, the cost price of the bicycle will be 1000 Rs.

Question no – (7) 

Solution : 

As per the question,

A businessman sold a flat making a profit of 16%.

He sold the flat for Rs 5,00,000 less, he would have suffered a loss of 16%.

Let, CP of flat = x Rs

∴ SP of flat in 16% profit

= x + x × 16/100

= x + 4x/25

In 16 % Loss, SP

= x – x × 16/100

= x – 4x/25

From question,

x + 4x/25 – x + 4x/25 = 500000

= 8x/25 = 5 Lakh

= x = 5 × 25/8 Lakh

= 500000 × 25/8

= 1562500 Rs

Therefore, the cost price of the flat 1562500 Rs.

Comparing Quantities Exercise 9.3 Solution :

Question no – (1)

Solution : 

(a) Discount,

= (2000 – 1500) = 500 Rs

∴ Discount Percentage,

= (500/2000 × 100)

= 25%

(b) Bedcover MRP = 800

Discount = 5%

∴ Selling Price = 800 – (800 × 5/100)

= 800 – 40

= 760

∴ Discount = (800 – 760) = 40 Rs

(c) MRP Cosmetics = 300

Discount = 1000

∴ Selling Price = (3000 – 1000) = 2000 Rs

Discount Percentage

= (1000/3000 × 100) %

= 33.33 %

(d) MRP = 10000

Discount = 2000

∴ Selling Price = (10000 – 2000) = 8000

∴ Discount Percentage,

= (2000/10000 × 100)%

= 20%

Question no – (2) 

Solution : 

(a) MP = 500

D = 20%

∴ Discount = (500 × 20/100)

= 100

∴ (500 – 100)

= 400%

Additional 5% discount

= (400 × 5/100)

= 20

∴ 6p = (400 – 20) = 380

(b) MP = 1000

Discount = (1000 × 20/100)

= 200

∴ (1000 – 200) = 800

Additional Discount, offer S.P

= 800 – (800 × 5/100)

= 800 – 40

= 760 Rs

(c) M.P = 1700

20% discount, = (1700 – 1700 × 20/100)

= (1700 – 340)

= 1360 Rs

Additional discount offer S.P

= (1360 – 1360 × 5/100)

= (1360 – 68)

= 1292 Rs

(d) MP = 600

Discount = 600 – (600 × 20/100)

= 600 – 120

= 480 Rs

After additional discount SP,

= 480 – (480 × 5/100)

= (480 – 24)

= 456 Rs

Question no – (3) 

Solution : 

(a) Mobile phone : Rs 9,000

= Let, MP = x Rs

∴ x × 90/100 × 80/100 = 9000

= x = 9000 × 100 × 100/90 × 80

= 12500 Rs

∴ MRP is 12500 Rs

(b) DVD player : Rs 1440

= Let, MP of player = x Rs

∴ x × 90/100 × 80/100 = 1440

= x = 1440 × 100 × 100/90 × 80

= 2000

(c) Monitor : Rs 5040

= Let, MP = x Rs

∴ x × 90/100 × 80/100 = 5040

= x = 5040 × 100 × 100/90 × 80

= 7000 Rs

(d) Pen drive : Rs 720

= Let, MP = x Rs

∴ x × 90/100 × 80/100 = 720

= x = 720 × 100 × 100/90 × 80

= 1000 Rs

Question no – (4) 

Solution : 

According to the question,

Cost price of a car is = Rs 5,00,000.

Salesman marks it at = 10% above the cost price.

Car is sold at a discount of = 5%.

∴ M.P. of car,

= 500000 + (500000 × 10/100)

= 500000 + 50000

= 550000

∴ 5% discount SP = 5

= 550000 – (550000 × 5/100)

= (550000 – 27500) Rs

= 522500

∴ Profit,

= (522500 – 500000)

= 22500 Rs.

Therefore, his profit will be 22500 Rs.

