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 👈