{"id":546210,"date":"2021-10-06T06:50:49","date_gmt":"2021-10-06T06:50:49","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546210"},"modified":"2021-10-07T05:49:41","modified_gmt":"2021-10-07T05:49:41","slug":"rd-sharma-solutions-for-class-8-maths-chapter-12-percentage","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-8-maths-chapter-12-percentage\/","title":{"rendered":"RD Sharma Solutions for Class 8 Maths Chapter 12\u2013Percentage"},"content":{"rendered":"\n

Class 8: Maths Chapter 12 solutions. Complete Class 8 Maths Chapter 12 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 12\u2013Percentage<\/h2>\n\n\n\n

RD Sharma 8th Maths Chapter 12, Class 8 Maths Chapter 12 solutions<\/p>\n\n\n\n

EXERCISE 12.1 PAGE NO: 12.2<\/p>\n\n\n\n

1. Write each of the following as percentage.
(i) 7\/25
(ii) 14\/625
(iii) 5\/8
(iv) 0.8
(v) 0.005
(vi) 3:25
(vii) 11: 80
(viii) 111: 125
(ix) 13: 75
(x) 15: 16
(xi) 0.18
(xii) 7\/125<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 7\/25<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 7\/25 \u00d7 100 = 28%<\/p>\n\n\n\n

(ii) 14\/625<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 14\/625 \u00d7 100 = 56\/25 = 2.24%<\/p>\n\n\n\n

(iii) 5\/8<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 5\/8 \u00d7 100 = 125\/2 = 62.5%<\/p>\n\n\n\n

(iv) 0.8<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 0.8 can be written as 8\/10<\/p>\n\n\n\n

so, 8\/10 \u00d7 100 = 80%<\/p>\n\n\n\n

(v) 0.005<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 0.005 can be written as 5\/1000<\/p>\n\n\n\n

so, 5\/1000 \u00d7 100 = 5\/10 = 0.5%<\/p>\n\n\n\n

(vi) 3:25<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 3\/25 \u00d7 100 = 3 \u00d7 4 = 12%<\/p>\n\n\n\n

(vii) 11 : 80<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 11\/80 \u00d7 100 = 55\/4 = 13.75%<\/p>\n\n\n\n

(viii) 111 : 125<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 111\/125 \u00d7 100 = 444\/5 = 88.8%<\/p>\n\n\n\n

(ix) 13 : 75<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 13\/75 \u00d7 100 = 52\/3 = 17.3%<\/p>\n\n\n\n

(x) 15 : 16<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 15\/16 \u00d7 100 = 375\/4 = 93.75%<\/p>\n\n\n\n

(xi) 0.18<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 0.18 can be written as 18\/100<\/p>\n\n\n\n

so, 18\/100 \u00d7 100 = 18%<\/p>\n\n\n\n

(xii) 7\/125<\/strong><\/p>\n\n\n\n

To change any number to percentage we have to multiply that number by 100,<\/p>\n\n\n\n

i.e., 7\/125 \u00d7 100 = 28\/5 = 5.6%<\/p>\n\n\n\n

2.<\/strong> Convert the following percentages to fractions and ratios.
(i) 25%
(ii) 2.5%
(iii) 0.25%
(iv) 0.3%
(v) 125%<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 25%<\/strong><\/p>\n\n\n\n

To convert percentage to fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 25\/100 = 1\/4 or 1:4<\/p>\n\n\n\n

(ii) 2.5%<\/strong><\/p>\n\n\n\n

To convert percentage to fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 2.5% can be written as 25\/10<\/p>\n\n\n\n

so, (25\/10)\/100 = 25\/1000 = 1\/40 or 1:40<\/p>\n\n\n\n

(iii) 0.25%<\/strong><\/p>\n\n\n\n

To convert percentage to fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 0.25% can be written as 25\/100<\/p>\n\n\n\n

so, (25\/100)\/100 = 25\/10000 = 1\/400 or 1:400<\/p>\n\n\n\n

(iv) 0.3%<\/strong><\/p>\n\n\n\n

To convert percentage to fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 0.3% can be written as 3\/10<\/p>\n\n\n\n

so, (3\/10)\/100 = 3\/1000 or 3:1000<\/p>\n\n\n\n

(v) 125%<\/strong><\/p>\n\n\n\n

To convert percentage to fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 125\/100 = 5\/4 or 5:4<\/p>\n\n\n\n

