RS Aggarwal 2019,2020 solution class 7 chapter 12 Simple Interest Exercise 12A

Exercise 12A

Page-165

Question 1:

Find the simple interest and the amount when:

Principal = Rs 6400, rate = 6% p.a. and time = 2 years

Answer 1:

$$\begin{gathered} \begin{array}{l}\mathrm{P}=\mathrm{Rs}. 6400, \mathrm{R}=\mathrm{6\%}, \mathrm{T}=2 \mathrm{years}\\ \mathrm{S}.\mathrm{I}. =\frac{\mathrm{P\times R\times T}}{\mathrm{100}}=\frac{6400\times 6\times 2}{100}\\ \mathrm{ }=\mathrm{Rs. }768\\ \mathrm{Amount=P+S.I}.\end{array} \\ =6400+768 \\ = \mathrm{Rs. 7168} \end{gathered}$$

Question 2:

Find the simple interest and the amount when:

Principal = Rs 2650, rate = 8% p.a. and time = $2\frac{1}{2}$ years

Answer 2:

$\text{P}=\text{Rs}.\text{ 265}0,\text{ R}=\text{8}\%,\text{ T=2}\frac{\text{1}}{\text{2}}\text{ years =}\frac{\text{5}}{\text{2}}\text{ years}$

$\text{S.I.}= \frac{\text{P×R×T}}{\text{100}}\text{=}\frac{\text{2650×8×5}}{\text{100×2}}$

                    $$\begin{gathered} \begin{array}{l}=\mathrm{Rs}.\mathrm{ }530\\ \mathrm{Amount}=\mathrm{P}+\mathrm{S}.\mathrm{I}.\end{array} \\ =2650+530 \\ =\mathrm{Rs}. 3180 \end{gathered}$$

Question 3:

Find the simple interest and the amount when:

Principal = Rs 1500, rate = 12% p.a. and time = 3 years 3 months.

Answer 3:

$$\begin{gathered} \begin{array}{l}\mathrm{P}=\mathrm{Rs}.1500, \mathrm{R}=\mathrm{12}\mathrm{\%},\mathrm{ T}=3+\frac{3}{12}=\frac{13}{4} \mathrm{years} \\ \mathrm{S}\mathrm{.I}.=\frac{\mathrm{P\times R\times T}}{\mathrm{100}}\mathrm{=}\frac{\mathrm{1500\times 12\times 13}}{\mathrm{100\times 4}}\\ \mathrm{ }=\mathrm{Rs}\mathrm{. 585}\\ \mathrm{Amount=P+S}\mathrm{.I}.\end{array} \\ =1500+585 \\ =Rs\mathrm{. 2085} \end{gathered}$$

Question 4:

Find the simple interest and the amount when:

Principal = Rs 9600, rate = $7\frac{1}{2}\%$ p.a. and time = 5 months.

Answer 4:

$$\begin{gathered} \mathrm{P}= \mathrm{Rs}. 9600 \\ \mathrm{R}=7\frac{1}{2}\% \\ \mathrm{T}=5 \mathrm{months} =\frac{5}{12} \mathrm{years} \\ \mathrm{S}.\mathrm{I}.=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{9600\times 15\times 5}{100\times 2\times 12} \\ =\mathrm{Rs}. 300 \\ \mathrm{Amount}= \mathrm{P}+ \mathrm{S}.\mathrm{I}. \\ =9600+300 \\ =\mathrm{Rs}. 9900 \end{gathered}$$

Question 5:

Find the simple interest and the amount when:

Principal = Rs 5000, rate = 9% p.a. and time = 146 days.

Answer 5:

$$\begin{gathered} \begin{array}{l}\mathrm{P}=\mathrm{Rs}.5000 , \mathrm{R}=\mathrm{9}\mathrm{\%} ,\mathrm{ T}=\mathrm{146 days}=\frac{146}{365} \mathrm{years}\\ \mathrm{S}.I.=\frac{\mathrm{P\times R\times T}}{\mathrm{100}}\mathrm{=}\frac{\mathrm{5000\times 9\times 146}}{\mathrm{100\times 365}}\\ \mathrm{ =Rs}\mathrm{. 180}\\ \mathrm{Amount=P+S}.I.\end{array} \\ =5000+180 \\ =Rs. 5180 \end{gathered}$$

Question 6:

Find the time when:

Principal = Rs 6400, SI = Rs 1152 and rate = 6% p.a.