Question no – (5) 

Solution : 

As per the question we know,

A shopkeeper gives successive discounts of 15% and 20%

Selling price is = Rs 8,160,

Now, let M.P. is = x Rs

x × 85/100 × 80/100 = 8160

= x = 8160 × 100 × 100/85 × 80

= 12000 Rs

Thus, the marked price of the TV will be 12000 Rs.

Question no – (6) 

Solution : 

Let, Cost Price is = 100

∴ MRP = 130

Discount = 15%

Selling Price = 130 – (130 × 15/100)

= (130 – 19.5) Rs

= 110.5 Rs.

Question no – (7) 

Solution : 

As per the question,

Mahesh wants to buy a vacuum cleaner that costs = Rs 8000.

The GST on the vacuum cleaner is = 18%

Amount give to shopkeeper,

= 8000 + (8000 × 18/100)

= 8000 + 1440

= 9440 Rs

Therefore, amount should he give to the shopkeeper will be 9440 Rs.

Question no – (8) 

Solution : 

According to the given question,

Reena bought a mobile phone for = Rs 16,240

Included a GST of = 12%.

Now, let price = x

x + x × 12/100 = 16240

= x + 12x/100 = 16240

= 100x +12x/100 = 16240

= x = 16240 × 100/112

= 14500 Rs

∴ Price is 14500 Rs

Therefore, the price of the mobile phone 14500 Rs before the GST was added.

Comparing Quantities Exercise 9.4 Solution :

Question no – (1) 

Solution : 

As per the given question,

A man paid Rs 8640 for a loan of Rs 8000 taken for 320 days.

1st, Simple Interest,

= (8640 – 8000)

= 640 Rs

2nd, Time,

= 320 day = 320/365 yr

= 64/73 yr

∴ Now the rate (r),

= 640 × 100 × 73/64 × 8000

= 9.12%

Thus, the rate of simple interest charged by the bank was 9.12%

Question no – (2) 

Solution : 

1st, Simple Interest of 8000 Rs

= 8000 × 8 × 3/100

= 1920 Rs

Now, let time ‘t’ of 3000 Rs

S.I = 3000 × 4 × t/100

= 1920 = 3000 × 4 × t/100

∴ t = 1920 × 100/3000 × 4

= 16 years

Therefore, the required time will be 16 years.

Question no – (3) 

Solution : 

Let, principal = ‘p’

t = 16 years

Amount = 3p

∴ Simple Interest = (3p – p) = 2p

Simple Interest = p r t/100

x × 12/100 ×2 = 7000

= 24x = 7000 × 100/24

and, 2p = p × r × 16/100

= r = 200/16 = 12.5%

Let, amount become 4 times in ‘t’ years

= 4p – p = 3p

= S.I

∴ 3p = p × 12.5 × t/100

∴ t = 300/12.5

= 24 years

Therefore, the time required by the same amount to become four times will be 24 years.

Question no – (4) 

Solution : 

In 2 years SI = (6 ×2) 12%

Amount = 7000

r = 6%

t = 2 years

∴ p (1 + 6.0/100) × 2 = 7000

= p × 1.12 = 7000

= p = 7000/1.12

= 6250 Rs

Therefore, the amount Surekha took as loan was 6250 Rs.

Question no – (5) 

Solution : 

As we know that,

Simple Interest = principal × rate × time/100

= 375 = 1500 × x × x/100

= x² = 375 × 100/1500

= 25

= x = 5

∴ R = 5%,

∴ T = 5 year

Therefore, the number of years will be 5 year.

Question no – (6) 

Solution : 

Given, R = 8/2 = 4%

t = 1 year 6 month

= 1 1/2 year

= 3/2 year

= 2 × 3/2

= 3 year

∴ Amount = p (1 + r/100)t

= 5000(1 + 4/100)3

= 5000 × (104/100)3

= 5624.32 Rs

Now, the Compound Interest,

= (5624.32 – 5000)

= 624.32 Rs

Therefore, the compound interest will be 624.32 Rs.