3.<\/strong> Express the following as decimal fractions.
(i) 27%
(ii) 6.3%
(iii) 32%<\/strong><\/p>\n\n\n\n

(iv) 0.25%
(v) 7.5%
(vi) 1\/8%<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 27%<\/strong><\/p>\n\n\n\n

To convert percentage to decimal fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 27% = 27\/100 = 0.27<\/p>\n\n\n\n

(ii) 6.3%<\/strong><\/p>\n\n\n\n

Here 6.3 can be written as 63\/10<\/p>\n\n\n\n

To convert percentage to decimal fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 6.3% = 63\/(10 \u00d7 100) =63\/1000 = 0.063<\/p>\n\n\n\n

(iii) 32%<\/strong><\/p>\n\n\n\n

To convert percentage to decimal fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 32% = 32\/100 = 0.32<\/p>\n\n\n\n

(iv) 0.25%<\/strong><\/p>\n\n\n\n

Here 0.25 can be written as 25\/100<\/p>\n\n\n\n

To convert percentage to decimal fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 0.25% = 25\/(100 \u00d7 100) = 25\/10000 = 0.0025<\/p>\n\n\n\n

(v) 7.5%<\/strong><\/p>\n\n\n\n

Here 7.5 can be written as 75\/10<\/p>\n\n\n\n

To convert percentage to decimal fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 7.5% = 75\/(10 \u00d7 100) =75\/1000 = 0.075<\/p>\n\n\n\n

(vi) 1\/8%<\/strong><\/p>\n\n\n\n

To convert percentage to decimal fractions we have to divide by 100,<\/p>\n\n\n\n

i.e., 1\/8% = 1\/(8 \u00d7 100) =1\/800 = 0.00125<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 12.2 PAGE NO: 12.9<\/p>\n\n\n\n

1.<\/strong> Find:
(i) 22% of 120
(ii) 25% of Rs 1000
(iii) 25% of 10 kg
(iv) 16.5% of 5000 metre
(v) 135% of 80 cm
(vi) 2.5% of 10000 ml<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 22% of 120<\/strong><\/p>\n\n\n\n

Here, 22% of 120 can be expressed as (22 \u00d7 120) \/ 100 = 2640\/100 = 26.40<\/p>\n\n\n\n

(ii) 25% of Rs 1000<\/strong><\/p>\n\n\n\n

Here, 25% of Rs 1000 can be expressed as (25 \u00d7 1000) \/ 100 = 25000\/100 = Rs 250<\/p>\n\n\n\n

(iii) 25% of 10 kg<\/strong><\/p>\n\n\n\n

Here, 25% of 10 Kg can be expressed as (25 \u00d7 10) \/ 100 = 250\/100 = 2.5 Kg<\/p>\n\n\n\n

(iv) 16.5% of 5000 metre<\/strong><\/p>\n\n\n\n

Here, 16.5% of 5000 metre can be expressed as (16.5 \u00d7 5000) \/ 100 = 16.5 \u00d7 50 = 825 m<\/p>\n\n\n\n

(v) 135% of 80 cm<\/strong><\/p>\n\n\n\n

Here, 135% of 80 cm can be expressed as (135 \u00d7 80) \/ 100 = (135 \u00d7 4) \/ 5 = 108 cm<\/p>\n\n\n\n

(vi) 2.5% of 10000 ml<\/strong><\/p>\n\n\n\n

Here, 2.5% of 10000 ml can be expressed as (2.5 \u00d7 10000) \/ 100 = 25000\/100 = 250 ml<\/p>\n\n\n\n

2.<\/strong> Find the number a, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3<\/strong><\/p>\n\n\n\n

(iii) \u00bd% of a is 50
(iv) 100% of a is 100<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 8.4% of a is 42<\/strong><\/p>\n\n\n\n

Here, 8.4% of a is 42 can be expressed as (8.4 \u00d7 a) \/ 100 = 42<\/p>\n\n\n\n

