RS Aggarwal 2019,2020 solution class 7 chapter 12 Simple Interest Test Paper 12

Test Paper 12

Page-168

Question 1:

Find the simple interest on Rs 6300 at 8% per annum for 8 months.

Answer 1:

$\begin{array}{l}\text{P}=\text{Rs 63}00,\text{ R}=\text{8}\% \text{, T}=\frac{8}{12} \text{years} \\ \therefore \text{S}\text{.I.}=\frac{\text{P×R×T}}{\text{100}}=\frac{\sout{6}\sout{3}\sout{0}\sout{0}\times 8\times {{\sout{8}}^{21}}^{{\sout{4}}^2}}{1\sout{0}\sout{0}\times \sout{1}{\sout{2}}_{{\sout{6}}_{{\sout{3}}_1}}}=\text{Rs }336\end{array}$

Question 2:

What sum will amount to Rs 6600 in 2 years at 10% per annum simple interest?

Answer 2:

$\begin{array}{l}\text{Let the sum be Rs }x.\\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}=\frac{x\times 1\sout{0}\times {\sout{2}}^1}{\sout{1}\sout{0}{\sout{0}}_5}=\frac{x}{5}\\ \\ \text{Now}, \text{A}=\text{P}+\text{S}.\text{I}\\ \text{A=}x+\frac{x}{5}=\frac{6x}{5}\\ \text{But the amount is Rs 66}00.\\ \therefore \frac{6x}{5}=6600\\ =>x=\frac{\sout{6}\sout{6}\sout{0}{\sout{0}}^{1100}\times 5}{\sout{6}}=5500\\ \text{Hence, the required sum is Rs 55}00.\end{array}$

Question 3:

At what rate per cent per annum simple interest will Rs 3625 amount to Rs 4495 in 2 years?

Answer 3:

$\begin{array}{l}\text{P}=\text{Rs 3625} ,\text{ A}=\text{Rs 4495,} \text{T}=\text{2 years}\\ \therefore \text{S}.\text{I.}=\text{A}―\text{P} \\ =\text{4495}―\text{3625} \\ =\text{Rs 87}0\\ \therefore \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}\\ =>870=\frac{3625\times \text{R}\times 2}{100}\\ =>\text{R}=\frac{870\times 100}{3625\times 2}\\ =>\text{R}=\frac{87000}{7250}=12\%\end{array}$

Question 4:

In what time will Rs 3600 amount to Rs 4410 at 9% per annum simple interest?

Answer 4:

$$\begin{gathered} \begin{array}{l}\text{P}=\text{Rs}\text{. 36}00,\text{ A}=\text{Rs}\text{. 441}0,\text{ R}=\text{9}\% \\ \text{S}.\text{I.}=\text{A}-\text{P }\\ =\text{441}0-\text{36}00 \\ =\text{Rs}\text{. 81}0\\ \therefore \text{T=}\frac{\text{S}\text{.I.×100}}{\text{P×R}}\end{array} \\ \text{=}\frac{\text{810×100}}{\text{3600×9}} \\ \text{=}\frac{\text{90}}{\text{36}} \\ \text{=2}\frac{\text{1}}{\text{2}}\text{ years} \end{gathered}$$

Question 5:

At what rate per cent per annum simple interest will a sum double itself in 12 years?

Answer 5:

$\begin{array}{l} \text{Let the sum be Rs }x. \\ \text{Amount }=\mathrm{Rs} \text{2}x \\ \therefore \text{S}.\text{I.}=(\text{2}x-x)=Rs x\\ \text{Time}=1\text{2 years}\\ \therefore \text{P}=x, S.I.=x, \text{T}=1\text{2 years}\\ \therefore \text{R=}\frac{\text{100×S}\text{.I.}}{\text{P×T}}=\frac{100\times x}{x\times 12}=8.3\%\end{array}$

Question 6:

A sum of money becomes $\frac{4}{3}$ of itself in 6 years at a certain rate of simple interest. Find the rate of interest.

