Read RD Sharma Solutions Class 7 Chapter 11 Percentage below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations
Exercise 11.1
Question :1. Express each of the following percents as fractions in the simplest forms:
(i) 45%
(ii) 0.25%
(iii) 150%
(iv) 6 1/4 %
Solution 1:
(i) 45% = 45/100
By simplifying the fraction, = 9/20
(ii) 0.25% = 0.25/100 = 25/10000
By simplifying the fraction, = 1/400
(iii) 150% = 150/100
By simplifying the fraction, = 3/2
(iv) 6 1/4 %
We written as 6.25 = 6.25/100
= 625/10000
By simplifying the fraction, = 1/16
Question :2. Express each of the following fractions as a percent:
(i) 3/4
(ii) 53/100
(iii) 1 3/5
(iv) 7/20
Solution 2:
(i) (3/4)
= (3/4) × 100 = 75%
(ii) (53/100)
= (53/100) × 100 = 53%
(iii) 1 3/5
Firstly, change the given mixed fraction into improper fraction
1 3/5 = (8/5)
= (8/5) × 100 = 160%
(iv) (7/20)
= (7/20) × 100 = 35%
Exercise 11.2
Question :1. Express each of the following ratios as per cents:
(i) 4: 5
(ii) 1: 5
(iii) 11: 125
Solution 1:
(i) 4: 5
We written as (4/5)
= (4/5) × 100 = 80%
(ii) 1: 5
We written as (1/5)
= (1/5) × 100 = 20%
(iii) 11: 125
We written as (11/125)
= (11/125) × 100 = (44/5) %
Question :2. Express each of the following percents as ratios in the simplest form:
(i) 2.5%
(ii) 0.4%
(iii) 13 3/4 %
Solution 2:
(i) 2.5%
= (2.5/100) = (25/1000) = (1/40)
(ii) 0.4%
= (0.4/100) = (4/1000) = (1/250)
(iii) 13 3/4 %
We have written as 13.75
= 13.75/100 = 1375/10000 = 11/80
Exercise 11.3
Question :1. Express each of the following precents as decimals:
(i) 12.5%
(ii) 75%
(iii) 128.8%
(iv) 0.05%
Solution 1:
(i) 12.5%
= 12.5/100 = 0.125
(ii) 75%
= 75/100 = 0.75
(iii) 128.8%
= 128.8/100 = 1.288
(iv) 0.05%
= 0.05/100 = 0.0005
Question :2. Express each of the following decimals as per cents:
(i) 0.004
(ii) 0.24
(iii) 0.02
(iv) 0.275
Solution 2:
(i) 0.004
We can write as (4/1000)
= (4/1000) × 100 = 0.4%
(ii) 0.24
We can write as (24/100)
= (24/100) × 100 = 24%
(iii) 0.02
We can write as (2/100)
= (2/100) × 100 = 2%
(iv) 0.275
We can write as (275/1000)
= (275/1000) × 100 = 27.5%
Question :3. Write each of the following as whole numbers or mixed numbers:
(i) 136%
(ii) 250%
(iii) 300%
Solution 3:
(i) 136%
= (136/100)
By simplifying = (34/25)
(ii) 250%
= (250/100)
By simplifying = (5/2)
(iii) 300%
= (300/100) = 3
Exercise 11.4
Question :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 1:
(i) 7% of Rs 7150
= (7/100) × 7150 = Rs 500.50
(ii) 40% of 400kg
= (40/100) × 400 = 160kg
(iii) 20% of 15.125liters
= (20/100) × 15.125 = 3.025liters
(iv) 3 1/3 % of 90km
We write 3 1/3 = (10/3)
= (10/300) × 90 = 3km
(v) 2.5% of 600 meters
= (2.5/100) x 600 = 15 meters
Question :2. Find the number whose 12 ½ % is 64.
Solution 2:
Let x be the required number
Then according to the given information, 12 ½ % × x = 64
= 12.5 % × x = 64
= (12.5/100) × x = 64
x = ((64 x 100))/12.5
x = 64 × 8 = 512
Therefore 512 is the number whose 12 ½ % is 64.
Question :3. What is the number, 6 ¼ % of which is 2?
Solution 3:
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
Thus 32 is the number whose 6 ¼ % is 32.