Which implies, a = (42 \u00d7 100) \/ 8.4 \u2014\u2014 (1)<\/p>\n\n\n\n

Here, 8.4 can been written as 84\/10<\/p>\n\n\n\n

Substituting the above in (1)<\/p>\n\n\n\n

a = (42 \u00d7 100 \u00d7 10) \/ 84<\/p>\n\n\n\n

a = 42000\/84<\/p>\n\n\n\n

a = 500<\/p>\n\n\n\n

(ii) 0.5% of a is 3<\/strong><\/p>\n\n\n\n

Here, 0.5 of a is 3 can be expressed as (0.5 \u00d7 a) \/ 100 = 3<\/p>\n\n\n\n

Which implies, a = (3 \u00d7 100) \/ 0.5 \u2014\u2014 (1)<\/p>\n\n\n\n

Here, 0.5 can been written as 5\/10<\/p>\n\n\n\n

Substituting the above in (1)<\/p>\n\n\n\n

a = (3 \u00d7 100 \u00d7 10) \/ 5<\/p>\n\n\n\n

a = 3000\/5<\/p>\n\n\n\n

a = 600<\/p>\n\n\n\n

(iii) \u00bd% of a is 50<\/strong><\/p>\n\n\n\n

Here, 0.5 of a is 50 can be expressed as (0.5 \u00d7 a) \/ 100 = 50<\/p>\n\n\n\n

Which implies, a = (50 \u00d7 100) \/ 0.5 \u2014\u2014 (1)<\/p>\n\n\n\n

Here, 0.5 can been written as 5\/10<\/p>\n\n\n\n

Substituting the above in (1)<\/p>\n\n\n\n

a = (50 \u00d7 100 \u00d7 10) \/ 5<\/p>\n\n\n\n

a = 50000\/5<\/p>\n\n\n\n

a = 10000<\/p>\n\n\n\n

(iv) 100% of a is 100<\/strong><\/p>\n\n\n\n

Here, 100% of a is 100 can be expressed as (100 \u00d7 a) \/ 100 = 100<\/p>\n\n\n\n

Which implies, a = (100 \u00d7 100) \/ 100<\/p>\n\n\n\n

a = 10000\/100<\/p>\n\n\n\n

a = 100<\/p>\n\n\n\n

3.<\/strong> x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given value x = 480<\/p>\n\n\n\n

And x is 5% of y<\/p>\n\n\n\n

Here, x is 5% of y can be expressed as (5 \u00d7 y) \/ 100 = x<\/p>\n\n\n\n

Which implies, x = y \u00d7 (5\/100)<\/p>\n\n\n\n

Substituting x = 480 in the above equation we get,<\/p>\n\n\n\n

480 = y \u00d7 (5\/100)<\/p>\n\n\n\n

Which implies, y = (480 \u00d7 100) \/ 5<\/p>\n\n\n\n

y = 48000\/5<\/p>\n\n\n\n

y = 9600<\/p>\n\n\n\n

It is also given that, y is 24% of z<\/p>\n\n\n\n

Which implies, y = z \u00d7 (24\/100)<\/p>\n\n\n\n

Substituting y = 9600 in the above equation we get,<\/p>\n\n\n\n

9600 = z \u00d7 (24\/100)<\/p>\n\n\n\n

Which implies, z = (9600 \u00d7 100) \/ 24<\/p>\n\n\n\n

z = 96000\/24<\/p>\n\n\n\n

z = 40000<\/p>\n\n\n\n

\u2234 y = 9600 and z = 40000<\/p>\n\n\n\n

4.<\/strong> A coolie deposits Rs 150 per month in his post office Savings Bank account. If this is 15% of his monthly income, find his monthly income.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the monthly income of a coolie be \u2018x\u2019<\/p>\n\n\n\n

Here, coolie deposits Rs 150 per month which is 15% of his monthly income<\/p>\n\n\n\n

From the above we can derive that,<\/p>\n\n\n\n

x \u00d7 (15\/100) =150<\/p>\n\n\n\n

Which implies, x = (150 \u00d7 100) \/ 15<\/p>\n\n\n\n

x = 15000\/15<\/p>\n\n\n\n

x = 1000<\/p>\n\n\n\n

\u2234 Monthly income is Rs 1000<\/p>\n\n\n\n

5.<\/strong> Asha got 86.875% marks in the annual examination. If she got 695 marks, find the total number of marks of the examination.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, Marks scored by Asha is 695<\/p>\n\n\n\n

And Percentage of marks Asha got is 86.875%<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Let the total marks be \u2018x\u2019<\/p>\n\n\n\n