Answer 6:

$$\begin{gathered} \begin{array}{l}\text{Let the sum be Rs }x.\\ \\ \text{Amount =}\frac{4}{3}x\\ \text{S}\text{.I}\text{.=A}-\text{P=}\frac{4}{3}x-x\\ \text{ =}\frac{x}{3}\\ \text{Let the rate be R}\%.\\ \\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}\\ =>\frac{x}{3}=\frac{x\times \text{R×6}}{100}\\ =>\text{R}=\frac{\sout{x}\times 100}{\sout{x}\times 6\times 3}=\frac{100}{18}=5.55\end{array} \\ \mathrm{Hence}, \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{interest} \mathrm{is} 5.55\%. \end{gathered}$$

Question 7:

Mark (✓) against the correct answer
At simple interest a sum becomes $\frac{49}{40}$ of itself in $2\frac{1}{2}$ years. The rate of interest per annum is
(a) 7%
(b) 8%
(c) 9%
(d) 12%

Answer 7:

(c) 9%
$\begin{array}{l}\text{Let the sum be Rs }x. \\ \\ \text{A}=\frac{49x}{40}\\ \text{We know:}\\ \text{A=P+S}\text{.I}\text{.}\\ \therefore \text{S}\text{.I. =A}-\text{P}\\ \text{ =}(\frac{49x}{40}-x)\\ \text{ =}\frac{49x-40x}{40}\\ \text{ =}\frac{9x}{40}\\ \text{Let the rate be R}\%\text{ per annum}. \\ \\ \text{S}\text{.I.=}\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =>\frac{9x}{40}=\frac{x\times \text{R}\times {\sout{5}}^1}{\sout{1}\sout{0}{\sout{0}}_{20}\times 2}\\ \text{=>R=}\frac{9\times \sout{2}{\sout{0}}^1\times {\sout{2}}^1}{\sout{4}{\sout{0}}_{{\sout{2}}_1}}\\ \text{=>R=9}\\ \end{array}$

Question 8:

Mark (✓) against the correct answer
A sum amounts to Rs 3626 in 219 days at 6% per annum simple interest. The sum is

(a) Rs 3000
(b) Rs 3200
(c) Rs 3500
(d) Rs 3600

Answer 8:

(c) 3500
$\begin{array}{l}\text{A=Rs 3626, R=6\%, T=219 days=}\frac{\text{219}}{\text{365}}\text{years}\\ \text{Let the required sum be Rs }x.\\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}=\frac{x\times 6\times 219}{100\times 365}=\frac{1314x}{36500}\\ \text{A}=\text{P}+\text{S.I.}\\ \text{ =}x+\frac{1314x}{36500}\\ =\frac{36500x+1314x}{36500}\\ =\frac{37814x}{36500}\\ \text{But the amount is Rs 3626.}\\ \therefore \frac{37814x}{36500}=3626\\ x=\frac{\sout{3}\sout{6}\sout{2}{\sout{6}}^{259}\times 36500}{\sout{3}\sout{7}\sout{8}\sout{1}{\sout{4}}_{2701}}=3500\\ \text{The required sum is Rs 35}00.\\ \end{array}$

Question 9:

Mark (✓) against the correct answer
In what time will Rs 6000 amount to Rs 6360 at 8% per annum simple interest?

