[5 marks sum]

Question 15 Marks

The compound interest, calculated yearly, on a certain sum of money for the second year is Rs. 1320 and for the third year is Rs. 1452. Calculate the rate of interest and the original sum of money.

Answer

Compound Interest for second year = ₹ 1320
Compound Interest for third year = ₹ 1452
r = ?, P = ?
For third year: P = ₹ 1320,
I = 1452 - 1320 = ₹ 132
T = 1 year
$\therefore I=\frac{ P \times R \times T }{100}$
$132=\frac{1320 \times R \times 1}{100}$
$R=\frac{132 \times 100}{1320}$
R = 10% p.a.
Now, let the principal for first year be x.
Then I $=\frac{ P \times R \times T }{100}=\frac{10 x}{100}=\frac{x}{100}$
Amount for first year = x + $\frac{x}{100}=\frac{11 x}{100}$

For second year:
$P=\frac{11 x}{10}$, R = 10% p.a., T = 1 year
I = $=\frac{\frac{11 x}{10} \times 10 \times 1}{100}=\frac{11 x}{100}$
From question,
$\frac{11 x}{100}=1320$
⇒ x = $\frac{1320 \times 100}{11}$
= 12000
Thus the original sum of money is ₹12,000

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Question 25 Marks

A sum of money amounts to ₹ 2,240 at 4 % p.a., simple interest in 3 years. Find the interest on the same sum for 6 months at $3 \frac{1}{2} \%$ p.a.

Answer

Let P = ₹ 100
S.I. $=\frac{\text { PRT }}{100}$
= ₹ $\frac{100 \times 4 \times 3}{100}$
= ₹ 12

A = P + S.I.
= ₹ 100 + ₹ 12
= ₹ 112If A is ₹ 112 then P is ₹ 100

If A is ₹ 1 then P is ₹ $\frac{100}{112}$

If A is ₹ 2,240 then P be $\frac{100}{112} \times 2,240$
= ₹ 100 x 20
= ₹ 2,000

So, principal is ₹ 2,000
Now, for second case
P = ₹ 2,000

Time = 6 months = $\frac{1}{2}$ year
$ R =3 \frac{1}{2}=\frac{7}{2} \% \text { p.a. }$
$2,000 \times \frac{7}{2} \times \frac{1}{2}$
∴ S.I. = ₹ $\frac{2,000 \times \frac{7}{2} \times \frac{1}{2}}{100}$
= ₹ $\frac{2,000 \times 7 \times 1}{100 \times 2 \times 2}$
= ₹ $\frac{20 \times 7}{4}$
= ₹ 35

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Question 35 Marks

A sum of money is lent out at compound interest for two years at 20% p.a., being reckoned yearly. If the same sum of the money was lent Gut at compound interest of the same rate of percent per annum C.I., being reckoned half yearly would have fetched Rs. 482 more by way of interest. Calculate the sum of money lent out.

Answer

Let the principal be ₹ 100
For first case r = 20% p.a.
n = 2 years

$A = P \left(1+\frac{r}{100}\right)^n$
= ₹ $\left(1+\frac{20}{100}\right)^2$
= ₹ $\left(1+\frac{1}{5}\right)^2$
= ₹ $\left(\frac{6}{5}\right)^2$
= ₹ $\frac{100 \times 6 \times 6}{5 \times 5}$
= ₹ $\frac{3,600}{25}$
= ₹ 144∴ C.I. = A - P
= ₹ 144 - ₹ 100
= ₹ 44

For second case
r = 20% p.a. $=\frac{20}{2} \%$ half yearly = 10% semi annual
Time = 2 years = 4 half years
$A = P \left(1+\frac{r}{100}\right)^n$
= ₹100 $\left(1+\frac{10}{100}\right)^4$
= ₹100 $\left(1+\frac{1}{10}\right)^4$
= ₹100 $\left(\frac{11}{10}\right)^4$
= ₹ $\frac{100 \times 11 \times 11 \times 11 \times 11}{10 \times 10 \times 10 \times 10}$
= ₹ $\frac{121 \times 121}{10 \times 10}$
= ₹ $\frac{14,641}{100}$
= ₹ 146·41

C.I. = A - P
= 146·41 - 100
= ₹ 46·41Difference between two interest
= ₹ 46·41 - ₹ 44·00
= ₹ 2·41
If difference is ₹2·41 then principal be ₹100
If difference is ₹482 then principal will be
= ₹$\frac{100}{2 \cdot 41} \times 482$
= ₹ $\frac{100 \times 482}{241} \times 100$

= ₹ 100 x 100 x 2
= ₹ 20,000
∴ Sum is ₹ 20,000

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Question 45 Marks

Find the difference between the simple interest and compound interest on 2,500 for 2 years at 4% p.a., compound interest being reckoned semi-annualy.

Answer

P = ₹ 2,500
r = 4% p.a. $=\frac{4}{2}$ = 2% half yearly
T = 2 year = 4 half years

$A = P \left(1+\frac{r}{100}\right)^n$
= ₹ 2,500 $\left(1+\frac{2}{100}\right)^4$
= ₹ 2,500 $\left(1+\frac{1}{50}\right)^4$
= ₹ 2,500 $\left(\frac{51}{50}\right)^4$
= ₹ $\frac{2,500 \times 51 \times 51 \times 51 \times 51}{50 \times 50 \times 50 \times 50}$
= ₹ $\frac{51 \times 51 \times 51 \times 51}{50 \times 50}$
= ₹ $\frac{2,601 \times 2,601}{2,500}$
= ₹ $\frac{67,65,201}{2,500}$
= ₹ 2,706.08

C.I. = A - P
= ₹ (2,706·08 - ₹ 2,500)
= ₹ 206.08
and S.I. = $\frac{ PRT }{100}$ = ₹ $\frac{2,500 \times 4 \times 2}{100}=$ ₹ 200

Difference between C.I. and S.I.
= C.I. - S.I.
= ₹ 206·08 - ₹ 200
= ₹ 6.08

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