Answer 6:

$$\begin{gathered} \begin{array}{l}\text{P}=\text{Rs}\text{. 64}00, \text{S}.\text{I. = Rs}\text{. 1152}, \text{R}=\text{6}\% \\ \text{T =}\frac{\text{S}\text{.I.×100}}{\text{P×R}}=\frac{1152\times \sout{1}\sout{0}\sout{0}}{64\sout{0}\sout{0}\times 6}\\ =\frac{1152}{384}\end{array} \\ =3 \text{years} \end{gathered}$$

Question 7:

Find the time when:

Principal = Rs 9540, SI = Rs 1908 and rate = 8% p.a.

Answer 7:

$\begin{array}{l}\text{P}=\text{Rs}.\text{ 954}0 , \text{S}.\text{I.}=\text{Rs}.\text{ 19}0\text{8, R}=\text{8}\%\\ \text{T = }\frac{\text{S}\text{.I.×100}}{\text{P×R}}\text{=}\frac{\text{1908×100}}{\text{9540×8}}\\ \text{ =}\frac{\text{10}}{\text{4}}\\ \text{ =2}\frac{\text{1}}{\text{2}}\text{ years}\end{array}$

Question 8:

Find the time when:

Principal = Rs 5000, amount = Rs 6450 and rate = 12% p.a.

Answer 8:

$\begin{array}{l}\text{P}=\text{Rs}\text{. 5}000,\text{ A=Rs}\text{. 645}0, \text{R}=\text{12}\%\\ \text{S}.\text{I.}=\text{A}-\text{P} \\ =\text{645}0-\text{5}000 \\ =\text{Rs}\text{. 145}0\\ \\ \text{T =}\frac{\text{S}\text{.I×100}}{\text{P×R}}\text{=}\frac{\text{1450×100}}{\text{5000×12}}\\ \text{ =}\frac{\text{29}}{\text{12}}\\ =2\frac{5}{12}\\ \text{ =2 years 5 months}\end{array}$

page-166

Question 9:

Find the rate when:

Principal = Rs 8250, SI = Rs 1100 and time = 2 years.

Answer 9:

$\begin{array}{l}\text{P }=\text{ Rs}.\text{ 825}0,\text{ S}.\text{I.}=\text{Rs}.\text{ 110}0, \text{T}=\text{2 years}\\ \text{R=}\frac{\text{S}\text{.I.×100}}{\text{P×T}}\text{=}\frac{\text{1100×100}}{\text{8250×2}}\\ \text{ =}\frac{\text{1100}}{\text{165}}\text{=6}\text{.67}\%\end{array}$

Question 10:

Find the rate when:

Principal = Rs 5200, SI = Rs 975 and time = $2\frac{1}{2}$ years.

Answer 10:

$$\begin{gathered} \begin{array}{l}\text{ P= Rs}\text{. 5200 , S}\text{.I.=Rs}\text{. 975 [ T=2}\frac{\text{1}}{\text{2}}\text{ years=}\frac{\text{5}}{\text{2}}\text{ years]}\\ \text{ R=}\frac{\text{S}\text{.I.×100}}{\text{P×T}}\text{=}\frac{\text{975×100×2}}{\text{5200×5}}\\ \text{ =}\frac{\text{195}}{\text{26}}\end{array} \\ \text{=7}\text{.5\%} \end{gathered}$$

Question 11:

Find the rate when:

Principal = Rs 3560, amount = Rs 4521.20 and time = 3 years.

