[3 marks sum]

Question 13 Marks

What is the rate of interest, if $₹3,750$ amounts to $₹4,650$ in $4$ years?

Answer

Principal $(P)= ₹ 3750$
Amount $(A)= ₹ 4650$
$\therefore$ S.I. $=A-P= ₹ 4650 - 3750 = ₹ 900$
$\text { Time }(T)=4 \text { years }$
$\therefore \text { Rate }=\frac{\text { S.I } \times 100}{P \times T}$
$=\frac{900 \times 100}{3750 \times 4}=6 \% \text { p.a. }$

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Question 23 Marks

In how many years will be ₹870 amount to ₹1,044, the rate of interest being 2 % p.a?

Answer

Principal $(P)=$₹ 870
Amount $(A)=$₹ 1044
$\therefore$ S.I. $=P-A=$ ₹ 1044 - ₹ 870= ₹ 174
Rate $(\mathrm{R})=2 \frac{1}{2}=\frac{5}{2} \%$ p.a.
$\therefore$ Time $=\frac{\text { S.I. } \times 100}{\mathrm{P} \times \mathrm{R}}$
$=\frac{174 \times 100 \times 2}{870 \times 5}=8$ years

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Question 33 Marks

In what time will a sum of money double itself at $8\%$ p.a $?$

Answer

Let the principal $(P)= ₹ 100$
$\therefore$ Amount $(A)= ₹ 100 \times 2 = ₹ 200$
$\therefore$ S.I. $=A-P= ₹ 200 - ₹ 100 = ₹ 100$
$\text { Rate }( R )=8 \% \text { p.a. } $
$\therefore \text { Time }=\frac{\text { S.I. } \times 100}{ P \times R } $
$ =\frac{100 \times 100}{100 \times 8} $
$ =\frac{25}{2}=12 \frac{1}{2} \text { years }$

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Question 43 Marks

Find the time in which Rs. 2,000 will amount to Rs. 2,330 at 11% p.a.?

Answer

Amount $(A)=$ Rs. 2330
Principal $(P)=$ Rs. 2000
$ \therefore \text { S.I. }=\mathrm{A}-\mathrm{P} $
= Rs. 2330 - Rs. 2000
$ =\text { Rs. } 330 $
$ R=11 \% \text { p.a. } $
$ \therefore \text { Time }=\frac{\text { S.I. } \times 100}{\mathrm{P} \times \mathrm{R}}=\frac{330 \times 100}{2000 \times 11} $
$=\frac{3}{2}=1 \frac{1}{2}$ years

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Question 53 Marks

Find the S.I. and amount on: Rs. 225 for 3 years 9 months at 16% p.a.

Answer

$\begin{aligned} & P=\text { Rs. } 225, R=16 \% \text { p.a. T }=3 \text { years } 9 \text { months } \\ & =3 \frac{9}{12}=3 \frac{3}{4} \text { years }=\frac{15}{4} \text { years } \\ & \therefore \text { S.I. }=\frac{\text { P.R.T. }}{100}=\frac{225 \times 16 \times 15}{100 \times 4}=\text { Rs. } 135 \\ & \therefore \text { Amount }=\text { P }+ \text { S.I. } \\ & =\text { Rs. } 225+\text { Rs. } 135 \\ & =\text { Rs. } 360\end{aligned}$

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Question 63 Marks

Find the S.I. and amount on: Rs. 3,380 for 30 months at $4 \frac{1}{2} \%$ p.a.

Answer

Principal $(P)=$ Rs. 3380
Rate $=4 \frac{1}{2} \%$ p.a. $=\frac{9}{2} \%$
Period $=30$ months $=\frac{30}{12}$ years
S.I. $=\frac{\text { PRT }}{100}=\frac{3380 \times 9 \times 30}{100 \times 2 \times 12}$
$=$ Rs. $\frac{1521}{4}=$ Rs. 380.25
Amount $=$ P + S.I.
$=$ Rs. $3380+380.25=$ Rs. 3760.25

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Question 73 Marks

What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years?

Answer

In second case, Principal $(P)=$ Rs. 900
Rate $(R)=4 \%$, Time $(T)=5$ years
$ \begin{aligned} & \therefore \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} \\ & =\frac{900 \times 4 \times 5}{100}=\text { Rs. } 180 \end{aligned} $
In first case, S.I. = Rs. 180
Rate $=5 \%$, Time $=3$ years
$ \begin{aligned} & \therefore \text { Sum }=\frac{\text { S.I. } \times 100}{\mathrm{R} \times \mathrm{T}} \\ & =\frac{180 \times 100}{5 \times 3}=\text { Rs. } 1200 \end{aligned} $

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Question 83 Marks

The interest on a sum of money at the end of $2 \frac{1}{2}$ years is $\frac{4}{5}$ of the sum. What is the rate per cent?

Answer

$\begin{aligned} & \mathrm{T}=2 \frac{1}{2} \text { years } \\ & =\frac{5}{2} \text { years } \\ & \text { S.I. }=\frac{4}{5} \mathrm{P} \\ & \mathrm{R}=\frac{\mathrm{S} . \mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}} \\ & =\frac{4 \mathrm{P} \times 100 \times 2}{5 \times \mathrm{P} \times 5} \\ & =32 \%\end{aligned}$

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MATHS [3 marks sum]

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