Simple Interest — Exercise 11

Find the S.I. and the amount on:

₹ 150 for 4 years at 5% per year.

Answer

By formula,

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.

Given,

P = ₹ 150, R = 5% and T = 4 years.

S.I. = $\dfrac{150 \times 5 \times 4}{100}$

= $\dfrac{3000}{100}$

= ₹ 30

Amount = P + S.I. = 150 + 30 = ₹ 180.

Hence, S.I. = ₹ 30 and Amount = ₹ 180.

Find the S.I. and the amount on:

₹ 350 for $3\dfrac{1}{2}$ years at 8% p.a.

Answer

By formula,

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.

Given,

P = ₹ 350, R = 8% and T = $3\dfrac{1}{2} \text{ years } = \dfrac{7}{2}$ years.

S.I. = $\dfrac{350 \times 8 \times 7}{100 \times 2}$

= $\dfrac{19600}{200}$

= ₹ 98

Amount = P + S.I. = 350 + 98 = ₹ 448.

Hence, S.I. = ₹ 98 and Amount = ₹ 448.

Find the S.I. and the amount on:

₹ 620 for 4 months at 8 paise per rupee per month.

Answer

By formula,

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.

Given,

P = ₹ 620, R = 8 paise per rupee per month = 8% per month and T = 4 months.

Since the rate is per month, the time is taken in months.

S.I. = $\dfrac{620 \times 8 \times 4}{100}$

= $\dfrac{19840}{100}$

= ₹ 198.40

Amount = P + S.I. = 620 + 198.40 = ₹ 818.40.

Hence, S.I. = ₹ 198.40 and Amount = ₹ 818.40.

Find the S.I. and the amount on:

₹ 3,380 for 30 months at $4\dfrac{1}{2}$% p.a.

Answer

By formula,

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.

Given,

P = ₹ 3,380, R = $4\dfrac{1}{2}$% = $\dfrac{9}{2}$% and T = 30 months = $\dfrac{30}{12}$ years = $\dfrac{5}{2}$ years.

S.I. = $\dfrac{3380 \times 9 \times 5}{100 \times 2 \times 2}$

= $\dfrac{152100}{400}$

= ₹ 380.25

Amount = P + S.I. = 3,380 + 380.25 = ₹ 3,760.25.

Hence, S.I. = ₹ 380.25 and Amount = ₹ 3,760.25.

Find the S.I. and the amount on:

₹ 600 from July 12 to Dec. 5 at 10% p.a.

Answer

By formula,

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.

Given,

P = ₹ 600, R = 10% p.a.

The starting date (July 12) is not included and the last date (Dec. 5) is included.

Time = (31 − 12) + 31 + 30 + 31 + 30 + 5 days

= 19 + 31 + 30 + 31 + 30 + 5 = 146 days = $\dfrac{146}{365}$ years = $\dfrac{2}{5}$ years.

S.I. = $\dfrac{600 \times 10 \times 2}{100 \times 5}$

= $\dfrac{12000}{500}$

= ₹ 24

Amount = P + S.I. = 600 + 24 = ₹ 624.

Hence, S.I. = ₹ 24 and Amount = ₹ 624.

Find the S.I. and the amount on:

₹ 850 from 10^th March to 3^rd August at $2\dfrac{1}{2}$% p.a.

Answer

By formula,

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.

Given,

P = ₹ 850, R = $2\dfrac{1}{2}$% = $\dfrac{5}{2}$% p.a.

The starting date (10^th March) is not included and the last date (3^rd August) is included.

Time = (31 − 10) + 30 + 31 + 30 + 31 + 3 days

= 21 + 30 + 31 + 30 + 31 + 3 = 146 days

= $\dfrac{146}{365} \text { years }$

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

S.I. = $\dfrac{850 \times 5 \times 2}{100 \times 2 \times 5}$

= $\dfrac{8500}{1000}$

= ₹ 8.50

Amount = P + S.I. = 850 + 8.50 = ₹ 858.50.

Hence, S.I. = ₹ 8.50 and Amount = ₹ 858.50.

Find the S.I. and the amount on:

₹ 225 for 3 years 9 months at 16% p.a.

Answer

By formula,

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.

Given,

P = ₹ 225, R = 16% and T = 3 years 9 months = $3\dfrac{9}{12} \text { years } = \dfrac{15}{4}$ years.

S.I. = $\dfrac{225 \times 16 \times 15}{100 \times 4}$

= $\dfrac{54000}{400}$

= ₹ 135

Amount = P + S.I. = 225 + 135 = ₹ 360.

