# Frank ICSE Class 8 Solutions Chapter 8

## Frank ICSE Mathematics Class 8 Solutions Chapter 8 Percentage and Its Applications

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 8, Percentage and Its Applications. Here students can easily find step by step solutions of all the problems for Percentage and Its Applications, Exercise 8.1 and 8.2 Also 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. Here in this post all the solutions are based on latest Syllabus.

Percentage and Its Applications Exercise 8.1 Solution :

Question no – (1)

Solution :

(a) 15 % Rs 240

= 240 × 15/100

= Rs 36

Therefore, 15 % Rs 240 is Rs 36

(b) 35% of Rs 4200

= 4200 × 35/100

= 42 × 35

= Rs 1470

Therefore, 35% of Rs 4200 is Rs 1470

(c) 75% Rs 1050

= 1050 × 75/100

= 15 × 105/2

= 1575/2

= 787.5

Therefore, 75% Rs 1050 is Rs 787.5

Question no – (2)

Solution :

(a) 15 × 100 × x

= 75

or, x = 75 × 100/15

= 500

Therefore, 15 × 100 × x = 75 is = 500

(b) 20/100 × y

= 75

or, y = 75 × 100/20

= 375

Therefore, If 20% of y is 75 find y is 375

(c) 20% of students failed. Then 80% students have passed

Suppose there are x students

= 80% of x

= 36

or, x × 80/100
= 36

or, x = 36 × 100/80

= 45

Therefore, The total number of students in the class is 45

(d) – (i) 15 × 100 × x = 75

or, x = 75 × 100/15

= 500

(ii) 20/100 × y = 75

or, y = 75 × 100/20

= 375

Question no – (3)

Solution :

Suppose total number of working days of school is = x

= x × 40/100

= 576

or, x = 192 × 100/75

= 256

Therefore, The total number of working days will be 256 days.

Question no – (4)

Solution :

= x × 40/100

= 576

or, x = 288/576 × 100/40

= 1440

Therefore, He have 1440 mangoes originally.

Question no – (5)

Solution :

Suppose the population in 2016 was x

= x × 105/100 × 105/100 = 132, 300

or, x 1, 32, 300 × 100/105 × 100/105

= 120000

The population was 120000.

Therefore, the population of the town will be 120000 in 2016.

Question no – (6)

Solution :

After 2 years the population will be

= 1, 80, 000 × 110/100 × 110/100

= 11 × 11 × 1800

= 217800

Therefore, the population of the town after 2 years will be 217800.

Question no – (7)

Solution :

Decrease of visitors

= 34000 – 32,300

= 1700

Decrease percentage

= (1700/34000 × 100)%

= 5%

Therefore, the decrease percent of visitors will be 5%

Question no – (8)

Solution :

Aman income after 2 years

= 30,000 × 109/100 × 109/100

= 35643

Therefore, His monthly income after 2 years will be 35643.

Question no – (9)

Solution :

Aman income after 2 years

= 30,000 × 109/100 × 109/100

= 35643

Therefore, His monthly income after 2 years will be 35643.

Question no – (10)

Solution :

The value of machine after 2 years

= 24000 × 95/100 × 95/100

= 12 × 95 × 19

= 21660 rupees

Therefore, the value of the machine will be Rs 21660 after 2 years.

Question no – (11)

Solution :

= 80,000 × 90/100 × 40/100

= 8 × 90 × 40

= 28800

Question no – (12)

Solution :

Suppose the number is x

= x × 120/100 × 80/100

= 24x/25

Since, 24x/25 < x therefore the number has decreased

And the decrease percentage

= (x – 24x/25/(x) × 100

= x/25/(x) × 100

= x/25 × 1/x × 100

= 4%

Therefore, the net decreased percent will be 4%

Question no – (13)

Solution :

Let, Alok salary is Rs 100 and Ajit salary is 125

Therefore,

= 125 – 100/125 × 100

=  25/125 × 100

=  20%

So, Alok’s salary is 20% less than Ajit’s salary.

Question no – (14)

Solution :

Kumar scored 960

So, Baljit scored

= 960 × 100/120

= 800

Ajit Scored

= 800 × 90/100

= 720

Therefore, Ajit Scored 720, Baljit  Scored  800

Question no – (15)

Solution :

Let, monthly income of Ameeta = x

∴ House rent =  x × 30/100

=  3x/10

Rest money,

=  x –  3x/10

= 10x – 3x/10

=  7x/10

Other Expenditure,

=  7x/10 × 60/100

=  21x/50

Rest money,

=  7x/10 – 21x/50

=  35x – 21x/50

= 14x/50

∴ 14x/50 = 7560

=  14x =  378000

= x =  378000/14

=  x = 27, 000

Therefore, Almeta’s monthly income will be Rs 27, 000

Question no – (16)

Solution :

A garden has total = 50000 trees

Guava trees are = 12%

Orange trees are = 15%

Coconut trees are = 10%

Rest are mango trees = ?

Total trees in the garden = 50000

Now, Guava trees,

= 50000 × 12/100

= 6000

Orange trees,

= 50000 × 15/100

= 7500

Coconut trees,

= 50000 × 10/100

= 5000

Therefore, total mango trees are,

– 50000

– 6000

– 7500

– 5000
————————–
31,500

Thus, the total number of mango trees will be 31,500

Question no – (17)

Solution :

Total students of a class = 35

Interested in mathematics = 60%

Not interested in mathematics = ?

Students interested in mathematics,

= 35 × 60/100

=  21

Now, Students not interested in mathematics,

= 35 – 21

=  14

Therefore, 14 students are not interested in mathematics.

