RD Sharma solution class 8 chapter 12 Percentage Exercise 12.2

Exercise 12.2

Page-12.9

Question 1:

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

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) 22\% \mathrm{of} 120=\frac{22}{100}\times 120 \\ =26.4 \\ \left(\mathrm{ii}\right) 25\% \mathrm{of} 1000=\frac{25}{100}\times 1000 \\ =250 \\ \left(\mathrm{iii}\right) 25\% \mathrm{of} 10 \mathrm{kg}=\frac{25}{100}\times 10 \mathrm{kg} \\ =\frac{25}{10}\mathrm{kg}=\frac{5}{2}\mathrm{kg} \\ =2.5 \mathrm{kg} \\ \left(\mathrm{iv}\right) 16.5\% \mathrm{of} 5000 \mathrm{m}=\frac{16.5}{100}\times 5000 \mathrm{m} \\ =16.5\times 50 \mathrm{m} \\ =825 \mathrm{m} \\ \left(\mathrm{v}\right) 135\% \mathrm{of} 80 \mathrm{cm}=\frac{135}{100}\times 80 \mathrm{cm} \\ =\frac{135\times 8}{10}\mathrm{cm} \\ =108 \mathrm{cm}=1.08 \mathrm{m} \\ \left(\mathrm{vi}\right) 2.5\% \mathrm{of} 10000 \mathrm{ml} = \frac{2.5}{100}\times 10000 \mathrm{ml} \\ =250 \mathrm{ml} \end{gathered}$$

Question 2:

Find the number a, if
(i) 8.4% of a is 42
(ii) 0.5% of a  is 3
(iii) $\frac{1}{2}\%$ of a is 50
(iv) 100% of a is 100

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) 8.4 \% of a \mathrm{is} 42. \\ \mathrm{i}.\mathrm{e}., \frac{8.4}{100}\times a=42 \\ \Rightarrow a=\frac{42\times 100}{8.4} \\ \therefore a=500 \\ \left(\mathrm{ii}\right) 0.5\% of a \mathrm{is} 3. \\ \mathrm{i}.\mathrm{e}., \frac{0.5}{100}\times a=3 \\ \Rightarrow a=\frac{3\times 100}{0.5} \\ \therefore a=600 \\ \left(\mathrm{iii}\right) \frac{1}{2}\% of a \mathrm{is} 50. \\ \mathrm{i}.\mathrm{e}., \frac{1}{2}\times \frac{1}{100}\times a=50 \\ \Rightarrow a=50\times 200 \\ \therefore a=10000 \\ \left(\mathrm{iv}\right) 100\% of a \mathrm{is} 100. \\ \mathrm{i}.\mathrm{e}., \frac{100}{100}\times a=100 \\ \Rightarrow 1\times a=100 \\ \therefore a=100 \end{gathered}$$

Question 3:

x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.

Answer 3:

$$\begin{gathered} \mathrm{Given}: \\ x=\frac{5}{100}y \\ \Rightarrow y=\frac{100}{5}x \\ \therefore y=20x \\ \mathrm{Also}, y=\frac{24}{100}z \\ \therefore z=\frac{100}{24}y \\ \mathrm{Putting} x=480 \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{equations}, \mathrm{w}e \mathrm{get}: \\ y=20\left(480\right)=9600 \\ z=\frac{100}{24}\times 9600=40000 \end{gathered}$$

Question 4:

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.

Answer 4:

$$\begin{gathered} \mathrm{Let} \mathrm{his} \mathrm{monthly} \mathrm{income} \mathrm{be} \mathrm{Rs} \mathrm{x} . \\ \mathrm{Then}, 15\% of x = 150 \\ \Rightarrow \frac{15}{100}x=150 \\ \Rightarrow x=\frac{150\times 100}{15} \\ =1000 \\ \therefore \mathrm{His} \mathrm{monthly} \mathrm{income} \mathrm{is} \mathrm{Rs} 1000. \end{gathered}$$

Question 5:

Asha got 86.875% marks in the annual examination. If she got 695 marks, find the total number of marks of the examination.

