[3 marks sum]

Question 13 Marks

In how many years will $Rs.2000$ amount to $Rs.2662$ at $10\% p.a.$ compound interest?

Answer

Here $ P=R s .2000, A=R s .2662, r=10 \%$
Now, $P\left(1+\frac{r}{100}\right)^t=A$
$\Rightarrow 2000\left(1+\frac{10}{100}\right)^{ t } $
$=2662 $
$\Rightarrow\left(\frac{11}{10}\right)^{ t } $
$=\frac{2662}{2000} $
$=\frac{1331}{1000} $
$=\left(\frac{11}{10}\right)^3$
By comparing powers, $t=3$
Hence time is $3$ years.

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Question 23 Marks

In what time will $Rs.15625$ amount to $Rs.17576$ at $4\% p.a.$ compound interest?

Answer

Here $P=\text { Rs. } 15625, A=17576, r=4 \%$
Now, $P\left(1+\frac{ r }{100}\right)^{ t }= A$
$\Rightarrow 15625\left(1+\frac{4}{100}\right)^{ t } $
$=17576 $
$\Rightarrow\left(\frac{26}{25}\right)^{ t } $
$=\frac{17576}{15625} $
$=\left(\frac{26}{25}\right)^3$
By comparing powers, $t=3$
Hence time is $3$ years.

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Question 33 Marks

Find compound interest on $Rs.31250$ for $3$ years, if the rates of interest for $1^{st},2^{nd}$ and $3^{rd}$ years be $8\%, 10\%$ and $12\%$ respectively.

Answer

Here $P=$ Rs. $31250, t=3$ years, $r=8 \%, 10 \%, 12 \%$ successively.
Now, Amount
$= P \left(1+\frac{ r _1}{100}\right)\left(1+\frac{ r _2}{100}\right)\left(1+\frac{ r _3}{100}\right)$
$=31250\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{12}{100}\right)$
$=31250\left(\frac{108}{100}\right)\left(\frac{110}{100}\right)\left(\frac{112}{100}\right)$
$=41580$
Hence, Amount $= Rs. 41580 .$

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Question 43 Marks

How much will $Rs.25000$ amount to in $2$ years at compound interest, if the rates for $1^{st}$ and $2^{nd}$ years be $4\%$ and $5\% p.a.$ respectively?

Answer

Here $P= Rs.25000, t=2$ years, $r=4 \%, 5 \%$ successively,
Now, Amount
$ = P \left(1+\frac{ r _1}{100}\right)\left(1+\frac{ r _2}{100}\right)$
$=25000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)$
$=25000\left(\frac{104}{100}\right)\left(\frac{105}{100}\right)$
$=27300$
Hence, Amount $= Rs. 27300.$

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Question 53 Marks

Find the amount on $Rs.36000$ in $2$ years $15\% p.a.$ compounded annually.

Answer

Here $P=R s .36000, t=2$ years, $r=15 \%$
Now, Amount
$= P \left(1+\frac{ r }{100}\right)^{ t } $
$=36000\left(1+\frac{15}{100}\right)^2 $
$=36000\left(\frac{115}{100}\right)^2 $
$=47610$
Hence, Amount $= Rs. 47610 .$

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Question 63 Marks

What principal will amount to $Rs.15729$ in two years, if the rate of interest for successive years are $5\%$ and $7\%$ respectively, the interest is being compounded annually.

Answer

Given :
Amount $=Rs. 15729, n=2$ years, $r_1=5$ and $r_2=7 \%$
$ A=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)$
$\Rightarrow 15729=P\left(1+\frac{5}{100}\right)\left(1+\frac{7}{100}\right)$
$\Rightarrow 15729=P\left(\frac{105}{100}\right)\left(\frac{107}{100}\right)$
$\Rightarrow P=\frac{15729 \times 100 \times 100}{105 \times 107}$
$\Rightarrow P=\text { Rs. } 14000 .$

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MATHEMATICS [3 marks sum]

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