5 Marks Questions

Question 15 Marks

Find the time when: A sum when reckoned at $7\frac12\%$ per annum amounts to $Rs. 3920$ in $3$ years. Find the sum.

Answer

Let the required sun be $Rs. xA = Rs. 3920,$
$\text{R}=7\frac12\%,$
$T = 3$ years
Now, $\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{\text{x}\times15\times3}{2\times100}$
$=\frac{9\text{x}}{40}$
$\text{A}=\text{P}+\text{S.I.}$
$=\text{x}+\frac{9\text{x}}{40}$
$=\frac{40\text{x}+\text{9x}}{40}$
$=\frac{49\text{x}}{40}$
But the Amount is Rs. 3920
$\Rightarrow\frac{49\text{x}}{40}=3920$
$\Rightarrow\text{x}=\frac{3920\times40}{49}$
$\Rightarrow\frac{156800}{49}=3200$
Hence, the required sum is $Rs. 3200$

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Question 25 Marks

Find the time when: 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

Amount in $3$ years = (Principal $+ S.I.$ for $3$ years) $= Rs. 837$
Amount in $2$ years = (Principal $+ S.I$. for $2$ years) $= Rs. 783$
On subtracting: S.I. for $1$ years $= (837 - 783) = Rs. 54$
$\text{S.I. for $2$ years}=\Big(\frac{54}{1}\times2\Big)=\text{Rs. }108$
$\therefore$ Sum = Amount for $2$ years $- S.I$. for $2$ years $=783-108$
$=\text{Rs. 675}$ $P = Rs. 675, S.I. = Rs. 108$ and $T = 2$ years
$\text{r}=\frac{\text{S.I.}\times100}{\text{P}\times\text{t}}$
$=\frac{108\times100}{675\times2}$
$=8\%$

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Question 35 Marks

Find the time when: A sum of money becomes $\frac85$ of itself in $5$ years at a certain rate of simple interest. Find the rate of interest.

Answer

Let the sum be Rs. x$\text{Amount}=\frac{8\text{x}}{5}$
$\therefore\text{S.I.}=\text{A}-\text{P}=\frac{\text{8x}}{\text{5}}-\text{x}$
$=\frac{3\text{x}}{5}$
Let the rate be $r\%$
$\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$\Rightarrow\frac{3\text{x}}{5}=\frac{\text{x}\times\text{r}\times5}{100}$
$\Rightarrow\text{3x}\times20$
$\Rightarrow\text{r}\times\text{x}\times5$
$\Rightarrow\text{r}=\frac{3\times\text{x}\times20}{\text{x}\times5}=12$
Hence, the rate of interest is $12\%.$

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Question 45 Marks

Find the time when: 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

In first case, Amount $(A) = Rs. 768$
Principal $(P) = Rs. 640 $
$S.I. = A - P = Rs. 768 - 640 = Rs. 128$
$\text{Time(t)}=2\text{ years}\ 6\text{ months}=2\frac12$
$=\frac52\text{ years}$
$\therefore\text{Rate}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{128\times100\times2}{640\times5}$
$=8\%\text{ p.a.}$ In second case, Principal $(P) = Rs. 850$ Rate $(r) = 8\%$ p.a. Time $(t) = 3$ years $\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{850\times8\times3}{100}$
$=\text{Rs. }204$
$\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }850+\text{Rs. }204$
$=\text{Rs. }1054$

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Question 55 Marks

Find the time when: 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

Amount $= Rs. 12122 $ Let principal $(P) = Rs. 100$
Rate (r) = 8% p.a. Time (t) = 2 years $\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{100\times8\times2}{100}$
$=\text{Rs. }16$ And amount = P \times S.I. = Rs. 100 + 16 $= \text{Rs. }116$ If amount is Rs. 116, then principal $= \text{Rs. }100$ And if amount is Rs. 1, then principal $=\text{Rs. }\frac{100}{116}$ And if amount is Rs. 12122, then principal $=\text{Rs. }\frac{100\times12122}{116}$
$=\text{Rs. }10450$
Now, rate (r) = 9% And time (t) = 2 years, 8 months $=2\frac23\text{ years}=\frac83\text{ years}$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\text{Rs. }\frac{10540\times9\times8}{100\times3}$
$=\text{Rs. }2508$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }10450+2508$
$=\text{Rs. }12958$

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Question 65 Marks

Find the time when: 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

Amount $= Rs. 4491$
Let principal $(P) = Rs. 100$ Rate $(r) =11\%$ p.a.
$\text{Time(t)}=2\text{ years}\ 3\text{ months}=2\frac14=\frac94\text{ years}$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{100\times11\times9}{100\times4}$
$=\text{Rs. }\frac{99}{4}$
$\therefore\text{Amount}\text{ A}=\text{P}+\text{S.I.}$
$=\text{Rs. }100+\text{Rs. }\frac{99}{4}=\text{Rs. }\frac{499}{4}$ If amount is $\text{Rs. }\frac{499}{4},$
then principal $= \text{Rs. }100$ If amount is $Rs. 1$, then principal $=\text{Rs. }\frac{100\times4}{499}$
$=\text{Rs. }3600$
Now, interest for $3$ years at the rate of $11\%$
$=\text{Rs. }=\frac{3600\times11\times3}{100}=\text{Rs. }1188$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }3600+\text{Rs. }1188$
$=\text{Rs. }4788$

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Question 75 Marks

Find the time when: 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 annual income is $Rs. 333.$

Answer

Total sum $= Rs. 3600$
Let first part $= Rs. x$
Then second part $= Rs. (3600 - x)$
Interest on first part for $1$ year at $9\%$ p.a. $=\frac{\text{x}\times9\times1}{100}=\frac{9\text{x}}{100}$
And interest on second part for $1$ years at $10\%$ p.a. $=\frac{(3600-\text{x})\times10\times1}{100}$
$=\frac{10(3600-\text{x})}{100}$ But total interst $= Rs. 333$
$\therefore\frac{9\text{x}}{100}+\frac{10(3600-\text{x})}{100}=333$
$\Rightarrow 9\text{x} + 10 (3600 - \text{x}) = 33300 $
$\Rightarrow 9\text{x} + 36000 - 10\text{x} = 33300 $
$\Rightarrow -\text{x} = 33300 - 36000 $
$\Rightarrow-\text{x} = - 2700 $
$\Rightarrow \text{x} = 2700$
First part $= Rs. 2700$ and
second part $= Rs. 3600 - 2700 = Rs. 900$

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Question 85 Marks

Find the time when: Divide $Rs. 3000$ into two parts so 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 $2$ years at $9%$ per annum.

Answer

Total sum $= Rs. 3000$
Let first part $= Rs. x$
Then second part $= Rs. (3000 - x)$
Now, interest on first part at the rate of $8\%\text{ for 4 years} =\frac{\text{x}\times8\times4}{100}=\frac{32}{100}\text{x}$ And
interest on the second part for $2$ years at $9\%=\frac{(3000-\text{x})\times9\times2}{100}$
$=\frac{18(3000-\text{x})}{100}$ But in both the case, interest is same.
$\therefore\frac{32}{100}\text{x}=\frac{18(3000-\text{x})}{100}$
$\Rightarrow32\text{x}=18(3000-\text{x)}$
$\Rightarrow\text{32x}=54000-18\text{x}$
$\Rightarrow32\text{x}+\text{18x}=54000$
$\Rightarrow50\text{x}=54000$
$\Rightarrow\text{x}=\frac{54000}{50}=1080$
$\therefore$ First part $= Rs. 1080$ And
second part $= Rs. (3000 - 1080) = Rs. 1920$

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