# Class 8 ICSE Maths Solutions Chapter 7

## Class 8 ICSE Maths Solutions Chapter 7 Percent and Percentage (Selina Concise)

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 7, Percent and Percentage. Here students can easily find step by step solutions of all the problems for Percent and Percentage, Exercise 7A, 7B and 7C 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 7 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Percent and Percentage Exercise 7(A) Solution :

Question no – (1)

Solution :

(i) 55% of 160 + 24% of 50 – 36% of 150

= 55/100 × 160 + 24/100 × 50 – 36/100 × 150

= 46

(ii) 9.3% of 500 – 4.8% of 250 – 2.5% of 240

= 9.3 × 500/100 – 4.8/100 × 250 – 2.5/100 × 240

= 9.3 × 5 – 12 × 10 – 0.5 × 12

= 28.5

Question no – (2)

Solution :

(i) As per the question,

Number is increased from 125 to 150

Increase,

= (150 – 125)

= 25

Increase percentage,

= 25/125 × 100

= 20%

Therefore, the percentage of increase will be 20%.

(ii) According to the question,

Number is increased from 125 to 100

Decrease percentage,

= (125 – 100/125 × 100)%

= (25/125 × 100)%

= 20%

Hence, the percentage of decrease will be 20%.

Question no – (3)

Solution :

(i) Given, 45 is what percent of 54

(45/54 × 100) %

= 83 1/3 %

So, 45 is 83 1/3% of 54.

(ii) Given, 2.7 is what percent of 18

(2.7/18 × 100) %

= 15%

Thus, 2.7 is 15% of 18.

Question no – (4)

Solution :

(i) Let, number ‘x’

252 = x × 35/100

= x × 7/20

= x = 252 × 20/7

= 720

Hence, the required number will be 720.

(ii) Let, number ‘x’

315 = x × 14/100

= x = 315 × 100/14

x = 2250

Therefore, the required number is 2250.

Question no – (5)

Solution :

(i) Given, Number is changed from 80 to 100

Percentage increase,

= (100 – 80/80 × 100)

= 20/80 × 100

= 25%

(ii) Given, Number is changed from 100 to 80

Percentage decrease,

= (100 – 80/100 × 100)

= 20/100 × 100

= 20%

(iii) Given, Number is changed from 6.25 to 7.50

Percentage increase,

= (1.25/6.25 × 100)

= 20%

Question no – (6)

Solution :

Selling Price of house = Rs. 2, 30, 500

Changes of the auctioneer,

= (8/100 × 2, 30, 500)

= 18, 440 Rs

Therefore, the charges of the auctioneer is Rs 18, 440.

Question no – (7)

Solution :

As per the question,

Number of oranges = 800

Rotten orange = 50

No of good oranges,

= (800 – 50)

= 750

Percentage of good oranges,

= (700/800 × 100)

= 93 3/4%

Therefore, the percentage of good oranges is 93 3/4%.

Question no – (8)

Solution :

According to the question,

Cistern contains = 5 thousand litres of water.

Leaked water % = 6%

Quantity of water leaked,

= (6/100 × 5000)

= 300 litres

Water left,

= (5000 – 300)

= 4700 litres

Thus, 4700 litres of water would be left in the cistern.

Question no – (9)

Solution :

Let, salary x

Expenditure.

= (87/100 × x)

= 87x/100

Saving = 325

x – 87x/100 = 325

= 13x/100 = 325

= x = 325 × 100/13

= 2500

Therefore, the salary will be Rs 2500

Question no – (10)

Solution :

(i) Error,

= (3.625 – 3.265)

= 0.360

Percentage error,

= (0.360/3.625 × 100)

= 9.93%

Therefore, the percentage error will be 9.93%.

(ii) Error,

= (5.87 × 10³ – 5.78 × 10³)

= 0.09 × 10³

Percentage error,

= (0.09 × 10³/5.78 × 10³ × 100)

= 1.56%

Therefore, the percentage error 1.56%.

Question no – (11)

Solution :

Losing candidate secured,

= (100 – 58)%

= (58 – 42)%

= 16%

= 18, 336

= 16/100 × votes polled = 1833 × 100/16

= 114600

= (58/100 × 114600)

= 66, 468

Secured by losing candidate,

= (42/100 × 1, 14, 600)

= 48, 132

Question no – (12)

Solution :

= (100 – 47)%

= 53%

Difference,

= (53 – 47)

= 6%

6% of votes polled = 12366

= 12366 × 100/6

= 2,06,100

= (53/100 × 2, 06, 100)

= 1, 09, 233.

Hence, votes secured by winning candidate are 1, 09, 233.

Question no – (13)

Solution :

(i) After one year.

Cost of scooter after 1 year.

= (100 – 15/100 × 800)

= (85/100 × 800)

= Rs 6800

(ii) After 2 years.