Question:4. If 6 is 50% of a number, what is that number?
Solution 4:
Let x be the required number
According to the question 50 % of x = 6
=(50/100) × x = 6
x = ((6 × 100))/50
x = 12
so, required number is 12
Exercise 11.5
Question :1. What percent 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 1:
(i) According to the given information required percentage = (6/24) × 100
= (100/4) = 25%
(ii) According to the given information required percentage = (10/125) × 100
= (1000/125) = 8%
(iii) According to the given information required percentage = (160/4) × 100
We know that 1km = 1000 meters
Thus 4km = 4000 meters
= (160/4000) × 100 = 4%
(iv) According to the given information required percentage = (25/8) × 100
We know that 1Rs = 100 paise
Thus 8Rs = 800 paise
= (25/800) × 100 = (25/8) = 3.125%
(v) As we know that 1day = 24 hours
1 hour = (1/24) day
8 hours = (8/24) day = (1/3) day
According to the given information required percentage = [(1/3)/2] × 100
= 100/6 = 16 2/3 %
(vi) We know that 1liter = 1000ml
According to the given information required percentage = (175/1000) × 100
= (17500/1000) = 17.50 %
Question :2. What percent is equivalent to (3/8)?
Solution 2:
=(3/8) = (3/8) × 100 = 37.5%
Question :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 3:
(i) Let x be the number
We know 4% of x = 8 by given information
(4/100) × x = 8
x = (800/4) = 200
(ii) Let the required number be x
We know 60% of x = 6 by given information
(60/100) × x = 6
x = (60/6) = 10
(iii) Let the required number be x
We know 30% of x = 6 by given information
(30/100) × x = 6
x = (((6 × 100))/30) = 20
(iv) Let the required number be x
We know 25% of x = 12 by given information
(25/100) × x = 12
x = ((12 × 100))/25 = 48
Question: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 4:
(i) 25 marks out of 30
Consider 25 marks out of 30 = (25/30) × 100
= (250/3) = 83.33%
Also 35 marks out of 40
Now, consider 35 marks out of 40 = (35/40) × 100
= 87.5%
Obviously 87.5 > 83.33
After converting into percentage 35 marks out of 40 = 87.5% is more
(ii) 100 runs scored off 110 balls
Consider 100 runs scored off 110 balls = (100/110) × 100
= 90.91%
Also 50 runs scored off 55 balls
Consider 50 runs scored off 55 balls = (50/55) × 100
= 90.91%
Both are equal here.
Question:5. Find 20% more than Rs.200.
Solution 5:
Given 20% of 200 = (20/100) × 200
= Rs 40
Therefore 20% more than Rs 200 = 200 + 40
= Rs 240
Question :6. Find 10% less than Rs.150
Solution 6:
Given 10% of 150 = (10/100) × 150
= Rs 15
Therefore 10% less than Rs 150 = 150 – 15
= Rs 135
Exercise 11.6
Question :1. Ashu had 24 pages to write. By the evening, he had completed 25% of his work. How many pages were left?
Solution 1:
The total number of pages Ashu had to write = 24 pages
Number of pages Ashu completed by the evening = 25% of 24
= (25/100) × 24 = (600/100) = 6
Thus number of pages left for completion = 24 – 6 = 18 pages
Question 2: A box contains 60 eggs. Out of which 16 2/3 % are rotten ones. How many eggs are rotten?
Solution 2:
The total number of eggs = 60 eggs
Number of eggs rotten = 16 2/3% of 60 eggs
= 16.66 % of 60 eggs = (16.66/100) × 60 = 10 eggs
Thus, number of eggs rotten = 10
Question 3: Rohit obtained 45 marks out of 80. What percent marks did he get?
Solution 3:
The total number of marks = 80
Marks which scored by Rohit = 45
The percentage obtained by Rohit = (45/80) × 100
= 56.25%
Question 4: Mr. Virmani saves 12% of his salary. If he receives Rs 15900 per month as salary, find his monthly expenditure.