From the above we can derive that,<\/p>\n\n\n\n

x \u00d7 (86.875\/100) =695<\/p>\n\n\n\n

Here, 86.875 can be expressed as 86875\/1000<\/p>\n\n\n\n

Which implies, x = (695 \u00d7 100 \u00d7 1000) \/ 86875<\/p>\n\n\n\n

x = 6950000\/86875 = 800<\/p>\n\n\n\n

Total no: of Marks, x = 800<\/p>\n\n\n\n

\u2234 Total number of marks is 800marks<\/p>\n\n\n\n

6. Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, number of days Deepti went to school = 216 days<\/p>\n\n\n\n

Deepti Attendance percentage is = 90%<\/p>\n\n\n\n

So, let the number of days when school remained opened be x days<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

(x \u00d7 90)\/100 = 216<\/p>\n\n\n\n

By using cross multiplication we get,<\/p>\n\n\n\n

x = (216\u00d7100)\/90<\/p>\n\n\n\n

= 240 days<\/p>\n\n\n\n

\u2234 Number of days the school remained opened for 240 days<\/p>\n\n\n\n

7. A garden has 2000 trees. 12% of these are mango trees 18% lemon and the rest are orange trees. Find the number of orange trees.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Total number of trees = 2000<\/p>\n\n\n\n

Number of mango trees = 12% of 2000<\/p>\n\n\n\n

= (12\/100) \u00d7 2000<\/p>\n\n\n\n

= 240 trees<\/p>\n\n\n\n

Number of lemon trees = 18% of 2000<\/p>\n\n\n\n

= (18\/100) \u00d7 2000<\/p>\n\n\n\n

= 360 trees<\/p>\n\n\n\n

Number of orange trees = 2000 \u2013 (Number of mango trees+ Number of lemon trees)<\/p>\n\n\n\n

= 2000 \u2013 (240+360)<\/p>\n\n\n\n

= 2000 \u2013 600<\/p>\n\n\n\n

= 1400 trees<\/p>\n\n\n\n

\u2234 Number of orange trees are 1400 trees<\/p>\n\n\n\n

8. Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The given details are,<\/p>\n\n\n\n

Amount of calorie daily needed = 2600 calorie<\/p>\n\n\n\n

Amount of protein needed = 12% of 2600<\/p>\n\n\n\n

= (12\/100) \u00d7 2600<\/p>\n\n\n\n

= 312 calorie<\/p>\n\n\n\n

Amount of fats needed = 25% of 2600<\/p>\n\n\n\n

= (25\/100) \u00d7 2600<\/p>\n\n\n\n

= 650 calorie<\/p>\n\n\n\n

Amount of carbohydrate needed = 63% of 2600<\/p>\n\n\n\n

= (63\/100) \u00d7 2600<\/p>\n\n\n\n

= 1638 calorie<\/p>\n\n\n\n

\u2234 Amount of calories required in protein is 312 calories, fat is 650 calories and carbohydrates is 1638 calories<\/p>\n\n\n\n

9. A cricketer scored a total of 62 runs in 96 balls. He hit 3 sixes, 8 fours, 2 two\u2019s and 8 singles. What percentage of the total runs came in
(i) Sixes
(ii) 4\u2019s
(iii) 2\u2019s
(iv) Singles<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The given details are,<\/p>\n\n\n\n

Total runs scored by cricketer = 62 runs<\/p>\n\n\n\n

(i)<\/strong> Runs scored in 3 sixes = 3\u00d76<\/p>\n\n\n\n

= 18<\/p>\n\n\n\n

Percentage of runs scored in sixes = (18\/62) \u00d7 100<\/p>\n\n\n\n

= 29.03%<\/p>\n\n\n\n

(ii)<\/strong> Runs scored in 8 fours = 8\u00d74<\/p>\n\n\n\n

= 32<\/p>\n\n\n\n

Percentage of runs scored in fours = (32\/62) \u00d7 100<\/p>\n\n\n\n

= 51.61%<\/p>\n\n\n\n

(iii)<\/strong> Runs scored in 2 two\u2019s = 2\u00d72<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

Percentage of runs scored in two\u2019s = (4\/62) \u00d7 100<\/p>\n\n\n\n

= 6.45%<\/p>\n\n\n\n

(iv)<\/strong> Runs scored in singles = 8 \u00d7 1<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