∴ Cost of scooter after 2 years,

= (85/100 × 100)

= Rs 5780

Question no – (14)

Solution :

Pass marks,

= (65 + 3)

= 68

% of pass marks = 68

Pass marks = 40%

Required maximum marks,

= (100/40 × 68)

= 170

Therefore, the maximum marks is 170.

Question no – (15)

Solution :

Pass marks = 125 + 15 = 140

Let, maximum marks = x

x × 35/100 = 140

= x = 100 × 140/35

= 400

Therefore, the maximum marks will be 400.

Question no – (16)

Solution :

(i) How many correct answers did each get

Number of correct answers got by John,

= (80/100 × 150)

= 120

No of correct answers by Mohan,

= (64/100 × 150)

= 96

= (96/120 × 100)

= 80%

Question no – (17)

Solution :

According to the given question,

Number = 8,000

Increased by = 205

Decreased by = 20%

The required number,

= (120/100 × 80/100 × 8000)

= 7680

Therefore, the resulting number is 7680

Question no – (18)

Solution :

Given in the question,

Number = 12,000

Decreased by = 25%

Increased by = 25%.

The required number,

= 12000 × 75/100 × 125/100

= 11,250

Therefore, the resulting number is 11,250.

Question no – (19)

Solution :

As per the question,

New cost = 120

305 decreased

= 120 × 30/100

= 36

New cost

= (120 – 36)

= 84

Overall change,

= (100 – 84)

= 16

Required percentage,

= (16/100 × 100)

= 16% decreased.

Question no – (20)

Solution :

After decrease new cost,

= (100 – 25)

= 75

Decrease by 40%

= 75 × 40/100

= 30

New cost,

= (75 – 30)

= 45

Overall change,

= (100 – 45)

= 85

Required percentage,

= (55/100 × 100)

= 55%

Therefore, the percentage change in the cost of the article is 55%

Percent and Percentage Exercise 7(B) Solution :

Question no – (1)

Solution :

Let, the number of oranges bought is 100.

Number of rotten oranges,

= 13/100 × 100

= 13

Oranges given in charity,

= (75/100 × 85)

= 261/4

Rest, 87 = 261/4

= 348 – 261/4

= 87/4

It the balance is 522 than number of oranges,

= (100 × 4/87 × 522)

= 2400

Therefore, the had bought 2400 oranges.

Question no – (2)

Solution :

Let, total number of people in town 100

Number of pupil died due to diseases,

= 5/100 × 100

= 5

Remaining = (100 – 5) = 96

Number pupil left town,

= (3/100 × 95)

= 57/20

∴ Remaining pupil,

= (95 – 57/20)

= 1900 – 57/20

= 1843/20

Now, It the remaining pupil in the town are 276450 than original number of pupil

= 100 × 20/1843 × 276450

= 300000

Therefore, the original number of pupil in the town is 300000.

Question no – (3)

Solution :

Candidate failed in English,

= (36 – 12)%

= 24%

Candidate failed only physics,

= (28 – 12)%

= 16%

Failed both subjects = 12%

Total failed

= (24 + 16 + 12)%

= 52%

(i) Percentage passed candidates,

= (100 – 52)%

= 48%

(ii) It failed candidate are 208, total candidate appeared,

= (100/52 × 208)

= 400

Question no – (5)

Solution :

Left, B’s income = 100

A’s income = 125

Difference = 25 Rs

If A’s income is Rs 100; then B’s income loss than A,

= (25/125 × 100)

= 20 Rs

Therefore, B’s income is less than A’s income 20%.

Question no – (6)

Solution :

Let, Neetu’s age = 100 year.

Mona’s age = 80 year.

Difference,

= (100 – 80)

= 20 years

Now, It Mona is 100 years, then Neetu is older Mona by,

= (20/80 × 100)

= 25%

Hence, Neetu 25% older than Mona.

Question no – (7)

Solution :

Let, original price = 100 Rs

Today’s price of sugar = 125 Rs

On Rs, 124; the price should be decreased = 25 Rs

On Rs 100, the price decreased by,

= (25/125 × 100)

= 20%

Therefore, it need to be 20% decreased.

Question no – (8)

Solution :

Let, required number x

x + 15x/100

= 391

= x = 391 × 100/115

= 340

Therefore, the number will be 340

Question no – (9)

Solution :

Left, number ‘x’

x – 23x/100 = 539

= 100x – 23x/100 = 539

= 77x = 100 × 539

= x = 100 × 539/77

= x = 700

Therefore, the number will be 700.

Question no – (10)

Solution :

Let, 3rd number = x

1st number,

= x + 20x/100

= 120/100

2nd number,

= x + 50x/100

= 150x/100

Required percentage,

= 150x/100/120x/100

= 125%

Therefore, 125% is the second of the first.

