Rd Sharma Solutions Class 8 Chapter 12 Percentage
Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 12, Percentage. Here students can easily find Exercise wise solution for chapter 12, Percentage. Students will find proper solutions for Exercise 12.1 and 12.2 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.
Percentage Exercise 12.1 Solution :
Question no – (1)
Solution :
(i) 7/25
= 7/25 × 100
= 28%
Therefore, 7/25 as percent = 28%
(ii) 14/625
= 14/625 × 100
= 56/25
= 2.24%
Hence, 14/625 as percent = 2.24%
(iii) 5/8
= 5/8 × 10
= 125/2
= 62.5%
Thus, 5/8 as percent = 62.5%
(iv) 0.8
= 8/10 × 100
= 80%
Hence, 0.8 as percent = 80%
(v) 0.005
= 5/1000 × 100
= 0.5%
Therefore, 0.005 as percent = 0.5%
(vi) 3 : 25
= 3/25 × 100
= 12%
Thus, 3 : 25 as percent = 12%
(viii) 11 : 80
= 11/80 × 100
= 13.75%
Thus, 11 : 80 as percent = 13.75%
(ix) 111 : 125
= 111/125 × 100
= 88.8%
Hence, 111 : 125 as percent = 88.8%
(x) 15 : 16
= 15/16 ×100
= 93.75%
Therefore, 15 : 16 as percent = 93.75%
(xi) 0.18
= 18/100 × 100
= 18%
Hence, 0.18 as percent = 18%
(xii) 7/125
= 7/125 × 100
= 5.6%
Thus, 7/125 as percent = 5.6%
Question no – (2)
Solution :
(i) 25%
= 25/100
= 1/4
= 1 : 4
Thus, 25% as fraction = 1/4 and as ratio = 1 : 4
(ii) 2.5%
= 25/10 × 1/100
= 1/40
= 1 : 40
Hence, 2.5% as fraction = 1/40 and as ratio = 1 : 40.
(iii) 0.25%
= 25/100 × 1/100
= 1/400
= 1 : 400
Hence, 0.25% as fraction = 1/400 and as ratio = 1 : 400
(iv) 0.3%
= 3/10 × 1/100
= 3/100
= 3 : 1000
Therefore, 0.3% as fraction = 3/100 and ratio = 3 : 100
(v) 125%
= 125/100
= 5/4
= 5 : 4
So, 25% as fraction = 5/4 and as ratio = 5 : 4
Question no – (3)
Solution :
(i) 27%
= 247/100
= 0.27
Note : To convert percent to decimal we have to take up the following point
Number 1, for the reason of percent the denominator is 100 is placed – 27/100
To make it decimal first, right 27 and count number of zero after one in denominator.
Here, is 2 zero.
So, the answer is 0.27.
(ii) 6.3%
= 63/10 × 1/100
= 0.063
Note : To convert percent to decimal we have to take up the following point
Number 1, for the reason of percent the denominator is 100 is placed – 63/100
For the reason of point in lower one 1 is placed and after that there is one digit 3 so 10 is placed,
Now it becomes 63/10 × 1/100
To make it decimal first, write 63 and count number of zero after 1 in denominator.
Here, is 3 zero
So, the answer is 0.063
(iii) 32%
= 32/100
= 0.32
Note : To convert percent to decimal we have to take up the following point
Number 1, for the reason of percent the denominator is 100 is placed – 32/100
To make it decimal first, right 32 and count number of zero after one in denominator.
Here, is 2 zero
So, the answer is 0.32
(iv) 0.25%
= 25/100 × 1/100
= 0.0025
Note : To convert percent to decimal we have to take up the following point
Number 1, for the reason of percent the denominator is 100 is placed – 0.25/100
For the reason of point in lower one 1 is placed and after that there is two digit 25 so 100 is placed, Now it becomes 25/100 × 1/100. To make it decimal first, write 25 and count number of zero after 1 in denominator.
Here, is 4 zero,
So, the answer is 0.0025
(v) 7.5%
= 75/10×100
= 0.075
Note : To convert percent to decimal we have to take up the following point
Number 1, for the reason of percent the denominator is 100 is placed – 75/100
For the reason of point in lower one 1 is placed and after that there is one digit 5 so 10 is placed, Now it becomes 75/10×100. To make it decimal first, write 75 and count number of zero after 1 in denominator.
Here, is 3 zero,
So, the answer is 0.075
(vi) 1/8%
= 1/8 × 100
= 0.00125
Therefore, 1/8% as decimal fraction = 0.00125
Percentage Exercise 12.2 Solution :
Question no – (1)
Solution :
(i) 22% of 120
= 22/100 × 120
= 132/5
= 26.40
(ii) 25% of Rs 1000
= 25/100 × 1000
= 250
Therefore, 25% of Rs 1000 is 250
(iii) 25% of 10 kg
25/100 × 10
= 2.5 kg
Hence, 25% of 10 kg is 2.5 kg
(iv) 16.5% of 5000 metre
= 165/10 × 100 × 5000
= 825 metre.
