Simple and Compound Interest — Exercise 8.1

Find the simple interest on ₹4,000 at 7.5% p.a. for 3 years 3 months. Also find the amount.

Answer

Given:

Principal (P) = ₹ 4,000

Rate of interest (R) = 7.5% p.a.

Time (T) = 3 years 3 months

= $3\dfrac{3}{12} \text{ years }= 3\dfrac{1}{4} \text{ years }= \dfrac{13}{4} \text{ years }$

As we know:

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$\Rightarrow \text{S.I.} = ₹ \dfrac{4000 \times 7.5 \times \dfrac{13}{4}}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{4000 \times 7.5 \times 13}{100 \times 4} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{390000}{400} \\[1em] \Rightarrow \text{S.I.} = ₹ 975$

Amount = Principal + Interest

= ₹ 4,000 + ₹ 975

= ₹ 4,975

Hence, simple interest = ₹ 975 and amount = ₹ 4,975.

What sum of money will yield ₹170.10 as simple interest in 2 years 3 months at 6% per annum?

Answer

Given:

Simple interest (I) = ₹ 170.10

Rate of interest (R) = 6% p.a.

Time (T) = 2 years 3 months

= $2\dfrac{3}{12} \text{ years }= 2\dfrac{1}{4} \text{ years }= \dfrac{9}{4} \text{ years }$

As we know:

$\text{P} = \dfrac{\text{I} \times 100}{\text{R} \times \text{T}}$

Substituting the values, we get

$\Rightarrow \text{P} = ₹ \dfrac{170.10 \times 100}{6 \times \dfrac{9}{4}} \\[1em] \Rightarrow \text{P} = ₹ \dfrac{170.10 \times 100 \times 4}{6 \times 9} \\[1em] \Rightarrow \text{P} = ₹ \dfrac{68040}{54} \\[1em] \Rightarrow \text{P} = ₹ 1{,}260$

Hence, the required sum of money = ₹ 1,260.

Find the rate of interest when ₹800 fetches ₹130 as simple interest in 2 years 6 months.

Answer

Given:

Principal (P) = ₹ 800

Simple interest (I) = ₹ 130

Time (T) = 2 years 6 months

= $2\dfrac{6}{12} \text{ years }= 2\dfrac{1}{2} \text{ years }= \dfrac{5}{2} \text{ years }$

As we know:

$\text{R} = \dfrac{\text{I} \times 100}{\text{P} \times \text{T}}$%

Substituting the values, we get

$\Rightarrow \text{R} = \dfrac{130 \times 100}{800 \times \dfrac{5}{2}}$%

$\Rightarrow \text{R} = \dfrac{130 \times 100 \times 2}{800 \times 5}$%

$\Rightarrow \text{R} = \dfrac{26000}{4000}$%

⇒ R = 6.5%

Hence, the rate of interest = 6.5% p.a.

Find the time when simple interest on ₹3.3 lakhs at 6.5% per annum is ₹75,075.

Answer

Given:

Principal (P) = ₹ 3.3 lakhs = ₹ 3,30,000

Rate of interest (R) = 6.5% p.a.

Simple interest (I) = ₹ 75,075

As we know:

$\text{T} = \dfrac{\text{I} \times 100}{\text{P} \times \text{R}}$

Substituting the values, we get

$\Rightarrow \text{T} = \dfrac{75075 \times 100}{330000 \times 6.5} \\[1em] \Rightarrow \text{T} = \dfrac{7507500}{2145000} \\[1em] \Rightarrow \text{T} = \dfrac{7}{2} \\[1em] \Rightarrow \text{T} = 3\dfrac{1}{2}$

Hence, the required time = $\mathbf{3\dfrac{1}{2}}$ years i.e. 3 years 6 months.

Find the sum of money when

simple interest at $7\dfrac{1}{4}$% p.a. for $2\dfrac{1}{2}$ years is ₹2,356.25

Answer

Given:

Simple interest (I) = ₹ 2,356.25

Rate of interest (R) = $7\dfrac{1}{4}$% p.a. = $\dfrac{29}{4}$% p.a.

Time (T) = $2\dfrac{1}{2}$ years

= $\dfrac{5}{2} \text{ years }$

As we know:

$\text{P} = \dfrac{\text{I} \times 100}{\text{R} \times \text{T}}$

Substituting the values, we get

$\Rightarrow \text{P} = ₹ \dfrac{2356.25 \times 100}{\dfrac{29}{4} \times \dfrac{5}{2}} \\[1em] \Rightarrow \text{P} = ₹ \dfrac{2356.25 \times 100 \times 4 \times 2}{29 \times 5} \\[1em] \Rightarrow \text{P} = ₹ \dfrac{1885000}{145} \\[1em] \Rightarrow \text{P} = ₹ 13{,}000$

Hence, the required sum of money = ₹ 13,000.