Answer 5:

$$\begin{gathered} \mathrm{Let} x \mathrm{be} \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{marks} \mathrm{of} \mathrm{the} \mathrm{examination}. \\ \mathrm{Then}, 86.875\% \mathrm{of} x=695 \\ \Rightarrow \frac{86.875}{100}x=695 \\ \Rightarrow x=\frac{695\times 100}{86.875} \\ =800 \\ \therefore \mathrm{The} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{marks} \mathrm{of} \mathrm{the} \mathrm{examination} \mathrm{is} 800. \end{gathered}$$

Question 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.

Answer 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{school} \mathrm{opened} \mathrm{for} x \mathrm{days} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{year}. \\ \mathrm{We} \mathrm{have}: \\ 90\% of x =216 \\ \Rightarrow \frac{90}{100}x=216 \\ \Rightarrow x=\frac{216\times 100}{90} \\ =240 \\ \therefore \mathrm{The} \mathrm{school} \mathrm{opened} \mathrm{for} 240 \mathrm{days} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{year}. \end{gathered}$$

Question 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.

Answer 7:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{orange} \mathrm{trees} \mathrm{be} x. \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{there} \mathrm{are} 2,000 \mathrm{trees}. \mathrm{While} 12\% \mathrm{of} \mathrm{them} \mathrm{are} \mathrm{mango} \mathrm{trees}, 18\% \mathrm{are} \mathrm{lemon} \mathrm{trees} \mathrm{and} \mathrm{the} \mathrm{rest} \mathrm{are} \mathrm{orange} \mathrm{trees}. \\ \mathrm{Now}, \left(12\% of 2000\right) +\left(18\% of 2000\right)+x=2000 \\ \Rightarrow \frac{12}{100}\times 2000+\frac{18}{100}\times 2000+x=2000 \\ \Rightarrow 240+360+x=2000 \\ \Rightarrow 600+x=2000 \\ \Rightarrow x=2000-600 \\ \Rightarrow x=1400 \\ \therefore \mathrm{There} \mathrm{are} 1,400 \mathrm{orange} \mathrm{tree}s. \end{gathered}$$

Page-12.10

Question 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.

Answer 8:

$$\begin{gathered} \mathrm{In} \mathrm{a} \mathrm{balanced} \mathrm{diet} \mathrm{of} 2600 \mathrm{calories}, 12\% \mathrm{is} \mathrm{protein}. \\ \therefore \mathrm{Amount} \mathrm{of} \mathrm{protein} \mathrm{in} \mathrm{food} \mathrm{intake} = 12\% \mathrm{of} 2600 \\ =\frac{12}{100}\times 2600 \\ =312 \\ \mathrm{Similarly}, \mathrm{a} \mathrm{balanced} \mathrm{diet} \mathrm{contain}s 25\% \mathrm{fats}. \\ \therefore \mathrm{Amount} \mathrm{of} \mathrm{fats} \mathrm{in} \mathrm{food} \mathrm{intake} = 25\% \mathrm{of} 2600 \\ =\frac{25}{100}\times 2600 \\ =650 \\ \mathrm{Similarly}, \mathrm{a} \mathrm{balanced} \mathrm{diet} \mathrm{contains} 63\% \mathrm{carbohydrates}. \\ \therefore \mathrm{Amount} \mathrm{of} \mathrm{carbohydrates} \mathrm{in} \mathrm{food} \mathrm{intake} = 63\% \mathrm{of} 2600 \\ =\frac{63}{100}\times 2600 \\ =1638 \end{gathered}$$

Question 9:

A cricketer scored a total of 62 runs in 96 balls. He hit 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in
(i) Sixes
(ii) fours
(iii) twos
(iv) singles