Hence, S.I. = ₹ 135 and Amount = ₹ 360.

On what sum of money does the S.I. for 10 years at 5% become ₹ 1,600?

Answer

S.I. = ₹ 1,600, T = 10 years and R = 5%.

$\text{ P }= \dfrac{\text{S.I.} \times 100}{\text{R} \times \text{T}}\\[1em] = \dfrac{1600 \times 100}{5 \times 10}\\[1em] = \dfrac{160000}{50}\\[1em] = ₹\ 3,200$

Hence, the required sum = ₹ 3,200.

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

Answer

P = ₹ 2,000, Amount = ₹ 2,330 and R = 11% p.a.

I = Amount − P = 2,330 − 2,000 = ₹ 330.

$\text{ T }= \dfrac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \dfrac{330 \times 100}{2000 \times 11}\\[1em] = \dfrac{33000}{22000}\\[1em] = \dfrac{3}{2} = 1\dfrac{1}{2}\ \text{years}$

Hence, the required time = $1\dfrac{1}{2}$ years.

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

Answer

Let the sum (principal) be ₹ P and R = 8% p.a.

When the money doubles itself, Amount = 2P.

So, I = Amount − P = 2P − P = ₹ P.

$\text{ T }= \dfrac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \dfrac{\text{P} \times 100}{\text{P} \times 8}\\[1em] = \dfrac{100}{8}\\[1em] = \dfrac{25}{2} = 12\dfrac{1}{2}\ \text{years}$

Hence, the required time = $12\dfrac{1}{2}$ years.

In how many years will ₹ 870 amount to ₹ 1,044, the rate of interest being $2\dfrac{1}{2}$% p.a.?

Answer

P = ₹ 870, Amount = ₹ 1,044 and R = $2\dfrac{1}{2}$% = $\dfrac{5}{2}$% p.a.

I = Amount − P = 1,044 − 870 = ₹ 174.

$\text{ T }= \dfrac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \dfrac{174 \times 100 \times 2}{870 \times 5}\\[1em] = \dfrac{34800}{4350}\\[1em] = 8\ \text{years}$

Hence, the required time = 8 years.

Find the rate percent, if the S.I. on ₹ 275 in 2 years is ₹ 22.

Answer

P = ₹ 275, T = 2 years and S.I. = ₹ 22.

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

$= \dfrac{22 \times 100}{275 \times 2}$%

$= \dfrac{2200}{550}$%

= 4%

Hence, the required rate = 4% p.a.

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

Answer

Let the sum be ₹ P, R = 8% p.a. and T = 5 years.

I = $\dfrac{\text{P} \times 8 \times 5}{100} = \dfrac{40\text{P}}{100} = \dfrac{2\text{P}}{5}$.

Amount = P + I

$\Rightarrow 700 = \text{P} + \dfrac{2\text{P}}{5}\\[1em] \Rightarrow 700 = \dfrac{7\text{P}}{5}\\[1em] \Rightarrow \text{P} = \dfrac{700 \times 5}{7} = ₹\ 500$

Hence, the required sum = ₹ 500.

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

Answer

P = ₹ 3,750, Amount = ₹ 4,650 and T = 4 years.

I = Amount − P = 4,650 − 3,750 = ₹ 900.

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

$= \dfrac{900 \times 100}{3750 \times 4}$%

$= \dfrac{90000}{15000}$%

= 6%

Hence, the required rate = 6% p.a.

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

Answer

First, find the rate of interest.

P = ₹ 6,000, Amount = ₹ 8,000 and T = 4 years.

I = Amount − P = 8,000 − 6,000 = ₹ 2,000.

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

$= \dfrac{2000 \times 100}{6000 \times 4}$%

$= \dfrac{200000}{24000}$%

= $\dfrac{25}{3}$%

Now, for the second case, P = ₹ 525, Amount = ₹ 700 and R = $\dfrac{25}{3}$%.

I = Amount − P = 700 − 525 = ₹ 175.

$\text{ T }= \dfrac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \dfrac{175 \times 100 \times 3}{525 \times 25}\\[1em] = \dfrac{52500}{13125}\\[1em] = 4\text{ years}$

Hence, the required time = 4 years.

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

Answer

Let the sum (principal) be ₹ P and T = $2\dfrac{1}{2} \text { years }= \dfrac{5}{2}$ years.

I = $\dfrac{4}{5}$ of the sum = ₹ $\dfrac{4\text{P}}{5}$.

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

$= \dfrac{\dfrac{4\text{P}}{5} \times 100 \times 2}{\text{P} \times 5}$%

$= \dfrac{4 \times 100 \times 2}{5 \times 5}$%

$= \dfrac{800}{25}$%

= 32%

Hence, the required rate = 32% p.a.

