[3 marks sum]

Question 13 Marks

Mr. Kumar borrowed $Rs. 15,000$ for two years. The rate of interest for the two successive years are $8\%$ and $10\%$ respectively. If he repays $Rs. 6,200$ at the end of the first year, find the outstanding amount at the end of the second year.

Answer

$P = ₹ 15,000$
Interest for $1^{st}$ year
$=\frac{15,000 \times 8 \times 1}{100}$
$= ₹ 1,200$
Amount after one year
$= ₹ (15,000 + 1,200)$
$= ₹ 16,200$
He repays ₹ $6,200$ at the end of the $1^{st}$ year
$\therefore$ Principal for $2^{nd}$ year
$= ₹ (16,200 - 6,200)$
$= ₹ 10,000$
Now interest for the 2nd year
$=\frac{10,000 \times 10 \times 1}{100}$
$= ₹1,000$
$\therefore $ Amount outstanding at the end of 2nd year
$= ₹ (10,000 + 1,000)$
$= ₹ 11,000$

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Question 23 Marks

The $S.I$. and $C.I$. on a sum of money for $2$ years is Rs. $200$ and $210$ respectively. If the rate of interest is the same. Find the sum and rate.

Answer

Let the sum be $P$ and rate of interest be $r \%$ then.
S.I. $=\frac{ P \times r \times n }{100}$
$200=\frac{ P \times r \times 2}{100}$
$P.r = 10000 ........(i)$
$C.I. = P \left[\left(1+\frac{r}{100}\right)^n-1\right]$
$\begin{aligned} & 210=\frac{10000}{r}\left[\left(1+\frac{r}{100}\right)^2-1\right] \\ & 210=\frac{10000}{r} \frac{(100+r)^2-100^2}{(100)^2}\end{aligned}$
$210=\frac{r^2+200 r}{r}$
$r^2 + 200 r = 210 r$
$r^2 = 10 r$
$r = 10%$
Using this in equation $(1)$, we get
$P=\frac{10000}{10}$
$= ₹ 1000$
$\therefore P = ₹ 1,000$
$r = 10%$

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Mathematics [3 marks sum]

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