Answer 9:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{The} \mathrm{cricketer} \mathrm{hits} 3 \mathrm{sixes}. \\ \mathrm{i}.\mathrm{e}., 3\times 6=18 \\ \therefore \frac{18}{62}\times 100=29.03\% \\ \left(\mathrm{ii}\right) \mathrm{The} \mathrm{cricketer} \mathrm{hits} 8 \mathrm{fours}. \\ \mathrm{i}.\mathrm{e}., 8\times 4=32 \\ \therefore \frac{32}{62}\times 100=51.61\% \\ \left(\mathrm{iii}\right) \mathrm{The} \mathrm{cricketer} \mathrm{hits} 2 \mathrm{twos}. \\ \mathrm{i}.\mathrm{e}., 2\times 2=4 \\ \therefore \frac{4}{62}\times 100=6.45\% \\ \left(\mathrm{iv}\right) \mathrm{The} \mathrm{cricketer} \mathrm{hits} 8 \mathrm{singles}. \\ \mathrm{i}.\mathrm{e}., 8\times 1=8 \\ \therefore \frac{8}{62}\times 100=12.90\% \end{gathered}$$

Question 10:

A cricketer hit 120 runs in 150 balls during a test match. 20% of the runs came in 6's, 30% in 4's, 25% in 2's and the rest in 1's. How many runs did he score in
(i) 6's
(ii) 4's
(iii) 2's
(iv) singles
What % of his shots were scoring ones?

Answer 10:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Let} \mathrm{the} \mathrm{cricketer} \mathrm{score} \mathrm{w} \mathrm{runs} \mathrm{in} 6\text{'}s. \\ \therefore 20\% of 120 =w \\ \Rightarrow \frac{20}{100}\times 120=w \\ \Rightarrow w=24 \\ \left(\mathrm{ii}\right) \mathrm{Let} \mathrm{the} \mathrm{cricketer} \mathrm{score} x \mathrm{runs} \mathrm{in} 4\text{'}s. \\ \therefore 30\% of 120 =x \\ \Rightarrow \frac{30}{100}\times 120=x \\ \Rightarrow x=36 \\ \left(\mathrm{iii}\right) \mathrm{Let} \mathrm{the} \mathrm{cricketer} \mathrm{score} y \mathrm{runs} \mathrm{in} 2\text{'}s. \\ \therefore 25\% of 120 =y \\ \Rightarrow \frac{25}{100}\times 120=y \\ \Rightarrow y=30 \\ \left(\mathrm{iv}\right) \mathrm{Let} \mathrm{the} \mathrm{cricket}er \mathrm{score} z \mathrm{runs} \mathrm{in} 1\text{'}\mathrm{s}. \\ \mathrm{Since}, \mathrm{the} \mathrm{cricketer} \mathrm{scores} \mathrm{the} \mathrm{rest} \mathrm{of} \mathrm{the} \mathrm{runs} \mathrm{in} 1\text{'}\mathrm{s}, \mathrm{we} \mathrm{have}: \\ 24+36+30+z=120 \\ \Rightarrow z+90=120 \\ \Rightarrow z=30 \end{gathered}$$

Question 11:

Radha earns 22% of her investment. If she earns Rs 187, then how much did she invest?

Answer 11:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{investment} \mathrm{be} \mathrm{Rs} x. \\ \mathrm{i}.\mathrm{e}., 22\% \mathrm{of} x = 187 \\ \Rightarrow \frac{22}{100}x=187 \\ \Rightarrow x=\frac{18700}{22} \\ =850 \\ \therefore \mathrm{Radha} \mathrm{invest}s \mathrm{Rs} 850. \end{gathered}$$

Question 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?

Answer 12:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{total} \mathrm{income} \mathrm{for} \mathrm{the} \mathrm{year} 1997 \mathrm{be} \mathrm{Rs} x. \\ \mathrm{i}.\mathrm{e}., 12\% of x =1440 \\ \Rightarrow \frac{12}{100}x=1440 \\ \Rightarrow x=\frac{144000}{12} \\ \Rightarrow x=12000 \\ \therefore \mathrm{Rohit}\text{'}\mathrm{s} \mathrm{total} \mathrm{income} \mathrm{was} \mathrm{Rs} 12,000. \end{gathered}$$

Question 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?

