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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
∴ Votes loses by the candidate
= 25540 × 45/100
= 11493
∴ Votes secured by winning candidate
= (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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