## RD Sharma Solutions for Class 7 Maths Chapter 11 Percentage Free Online

Exercise 11.1 Page No: 11.3

**1. Express each of the following per cents as fractions in the simplest forms:**

**(i) 45%**

**(ii) 0.25%**

**(iii) 150%**

**(iv) 6 (1/4) %**

**Solution:**

(i) Given 45%

= (45/100)

On simplifying the above fraction we get

= (9/20)

(ii) Given 0.25%

= (0.25/100)

= (25/10000)

On simplifying the above fraction we get

= (1/400)

(iii) Given 150%

= (150/100)

On simplifying the above fraction we get

= (3/2)

(iv) Given 6 (1/4) %

We can write 6 (1/4) as 6.25

= (6.25/100)

= (625/1000)

= (1/16)

**2. Express each of the following fractions as a per cent:**

**(i) (3/4)**

**(ii) (53/100)**

**(iii) 1 (3/5)**

**(iv) (7/20)**

**Solution:**

(i) Given (3/4)

= (3/4) × 100

= 75%

(ii) Given (53/100)

= (53/100) × 100

= 53%

(iii) Given 1 (3/5)

Convert the given mixed fraction into improper fraction

1 (3/5) = (8/5)

= (8/5) × 100

= 160%

(iv) Given (7/20)

= (7/20) × 100

= 35%

Exercise 11.2 Page No: 11.4

**1. Express each of the following ratios as per cents:**

**(i) 4: 5**

**(ii) 1: 5**

**(iii) 11: 125**

**Solution:**

(i) Given 4: 5

4: 5 can be written as (4/5)

= (4/5) × 100

= 80%

(ii) Given 1: 5

1: 5 can be written as (1/5)

= (1/5) × 100

= 20%

(iii) Given 11: 125

11: 125 can be written as (11/125)

= (11/125) × 100

= (44/5) %

**2. Express each of the following per cents as ratios in the simplest form:**

**(i) 2.5%**

**(ii) 0.4%**

**(iii) 13 (3/4) %**

**Solution:**

(i) Given 2.5%

= (2.5/100)

= (25/1000)

= (1/40)

(ii) Given 0.4%

= (0.4/100)

= (4/1000)

= (1/250)

(iii) Given 13 (3/4) %

13 (3/4) = 13.75

= 13.75/100

= 1375/10000

= 11/80

Exercise 11.3 Page No: 11.5

**1. Express each of the following per cents as decimals:**

**(i) 12.5%**

**(ii) 75%**

**(iii) 128.8%**

**(iv) 0.05%**

**Solution:**

(i) Given 12.5%

= (12.5/100)

= 0.125

(ii) Given 75%

= (75/100)

= 0.75

(iii) Given 128.8%

= (128.8/100)

= 1.288

(iv) Given 0.05%

= (0.05/100)

= 0.0005

**2. Express each of the following decimals as per cents:**

**(i) 0.004**

**(ii) 0.24**

**(iii) 0.02**

**(iv) 0.275**

**Solution:**

(i) Given 0.004

0.004 can be written as 4/1000

= (4/1000) × 100

= 0.4%

(ii) Given 0.24

0.24 can be written as (24/100)

= (24/100) × 100

= 24%

(iii) Given 0.02

0.02 can be written as (2/100)

= (2/100) × 100

= 2%

(iv) Given 0.275

0.275 can be written as (275/1000)

= (275/1000) × 100

= 27.5%

**3. Write each of the following as whole numbers or mixed numbers:**

**(i) 136%**

**(ii) 250%**

**(iii) 300%**

**Solution:**

(i) Given 136%

= (136/100)

On simplifying we get

= (34/25)

(ii) Given 250%

= (250/100)

On simplifying

= (5/2)

(iii) Given 300%

= (300/100)

= 3

Exercise 11.4 Page No: 11.7

**1. Find each of the following:**

**(i) 7% of Rs 7150**

**(ii) 40% of 400kg**

**(iii) 20% of 15.125liters**

**(iv) 3 (1/3) % of 90km**

**(v) 2.5% of 600meters**

**Solution:**

(i) Given 7% of Rs 7150

= (7/100) × 7150

= Rs 500.50

(ii) Given 40% of 400kg

= (40/100) × 400

= 160kg

(iii) Given 20% of 15.125liters

= (20/100) × 15.125

= 3.025liters

(iv) Given 3 (1/3) % of 90km

We know that 3 (1/3) = (10/3)

= (10/300) × 90

= 3km

**2. Find the number whose 12 ½ % is 64.**

**Solution:**

Let the required number be x

Then according to the question, 12 ½ % × x = 64

= 12.5 % × x = 64

= (12.5 × 100) × x = 64

x = (64/100)/ 12.5

x = 64 × 8 = 512

Therefore 512 is the number whose 12 ½ % is 64.