Solution 4:
Mr. Virmani’s salary per month = Rs. 15900
Mr. Virmani’s savings = 12% of Rs. 15900
= (12/100) × 15900 = Rs. 1908
Mr Virmani’s monthly expenditure = salary – savings
= Rs. (15900 – 1908) = Rs. 13992
Question: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 5:
Total amount with lawyer = Rs. 250000
Inheritance of first son’s = 30% of 250000
= (30/100) × 250000 = (7500000/100) = Rs. 75000
Inheritance of second son’s = 45% of 250000
= (45/100) × 250000= (11250000/100) = Rs. 112500
Inheritance of third son’s = 25% of 250000
= (25/100) × 250000 = (6250000/100) = Rs. 62500
Question:6. Rajdhani College has 2400 students, 40% of whom are girls. How many boys are there in the college?
Solution 6:
In Rajdhani College total number of students = 2400
Number of girls = 40% of 2400
= (40/100) × 2400 = (96000/100) = 960
Number of boys = total number of students – number of girls
= 2400 – 960 = 1440 boys
Question:7. Aman obtained 410 marks out of 500 in the CBSE XII examination while his brother Anish gets 536 marks out of 600 in the IX class examination. Find whose performance is better?
Solution 7:
Aman’s marks in CBSE XII = (410/500)
Percentage of marks obtained by Aman = (410/500) × 100 = 82%
Anish’s marks in CBSE IX = (536/600)
Percentage of marks obtained by Anish = (536/600) × 100 = 89.33%
Obviously 89.33 > 82
Thus, Anish’s performance is better than Aman’s
Question:8. Rahim obtained 60 marks out of 75 in Mathematics. Find the percentage of marks obtained by Rahim in Mathematics.
Solution 8:
Marks obtained by Rahim in Mathematics = (60/75)
Percentage of marks obtained by Rahim = (60/75) × 100
= 80%
Question:9. In an orchard, 16 2/3 % of the trees are apple trees. If the number of trees in the orchard is 240, find the number of other types of trees in the orchard.
Solution 9:
Let x be the number of apple trees
Number of trees in the orchard = 240
Number of apple trees = 16 2/3 %
According to the given information condition, 16 (2/3) % of 240 = x
= 16.66 % of 240 = x
x = (16.66/100) × 240 = 40 trees
Number of other types of trees = Total number of trees – number of apple trees
= 240 – 40
= 200 trees
Question:10. Ram scored 553 marks out of 700 and Gita scored 486 marks out of 600 in science. Whose performance is better?
Solution 10:
Marks scored by Ram = (553/700)
By Ram percentage of marks scored = (553/700) × 100
= 0.79 × 100 = 79%
Marks scored by Gita = (486/600)
By Gita percentage of marks scored = (486/600) × 100
= 0.81 × 100 = 81
Gita’s performance (81%) is better than Ram’s (79%).
Question:11. Out of an income of Rs 15000, Nazima spends Rs 10200. What percent of her income does she save?
Solution 11:
Nazima’s total income = Rs 15000
Amount Nazima spends = Rs 10200
Amount Nazima saves = 15000 – 10200 = Rs 4800
Percentage of income Nazima saves = (4800/15000) × 100
= (480000/15000) = 32%
Nazima saves 32% of her income.
Question: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 12:
Total number of students in the school = 880
Number of boys in the school = 45% of 880
= (45/100) × 880 = (39600/100) = 396
Number of boys = 396
Number of girls in the school = total number of students – number of boys
= 880 – 396
Number of girls = 484
Question:13. Mr. Sidhana saves 28% of his income. If he saves as 840 per month, find his monthly income.
Solution 13:
Let be x is Mr. Sidhana’s monthly income
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
So, Mr. Sidhana’s monthly income = Rs 3000
Question: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 14:
Total number of appeared students for the examination = 1650
Number of failed students = 8% of 1650
= (8/100) ×1650 = ((8 × 1650))/100 = 13200/100
Number of students failed = 132
Number of students passed = 1650 – 132
= 1518
Percentage of students passed =(1518/1650) × 100
= 0.92 × 100 = 92%
92% of the students passed the examination.
Question:15. In an examination, 92% of the candidates passed and 46 failed. How many candidates appeared?
Solution 15:
Let be x the total number of candidates
Number of failed candidates = 46
Number of passed candidates who = 92% of x
According to the given condition
92% of x = x – 46
⇒ (92/100) x = x – 46
⇒ 92x = 100x – 4600
⇒ -8x = – 4600
⇒ x = 4600/8 = 575
Number of candidates who appeared for the examination = 575