Percentage of runs scored in singles = (8\/62) \u00d7 100<\/p>\n\n\n\n

= 12.9%<\/p>\n\n\n\n

10. A cricketer hit 120 runs in 150 balls during a test match. 20% of the runs came in 6\u2019s, 30% in 4\u2019s, 25% in 2\u2019s and the rest in 1\u2019s. How many runs did he score in
(i) 6\u2019s<\/strong><\/p>\n\n\n\n

(ii) 4\u2019s
(iii) 2\u2019s<\/strong><\/p>\n\n\n\n

(iv) singles
What % of his shots were scoring ones?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The given details are,<\/p>\n\n\n\n

Total number of runs scored by cricketer = 120<\/p>\n\n\n\n

(i)<\/strong> 20% of Runs scored in 6\u2019s = (20\/100) \u00d7 120 = 24 runs<\/p>\n\n\n\n

(ii)<\/strong> 30% of Runs scored in 4\u2019s = (30\/100) \u00d7 120 = 36 runs<\/p>\n\n\n\n

(iii)<\/strong> 25% of Runs scored in 2\u2019s = (25\/100) \u00d7 120 = 30 runs<\/p>\n\n\n\n

(iv)<\/strong> Runs scored in singles = 120 \u2013 (24+36+30)<\/p>\n\n\n\n

= 120 \u2013 90 = 30 runs<\/p>\n\n\n\n

Percentage of shots scoring ones = (Runs came in singles\/Total runs scored) \u00d7 100<\/p>\n\n\n\n

= (30\/120) \u00d7 100<\/p>\n\n\n\n

= 25%<\/p>\n\n\n\n

11. Radha earns 22% of her investment. If she earns Rs 187, then how much did she invest?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, percentage Radha earns = 22% of investment<\/p>\n\n\n\n

So, let us consider total investment be Rs x<\/p>\n\n\n\n

By calculating, (x\/100) \u00d7 22 = 187<\/p>\n\n\n\n

By cross multiplying we get,<\/p>\n\n\n\n

(x\/100) = 187\/22<\/p>\n\n\n\n

x = (187\u00d7100)\/22<\/p>\n\n\n\n

= 850<\/p>\n\n\n\n

\u2234 Radha\u2019s total investment is Rs 850<\/p>\n\n\n\n

12. Rohit deposits 12% of his income in a bank. He deposited Rs 1440 in the bank during 1997. What was his total income for the year 1997?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The given details are,<\/p>\n\n\n\n

Percentage Rohit deposited in bank = 12% of total income<\/p>\n\n\n\n

Rohit deposited money during the year 1997 = Rs 1440<\/p>\n\n\n\n

So, let us consider the total income of Rohit as Rs x<\/p>\n\n\n\n

By calculating,<\/p>\n\n\n\n

(x\/100) \u00d7 12 = 1440<\/p>\n\n\n\n

By cross multiplying<\/p>\n\n\n\n

x = (1440\u00d7100)\/22<\/p>\n\n\n\n

= 12000<\/p>\n\n\n\n

\u2234 Rohit\u2019s total income for the year 1997 is Rs 12000<\/p>\n\n\n\n

13. Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Percentage of nitre in gunpowder = 75%<\/p>\n\n\n\n