Answer 11:

$\begin{array}{l}\text{P}=\text{Rs}\text{. 356}0 ,\text{ A}=\text{Rs}\text{. 4521}.\text{2}0 \text{, T}=\text{3 years}\\ \text{S}.\text{I.}=\text{ A}-\text{P} =\text{4521}.\text{2}0-\text{356}0\\ =\text{Rs}\text{. 961}.\text{2}0 \\ \text{R =}\frac{\text{S}\text{.I.×100}}{\text{P×T}}\text{=}\frac{\text{961}\text{.20×100}}{\text{3560×3}}\\ \text{ =}\frac{\text{96120×100}}{\text{100×3560×3}}\\ \text{ =9\%}\end{array}$

Question 12:

Shanta borrowed Rs 6000 from the State Bank of India for 3 years 8 months at 12% per annum. What amount will clear off her debt?

Answer 12:

$\begin{array}{l}\text{P}=\text{Rs 6}000, \text{R}=\text{12}\%,\text{ T=}3 \text{years 8 months = 3}\frac{\text{8}}{\text{12}}\text{=}\frac{\text{44}}{\text{12}}\text{years}\\ \text{S}\text{.I. =}\frac{\text{P×R×T}}{\text{100}}\text{=}\frac{\text{6000×12×44}}{\text{100×12}}\text{= Rs 2640}\\ \text{A}= \text{P}+\text{S}.\text{I.} \\ = \text{6}000+\text{264}0 \\ = \text{Rs 864}0\end{array}$

Question 13:

Hari borrowed Rs 12600 from a moneylender at 15% per annum simple interest. After 3 years, he paid Rs 7070 and gave a goat to clear off the debt. What is the cost of the goat?

Answer 13:

$$\begin{gathered} \begin{array}{l}\text{P}=\text{Rs}\text{. 126}00 \text{R}=\text{15}\% \text{T}=\text{3 years}\\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}\text{=}\frac{\text{12600×15×3}}{\text{100}}\\ \text{ =Rs}\text{. 5670}\\ \text{A}=\text{Rs}\text{. 126}00+\text{Rs}\text{. 567}0=\text{Rs}\text{. 1827}0 \\ \text{ Hari had to pay Rs}\text{. 1827}0\text{ to the money lender, but he paid Rs}\text{. 7}0\text{7}0\text{ and a goat}.\\ \therefore \text{Cost of the goat }=\text{Rs}\text{. 1827}0-\text{Rs}\text{. 7}0\text{7}0\end{array} \\ =\text{Rs}\text{. 112}00 \end{gathered}$$

Question 14:

The simple interest on a certain sum for 3 years at 10% per annum is Rs 829.50. Find the sum.

Answer 14:

$$\begin{gathered} \begin{array}{l}\text{Let the sum be Rs. P.}\\ \text{S}\text{.I.}=\text{Rs}\text{. 829}.\text{5}0,\text{ T}=3 \text{years, R}=\text{1}0\%\\ \\ \mathrm{Now, }\text{P=}\frac{\text{S}\text{.I×100}}{\text{R×T}}\\ \text{ =}\frac{\text{829}\text{.50×100}}{\text{10×3}}\\ \text{ =}\frac{\text{8295}}{3}\end{array} \\ \text{= 2765} \\ \mathrm{Hence}, \mathrm{the} \mathrm{sum} \mathrm{is} \mathrm{Rs}. 2765. \end{gathered}$$

Question 15:

A sum when reckoned at $7\frac{1}{2}\%$ per annum amounts to Rs 3920 in 3 years. Find the sum

Answer 15:

$\begin{array}{l} \text{Let the required sum be Rs. }x. \\ \text{A}=\text{Rs}\text{. 392}0, \text{R}=7\frac{1}{2}\%, \text{T}=\text{3 years}\\ \text{Now}, \\ \text{Now}, \text{S}\text{.I}\text{.=}\frac{\text{P×R×T}}{\text{100}}=\frac{x\times 15\times 3}{2\times 100}=\frac{9x}{40}\\ \text{A}=\text{P}+\text{S}.\text{I.} \\ \text{ =} x+\frac{9x}{40}=\frac{40x+9x}{40} =\frac{49x}{40} \\ \text{But the amount is Rs}\text{. 392}0.\\ =>\frac{49x}{40}=3920\\ =>x=\frac{3920\times 40}{49}=\frac{156800}{49}=3200\\ \text{Hence, the required sum is Rs}\text{. 32}00.\end{array}$

Question 16:

A sum of money put at 11% per annum amounts to Rs 4491 in 2 years 3 months. What will it amount to in 3 years at the same rate?