Question no – (18)

Solution :

Let, match played by them = x

Won match,

=  x × 60/100

=  3x/5

Lost match,

=  x – 3x/5

=  5x – 3x/5

=  2x/5

Therefore, 2x/5 =  24

= 2x = 5 × 24

= x = 5 × 24/2

= x = 60

So, the total number of matches played by them will be 60.

Question no – (19)

Solution :

Let, he had originally Rs = x

Spent money,

= x × 85/100

=  17x =20

Rest money

= x – 17x/20

=  20x  – 17x/20

=  3x/20

Therefore, 3x/20 =  810

= 3x =  810 × 20

= x = 810 × 20/3

= x = 5400

Thus, he had originally Rs 5400

Question no – (20)

Solution :

A man travels the distance by air = 50%

Travels the distance train = 35%

Travels the distance by bus = 10%

Travels the distance by taxi = 144 km

Let, total distance travelled by man is x

Therefore,

By Air,

= x × 30/100

= x/2

By Train,

= x × 35/100

= 7x/20
By Bus,

= x × 10/100

= x/10

Now remaining distance,

= x – x/2 – 7x/20 – x/10

= 20x – 10x – 7x – 2x/20

= x/20

Therefore, x/20 = 144

= x = 2880

Air = 2880/2

= 1440 KM

Train = 7 × 2880/20

= 1008 km

Bus = 2880/10

= 288 Km

Question no – (21)

Solution :

Let, total family be x

Therefore,

Times of India,

= x × 65/100

= 13x/20

Indian Express,

= x × 75/100

= 3x/4

Neither,

= x × 15/100

= 3x/20

Therefore,

= x – 3x/20

= 20x – 3x/20

= 17x/20

Now total families lived in that locality

= 13x/20 + 3x/4 + 3x/20 = x 330

= 13x + 15x 3x/20 = x + 330

= 31x/20 = x + 330

= 31x = 20x + 6600

= 31x – 20x = 6600

= 11x = 660

= x = 6600/11

= x = 600

Therefore, a total of 600 families live in that locality.

Revision Exercise Questions Solution :

Question no – (1)

Solution :

= Total number of working days

= 186 × 100/75

= 248

Therefore, the total number of working days of the school will be 248.

Question no – (2)

Solution :

Annual income,

= 28000 × 100/7

= 400000 rupees

Hence, the annual income of Mohit will be 400000 rupees

Question no – (3)

Solution :

The initial salary

= 36300 × 100/110

= 33000 rupees

Therefore, his salary was 33000 rupees before the increase

Question no – (4)

Solution :

Final Value,

= 12500 × 95/100 × 108/100

= 25 × 19 × 27

= 25 × 513

= 12825

Thus, the value of the land will be Rs 12825 after 2 years

Question no – (5)

Solution :

Final Charge

= 1500 × 90/100 × 98/100

= 27 × 49

= 1323

So, Ramesh will charge Seema Rs 1323

Question no – (6)

Solution :

Cost price of 5 kg mixture

= (3 × 240) + (2 × 180)

= 270 + 360

= 1080

Selling price of 5 kg

= 270 × 5

= 1350

Profit percent,

= {(1350 – 1080/1350) × 100}%

= {270/1350 × 100}%

= 20%

That’s why, his profit will be 20%

Question no – (7)

Solution :

Cost of each banana

= 42/12

= 7/2 rupees

In 98 rupees Reena can be buy

= 98/7/2

= 98 × 2/7

= 28 bananas

Hence, Reena can buy 28 bananas

Question no – (8)

Solution :

(a) S.P = 8700 × 80/100

= 6960 rupees

(b) C.P = 6960

= 100/116

= 6000 rupees

Question no – (9)

Solution :

Let, the cost price x rupees

= x + 12x/100

= 840

or, 112x/100

= 840

or, x = 840 × 100/112

= 750

Therefore, the price will be RS 750 before the tax.

Objective Type Questions Solutions :

Question no – (1)

Solution :

C.P of 12 articles = S.P of 10 articles

or, C.P of (12 + 2) articles – S.P of 10 articles

or, S.O of 10 articles

= C.P of 10 articles

= C.P of 2 articles

Profit in selling 10 articles

= C.P of 2 articles

Percentage of profit

= (2/10 × 100)%

= 20%

Therefore, the profit percent will be 20%

Question no – (2)

Solution :

From the question we know,

Cost price of the shoes is = 2500

and GST is = 18%

Now, selling Price of the shoes,

S.P = 250 × 118/100

= 2950

Therefore, the selling shoes 2950.

Question no – (3)

Solution :

From question we know,

Cost price of the article is = 6400

and the profit percent is 6%

Now, selling price will be

S.P = 6400 × 106/100

= 6784

Therefore, selling price of the article 6784.

Question no – (4)

Solution :

From question we know,

Selling price of the article is = 2750

and the profit percent is 10%

Now, cost price will be

C.P = 2750 × 100/110

= 2500

Therefore, cost price of article is 2500.

Question no – (5)

Solution :

From question we know,

MRP of the article is = 1500

and the discount percent is 10%

Now, selling price will be

S.P = 1500 × 90/100

= 1350

Therefore, the selling price of the article is 1350.

Question no – (6)

Solution :

From question we know,

Selling price of the article is = 380

and the discount percent is = 5%

Now, marked price will be

= 3800 × 100/95

= 400

Therefore, the marked price of the article 400.

Question no – (7)

Solution :

Marked price of the commodity = 2000

Discount percent = 12% and 5%

Selling price of the commodity

S.P = 2000 × 88/100 × 95/100

= 1672

Therefore, The selling price of the commodity will be Rs 1672.

Next Chapter Solution :

Updated: June 24, 2023 — 11:33 am