Answer 13:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{gunpowder} \mathrm{that} \mathrm{contains} 9 \mathrm{kg} \mathrm{nitre} \mathrm{be} \mathrm{x} \mathrm{kg}. \\ \mathrm{i}.\mathrm{e}., \frac{75}{100}\times \mathrm{x}=9 \\ \Rightarrow \frac{100}{75}=\frac{\mathrm{x}}{9} \\ \Rightarrow \mathrm{x}=\frac{900}{75} \\ \Rightarrow \mathrm{x}=12 \\ \therefore \mathrm{The} \mathrm{amount} \mathrm{of} \mathrm{gunpowder} \mathrm{containing} 9 \mathrm{kg} \mathrm{nitre} \mathrm{is} 12 \mathrm{kg}. \\ \mathrm{Let} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{gunpowder} \mathrm{containing} 2.3 \mathrm{kg} \mathrm{sulphur} \mathrm{be} \mathrm{y} \mathrm{kg}. \\ \mathrm{i}.\mathrm{e}., \frac{10}{100}\times \mathrm{y}=2.3 \\ \Rightarrow \frac{100}{10}=\frac{\mathrm{y}}{2.3} \\ \Rightarrow \mathrm{y}=\frac{230}{10} \\ \Rightarrow \mathrm{y}=23 \\ \therefore \mathrm{The} \mathrm{amount} \mathrm{of} \mathrm{gunpowder} \mathrm{containing} 2.3 \mathrm{kg} \mathrm{sulphur} \mathrm{is} 23 \mathrm{kg}. \end{gathered}$$

Question 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?

Answer 14:

$$\begin{gathered} \mathrm{Composition} \mathrm{of} \mathrm{the} \mathrm{alloy} = 15 \mathrm{parts} \mathrm{of} \mathrm{tin} + 105 \mathrm{parts} \mathrm{of} \mathrm{copper} \\ =120 \mathrm{parts} \\ \therefore \mathrm{Percentage} \mathrm{of} \mathrm{tin}= \frac{15}{120}\times 100 \\ =12.5\% \\ \mathrm{Also}, \mathrm{percentage} \mathrm{of} \mathrm{copper}= \frac{105}{120}\times 100 \\ =87.5\% \end{gathered}$$

Question 15:

An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.

Answer 15:

$$\begin{gathered} \mathrm{Percentage} \mathrm{of} \mathrm{copper} \mathrm{in} \mathrm{the} \mathrm{alloy}=32 \\ \mathrm{Percentage} \mathrm{of} \mathrm{nickel} \mathrm{in} \mathrm{the} \mathrm{alloy}=40 \\ \mathrm{Percentage} \mathrm{of} \mathrm{zinc} \mathrm{in} \mathrm{the} \mathrm{alloy}=100-32-40=28 \\ \therefore \mathrm{Amount} \mathrm{of} \mathrm{zinc} \mathrm{in} 1 \mathrm{kg} \mathrm{of} \mathrm{the} \mathrm{alloy} = \left(0.28\times 1\right) \mathrm{kg} \\ =280 \mathrm{gm} \end{gathered}$$

Question 16:

A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride?

Answer 16:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{total} \mathrm{ride} \mathrm{be} x \mathrm{km}. \\ \mathrm{Then}, 10\% \mathrm{of} x = 122 \\ \Rightarrow \frac{10}{100}x=122 \\ \Rightarrow x=1220 \\ \therefore \mathrm{The} \mathrm{length} \mathrm{of} \mathrm{total} \mathrm{ride} \mathrm{is} 1,220 \mathrm{km}. \end{gathered}$$

Question 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?

Answer 17:

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{female} \mathrm{student}s = 300-142 \\ =158 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{female} \mathrm{teachers} = 30-12 \\ =18 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{females}= 158+18 \\ =176 \\ \mathrm{Total} \mathrm{population} \mathrm{of} \mathrm{the} \mathrm{school}= 300+30 \\ =330 \\ \mathrm{Percentage} \mathrm{of} \mathrm{females} = \frac{176}{330}\times 100 \\ =\frac{160}{3}\% \mathrm{or} 53.33\% \end{gathered}$$

Question 18:

Aman's income is 20% less than that of Anil. How much percent is Anil's income more than Aman's income?