Hence, 16.5% of 5000 metre is 825 metre.
(v) 135% of 80 cm
= 135/100 × 80
= 108 cm
Thus, 135% of 80 cm is 108 cm.
(vi) 2.5% of 10000 ml
= 25/10 × 100 × 10000
= 250 ml
Therefore, 2.5% of 10000 ml is 250 ml.
Question no – (2)
Solution :
(i) 8.4 of a is 42
or, 84/10 × 100 × a = 42
or, a = 42 × 10 × 100/84
= 500
Hence, the number a will be 500
(ii) 0.5% of a is 3
or, 5/10 × 100 × a = 3
or, a = 3 × 10 × 100/5
= 600
Thus, the number a will be 600.
(iii) 1/2% of a is 50
or, 1/2× 100 × a = 50
or, a = 50 × 2 × 100
= 10000
So, the number a will be 10000
(iv) 100% of a is 100
or, 100/100 × a = 100
or, a = 100
Therefore, the number a will be 100.
Question no – (3)
Solution :
Given, x = 5% of y
or, 480 = 5/100 × y ….. [∵ x = 480]
or, y = 480× 100/5 = 9600
Now, y = 24% of z
or, 24/100 × z = 9600 ….. [∵ y = 9600]
or, z = 9600 × 100/24 = 40000
∴ y = 9600, z = 40000
Therefore, the values of y will be 9600 and z will be 40000.
Question no – (4)
Solution :
Let, monthly income = x
Now, according to question,
= x × 15/100 = 150
or, x = 150 × 100/15
= x = 1000
Therefore, his monthly income will be 1000 Rs.
Question no – (5)
Solution :
Let, Total number of marks = x
According to question,
= x × 86.875% = 695
or, x × 86875/1000 × 100 = 695
or, x = 695 × 1000 × 100/86875 = 800
Therefore, total number of marks of the examination is 800.
Question no – (6)
Solution :
Let, number of days an which the school was opened = x
According to question,
= x × 90/100 = 216
or, x = 216 × 100/90 = 240
Therefore, the school was opened for 240 days.
Question no – (7)
Solution :
Total trees = 2000 trees
Number of mango tree,
= 200× 12/100
= 240 trees
Number of Lemon tree,
= 2000 × 18/100
= 360 trees
Number of Orange tree,
= 200 – (240 + 360)
= 2000 – 600
= 1400 trees.
Therefore, the number of orange trees are 1400 tree.
Question no – (8)
Solution :
Amount of protein,
= (26000 × 12/100)
= 312 cal
∴ Calories in protein = 312
Amount of fat,
= (2600 × 25/100)
= 650 cal.
∴ Calories in fat = 650
Amount of carbohydrate,
= (2600 × 3/100)
= 1638 cal
∴ Calories in carbohydrate = 1638
Question no – (9)
Solution :
As per the question,
Total runs = 62
(i) He hit 3 sixes :
∴ Run came in sixes
= 3 × 6
= 18 run
∴ Run percentage in sixes
= 18/62 × 100
= 29.03%
(ii) he hit 8 fours :
∴ Run came in four
= 8 × 4
= 32 run
∴ Run% came in four
= 32/62 × 100
= 51.62%
(iii) he hit 2 tows :
∴ Run came in two
= 2 × 4
= 4 run
∴ Run% came in two
= 4/62 × 100
= 6.45%
(iv) he hit 8 singles :
∴ Run came in singles
= 8 × 1
= 8 run
∴ Run% came to in singles
= 8/62 × 100
= 12.90%
Question no – (10)
Solution :
According to the question,
Total run = 120 run
(i) 20% run came in 6’s
∴ Total run came in 6’s
= 120 × 20/100
= 24 run
(ii) 30% total run came in 2’s
∴ Total run came in,
= 120 × 30/100
= 36 run
(iii) 25% run came in 4’s 30
∴ Total run came in 11
= 120 × 25/100
= 30 run
∴ Total run came in singles
= 120 – (24 + 36 + 30)
= 120 – 90
= 30 run
Question no – (11)
Solution :
Let she invest = x
According to question,
= x × 22/100 = 187
or, x = 187 × 100/22
= 850 Rs.
Therefore, Radha was invested 850 Rs.
Question no – (12)
Solution :
Let, his total income = x
According to question,
= x × 12/100 = 1440
or, x = 1440 × 100/12
= 12000
Therefore, Rohit’s total income for the year 1997 is 12000.
Question no – (13)
Solution :
Let, Amount of gunpowder x containing 9 kg of nitre.
∴ x × 75/100 = 9.
= 75x = 900
= x = 900/75
= 12 kg.
Now Let, x kg be the gunpowder containing 2.5 kg of Sulphur.
∴ x× 10/100 = 2.5
= 10x = 250.
= x = 250/10
= 25 kg.
