# Joy of Mathematics Class 8 Solutions Chapter 8

## Joy of Mathematics Class 8 Solutions Chapter 8 Percentage

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 8 Percentage. Here students can easily find step by step solutions of all the problems for Percentage. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 8

Percentage Exercise 8 Solution

Question no – (1)

Solution :

(a) Given, 25%

= 25/100

= 1/4

(b) Given, 125%

= 125/100

= 5/4

(c) Given, 4 1/5%

= 21/5%

= 21/5 × 1/100

= 21/500

(d) Given, 0.25%

= 25/100 × 1/100

= 1/400

Question no – (2)

Solution :

(a) Given, 3/8

= 3 × 100/8 (b) Given, 1/16

= 1 × 100/16 (c) Given, 4 1/5

= 21 × 100/5

= 2100/5 (d) Given, 1 1/4

= 5/4

= 5 × 100/4

= 125%

(f) Given,10.375

= 10375 × 100/1000

Question no (3)

Solution :

(a) 20% of Rs 1250

= 20/100 × 1250

= Rs 250

(b) 18% of 990 km

= 18/100 × 990 km

= 891/5 (c) 12% of 1.512 kg

= 12/100 × 1.512/1000

= 18184/100000

= 0.18184 kg

(d) 8% of 5L

= 8/100 × 5

= 2/5

Question no – (4)

Solution :

(a) 12% of x is 144

=> 12/100 × x = 144

=> x = 144 × 100/12

=> x = 1200

(b) 3.5% of x is 287

=> 35/10 × 100 × x = 287

=> x = 287 × 1000/35

=> x = 8200

(c) x% of 1100 is 132

=> x/100 × 1100 = 132

=> x = 132/11

= 12

(d) x% of 1250 is 25

=> x/100 × 1250 = 25

=> x = 25 × 2/25

=> x = 2

Question no – (5)

Solution :

The price of diesel is Rs 55 per liter

Now price increased by 15%

New price of diesel per liter,

= 55 × 115/100

= 6325/100

= 63.25

New price of diesel Rs 63.25 per liter.

Question no – (6)

Solution :

Let, the total number of students in the school = 100%

60% students in the school are boys.

(100 – 60)% = 40% students in the school are girls.

Now, 40% of total number of student = 240

=> 40/100 × total number of student = 240

=> Total number of student

= 240 × 100/40

= 600

Total number of students in the school 600.

Question no – (7)

Solution :

Let, the salary of Amit = 100%

After 10% increase, salary = (100 + 10)%

= 110%

His new salary = Rs 13585

Now, 110% = 13585

=> 1% = 13585/110

=> 1% = 123.5

Amit salary before the increase

= 123.5 × 100

= 1235 × 100/10

= 12350

Amit salary before the increase Rs 12350.

Question no – (8)

Solution :

Let, the total income be = Rs 100

20% spends house rent

= 100 × 20/100

= 20

Remaining

= (100 – 20)

= 80

Now, 60% of remaining

80 × 60/100 = 48

Savings

= 100 – (48 + 20)

= 100 – 68

= 32

According to question

32 → 2400

1 → 75

The total income

= 75 × 100

= 7500

Question no – (9)

Solution :

Let,

C’s share is = 100

B’s share is = 100 × 50/100

A;s share is = 50 × 50/100

= 25

Hence, A : B : C = 25 : 50 : 100

= 1 : 2 : 4

Divided Rs 3500 among A, B, C

A’s share = 3500 × 1/7 = 500

B’s share = 3500 × 2/7 = 1000

C’s share = 3500 × 4/7 = 2000

Question no – (10)

Solution :

Let, the number is = 100

After increase 20% the number = 120

After decrease 20% the number

= 80/100 × 120

= 96

∴ Net increase or decrease = change in number/original number × 100

= 100 – 96/100 × 100

= 4/100 × 100

= 4%

Thus the number is decreased by 4%.

Question no – (11)

Solution :

Let, the total number of votes = 100

A candidate secured = 55% of the votes

Lose by = (100 – 55)% = 45%

Now, won by = (55 – 45)% = 10%

10% = 2554

1% = 2554/10

100% = 2554/10 × 100

= 25540

Total number of votes = 25540

= 25540 × 45/100

= 11493

= (25540 – 11493)

= 14047.

Question no – (12)

Solution :

Let, the maximum marks = x

A student secured 43% marks and got 24% marks more

= x × 43/100 – 24

Now, x × 36/100 + 32 = x × 43/100 – 24

=> 36x + 3200/100 = 43x – 2400/100

=> 36x + 3200 = 43x – 2400

=> – 7x = – 2400 – 3200

=> – 7x = – 5600

=> x = – 5600/- 7

=> x = 800

Maximum marks = 800

Question no – (13)

Solution :

A candidate got 48 marks and failed by 12 marks.