Answer 18:

$$\begin{gathered} \mathrm{Let} \mathrm{Anil}\text{'}\mathrm{s} \mathrm{income} \mathrm{be} x. \\ \mathrm{Then}, \mathrm{Aman}\text{'}\mathrm{s} \mathrm{income} = x-\frac{20x}{100} \\ =\frac{8x}{10} \\ \mathrm{Difference} \mathrm{in} \mathrm{the} \mathrm{incomes} \mathrm{of} \mathrm{Anil} \mathrm{and} \mathrm{Aman} = x-\frac{8x}{10} \\ =\frac{2x}{10} \\ \therefore \mathrm{Percentage} \mathrm{of} \mathrm{the} \mathrm{difference} \mathrm{in} \mathrm{the} \mathrm{incomes} \mathrm{of} \mathrm{Anil} \mathrm{and} \mathrm{Aman} \mathrm{to} \mathrm{that} \mathrm{of} \mathrm{Aman}\text{'}\mathrm{s} \mathrm{incom}e = \left(\frac{2x}{10}\right)/\left(\frac{8x}{10}\right)\times 100 \\ =\frac{1}{4}\times 100 \\ =25\% \end{gathered}$$

Question 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?

Answer 19:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{depreciates} \mathrm{by} 5\% \mathrm{every} \mathrm{year}. \\ \mathrm{Present} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{machine} = \mathrm{Rs} 100000 \\ \therefore \mathrm{For} \mathrm{the} \mathrm{first} \mathrm{year}, 5\% \mathrm{of} 100000 = \frac{5}{100}\times 100000 \\ =\mathrm{Rs} 5000 \\ \therefore \mathrm{Value} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{after} \mathrm{one} \mathrm{year} = 100000-5000=\mathrm{Rs} 95000 \\ \mathrm{Value} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{in} \mathrm{the} \mathrm{second} \mathrm{year} = \mathrm{Rs} 95000 \\ \therefore 5\% \mathrm{of} 95000=\frac{5}{100}\times 95000 \\ =\mathrm{Rs} 4750 \\ \therefore \mathrm{Value} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{after} 2 \mathrm{years} =95000-4750=\mathrm{Rs} 90250 \\ \therefore \mathrm{After} \mathrm{two} \mathrm{years}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{machine} \mathrm{will} \mathrm{be} \mathrm{Rs} 90,250. \end{gathered}$$

Question 20:

The population of a town increases by 10% annually. If the present population is 60000, what will be its  population after 2 years?

Answer 20:

$$\begin{gathered} \mathrm{Present} \mathrm{population} = 60000 \\ \mathrm{It} \mathrm{increases} \mathrm{by} 10\%\hspace{0.17em}\mathrm{annually}. \\ \therefore \mathrm{Increase} \mathrm{in} \mathrm{the} \mathrm{population} \mathrm{in} \mathrm{the} \mathrm{first} \mathrm{year} = 10\% \mathrm{of} 60000 =\frac{10}{100}\times 60000 \\ =6000 \\ \therefore \mathrm{Population} \mathrm{after} 1 \mathrm{year} = 60000+6000=66000 \\ \mathrm{Increase} \mathrm{in} \mathrm{the} \mathrm{population} \mathrm{in} \mathrm{the} \mathrm{second} \mathrm{year}, 10\% \mathrm{of} 66000 =\frac{10}{100}\times 66000 \\ =6600 \\ \mathrm{Thus}, \mathrm{population} \mathrm{after} 2 \mathrm{years} = 66000+6600=72600 \end{gathered}$$

Question 21:

The population of a town increases by 10% annually. If the present population is 22000, find its population a year ago.