Amount of nitre in gunpowder = 9 kg<\/p>\n\n\n\n

Let us consider the amount of gunpowder be \u2018x\u2019 kg<\/p>\n\n\n\n

So, by calculating (x\/100) \u00d7 75 = 9<\/p>\n\n\n\n

By cross multiplying<\/p>\n\n\n\n

x\/100 = 9\/75<\/p>\n\n\n\n

x = (9\u00d7100)\/75<\/p>\n\n\n\n

= 12kg<\/p>\n\n\n\n

Percentage of sulphur in gunpowder = 10%<\/p>\n\n\n\n

Amount of sulphur in gunpowder = 2.3 kg<\/p>\n\n\n\n

Let us consider amount of gunpowder be \u2018x\u2019 kg<\/p>\n\n\n\n

So, by calculating (x\/100) \u00d7 10 = 2.3<\/p>\n\n\n\n

By cross multiplying<\/p>\n\n\n\n

x\/100 = 2.3\/10<\/p>\n\n\n\n

x = (2.3\u00d7100)\/10<\/p>\n\n\n\n

= 23kg<\/p>\n\n\n\n

\u2234 The amount of gunpowder in nitre is 12kg<\/p>\n\n\n\n

The amount of gunpowder in sulphur is 23kg<\/p>\n\n\n\n

14. An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Amount of tin in an alloy = 15 parts<\/p>\n\n\n\n

Amount of copper in an alloy = 105 parts<\/p>\n\n\n\n

So, total weight of alloy = 15 + 105 = 120 parts<\/p>\n\n\n\n

Now, by calculating<\/p>\n\n\n\n

Percentage of copper in alloy = (105\/120) \u00d7 100<\/p>\n\n\n\n

= 525\/6<\/p>\n\n\n\n

= 87.50%<\/p>\n\n\n\n

\u2234 Percentage of copper in an alloy is 87.50%<\/p>\n\n\n\n

15. An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Alloy contains, 32% of copper<\/p>\n\n\n\n

40% of nickel<\/p>\n\n\n\n

Remaining zinc<\/p>\n\n\n\n

Mass of alloy = 1kg = 1000 grams<\/p>\n\n\n\n

Mass of copper in alloy = (1000\/100) \u00d7 32<\/p>\n\n\n\n

= 320 grams<\/p>\n\n\n\n

Mass of nickel in alloy = (1000\/100) \u00d7 40<\/p>\n\n\n\n

= 400 grams<\/p>\n\n\n\n

So, mass of zinc in alloy = 1000 \u2013 (320 + 400)<\/p>\n\n\n\n

= 1000 \u2013 720<\/p>\n\n\n\n

= 280 grams<\/p>\n\n\n\n

\u2234 Mass of zinc in 1kg of alloy is 280 grams<\/p>\n\n\n\n

16. A motorist travelled 122 kilometers before his first stop. If he had 10%of his journey to complete at this point, how long was the total ride?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Motorist total distance travelled before first stop = 122 km<\/p>\n\n\n\n

Journey completed at first stop = 10 %<\/p>\n\n\n\n

Let us consider total ride to be travelled be \u2018x\u2019 km<\/p>\n\n\n\n

So, by calculating<\/p>\n\n\n\n

(x\/100) \u00d7 10 = 122<\/p>\n\n\n\n

By cross multiplying we get,<\/p>\n\n\n\n

x\/100 = 122\/10<\/p>\n\n\n\n

x = (122 \u00d7 100)\/10<\/p>\n\n\n\n

= 1220 km<\/p>\n\n\n\n

\u2234 Motorist total ride is 1220 km<\/p>\n\n\n\n

17. A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The given details are,<\/p>\n\n\n\n

In a school, number of students are = 300<\/p>\n\n\n\n

Number of boys = 142<\/p>\n\n\n\n

Number of girls = 300 \u2013 142 = 158<\/p>\n\n\n\n

In a school, number of teachers are = 30<\/p>\n\n\n\n

Number of male teachers are = 12<\/p>\n\n\n\n

Number of female teachers are = 30 \u2013 12 = 18<\/p>\n\n\n\n

Total number of students and teachers is = 300+30 = 330<\/p>\n\n\n\n

Total numbers of female in the school is = 158+18 = 176<\/p>\n\n\n\n

Percentage of female in the school = (176\/330) \u00d7 100<\/p>\n\n\n\n

= 160\/3%<\/p>\n\n\n\n

\u2234 Total of 160\/3% are female in the school.<\/p>\n\n\n\n

18. Aman\u2019s income is 20% less than that of Anil. How much percent is Anil\u2019s income more than Aman\u2019s income?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given Aman\u2019s income is 20% less than Anil\u2019s income<\/p>\n\n\n\n

Let us consider Aman\u2019s and Anil\u2019s income as Rs x<\/p>\n\n\n\n

Aman\u2019s income = x \u2013 x \u00d7 (20\/100)<\/p>\n\n\n\n

= x \u2013 x \u00d7 (1\/5)<\/p>\n\n\n\n

= x \u2013 x\/5<\/p>\n\n\n\n

= (5x-x)\/5<\/p>\n\n\n\n

= 4x\/5<\/p>\n\n\n\n

Let us find the difference between Anil\u2019s and Aman\u2019s income = x \u2013 4x\/5<\/p>\n\n\n\n

= (5x-4x)\/5<\/p>\n\n\n\n

= x\/5<\/p>\n\n\n\n

When, Anil\u2019s income is more than Aman\u2019s income the percentage is = (x\/5)\/(4x\/5) \u00d7 100<\/p>\n\n\n\n