(a) 9 months
(b) 8 months
(c) $1\frac{1}{4}$  years
(d) $1\frac{1}{2}$ years

Answer 9:

(a) 9 months

$\begin{array}{l}\text{P}=\text{Rs 6}000, \text{A}=\text{Rs 636}0, \text{R}=\text{8}\% \\ \text{S}.\text{I.}=\text{A}-\text{P} \\ =\text{636}0-\text{6}000 \\ =\text{Rs}\text{. 36}0\\ \\ \text{T=}\frac{\text{S}\text{.I.×100}}{\text{P×T}}\text{=}\frac{\text{360×100}}{\text{6000×8}}\\ \text{ =}\frac{6}{8}\text{ years}\\ \text{ =}\frac{\text{6}}{\text{8}}\text{×12 months}\\ \text{ =}\frac{\text{72}}{\text{8}}\\ \text{ =9 months}\end{array}$

Question 10:

Mark (✓) against the correct answer
The simple interest on a sum for 5 yars is $\frac{3}{5}$ of the sum. The rate per cent per annum is
(a) 8%
(b) 10%
(c) 12%
(d) $12\frac{1}{2}\%$

Answer 10:

$$\begin{gathered} \left(c\right) 12\% \\ \begin{array}{l} \text{Let the sum be Rs }x.\\ \text{S}\text{.I}\text{.=Rs }\frac{3}{5} \\ \text{Let the rate be R}\%\text{ per annum}.\\ \text{S}\text{.I.=}\frac{\text{P×R×T}}{\text{100}}\text{=}\frac{x\text{×R×5}}{\text{100}}\text{=}\frac{\text{R}x}{\text{20}}\\ \therefore \frac{3x}{5}=\frac{\text{R}x}{20}\\ =>\text{R}=\frac{3x\times 20}{5x}=12\%\\ \end{array} \end{gathered}$$

Question 11:

Mark (✓) against the correct answer
The simple interest at x% per annum for x years will be Rs x on a sum of

(a) Rs x
(b) Rs 10 x
(c) Rs 100 x
(d) Rs $\left(\frac{100}{x}\right)$

Answer 11:

$$\begin{gathered} \left(d\right) \mathrm{Rs} \frac{100}{\mathrm{x}} \\ \begin{array}{l}\text{Let the sum be Rs P.}\\ \text{Time=}\mathrm{x}\text{ years}\\ \text{Rate of interest}, \text{r}=\mathrm{x}\mathrm{\%}\text{ per annum}\\ \text{S}\text{.I}\text{.=Rs }\mathrm{x}\\ \text{Simple Interest }=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ \text{ =>}\mathrm{x}\text{ =}\frac{\text{P}\times \mathrm{x}\times \mathrm{x}}{100}\\ \text{ =>P=}\frac{\sout{\mathrm{x}}\times 100}{\sout{\mathrm{x}}\times \mathrm{x}}\\ \therefore \text{P=}\left(\frac{100}{\mathrm{x}}\right)\\ \end{array} \end{gathered}$$

Question 12:

Mark (✓) against the correct answer
At what rate per cent per annum simple interest will a sum double itself in 10 years?

(a) 8%
(b) 10%
(c) 12%
(d) $12\frac{1}{2}\%$

Answer 12:

$$\begin{gathered} \text{(b) 10\%} \\ \begin{array}{l}\text{Let the sum be Rs }x.\\ \text{Amount=Rs 2}x\\ \text{Time=10 years}\\ \therefore \text{S}\text{.I}\text{.=A}-\text{P=2}x-x\\ \text{ =Rs }x\\ \therefore \text{S}\text{.I.=}\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =>x=\frac{x\times \text{R}\times \text{10}}{100}\\ =>\text{R=}\frac{\sout{x}\times \sout{1}\sout{0}\sout{0}}{\sout{x}\times \sout{1}\sout{0}}\end{array} \\ =>\mathrm{R}=10 \end{gathered}$$

Question 13:

Fill in the blanks.

(i) $P=\frac{100\times (......)}{R\times T}$
(ii) $R=\frac{100\times \mathrm{SI}}{(......)\times T}$
(iii) At (......)% per annum simple interest a sum doubles itself in 10 years.
(iv) At simple interest a sum becomes $\frac{6}{5}$ of itself in $2\frac{1}{2}$ years. The rate of interest is (......)% per annum.

