[4 marks sum]

Question 14 Marks

In 4 years, ₹6,000 amount to ₹8,000. In what time will ₹525 amount to ₹700 at the same rate?

Answer

In first case, principal $(P)=$₹ 6000
Amount $(A)$= ₹ 8000
$ \therefore \text { S.I. }=A-P=$ ₹ 8000 - ₹ 6000= ₹ 2000
Time $(T)=4$ years
$ \begin{aligned} & \therefore \mathrm{R}=\frac{\text { S.I. } \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{2000 \times 100}{6000 \times 4} \\ & =\frac{25}{3} \%=8 \frac{1}{3} \% \text { p.a. } \end{aligned} $
In second case, Principal $(P)=$ ₹ 525
Amount $(A)=$ ₹ 700
$ \therefore \text { S.I. }=A-P=$ ₹ 700 - ₹ 525= ₹ 175
Rate $(R)=\frac{25}{3} \%$ of p.a.
$ \begin{aligned} & \therefore \text { Time }=\frac{\text { S.I. } \times 100}{\mathrm{P} \times \mathrm{R}} \\ & =\frac{\text { Rs. } 175 \times 100 \times 3}{525 \times 25}=4 \text { years } \end{aligned} $

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Question 24 Marks

Find the sum which will amount to ₹700 in 5 years at 8% rate p.a.

Answer

Amount ₹ 700, Rate $(R)=8 \%$ p.a.
Time $(\mathrm{T})=5$ years
Let principal $(P)=$ ₹ 100
then S.I. $=\frac{\text { P.R.T. }}{100}=\frac{100 \times 8 \times 5}{100}=$ ₹ 40
$\therefore$ Amount $(A)=P+$ S.I.
= ₹ 100 $+ 40=$ ₹ 140
If amount is ₹ 140 , then principal = ₹ $100$
and, if amount is Rs. 700 , then principal
= ₹$\frac{100 \times 700}{140}= $₹ 500

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Question 34 Marks

Find the S.I. and amount on: Rs. 850 from 10th March to 3rd August at $2 \frac{1}{2} \%$ p.a.

Answer

$ P=\text { Rs. } 850, R=2 \frac{1}{2} \%=\frac{5}{2} \% \text { p.a. } $
$T=10$ th march to 3rd Aug.

March	= 21 Days
April	= 30 Days
May	= 31 Days
June	= 30 Days
July	= 31 Days
Aug.	= 03 Days
Total	146 Days

$\begin{aligned} & =\frac{146}{365} \text { years }=\frac{2}{5} \text { years } \\ & \therefore \text { S.I. }=\frac{\text { P.R.T }}{100}=\frac{850 \times 5 \times 2}{100 \times 2 \times 5}=\frac{850}{100}\end{aligned}$
$\therefore$ Amount $=\mathrm{P}+$ S.I.
$=$ Rs. $850+$ Rs. 8.50
$=$ Rs. 858.50

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Question 44 Marks

Find the S.I. and amount on: 600 from July 12 to Dec. 5 at 10% p.a.

Answer

$\begin{aligned} & P=\text { Rs. } 600, R=10 \% \text { p.a. } \\ & T=\text { July } 12 \text { to } D \text { ec. } 5\end{aligned}$

July	= 19 Days
Aug.	= 31 Days
Sep.	= 30 Days
Oct.	= 31 Days
Nov.	= 30 Days
Dec.	= 05 Days
Total	146 Days

$\begin{aligned} & =\frac{146}{365} \text { years }=\frac{2}{5} \text { years } \\ & \therefore \text { S.I. }=\frac{\text { P.R.T }}{100}=\frac{600 \times 10 \times 2}{100 \times 5}=\text { Rs. } 24 \\ & \therefore \text { Amount }=\text { P + S.I. } \\ & =\text { Rs. } 600+\text { Rs. } 24 \\ & =\text { Rs. } 624\end{aligned}$

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Question 54 Marks

P and Q invest Rs. 36,000 and Rs. 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets Rs. 3,080 more interest than Q, find the rate of interest.

Answer

P's investment $\left(P_1\right)=$ Rs. 36000
and Q's investment $\left(P_2\right)=$ Rs. 25000
Period $(T)=4$ years, Let rate of interest $=x \%$
P's interest $=$ Rs. $\frac{36000 \times \mathrm{x} \times 4}{100}$
$=$ Rs. $1440 x \quad\left(\because\right.$ S.I. $\left.=\frac{\text { PRT }}{100}\right)$
$ \text { and Q's interest }=\frac{25000 \times \mathrm{x} \times 4}{100}=\text { Rs. } 1000 \mathrm{x} $
Difference in their interest
$ \text { = Rs. }(1440-1000) x=\text { Rs. } 440 x $
But the difference $=$ Rs. 3080
$ \therefore 440 x=3080 \Rightarrow x=\frac{3080}{440} \Rightarrow x=7 \% $
$\therefore$ Rate of interest $=7 \%$ p.a.

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Question 64 Marks

A sum of Rs. 1,780 become Rs. 2,136 in 4 years, Find:
(i) the rate of interest.
(ii) the sum that will become Rs. 810 in 7 years at the same rate of interest?

Answer

(i) In the first case, Principal $(P)=$ Rs. 1780
Amount $(A)=$ Rs. 2136
$\therefore$ S.I. $=$ A $-\mathrm{P}=$ Rs. $2136-1780=$ Rs. 356
Time $(T)=4$ years
$\therefore$ Rate $=\frac{\text { S.I. } \times 100}{\mathrm{P} \times \mathrm{T}}$
$=\frac{356 \times 100}{1780 \times 4}=5 \%$ p.a
(ii) In second case, Let principal $(P)=$ Rs. 100
Rate $(R)=5 \%$ p.a., Time $(T)=7$ years
$ \begin{aligned} & \therefore \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} \\ & =\frac{100 \times 5 \times 7}{100}=\text { Rs. } 35 \end{aligned} $
$\therefore$ Amount $=$ P + S.I. $=$ Rs. $100+35=$ Rs. 135
If the amount is Rs. 135 , then the principal = Rs. 100 and if the amount is Rs. 810 , then the principal.
$ =\text { Rs. } \frac{100 \times 810}{135}=\text { Rs. } 600 $

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Question 74 Marks

A sum amounts to Rs. 2,652 in 6 years at 5% p.a. simple interest. Find:
(i) the sum
(ii) the time in which the same sum will double itself at the same rate of interest.

Answer

(i) In First case, Let principal $(P)=$ rs. 100
Rate $(R)=5 \%$ p.a., Time $(T)=6$ years
$ \begin{aligned} & \therefore \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} \\ & =\frac{100 \times 5 \times 6}{100}=\text { Rs. } 30 \end{aligned} $
and, amount $=$ Rs. $100+$ Rs. $30=$ Rs. 130
If amount is Rs. 130 , then principal $=$ Rs. 100
and, if amount is Rs. 2652, then principal
$ =\frac{100 \times 2652}{130}=\text { Rs. } 2040 $
In second case, Let sum $(P)=$ Rs. 100
Amount $(A)=$ Rs. $100 \times 2=$ Rs. 200
$ \text { S.I. }=\text { A - P = Rs. } 200-100=\text { Rs. } 100 $
Rate $=5 \%$ p.a.
$ \text { Time }=\frac{\text { S.I. } \times 100}{P \times R}=\frac{100 \times 100}{100 \times 5}=20 \text { years } $

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