= 25%<\/p>\n\n\n\n

\u2234 25% of Anil\u2019s income is more than Aman\u2019s income.<\/p>\n\n\n\n

19. The value of a machine depreciates every year by 5%. If the present value of the machine be Rs 100000, what will be its value after 2 years?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Present value of machine is = Rs 100000<\/p>\n\n\n\n

Every year the depreciation in price is = 5%<\/p>\n\n\n\n

So, value after two years = 100000 \u00d7 (100-5)\/100 \u00d7 (100-5)\/100<\/p>\n\n\n\n

= 100000 \u00d7 95\/100 \u00d7 95\/100<\/p>\n\n\n\n

= 90250<\/p>\n\n\n\n

\u2234 Value of machine after two years is Rs 90250<\/p>\n\n\n\n

20. The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Present population of town is = 60000<\/p>\n\n\n\n

Annually population increases by = 10%<\/p>\n\n\n\n

So, Population after 2 years = present population \u00d7 [(100 + increased %)\/100] years<\/p>\n\n\n\n

= 60000 \u00d7 (100+10)\/100 \u00d7 (100+10)\/100<\/p>\n\n\n\n

= 60000 \u00d7 110\/100 \u00d7 110\/100<\/p>\n\n\n\n

= 60000 \u00d7 11\/10 \u00d7 11\/10<\/p>\n\n\n\n

= 72600<\/p>\n\n\n\n

\u2234 After 2 years population will be 72600<\/p>\n\n\n\n

21. The population of a town increases by 10% annually. If the present population is 22000, find its population a year ago.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Present population of town is = 22000<\/p>\n\n\n\n

Let the population of town be 100 a year ago.<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Annual population increase is = 10% of 100 = 10<\/p>\n\n\n\n

The present population = 100 + 10 = 110<\/p>\n\n\n\n

If present population is 110, population year ago = 100<\/p>\n\n\n\n

If present population is 1, population year ago = 100\/110<\/p>\n\n\n\n

If present population is 22000, population year ago = 100\/110 \u00d7 22000<\/p>\n\n\n\n

= 10\/11 \u00d7 22000<\/p>\n\n\n\n

= 20000<\/p>\n\n\n\n

\u2234 1 year ago population was 20000<\/p>\n\n\n\n

22. Ankit was given an increment of 10% on his salary. His new salary is Rs 3575. What was his salary before increment?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the salary of Ankit before increment be = Rs x<\/p>\n\n\n\n

New salary of Ankit = Rs 3575<\/p>\n\n\n\n

Increase in salary is = 10% of 100 = 10<\/p>\n\n\n\n

Present salary = 100 + 10 = 110<\/p>\n\n\n\n

So, Salary of Ankit before increment is x \u00d7 110\/100 = 3575<\/p>\n\n\n\n

By calculating for x we get,<\/p>\n\n\n\n

x \u00d7 110 = 3575 \u00d7 100<\/p>\n\n\n\n

x = (3575 \u00d7 100)\/110<\/p>\n\n\n\n

= 3250<\/p>\n\n\n\n

\u2234 Salary of Ankit before increment is Rs 3250<\/p>\n\n\n\n

23. In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given details are,<\/p>\n\n\n\n

Increase in petrol price by = 10%<\/p>\n\n\n\n

Reduction in consumption while having same expenditure =<\/p>\n\n\n\n

(increase%)\/(100+increase%) \u00d7 100<\/p>\n\n\n\n

= 10\/(100+10) \u00d7 100<\/p>\n\n\n\n

= 1000\/110<\/p>\n\n\n\n

= 100\/11<\/p>\n\n\n\n

= 91119111%<\/p>\n\n\n\n

\u2234 at the cost of same expenditure one can reduce 91119111% of consumption of petrol.<\/p>\n\n\n\n

24. Mohan\u2019s income is Rs 15500 per month. He saves 11% of his income. If his income increases by 10%, then he reduces his saving by 1%, how much does he save now?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Mohan monthly income is = Rs 15500<\/p>\n\n\n\n