Answer 21:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{population} \mathrm{of} \mathrm{the} \mathrm{town} \mathrm{one} \mathrm{year} \mathrm{ago} = x \\ \mathrm{Now}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{population} \mathrm{of} \mathrm{the} \mathrm{town} \mathrm{increases} \mathrm{by} 10\% \mathrm{annually}. \\ \mathrm{Present} \mathrm{population} = x + 10\% \mathrm{of} x \\ =x + \frac{10}{100}\times x \\ =x + \frac{x}{10} \\ =\frac{11x}{10} \\ \mathrm{But} \mathrm{present} \mathrm{population} \mathrm{of} \mathrm{the} \mathrm{town} = 22000 \\ \mathrm{So}, \frac{11x}{10} = 22000 \\ \Rightarrow 11x = 220000 \\ \Rightarrow x = \frac{220000}{11} \\ \Rightarrow x = 20000 \\ \mathrm{So}, \mathrm{population} \mathrm{of} \mathrm{the} \mathrm{town} 1 \mathrm{year} \mathrm{ago} = 20000 \end{gathered}$$

Question 22:

Ankit was given an increment of 10% on his salary. His new salary is Rs 3575. What was his salary before increment?

Answer 22:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{initial} \mathrm{salary} \mathrm{be} \mathrm{Rs} x. \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ \mathrm{Salary} \mathrm{before} \mathrm{increment} + \mathrm{increment} \mathrm{given} \mathrm{on} \mathrm{salary} = \mathrm{new} \mathrm{salary} \\ \therefore x + 10\% of x = 3575 \\ \Rightarrow x+\frac{10}{100}x=3575 \\ \Rightarrow 100x+10x=357500 \\ \Rightarrow 110x=357500 \\ \Rightarrow x=\frac{357500}{110} \\ \Rightarrow x=3250 \\ \therefore \mathrm{Salary} \mathrm{before} \mathrm{increment} = \mathrm{Rs} 3,250 \end{gathered}$$

Question 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?

Answer 23:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{reduce} \mathrm{the} \mathrm{consumption} \mathrm{such} \mathrm{that} \mathrm{the} \mathrm{expenditure} \mathrm{does} \mathrm{not} \mathrm{increase}. \\ \mathrm{For} \mathrm{this}, \mathrm{we} \mathrm{use} \mathrm{the} \mathrm{following} \mathrm{formula}: \\ \left(\frac{r}{r+100}\right)\times 100, \mathrm{where} \mathrm{r} = \mathrm{percentage} \mathrm{rise} \mathrm{in} \mathrm{the} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{commodity} \\ \therefore \mathrm{Percentage} \mathrm{reduction} \mathrm{in} \mathrm{the} \mathrm{consumption} = \left(\frac{10}{10+100}\right)\times 100=\left(\frac{10}{110}\right)\times 100 (\because r = 10\%) \\ =9\frac{1}{11} \end{gathered}$$

Question 24:

Mohan's 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?

Answer 24:

$$\begin{gathered} \mathrm{Mohan}\text{'}\mathrm{s} \mathrm{saving} = 11\% \mathrm{of} 15500 \\ =\frac{11}{100}\times 15500 \\ =\mathrm{Rs}. 1705 \end{gathered}$$

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{Mohan}\text{'}\mathrm{s} \mathrm{income} \mathrm{increases} \mathrm{by} 10\%. \\ \therefore \mathrm{Increase} \mathrm{in} \mathrm{income} =\frac{10}{100}\times 15500=\mathrm{Rs} 1550 \\ \mathrm{Increased} \mathrm{income} =\mathrm{Rs} (15500+1550)=\mathrm{Rs} 17050 \\ \mathrm{Now}, \mathrm{percentage} \mathrm{of} \mathrm{saving}=(11-1)=10\% \\ \therefore \mathrm{Saving} = \frac{10}{100}\times 17050=\mathrm{Rs} 1705 \\ \mathrm{Thus}, \mathrm{the} \mathrm{amounts} \mathrm{of} \mathrm{his} \mathrm{present} \mathrm{and} \mathrm{earlier} \mathrm{savings} \mathrm{are} \mathrm{the} \mathrm{same}. \end{gathered}$$

Question 25:

Shikha's income is 60% more than that of Shalu. What percent is Shalu's income less than Shikha's?