**3. What is the number, 6 ¼ % of which is 2?**

**Solution:**

Let the required number be x

Then according to the question, 6 ¼ % × x = 64

= 6.25 % × x = 2

= (6.25/100) × x = 2

x = (2 × 100)/ 6.25

x = 2 × 16 = 32

Therefore 32 is the number whose 12 ½ % is 64.

**4. If 6 is 50% of a number, what is that number?**

**Solution:**

Let the required number be x

Given that 50 % of x = 6

(50/100) × x = 6

x = (300 × 2)/ 50

x = 12

The required number is 12

Exercise 11.5 Page No: 11.9

**1. What per cent of**

**(i) 24 is 6?**

**(ii) Rs 125 is Rs 10?**

**(iii) 4km is 160 meters?**

**(iv) Rs 8 is 25 paise?**

**(v) 2 days is 8 hours?**

**(vi) 1 liter is 175ml?**

**Solution:**

(i) According to the question required percentage = (6/24) × 100

= (100/4)

= 25%

(ii) According to the question required percentage = (10/125) × 100

= (1000/125)

= 8%

(iii) According to the question required percentage = (160/4) × 100

We know that 1km = 1000 meters

Therefore 4km = 4000 meters

= (160/4000) × 100

= 4%

(iv) According to the question required percentage = (25/8) × 100

We know that 1Rs = 100 paise

Therefore 8Rs = 800 paise

= (25/800) × 100

= (25/8)

= 3.125%

(v) We know that 1day = 24 hours

1 hour = (1/24) day

8 hours = (8/24) day = (1/3) day

According to the question required percentage = (1/3)/2 × 100

= 100/6

= 16 (2/3) %

(vi) We know that 1liter = 1000ml

According to the question required percentage = (175/1000) × 100

= 17500/1000

= 17.50 %

**2. What per cent is equivalent to (3/8)?**

**Solution:**

Given (3/8)

= (3/8) × 100

= 37.5%

**3. Find the following:**

**(i) 8 is 4% of which number?**

**(ii) 6 is 60% of which number?**

**(iii) 6 is 30% of which number?**

**(iv) 12 is 25% of which number?**

**Solution:**

(i) Let x be the required number

Given that 4% of x = 8

(4/100) × x = 8

x = (800/4)

x = 200

(ii) Let the required number be x

Given that 60% of x = 6

(60/100) × x = 6

x = (60/6)

x = 10

(iii) Let the required number be x

Given that 30% of x = 6

(30/100) × x = 6

x = (6 × 100)/30

x = 20

(iv) Let the required number be x

Given that 25% of x = 12

(25/100) × x = 12

x = (12 × 100)/25

x = 48

**4. Convert each of the following pairs into percentages and find out which is more?**

**(i) 25 marks out of 30, 35 marks out of 40**

**(ii) 100 runs scored off 110 balls, 50 runs scored off 55 balls**

**Solution:**

(i) Given 25 marks out of 30

Consider 25 marks out of 30 = (25/30) × 100

= (250/3)