Answer 13:

$$\begin{gathered} \begin{array}{l}\left(i\right)P=\frac{100\times S.I.}{\mathrm{R\times T}}\\ \left(\mathrm{ii}\right)\mathrm{R=}\frac{\mathrm{100\times S}.I.}{\mathrm{(P)\times T}}\end{array} \\ \left(\mathrm{iii}\right)10\% \\ \mathrm{Let} \mathrm{the} \mathrm{sum} \mathrm{be} \mathrm{Rs} x. \\ \mathrm{Amount}= \mathrm{Rs} 2x \\ \mathrm{S}.\mathrm{I}.=\mathrm{A}-\mathrm{P} \\ =2x-x \\ =\mathrm{Rs} x \\ \mathrm{Let} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{inetrset} \mathrm{be} R\%. \\ \mathrm{Time}=10 \mathrm{yrs} \\ S\mathit{.}I.=\frac{P\mathit{\times }R\mathit{\times }T}{100} \\ \Rightarrow x=\frac{x\times R\times 10}{100} \\ \Rightarrow R=\frac{100}{10} \\ \Rightarrow R =10 \\ \left(iv\right) 8\% \\ \mathrm{Let} \mathrm{the} \mathrm{sum} \mathrm{be} \mathrm{Rs} x. \\ \mathrm{Amount}= \mathrm{Rs} \frac{6}{5}x \\ \mathrm{S}.\mathrm{I}.=\mathrm{A}-\mathrm{P} \\ =\frac{6}{5}x-x \\ =\mathrm{Rs} \frac{x}{5} \\ \mathrm{Let} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{inetrset} \mathrm{be} R\%. \\ \mathrm{Time}=2.5 \mathrm{yrs} \\ S\mathit{.}I.=\frac{P\mathit{\times }R\mathit{\times }T}{100} \\ \Rightarrow \frac{x}{5}=\frac{x\times R\times 2.5}{100} \\ \Rightarrow R=\frac{100}{2.5\times 5} \\ \Rightarrow R =8 \end{gathered}$$

Question 14:

Write 'T' for true and 'F' for false

(i) Simple interest on Rs x for x years is Rs x. Then, the rate of interest is x% per annum.
(ii) Rate = $\frac{100\times \mathrm{SI}}{P\times T}$.
(iii) A sum doubles itself at simple interest at 10% per annum in 10 years.
(iv) Simple interest on Rs 1000 at 5% per annum for 73 days is Rs 10.

Answer 14:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{False} \\ S.I.=\frac{P\times R\times T}{100} \\ \Rightarrow x=\frac{x\times R\times x}{100} \\ \Rightarrow R=\frac{100\times x}{x\times x} \\ \Rightarrow R=\frac{100}{x}\% \\ \left(\mathrm{ii}\right) \mathrm{True} \\ \left(\mathrm{iii}\right) \mathrm{True} \\ \mathrm{Let} \mathrm{the} \mathrm{sum} \mathrm{be} \mathrm{Rs} x. \\ \mathrm{S}.\mathrm{I}.=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \mathrm{S}.\mathrm{I}.=\frac{\mathrm{x}\times 10\times 10}{100} \\ \mathrm{S}.\mathrm{I}.=\mathrm{Rs} x \\ \mathrm{Amount}= \mathrm{S}.\mathrm{I}.+\mathrm{P} \\ =\mathrm{x}+\mathrm{x} \\ =\mathrm{Rs} 2\mathrm{x} \\ \left(\mathrm{iv}\right) \mathrm{True} \\ \mathrm{P}=\mathrm{Rs} 1000 , \mathrm{R}=5\%, \mathrm{T}=73 \mathrm{days}=\frac{73}{365} \mathrm{yrs} \\ \mathrm{S}.\mathrm{I}.=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{1000\times 5\times 73}{100\times 365} \\ =\mathrm{Rs} 10 \end{gathered}$$