Question no – (7) 

Solution : 

Principal = 20000

rate= 20/2%

= 10%

time = 3/2 × 2

= 3 year

Amount = 20000(1 + 10/100)3

= 20000 × 110 × 110 × 110/100 × 100 × 100

= 26620 Rs

Compound Interest,

= (26620 – 20000)

= 6620 Rs

Simple Interest,

= 20000 × 20 × 1.5/100

= 6000

∴ CI – SI = (6620 – 6000) Rs

= 620 Rs

Question no – (8) 

Solution : 

1st, the Simple Interest,

= p × r × t/100

= 2000 × 8 × 3/100

= 480 Rs

2nd, the Amount = P(1 + R/100)t

= 2000 (1 + 8/100)3

= 2000 × (108/100)3

= 2000 × 27/25 × 27/25 × 27/25

= 2519.42 Rs

∴ Compound Interest,

= (2519.42 – 2000)

= 519.42 Rs

Now, C.I – S.I

= (519.42 – 480)

= 39.42

Therefore, the difference between CI and SI will be Rs. 39.42

Question no – (9) 

Solution : 

As per the given question we know,

Principal = 12000,

time = 2

rate = 14/2 = 7%

∴ Amount = 12000 (1 + 7/100)²

= 12000 (107/100)²

= 12000 × 107 × 107/100 × 100

= 13738.80

∴ Compound Interest,

= (13738.80 – 12000)

= 1738.80 Rs

Thus, the Compound interest will be 1738.80 Rs.

Question no – (10) 

Solution : 

As per the given question we know,

Principal = 8000

Rate = 10%

time = 1 1/2 year = 6 quarter

∴ Amount,

= 8000 (1 + 10/100)6

= 8000 × 110/100 × 110/100 × 110/100

= 14160 Rs

∴ Compound Interest,

= (14160 – 8000) Rs

= 6160 Rs.

Therefore, the amount is 14160 and the compound interest is 6160 Rs.

Question no – (11) 

Solution : 

Let, time taken ‘t’

∴ 5618 = 5000 (1 + 6/100)^t

= 5618/5000 = (1 + 6/100)^t

= (106/100)^t

= 2809/2500

= (53/50)^t

= (53/50)²

= (53/50)^t

∴ t = 2 yr

Therefore, Rs 5000 to become Rs 5618 pit will take 2 years.

Question no – (12) 

Solution : 

Let, Rate of interest = R

∴ 13310 = 10000 (1 + R/100)³

= 13310/10000 = (1 + R/100)³

= (11/10)³ = (1 + R/100)³

= R/100 = 11/10 – 1

= 11 – 10/10

= 1/10

∴ Rate = 100/10 = 10%

Therefore, the rate of interest will be 10%

Question no – (13) 

Solution : 

Let, principal = P

∴ 13225 = p (1 + 15/100)²

= p (115/100)²

= p (115/100)² = 13225

= p = 13225/(115/100)²

= 13225 × 100 × 100/115 × 115

= 10000 Rs

Therefore, the sum of money invested by Seema is 10000 Rs.

Question no – (14) 

Solution : 

Let, principal = p

∴ 6655 = p(1 + 10/100)³

= p (110/100)³

= p × (11/10)³

= p × (11/10)³ = 6655

∴ p = 6655 × 10 × 10 × 10/11 × 11 × 11

= 5000 Rs

Therefore, the required sum of money will be 5000 Rs.

Question no – (15) 

Solution : 

Difference will be,

= 10000 × 6.5 × 6.5/100 × 100 Rs

= 42.25 Rs

Therefore, the difference between the Compound Interest (C.I) and Simple Interest (S.I) will be 42.25 Rs.

Question no – (16) 

Solution : 

As per the given question,

Rate = 2%

Time = 3 yr

∴ Compound Interest = p (1 + r/100)t – p

= 306.04 = p (1 + 2/100)3 – p

= p [1.061208 – 1]

= p = 306.04/0.061208

= 5000 Rs

Now, Simple Interest,

= 5000 × 2 × 3/100

= 300 Rs

Therefore, the Simple Interest (S.I) will be 300 Rs.