= 83.33%

Also given that 35 marks out of 40

Now consider 35 marks out of 40 = (35/40) × 100

= 87.5%

Clearly 87.5 > 83.33

After converting into percentage 35 marks out of 40 = 87.5% is more

(ii) Given 100 runs scored off 110 balls

Consider 100 runs scored off 110 balls = (100/110) × 100

= 90.91%

Also given that 50 runs scored off 55 balls

Consider 50 runs scored off 55 balls = (50/55) × 100

= 90.91%

Here both are equal

**5. Find 20% more than Rs.200.**

**Solution:**

Consider 20% of 200 = (20/100) × 200

= Rs 40

Therefore 20% more than Rs 200 = 200 + 40

= Rs 240

**6. Find 10% less than Rs.150**

**Solution:**

Consider 10% of 150 = (10/100) × 150

= Rs 15

Therefore 10% less than Rs 150 = 150 – 15

= Rs 135

Exercise 11.6 Page No: 11.13

**1. Ashu had 24 pages to write. By the evening, he had completed 25% of his work. How many pages were left?**

**Solution:**

Given total number of pages Ashu had to write = 24

Number of pages Ashu completed by the evening = 25% of 24

Number of pages Ashu completed by the evening = 25% of 24

= (25/100) × 24

= 600/100

= 6

Therefore number of pages left for completion = 24 – 6 = 18 pages

Therefore number of pages left for completion = 24 – 6 = 18 pages

**2. A box contains 60 eggs. Out of which 16(2/3) % are rotten ones. How many eggs are rotten?**

**Solution:**

Given that total number of eggs = 60

Number of eggs rotten = 16 (2/3) % of 60 eggs

= 16.66 % of 60 eggs

= (16.66/100) × 60

= 10 eggs

Therefore number of eggs rotten = 10

**3. Rohit obtained 45 marks out of 80. What per cent marks did he get?**

**Solution:**

Given total number of marks = 80

Marks scored by Rohit = 45

Percentage obtained by Rohit = (45/80) × 100

= 56.25%

**4. Mr Virmani saves 12% of his salary. If he receives Rs 15900 per month as salary, find his monthly expenditure.**

**Solution:**

Given Mr Virmani’s salary per month = Rs. 15900

Mr Virmani’s savings = 12% of Rs. 15900

Mr Virmani’s savings = 12% of Rs. 15900

= (12/100) × 15900

= Rs. 1908

Mr Virmani’s monthly expenditure = salary – savings

Mr Virmani’s monthly expenditure = salary – savings

= Rs. (15900 – 1908)

= Rs. 13992

**5. A lawyer willed his 3 sons Rs 250000 to be divided into portions 30%, 45% and 25%. How much did each of them inherit?**

**Solution:**

Given total amount with the lawyer = Rs. 250000

First son’s inheritance = 30% of 250000

First son’s inheritance = 30% of 250000

= (30/100) × 250000

= 7500000/100

= Rs. 75000

Second son’s inheritance = 45% of 250000

Second son’s inheritance = 45% of 250000

= (45/1000) × 250000

= 11250000/100

= Rs. 112500

Third son’s inheritance = 25% of 250000

Third son’s inheritance = 25% of 250000

= (25/100) × 250000

= 6250000/100

= Rs. 62500

**6. Rajdhani College has 2400 students, 40% of whom are girls. How many boys are there in the college?**

**Solution:**

Given total number of students in Rajdhani College = 2400

Number of girls = 40% of 2400

Number of girls = 40% of 2400

= (40/100) × 2400

= 96000/100

= 960

Number of boys = total number of students – number of girls

Number of boys = total number of students – number of girls

= 2400 – 960 = 1440 boys

**7. Aman obtained 410 marks out of 500 in CBSE XII examination while his brother Anish gets 536 marks out of 600 in IX class examination. Find whose performance is better?**

**Solution:**

Given Aman’s marks in CBSE XII = 410/500

Percentage of marks obtained by Aman = (410/500) × 100

Percentage of marks obtained by Aman = (410/500) × 100

= 82%

Given that Anish’s marks in CBSE IX = 536/600

Percentage of marks obtained by Anish = (536/600) × 100

Percentage of marks obtained by Anish = (536/600) × 100

= 89.33%

Clearly 89.33 > 82

Therefore, Anish’s performance is better than Aman’s

Therefore, Anish’s performance is better than Aman’s

**8. Rahim obtained 60 marks out of 75 in Mathematics. Find the percentage of marks obtained by Rahim in Mathematics.**

**Solution:**

Given marks obtained by Rahim in mathematics = 60/75

Percentage of marks obtained by Rahim = (60/75) × 100

Percentage of marks obtained by Rahim = (60/75) × 100

= 80%

**9. In an orchard, 16 (2/3) % of the trees are apple trees. If the number of trees in the orchard is 240, 3 find the number of other type of trees in the orchard.**