Find the sum of money when

final amount is ₹11,300 at 4% p.a. for 3 years 3 months

Answer

Given:

Amount (A) = ₹ 11,300

Rate of interest (R) = 4% p.a.

Time (T) = 3 years 3 months

= $3\dfrac{3}{12} \text{ years }= 3\dfrac{1}{4} \text{ years }= \dfrac{13}{4} \text{ years }$

As we know:

$\text{A} = \Big(1 + \dfrac{\text{R} \times \text{T}}{100}\Big)\text{P}$

Substituting the values, we get

$\Rightarrow 11300 = \Big(1 + \dfrac{4 \times \dfrac{13}{4}}{100}\Big)\text{P} \\[1em] \Rightarrow 11300 = \Big(1 + \dfrac{4 \times 13}{100 \times 4}\Big)\text{P} \\[1em] \Rightarrow 11300 = \Big(1 + \dfrac{13}{100}\Big)\text{P} \\[1em] \Rightarrow 11300 = \dfrac{113}{100}\text{P} \\[1em] \Rightarrow \text{P} = \dfrac{11300 \times 100}{113} \\[1em] \Rightarrow \text{P} = ₹ 10{,}000$

Hence, the required sum of money = ₹ 10,000.

How long will it take a certain sum of money to triple itself at $13\dfrac{1}{3}$% per annum simple interest?

Answer

Given:

Rate of interest (R) = $13\dfrac{1}{3}$% p.a. = $\dfrac{40}{3}$% p.a.

Let the sum of money (principal) be ₹ P.

As the sum of money triples itself,

Amount (A) = ₹ 3P

Interest = Amount - Principal

= ₹ 3P - ₹ P

= ₹ 2P

As we know:

$\text{T} = \dfrac{\text{I} \times 100}{\text{P} \times \text{R}}$

Substituting the values, we get

$\Rightarrow \text{T} = \dfrac{2\text{P} \times 100}{\text{P} \times \dfrac{40}{3}} \\[1em] \Rightarrow \text{T} = \dfrac{2\text{P} \times 100 \times 3}{\text{P} \times 40} \\[1em] \Rightarrow \text{T} = \dfrac{600}{40} \\[1em] \Rightarrow \text{T} = 15$

Hence, the required time = 15 years.

At a certain rate of simple interest ₹4,050 amounts to ₹4,576.50 in 2 years. At the same rate of simple interest, how much would ₹1 lakh amount to in 3 years?

Answer

Given:

Principal (P) = ₹ 4,050

Amount (A) = ₹ 4,576.50

Time (T) = 2 years

Interest = Amount - Principal

= ₹ 4,576.50 - ₹ 4,050

= ₹ 526.50

As we know:

$\text{R} = \dfrac{\text{I} \times 100}{\text{P} \times \text{T}}$%

Substituting the values, we get

$\Rightarrow \text{R} = \dfrac{526.50 \times 100}{4050 \times 2}$%

$\Rightarrow \text{R} = \dfrac{52650}{8100}$%

⇒ R = 6.5%

So, the rate of interest = 6.5% p.a.

Now, for the second case :

Principal (P) = ₹ 1 lakh = ₹ 1,00,000

Rate of interest (R) = 6.5% p.a.

Time (T) = 3 years

As we know:

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$\Rightarrow \text{S.I.} = ₹ \dfrac{100000 \times 6.5 \times 3}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{1950000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 19{,}500$

Amount = Principal + Interest

= ₹ 1,00,000 + ₹ 19,500

= ₹ 1,19,500

Hence, ₹ 1 lakh will amount to ₹ 1,19,500 in 3 years.

What sum of money invested at 7.5% p.a. simple interest for 2 years produces twice as much interest as ₹9,600 in 3 years 6 months at 10% p.a. simple interest?

Answer

Given:

For the second sum :

Principal (P) = ₹ 9,600

Rate of interest (R) = 10% p.a.

Time (T) = 3 years 6 months

= $3\dfrac{6}{12} \text{ years }= 3\dfrac{1}{2} \text{ years }= \dfrac{7}{2} \text{ years }$

As we know:

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$\Rightarrow \text{S.I.} = ₹ \dfrac{9600 \times 10 \times \dfrac{7}{2}}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{9600 \times 10 \times 7}{100 \times 2} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{672000}{200} \\[1em] \Rightarrow \text{S.I.} = ₹ 3{,}360$

Twice of this interest = 2 × ₹ 3,360 = ₹ 6,720

Let the money invested at 7.5% p.a. for 2 years be ₹ $x$.

According to given information,

$\Rightarrow \dfrac{x \times 7.5 \times 2}{100} = 6720 \\[1em] \Rightarrow \dfrac{15x}{100} = 6720 \\[1em] \Rightarrow x = \dfrac{6720 \times 100}{15} \\[1em] \Rightarrow x = \dfrac{672000}{15} \\[1em] \Rightarrow x = 44{,}800$

Hence, the required sum of money = ₹ 44,800.