Question no – (11)

Solution :

Let, 3rd number 100

1st number = 20

2nd number = 50

2nd number as the percent of the first,

= (50/20 × 100)%

= 250%

Therefore, the percent will be 250%

Question no – (12)

Solution :

Let, 3rd numbers = x

1st number,

= x – 30x/100

= 70x/100

= 7x/10

2nd number,

= x – 40x/100

= 60x/100

= 6x/10

Required %,

= 6x/10/7x/10 × 100

= 6x/7x × 100

= 85 5/7%

Therefore, the required percent will be 85 5/7%

Percent and Percentage Exercise 7(C) Solution :

Question no – (1)

Solution :

Here, total ball,

= (8 + 11 + 6)

= 25

Blue balls = 11

Required percentage,

= (11/25 × 100)

= 44%

Therefore, the percentage of blue balls in the bag will be 44%.

Question no – (2)

Solution :

Total money,

= (1350 + 650)

= 2000 Rs

Rohit required = 650 Rs

Required percentage,

= 650/2000 × 100

= 32.5%

Therefore, Mohan will get 32.5% from Rohit.

Question no – (3)

Solution :

As per the question we know,

Monthly income of a man = 16,000

Income-tax = 15%

Income tax,

= (16000 × 15/100)

= 2400 Rs

Rest = (16000 – 2400)

= 13600 Rs

Amount spent to rent, food, clothing,

= (13,600 × 75/100)

= 10,200 Rs

Remaining balance,

= (13600 – 10200)

= 3400 Rs

Therefore, Rs  3400 is still left whit the man.

Question no – (4)

Solution :

Let, original number = 100

New number = 120

Decrease 10%

120 × 10/100

= 12

New number,

= (120 – 12)

= 108

Required change percentage,

= (8/100 × 100)%

= 8%

Therefore, the overall change in the number as the percent is 8%.

Question no – (5)

Solution :

Let, number = 100

New number = 110

Increase mot,

= (110 × 20/100)

= 22

New formed number,

= (110 + 22)

= 132

Total change,

= (132 – 100)

= 32

Increase %,

= (32/100 × 100)

= 32%

Therefore, the overall change percentage in the number will be 32%.

Question no – (6)

Solution :

Let, starting production = 100

Decrease = 25%

∴ New production,

= (100 – 25)

= 75

In 2004, increased by 40%,

= (75 × 40/100)

= 30

∴ New production,

= (75 + 30)

= 105

Change in 2 years,

= (105 – 100)

= 5

∴ Increase percent,

= (5/100 × 100)

= 5%

Hence, resulting change of in production during these two years will be 5%

Question no – (7)

Solution :

1 orange price = 24/12 = 2 Rs

New price,

= 50/20

= 2.50 Rs

Increase in price,

= (2.5 – 2.0)

= 0.50 Rs

% increase in price,

= (0.50/2 × 100)

= 25%

Therefore, the percentage change in the price of oranges will be 25%.

Question no – (8)

Solution :

First, Math percentage,

= (120/150 × 100)

= 80%

Second, English percentage,

= (136/200 × 100)

= 68%

Third, Science,

= (108/150 × 100)

= 72%

Total number,

= (120 + 136 + 108)

= 364

Max mark,

= (150 + 150 + 200)

= 500

Whole %,

= (364/500 × 100)

= 72.8%

Therefore, the whole percentage of his score will be 72.8%

Question no – (9)

Solution :

Let, age of B = 100 years

Age of A

= (100 + 100 × 25/100)

= 125 years

Different = (125 – 100)

= 25 years

B is younger than A,

= (25/125 × 100)

= 20%

Thus, B younger than A by 20%.

Question no – (11)

Solution :

= (29200 + 58800 + 72000)

= 160,000

Percentage of vote scored by winning candidate,

= (72000/160000 × 100)

= 45%

Therefore, the percentage of votes scored by the winning candidate is 45%

Question no – (12)

Solution :

(i) Let, number = x

Here, x + 23x/100 = 861

= x = 100 × 861/123

= 700

Hence, The required number will be 700.

(ii) Let, number 100,

x – 16x/100 = 798

= x = 798 × 100/84

= 950

Therefore, the required number is 950.

Question no – (13)

Solution :

Let, price x kg sugar = 100Rs

After increase new price = 120 Rs

For Rs 120 sugar obtained x kg

For Rs 100, sugar sot = (x/120 × 100)

= 5x/6 kg

Decrease in consumption,

= (x – 5x/6)

= x/6 kg

Required percentage decrease,

= x/6/x × 100

= 16 2/3%

Therefore, the consumption of sugar must be decreased by 16 2/3%

Next Chapter Solution :

Updated: June 20, 2023 — 2:59 pm