Comparing Quantities Exercise 9.5 Solution :

Question no – (1) 

Solution : 

As per the given question,

Cost of an antique vase increases by = 6% every year.

Current price is = Rs 2500

Price after 3 years will be,

= p(1 + r/100)³

= 2500(1 + 6/100)³

= 2500 × (106/100)³

= 2500 × 53/50 × 53/50 × 53/50

= 2977.54 Rs

Therefore, the price of antique vase after 3 years will be 2977.54 Rs.

Question no – (2) 

Solution : 

According to the question,

Car was bought in current 2015 for Rs 3,00,000.

Value of the car depreciates by 5%

∴ Worth of car in 2018 will be,

= 300000 (1 – 5/100)³

= 300000 (95/100)³

= 300000 × 95 × 95 × 95/100 × 100 × 100

= 257212.5 Rs

Therefore, the value of car in the year 2018 will be 257212.5 Rs.

Question no – (3) 

Solution : 

As per the question we know,

Engineering college has = 5,000 students in 2016.

strength of the students increases at a rate of = 4% every year

∴ Number of students in 2018 will be,

5000 (1 + 4/100)²

= 5000 × 104/100 × 104/100

= 5408

Therefore, the number of students in 2018 will be 5408.

Question no – (4) 

Solution : 

According to the question,

Tourists who visited a hill station in 2016 was = 50,00,000

It increased by every year following 2016 = 4%

∴ No of tourists in 2019 visited,

= 5000000 (1 + 4/100)³

= 5000000 (104/100)³

= 5000000 (52/50)³

= 5000000 × 26/25 × 26/25 × 26/25

= 5624320 tourists

Thus, the number of tourists visiting the hill station in 2019 is 5624320 tourists.

Question no – (5) 

Solution : 

According to the question,

Mobile phone Depreciates at a rate of 6% every year.

Current cost is = Rs 20,000

∴ Price after 4 years,

= p (1 – 6/100)⁴

= 20000 (1 – 6/100)⁴

= 20000 (94/100)⁴

= 20000 × 94 × 94 × 94 × 94/100 × 100 × 100 × 100

= 15614.98

Therefore, the price of the mobile phone after 4 years will be 15614.98 Rs.

Question no – (6) 

Solution :

As per the question we know,

Fare for a certain route of buses increases at a rate of = 10% each year.

Current fare is = Rs 5

∴ Fare after 2 years,

= 5 (1 + 10/100)²

= 5 × 110 × 110/100 × 100

= 6.05 Rs

Therefore, the fare after 2 years will be 6.05 Rs.

Question no – (7) 

Solution : 

According to the give question,

Population of bacteria in a culture is increasing at the rate of 3% per hour.

Initially it was = 4,50,000.

∴ After 2 hour population of bacteria,

= 450000 (1 + 3/100)2

= 450000 (103/100)2

= 450000 × 103 × 103/100 × 100

= 477405

Hence, approximate population of bacteria at the end of 2 hours will be 477405.

Question no – (8) 

Solution : 

As per the given question,

Value of a residential flat constructed at a cost of = Rs 10,00,000

Cost appreciating at the rate of = 10% every year

∴ Approximate value after 4 years,

= 1000000 (1 + 10/100)⁴

= 1000000 (100/100)⁴

= 1000000 × 110 × 110 × 110 × 110/100 × 100 × 100 × 100

= 1464100 Rs

Therefore, its approximate value after 4 years will be 1464100 Rs.

Question no – (9) 

Solution : 

According to the question,

rate of 5% per year consecutively for 2 years

rate of 1.5% for the next 3 years

Population 5 years ago = 23,45,000,

∴ Current population of that town,

= 2345000 (1 + 5/100)² (1 – 1.5/100)³

= 2345000 × (105/100)² × (100 – 1.5/100)³

= 2345000 × 105 × 105/100 × 100 × 98.5 × 98.5 × 98.5/100 × 100 × 100

= 2470758

Therefore, the current population of that town will be 2470758.

Next Chapter Solution : 

👉 Chapter 10 👈