Answer 16:

$\begin{array}{l}\text{Given: R=11\%, T=2 years 3 months = 2+}\frac{3}{12}=\frac{27}{12} \text{years}\\ \text{Let the required sum be Rs. }x.\\ \\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}\text{=}\frac{x\times 11\times \sout{2}{\sout{7}}^9}{100\times \sout{1}{\sout{2}}_4}=\frac{99x}{400}\\ \text{A}=\text{P}+\text{S}.\text{I.}\\ =x+\frac{99x}{400}=\frac{400x+99x}{400}=\frac{499x}{400}\\ \text{But the amount is Rs}\text{. 4491.}\\ =>\frac{499x}{400}=4491\\ =>x=\frac{4491\times 400}{499}=\frac{1796400}{499}=3600\\ \text{Hence, the required sum is Rs}\text{. 3600.}\\ \text{∴ S}\text{.I}\text{.=}\frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{3600\times 11\times 3}{100}=\text{Rs}\text{. 1188}\\ \therefore \text{Amount=P+S}\text{.I.=3600+1188}\\ \text{ =Rs}\text{. 4788}\end{array}$

Question 17:

A sum of money invested at 8% per annum amounts to Rs 12122 in 2 years. What will it amount to in 2 years 8 months at 9% per annum?

Answer 17:

$\begin{array}{l}\text{Let the required sum be Rs. }x.\\ \\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}=\frac{x\times 8\times 2}{100}=\frac{16x}{100}\\ \text{A}=\text{P}+\text{S}.\text{I.}\\ =x+\frac{16x}{100}=\frac{100x+16x}{100}=\frac{116x}{100}\\ \text{But the amount is Rs}\text{. 12122.}\\ =>\frac{116x}{100}=12122\\ =>x=\frac{12122\times 100}{116}=10450\\ \\ \mathrm{Now, }\text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}=\frac{1045\sout{0}\times {\sout{9}}^3\times \sout{3}{\sout{2}}^8}{10\sout{0}\times \sout{1}{\sout{2}}_{4_1}}=\text{Rs}. 2508\\ \therefore \text{A=P+S}\text{.I.}\\ \text{ = Rs}\text{. 10450 + Rs}\text{. 2508}\\ \text{ = Rs}\text{. 12958}\end{array}$

Question 18:

At what rate per cent per annum will Rs 3600 amount to Rs 4734 in $3\frac{1}{2}$ years?

Answer 18:

$$\begin{gathered} \begin{array}{l}\text{P}=\text{Rs}\text{. 36}00 \text{A}=\text{Rs}\text{. 4734} \text{ T=} 3\frac{1}{2}=\frac{7}{2} \text{years} \\ \text{ S}.\text{I.}=\text{A}-\text{P} \\ =\text{4734}-\text{36}00 \\ =\text{Rs}\text{. 1134} \\ \text{R=}\frac{\text{S}\text{.I.×100}}{\text{P×T}}\\ \end{array} \\ \text{=}\frac{\text{1134×100×2}}{\text{3600×7}} \\ \text{=9\%} \end{gathered}$$

Question 19:

If Rs 640 amounts to Rs 768 in 2 years 6 months, what will Rs 850 amount to in 3 years at the same rate per cent per annum?

Answer 19:

$\begin{array}{l}\text{P}=\text{Rs}\text{. 64}0,\text{ A}=\text{Rs}\text{. 768, T=}2 \text{years 6 months =}\frac{\text{5}}{\text{2}}\text{ years} \\ \text{ S}.\text{I.}=\text{A}-\text{P} \\ =\text{768}-\text{64}0 \\ =\text{Rs}\text{. 128} \\ \text{R=}\frac{\text{S}\text{.I.×100}}{\text{P×T}}\text{=}\frac{\text{128×100×2}}{\text{640×5}}\text{=8\%}\\ \\ \text{P}=\text{Rs}\text{. 85}0,\text{ R}=\text{8}\%\text{, T}=\text{3 years}\\ \therefore \text{S}\text{.I. =}\frac{\text{P×R×T}}{\text{100}}\text{=}\frac{\text{850×8×3}}{\text{100}}\text{=}\frac{\text{2040}}{\text{10}}\text{=Rs}\text{. 204}\\ \therefore \text{A}=\text{P}+\text{S}.\text{I.} \\ =\text{85}0+\text{2}0\text{4}\\ =\text{Rs}.\text{ 1}0\text{54}\end{array}$

Question 20:

In what time will Rs 5600 amount to Rs 6720 at 8% per annum?