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

Answer

First, find the interest on ₹ 900 at 4% for 5 years.

$\text{ I }= \dfrac{900 \times 4 \times 5}{100}\\[1em] = \dfrac{18000}{100}\\[1em] = ₹\ 180$

Now, let the required sum be ₹ P, lent at R = 5% for T = 3 years, giving I = ₹ 180.

$\text{ P }= \dfrac{\text{I} \times 100}{\text{R} \times \text{T}}\\[1em] = \dfrac{180 \times 100}{5 \times 3}\\[1em] = \dfrac{18000}{15}\\[1em] = ₹\ 1,200$

Hence, the required sum = ₹ 1,200.

A sum of ₹ 1,780 becomes ₹ 2,136 in 4 years. Find:

(i) the rate of interest.

(ii) the sum that will become ₹ 810 in 7 years at the same rate of interest.

Answer

(i) P = ₹ 1,780, Amount = ₹ 2,136 and T = 4 years.

I = Amount − P = 2,136 − 1,780 = ₹ 356.

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

$= \dfrac{356 \times 100}{1780 \times 4}$%

$= \dfrac{35600}{7120}$%

= 5%

Hence, the rate of interest is 5% p.a.

(ii) Let the required sum be ₹ P, R = 5% and T = 7 years.

I = $\dfrac{\text{P} \times 5 \times 7}{100} = \dfrac{35\text{P}}{100} = \dfrac{7\text{P}}{20}$.

Amount = P + I

$\Rightarrow 810 = \text{P} + \dfrac{7\text{P}}{20}\\[1em] \Rightarrow 810 = \dfrac{27\text{P}}{20}\\[1em] \Rightarrow \text{P} = \dfrac{810 \times 20}{27} = ₹\ 600$

Hence, the required sum is ₹ 600.

A sum amounts to ₹ 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) Let the sum be ₹ P, R = 5% p.a. and T = 6 years.

I = $\dfrac{\text{P} \times 5 \times 6}{100} = \dfrac{30\text{P}}{100} = \dfrac{3\text{P}}{10}$.

Amount = P + I

$\Rightarrow 2652 = \text{P} + \dfrac{3\text{P}}{10}\\[1em] \Rightarrow 2652 = \dfrac{13\text{P}}{10}\\[1em] \Rightarrow \text{P} = \dfrac{2652 \times 10}{13} = ₹\ 2,040$

Hence, the sum is ₹ 2,040.

(ii) When the sum doubles itself, I = Amount − P = 2P − P = ₹ P and R = 5%.

$\text{ T }= \dfrac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \dfrac{\text{P} \times 100}{\text{P} \times 5}\\[1em] = \dfrac{100}{5}\\[1em] = 20\ \text{years}$

Hence, the time in which the sum will double itself is 20 years.

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

Answer

Let the rate of interest be R% p.a. and T = 4 years.

For P : Principal = ₹ 36,000, Time = 4 years

I = $\dfrac{36000 \times \text{R} \times 4}{100} = ₹\ 1,440\text{R}$.

For Q : Principal = ₹ 25,000, Time = 4 years

I = $\dfrac{25000 \times \text{R} \times 4}{100} = ₹\ 1,000\text{R}$.

Given, P gets ₹ 3,080 more interest than Q.

⇒ 1,440 R - 1,000 R = 3,080

⇒ 440 R = 3,080

$\Rightarrow \text{R} = \dfrac{3080}{440}$

⇒ R = 7

⇒ R% = 7%

Hence, the required rate of interest = 7% p.a.

A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.

Answer

Let the sum (money lent) be ₹ P and T = 5 years.

Interest (I) = one-fourth of the money lent = ₹ $\dfrac{\text{P}}{4}$.

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

R% $= \dfrac{\dfrac{\text{P}}{4} \times 100}{\text{P} \times 5}$%

R% $= \dfrac{100}{4 \times 5}$%

R% $= \dfrac{100}{20}$%

R% = 5%

R = 5

Hence, R = 5 and rate of interest = 5%.

The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.

Answer

Let the sum be ₹ P and T = 5 years.

I = 30% of the sum = $\dfrac{30}{100} \times \text{P} = \dfrac{3\text{P}}{10}$.

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

$= \dfrac{\dfrac{3\text{P}}{10} \times 100}{\text{P} \times 5}$%

$= \dfrac{3 \times 100}{10 \times 5}$%

$= \dfrac{300}{50}$%

= 6%

Hence, the required rate of interest = 6% p.a.