**Solution:**

Let the number of apple trees be x

Number of trees in the orchard = 240

Number of trees in the orchard = 240

Number of apple trees = 16 (2/3) %

According to the given condition, 16 (2/3) % of 240 = x

= 16.66 % of 240 = x

x = (16.66/100) × 240

x = 40 trees

Number of other types of trees = Total number of trees – number of apple trees

= 240 – 40

= 200 trees

**10. Ram scored 553 marks out of 700 and Gita scored 486 marks out of 600 in science. Whose performance is better?**

**Solution:**

Given marks scored by Ram = 553/700

Percentage of marks scored by Ram = (553/700) × 100

Percentage of marks scored by Ram = (553/700) × 100

= 0.79 × 100 = 79%

Also given that marks scored by Gita = (486/600)

Also given that marks scored by Gita = (486/600)

Percentage of marks scored by Gita = (486/600) × 100

= 0.81 × 100 = 81

Gita’s performance (81%) is better than Ram’s (79%).

Gita’s performance (81%) is better than Ram’s (79%).

**11. Out of an income of Rs 15000, Nazima spends Rs 10200. What per cent of her income does she save?**

**Solution:**

Given Nazima’s total income = Rs 15000

Amount Nazima spends = Rs 10200

Amount Nazima saves = 15000 – 10200

Amount Nazima spends = Rs 10200

Amount Nazima saves = 15000 – 10200

= Rs 4800

Percentage of income Nazima saves = (4800/10200) × 100

Percentage of income Nazima saves = (4800/10200) × 100

= 480000/10200

= 32%

Nazima saves 32% of her income.

Nazima saves 32% of her income.

**12. 45% of the students in a school are boys. If the total number of students in the school is 880, find the number of girls in the school.**

**Solution:**

Given total number of students in the school = 880

Number of boys in the school = 45% of 880

= (45/100) × 880

Number of boys in the school = 45% of 880

= (45/100) × 880

= 39600/100

Number of boys = 396

Number of girls in the school = total number of students – number of boys

Number of girls in the school = total number of students – number of boys

= 880 – 396

Number of girls = 484

**13. Mr. Sidhana saves 28% of his income. If he saves as 840 per month, find his monthly income.**

**Solution:**

Let Mr. Sidhana’s monthly income be x

Monthly savings of Mr. Sidhana’s = Rs 840

28% of x

⇒ (28/100) × x

⇒ 28x = Rs 84000

⇒ x = (84000/28) = Rs 3000

Mr. Sidhana’s monthly income = Rs 3000

Monthly savings of Mr. Sidhana’s = Rs 840

28% of x

*=*Rs 840⇒ (28/100) × x

*= Rs 840*⇒ 28x = Rs 84000

⇒ x = (84000/28) = Rs 3000

Mr. Sidhana’s monthly income = Rs 3000

**14. In an examination, 8% of the students fail. What percentage of the students pass? If 1650 students appeared in the examination, how many passed?**

**Solution:**

Given total number of students who appeared for the examination = 1650

Number of students who failed = 8% of 1650

= (8/100) ×1650

= (8 × 1650)/100

= 13200/100

= (8 × 1650)/100

= 13200/100

Number of students who failed = 132

Number of students who passed = 1650 – 132

Number of students who passed = 1650 – 132

= 1518

Percentage of students who passed = (1518/1650) × 100

Percentage of students who passed = (1518/1650) × 100

= 0.92 × 100 = 92%

92% of the students passed the examination.

**15. In an examination, 92% of the candidates passed and 46 failed. How many candidates appeared?**

**Solution:**

Let the total number of candidates be x

Number of candidates who failed = 46

Number of candidates who passed = 92% of x

According to the given condition

92% of x = x –

⇒ (92/100) x = x – 6

⇒ 92x = 100x –

⇒ -8x = – 4600

⇒ x = 4600/8 = 575

Number of candidates who appeared for the examination = 575

Number of candidates who failed = 46

Number of candidates who passed = 92% of x

According to the given condition

92% of x = x –

*46*⇒ (92/100) x = x – 6

⇒ 92x = 100x –

*4600*⇒ -8x = – 4600

⇒ x = 4600/8 = 575

Number of candidates who appeared for the examination = 575