Passing marks,

= 48 + 12

= 60

Passing percentage = 40%

40% → 60

1% → 60/40

100% → 60/40 × 100 = 150

Maximum marks 150

Question no – (14)

Solution :

Let, the original rent be = x

Rental after 1st raise

Raise = 10% × x

= 10/100 × x

= x/10

= 0.1x

Rental = x + 0.1x = 1.1x

2nd raise

Raise = 10% of 1.1x

= 10/100 × 1.1x/10

= 11x/100

= 0.11x

Total Rental,

= 1.1x + 0.11x

= 1.21x

Total increase,

= 1.21x – x

= 0.21x

Percentage increase,

= 0.21x/x × 100 × 100

= 21%

Question no – (15)

Solution :

1st race = 2 hours 13 min

= (2 × 60) min + 13 min

= 120 + 13)min

= 133 min

Wrongly announced = 2 hours 31 minutes

= (2 × 60) min + 31 min

= (120 + 31) min

= 151 min

Error = (151 – 133) min = 18 min

Percentage error = 18 error/original raise × 100

= 18/133 × 100

= 1800/133

= 13.53%

Percentage error is 13.53%

Question no – (16)

Solution :

Let, the total income be = 100

60% of his income on household expenses

= 100 × 60/100

= 60

Remaining

= (100 – 60)

= 40

20% of remaining on re-creator

= 40 × 20/100

= 8

Now, Remaining = (40 – 8) = 32

20% of remaining o petrol

= 32 × 20/100

= 32/5

Savings = Income – Expenditure

= 100 – (60 + 8 + 32/5)

= 100 – (300 + 40 + 32/5)

= 100 – 372/5

= 500 – 372/5

= 128/5

According to the question,

128/5 → 2560

1 → 2560 × 5/128

= 100

100 → 100 × 100 = 10000

Therefore the total income = 10,000

Question no – (17)

Solution :

Let, the population of the town in the beginning = 100

5% population of a town died

= 100 × 5/100

= 5

Remaining population

= (100 – 5)

= 95

Now, 15% of the remaining migrated from the town

= 95 × 15/100

= 57/4

Question no – (18)

Solution :

Let, B’s income is = 100

A’s income is = 110

B’s income less than A’s income

= (A – B)’s income/A’s income × 100

= 10/110 × 100

= 100/11 Question no – (19)

Solution :

Let, the original price = 100

After increasing 20% the price = 120

Reduced percentage of price

= 120 – 100/120 × 100

= 20/120 × 100

= 200/12 Reduced price = 16 2/3

Question no – (20)

Solution :

The sides of rectangle are 40 cm by 30 cm

∴ Area of rectangle = (40 × 30) cm2

= 1200 cm2

Now, each side in increased by 10%

Then, length = (40 + 40 × 10/100)

= 40 + 4

= 44

Breadth = (30 + 30 × 10/100)

= 30 + 3

= 33

Now, Area,

= 44 × 33

= 1452

Percentage increase in the area

= difference of area/original area × 100

= 1452 – 1200/1200 × 100

= 252/1200 × 100

= 21%

Hence, percentage increase in the area 21%.

Question no – (21)

Solution :

Total marks scored = 62 + 40 + 50 + 48 + 32

= 232

Total marks

= 80 + 60 + 60 + 50 + 40

= 290

Aggregate percentage

= 232/290 × 100%

Percentage score in English

= 62/80 × 100

= 77.5

Percentage score in Hindi

= 40/60 × 100

= 66.67

Percentage score in science

= 50/60 × 100

= 83.34

Percentage score in mathematics

= 48/50 × 100

= 96

Percentage score in social studies

= 32/40 × 100

= 80

The best performance is in mathematics.

Question no – (22)

Solution :

Let, the original price = 100

After increase 20% the price

= 100 + 20

= 120

Again, increase 20% of the price

= 120 × 20/100

= 24

Total price,

= (120 + 24)

= 144

Percentage

= 144 – 100/100 × 100

= 44/100 × 100

= 44%

Question no – (23)

Solution :

Total number of students = 2000

Total number of girls = 800

Total number of boys

= (2000 – 800)

= 1200

80% of the girls passed

= 800 × 80/100

= 640

Failed girls student

= (800 – 640)

= 160

70% of the boys passed

= 1200 × 70/100

= 840

Failed boys student,

= (1200 – 840)

= 360

Total number of failed student,

= 160 + 360

= 520

Percentage of failed student,

= 520/2000 × 100

= 26%

Percentage of failed students 26%.

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Updated: May 30, 2023 — 7:22 am