Answer 25:

$$\begin{gathered} \mathrm{Let} \mathrm{Shalu}\text{'}\mathrm{s} \mathrm{income} \mathrm{be} \mathrm{Rs} x. \\ \therefore \mathrm{Shikha}\text{'}\mathrm{s} \mathrm{income}=\mathrm{Rs} \left(\mathrm{x}+\frac{60x}{100}\right)=\mathrm{Rs} \frac{160x}{100}=\mathrm{Rs} \frac{16x}{10} \\ \mathrm{Difference} \mathrm{in} \mathrm{the} \mathrm{incomes} \mathrm{of} \mathrm{Shikha} \mathrm{and} \mathrm{Shalu} =\frac{16x}{10}-x=\frac{16x-10x}{10}=\mathrm{Rs} \frac{6x}{10} \\ \mathrm{Percentage} \mathrm{of} \mathrm{the} \mathrm{difference} \mathrm{in} \mathrm{the} \mathrm{incomes} \mathrm{of} \mathrm{Shikha} \mathrm{and} \mathrm{Shalu} \mathrm{to} \mathrm{that} \mathrm{of} \mathrm{Shikha}\text{'}\mathrm{s} \mathrm{income}=\frac{\frac{6x}{10}}{\frac{16x}{10}}\times 100=\frac{600}{16}=37.5\% \\ \therefore \mathrm{The} \mathrm{income} \mathrm{of} \mathrm{Shalu} \mathrm{is} \mathrm{less} \mathrm{than} \mathrm{that} \mathrm{of} \mathrm{Shikha} \mathrm{by} 37.5\%. \end{gathered}$$

Question 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?

Answer 26:

$$\begin{gathered} \mathrm{Let} x, y and z \mathrm{be} \mathrm{the} \mathrm{amounts} \mathrm{received} \mathrm{by} \mathrm{the} \mathrm{first}, \mathrm{second} \mathrm{and} \mathrm{the} \mathrm{third} \mathrm{person}, \mathrm{respectively}. \\ \mathrm{We} \mathrm{have}: \\ x=50\% \mathrm{of} y \\ =\frac{50}{100}y=\frac{1}{2}y \\ \therefore y=2x \\ \mathrm{Again}, y=50\% \mathrm{of} z \\ =\frac{1}{2}z \\ \therefore z=2y \\ =2\left(2x\right) \\ =4x \\ \mathrm{Also}, x+y+z=3500 \\ \mathrm{Substituting} \mathrm{the} \mathrm{values} \mathrm{of} z and y, \mathrm{we} \mathrm{get}: \\ x+2x+4x=3500 \\ \Rightarrow 7x=3500 \\ \Rightarrow x=500 \\ \therefore y=2x=2\left(500\right) \\ =1000 \\ \therefore z=4x=4\left(500\right) \\ =2000 \\ \mathrm{Thus}, \mathrm{the} \mathrm{amounts} \mathrm{received} \mathrm{by} \mathrm{the} \mathrm{first}, \mathrm{second} \mathrm{and} \mathrm{the} \mathrm{third} \mathrm{person} \mathrm{are} \mathrm{Rs} 500, \mathrm{Rs} 1000 \mathrm{and} \mathrm{Rs} 2000, \mathrm{respectively}. \end{gathered}$$

Question 27:

After a 20% hike, the cost of Chinese Vase is Rs 2000. What was the original price of the object?

Answer 27:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{original} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{object} \mathrm{be} x. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}: \\ 20\%\mathrm{of} x +x = 2000 \\ \Rightarrow \frac{20x}{100}+x=2000 \\ \Rightarrow \frac{120}{100}x=2000 \\ \Rightarrow x=\frac{200000}{120} \\ \Rightarrow x=1666.67 \\ \mathrm{Thus}, \mathrm{the} \mathrm{original} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{object} \mathrm{was} \mathrm{Rs} 1,666.67. \end{gathered}$$