Therefore, 25 kg gunpowder contains 2.5 kg of Sulphur.
Question no – (14)
Solution :
Total no of part Alloy
= (15 + 105)
= 120
Now No of part of copper,
= 105
∴ Percentage of copper in alloy,
= 105/120 × 100
= 87.5%
Therefore, the percentage of copper in the alloy is 87.5%
Question no – (15)
Solution :
Total alloy :
= 1 kg
= 1000 gm
Copper contains :
= 1000 × 32/100
= 320 kg
Nickel contains :
= 1000 × 40/100
= 400 kg
∴ Zinc contains :
= 1000 – (320 + 400)
= 1000 – 720
= 280 gm
Therefore, the mass of the zinc in 1 kg of the alloy is 280 gm.
Question no – (16)
Solution :
Let, Total ride = x km
According to question,
= x × 10/100 = 122
or, x = 1220 km
Therefore, the total ride will be 1220 km.
Question no – (17)
Solution :
Total students = 300
Boys = 142
∴ Girls students
= 300 – 142
= 158
∴ Girls students%
= 158/300 × 100
Total teachers = 30
∴ Men teachers = 12
Female = 30 – 12 = 8
∴ Female teachers,
=- 8/30 × 100
= 8/3 = 8/3
∴ Total girls students + female students%
= (158/3 + 8/3)
= 166/3%
Question no – (18)
Solution :
Let, Anil’s income = x
∴ Aman’s income,
= x – 20x/100
= 100 – 20x/100
= 80x/100
= 8x /10
Difference between Anil and Aman’s income,
= x – 8x/10
= 10x – 8x/10
= 2x/10
∴ Percentage between Anil and Aman’s income,
= (2x/10/8x/10) × 100
= 2x × 10/8x × 10 × 100
= 25%
Question no – (19)
Solution :
Present value of machine = Rs 100000
Every year of machine depreciate = 5%
∴ After 1 year machine value :
= 10000 × (100 – 5)/100
= 100000 × 95/100
= 95000 Rs.
Again After 2 year machine value :
= 95000 × (100 – 5) /100
= 95000 × 95/100
= 90250 Rs.
Therefore, its value after 2 years will be 90250 Rs.
Question no – (20)
Solution :
As per the given question,
Present population = 60000
After increasing,
= (60,000 × 10/100)
= 6000
Total population,
= (60,000 + 6000)
= 66000
Increase after 2 years
= 10 % of 66,000
= 10/100 × 66,000
= 6600
Hence, population after 2nd year,
= 66000 + 600
= 72600
Therefore, the population of the town after 2 years will be 72,600.
Question no – (21)
Solution :
According to the given question,
Present population = 22000
Let, before 1 year population = x
According to question,
= x × 110/100 = 22000
or, x = 22000 × 10/11 = 20000
Therefore, a year ago the population of the town was 20,000.
Question no – (23)
Solution :
Let, price of petrol = 100
After incoming price of petrol
= 100 + (100 × 10/100)
= 110
Let, at the same expenditure the
consumption should be reduce by x %
According to question,
x = 110 – 100/110 × 100
or, x = 10/110 × 10
= 100/11
= 9 1/11%
Therefore, consumption need to reduce by 9 1/11%, then the expenditure does not increase.
Question no – (24)
Solution :
Mohan’s income = 15500
Mohan’s savings
= 15500 × 11/10
= 1705
Increasing income :
= 15500 × 110/100
= 17050
Now, Mohan’s saving :
= 17050 × (11 – 1)/100
= 17050 × 10/100
= 1705
Therefore, Both saving are same.
Question no – (25)
Solution :
Let, Shalu’s income = x
Shikha income,
= x + 60x/100
= 100 + 60x/100
= 1600x/100
= 8x/5
Difference between Shalu and Shikhas income,
= 8x – x/5
= 8x – 5x/5
= 3x/5
∴ Percentage of income Shalu and Shikha,
= 3x/5/8x/5 × 100
= 75/2 %
= 37.5 %
Question no – (26)
Solution :
Let, 1st persons share = x
∴ 2nd persons share,
= x × 50/100
= x/2
∴ 3rd persons share,
= x/2 × 50/100
= x/4
Now, according to question,
x + x/2 + x/4 = 3500
or, 4x + 2x+x/4 = 3500
or, 7x = 3500 × 4/7 = 2000
∴ 1st persons shared = 2000
∴ 2nd persons shared,
= 2000/2
= 1000
∴ 3rd persons sheared,
= 2000/4
= 500
Therefore, each of the them will get, 2000, 1000, and 500
Question no – (27)
Solution :
Let, original price = x
After hike,
= 20x/100
Now, as per the question,
x + 20x/100 = 2000
or, 100x + 20x/100 = 2000
or, 2000 ×100
or, x = 2000 × 100/120
= 500/3
= 1666.67 Rs.
Therefore, the original price of the object will be 1666.67 Rs.
Next Chapter Solution :
👉 Chapter 13 👈