Answer 20:

$$\begin{gathered} \begin{array}{l}\text{P}=\text{Rs}\text{. 56}00, \text{A}=\text{Rs}\text{. 672}0,\text{ R}=\text{8}\%\\ \text{ S}.\text{I.}=\text{A}-\text{P} \\ =\text{672}0-\text{56}00 \\ =\text{Rs}\text{. 112}0\\ \text{T=}\frac{\text{S}\text{.I.×100}}{\text{P×R}}\end{array} \\ \text{=}\frac{\text{1120×100}}{\text{5600×8}} \\ \text{=}\frac{\text{1120}}{\text{448}} \\ \text{=2}\frac{\text{1}}{\text{2}}\text{ years} \end{gathered}$$

Question 21:

A sum of money becomes $\frac{8}{5}$ of itself in 5 years at a certain rate of simple interest. Find the rate of interest.

Answer 21:

$$\begin{gathered} \begin{array}{l}\text{Let} \text{the sum be Rs. }x.\\ \text{ Amount=}\frac{8x}{5}\\ \therefore \text{S}\text{.I}\text{.=A}-\text{P=}\frac{8x}{5}-x\\ \text{ =}\frac{3x}{5}\\ \text{Let the rate be R}\%.\\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}\\ =>\frac{3x}{5}=\frac{x\times \text{R}\times {\sout{5}}^1}{\sout{1}\sout{0}{\sout{0}}_{20}}\\ =>3x\times 20=\text{R}\times x\times 5\\ =>\text{R}=\frac{3\times \cancel{x}\times \sout{2}{\sout{0}}^4}{\cancel{x}\times \sout{5}}=12\end{array} \\ \mathrm{Hence}, \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{interest} \mathrm{is} 12\%. \end{gathered}$$

Question 22:

A sum of money lent at simple interest amounts to Rs 783 in 2 years and to Rs 837 in 3 years. Find the sum and the rate per cent per annum.

Answer 22:

$\begin{array}{l}\text{Amount in 3 years}=(\text{Principal}+\text{ S}.\text{I}.\text{ for 3 years})=\text{Rs}.\text{ 837} \\ \text{Amount in 2 years}=(\text{Principal}+\text{ S}.\text{I. for 2 years})=\text{Rs}.\text{ 783} \\ \text{On subtracting}: \\ \text{S}.\text{I. for} 1\text{ year}=(\text{837}-\text{783})=\text{Rs}.\text{ 54}\\ \text{S}.\text{I. for 2 years=}(\frac{54}{1}\times 2)=\text{Rs}\text{. }108\\ \therefore \text{Sum}=\text{Amount for 2 years}-\text{S}.\text{I. for 2 years}\\ =\text{783}-\text{1}0\text{8} \\ =\text{Rs}.\text{ 675}\\ \\ \text{P}=\text{Rs}\text{. 675}, \text{S}.\text{I.}=\text{Rs}\text{. 1}0\text{8 and} \text{ T}=\text{2 years}\\ \text{R}=\frac{\text{S}\text{.I}\text{.}\times \text{100}}{\text{P}\times \text{T}}\\ =\frac{108\times \sout{1}\sout{0}{\sout{0}}^{\sout{5}{\sout{0}}^2}}{\sout{6}\sout{7}{\sout{5}}_{27}\times {\cancel{2}}_1}\\ =8\%\end{array}$

Question 23:

A sum of money lent at simple interest amount to Rs 4745 in 3 years and to Rs 5475 in 5 years. Find the sum and the rate per cent per annum.