Mohan savings is = 11% of 15500<\/p>\n\n\n\n

= 15500 \u00d7 11\/100<\/p>\n\n\n\n

= Rs 1705<\/p>\n\n\n\n

Monthly income increases by = 10%<\/p>\n\n\n\n

New monthly income is = 15500 + 10\/100 \u00d7 15500<\/p>\n\n\n\n

= 15500 + 1550<\/p>\n\n\n\n

= Rs 17050<\/p>\n\n\n\n

When savings reduced by 1% will result in = 11 \u2013 1 = 10% of 17050<\/p>\n\n\n\n

New savings = (10\/100) \u00d7 17050<\/p>\n\n\n\n

= Rs 1705<\/p>\n\n\n\n

\u2234 Savings is Rs 1705, which remains the same even after increment.<\/p>\n\n\n\n

25. Shikha\u2019s income is 60% more than that of Shalu. What percent is Shalu\u2019s income less than Shikha\u2019s?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider Shikha\u2019s and Shalu\u2019s income be Rs x<\/p>\n\n\n\n

So, Shikha\u2019s income is 60% more of Shalu\u2019s = x + x \u00d7 60\/100<\/p>\n\n\n\n

= x + 3x\/5<\/p>\n\n\n\n

= (5x+3x)\/5<\/p>\n\n\n\n

= 8x\/5<\/p>\n\n\n\n

Difference between Shikha\u2019s and Shalu\u2019s income will be = 8x\/5 \u2013 x<\/p>\n\n\n\n

= (8x-5x)\/5<\/p>\n\n\n\n

= 3x\/5<\/p>\n\n\n\n

When Shalu\u2019s income is less than Shikha\u2019s income (in %) = (3x\/5)\/(8x\/5) \u00d7 100<\/p>\n\n\n\n

= 3x\/8x \u00d7 100<\/p>\n\n\n\n

= 300\/8<\/p>\n\n\n\n

= 37.5%<\/p>\n\n\n\n

\u2234 By 37.5%, Shalu\u2019s income is less than Shikha\u2019s income.<\/p>\n\n\n\n

26. Rs 3500 is to be shared among three people so that the first person gets 50% of the second, who in turn gets 50% of the third. How much will each of them get?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the total money to be shared is = Rs 3500<\/p>\n\n\n\n

Let us consider third person get = Rs x<\/p>\n\n\n\n

So, second person gets (50% of third) = 50% of x<\/p>\n\n\n\n

= 50\/100 \u00d7 x<\/p>\n\n\n\n

= Rs x\/2<\/p>\n\n\n\n

Now, first person gets (50% of second) = 50% of x\/2<\/p>\n\n\n\n

= 50\/100 \u00d7 x\/2<\/p>\n\n\n\n

= Rs x\/4<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

x\/4 + x\/2 + x = 3500<\/p>\n\n\n\n

by taking 4 as LCM<\/p>\n\n\n\n

(x+2x+4x)\/4 = 3500<\/p>\n\n\n\n

By cross multiplying<\/p>\n\n\n\n

x+2x+4x = 3500 \u00d7 4<\/p>\n\n\n\n

7x = 14000<\/p>\n\n\n\n

x = 14000\/7<\/p>\n\n\n\n

= 2000<\/p>\n\n\n\n

\u2234 Each of the person gets,<\/p>\n\n\n\n

First person (x\/4) gets = x\/4 = 2000\/4 = Rs 500<\/p>\n\n\n\n

Second person (x\/2) gets = x\/2 = 2000\/2 = Rs 1000<\/p>\n\n\n\n

Third person (x) gets = x = Rs 2000<\/p>\n\n\n\n

27. After a 20% hike, the cost of Chinese Vase is Rs 2000. What was the original price of the object?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let cost price of Chinese Vase before hike be = Rs x<\/p>\n\n\n\n

The hike is = 20% of 100 = 20\/100<\/p>\n\n\n\n

The cost price of Chinese Vase after hike is = Rs 2000<\/p>\n\n\n\n

So, let\u2019s calculate for x,<\/p>\n\n\n\n

x + x\u00d720\/100 = 2000<\/p>\n\n\n\n

x + x\/5 = 2000<\/p>\n\n\n\n

(5x+x)\/5 = 2000<\/p>\n\n\n\n

6x = 2000\u00d75<\/p>\n\n\n\n

x = 10000\/6<\/p>\n\n\n\n

= 1666.6667<\/p>\n\n\n\n

\u2234 Original price of Chinese Vase is = Rs. 1666.67<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 12: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 12\u2013Percentage<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 8 Maths Chapter 12\u2013Percentage PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n