Answer 23:

$\begin{array}{l}\text{Amount in 5 years}=(\text{Principal}+\text{ S}.\text{I}.\text{ for 5 years})=\text{Rs}.\text{ 5475} \\ \text{Amount in 3 years}=(\text{Principal}+\text{ S}.\text{I. for 3 years})=\text{Rs}.\text{ 4745} \\ \text{On subtracting}: \\ \text{S}.\text{I. for} \text{2 years}=(\text{5475}-\text{4745})=\text{Rs}\text{. 73}0\\ \text{S}.\text{I. for 3 years=}(\frac{730}{2}\times 3)=\text{Rs}\text{. }1095\\ \therefore \text{Sum}=\text{Amount for 3 years}-\text{S}.\text{I. for 3 years}\\ =\text{4745}-\text{1}0\text{95} \\ =\text{Rs}.\text{ 365}0\\ \text{ P=Rs}\text{. 3650, S}\text{.I}\text{.=Rs}\text{. 1095, T=3 years}\\ \text{R=}\frac{\text{S}\text{.I}\text{.}\times \text{100}}{\text{P}\times \text{T}}\\ \text{ =}\frac{1095\times 100}{3650\times 3}\\ \text{ = 10\%}\\ \end{array}$

Question 24:

Divide Rs 3000 into two parts such that the simple interest on the first part for 4 years at 8% per annum is equal to the simple interest on the second part for 2 years at 9% per annum.

Answer 24:

$$\begin{gathered} \begin{array}{l}\text{Let the first part be Rs. }x.\\ \text{Second part }=(\text{3}000-x)\\ \therefore \text{S}.\text{I. on x at 8}\% \text{per annum for 4 years}=\frac{x\times 8\times {\sout{4}}^{{\sout{2}}^1}}{\sout{1}\sout{0}{\sout{0}}_{\sout{5}{\sout{0}}_{25}}}=\frac{8x}{25}\\ \text{S}.\text{I. on }(\text{3}000-x)\text{ at 9}\% \text{per annum }=\frac{(3000-x)\times 9\times {\sout{2}}^1}{\sout{1}\sout{0}{\sout{0}}_{50}}\\ =\frac{27000-9x}{50}\\ \therefore \frac{8x}{25}=\frac{27000-9x}{50}\\ =>8x=\frac{(27000-9x)\times \sout{2}{\sout{5}}^1}{\sout{5}{\sout{0}}_2}\\ =>16x=27000-9x\\ =>16x+9x=27000\\ =>x=\frac{\sout{2}\sout{7}\sout{0}\sout{0}{\sout{0}}^{1080}}{\sout{2}{\sout{5}}_1}=1080\\ \therefore \mathrm{F}\text{irst part = Rs}\text{.1}0\text{8}0\end{array} \\ \mathrm{Second part }=(\text{3}000-\text{1}0\text{8}0)=\text{Rs}\text{. 192}0 \end{gathered}$$

Question 25:

Divide Rs 3600 into two parts such that if one part be lent at 9% per annum and the other at 10% per annum, the total annum income is Rs 333.

Answer 25:

$$\begin{gathered} \begin{array}{l}\text{Let the first part be Rs. }x. \\ \text{Second part }=(\text{36}00-x)\\ \therefore \text{S}.\text{I. on x at 9}\% \text{per annum for 1 years}=\frac{x\times 9\times 1}{100}=\frac{9x}{100}\\ \text{And, S}.\text{I. on }(\text{36}00-x)\text{ at} \text{1}0\% \text{per annum =}\frac{(3600-x)\times 1\times \sout{1}{\sout{0}}^1}{10\sout{0}}=\frac{3600-x}{10}\\ \therefore \frac{9x}{100}+\frac{3600-x}{10}=333\\ =>\frac{9x+36000-10x}{100}=333\\ =>-x+36000=33300\\ =>-x=33300-36000\\ =>-x=-2700\\ =>x=2700\\ \text{First part = Rs}\text{. 2700 }\end{array} \\ S\mathrm{econd part }=(\text{36}00-2700)=\text{Rs}\text{. }900 \end{gathered}$$