Simple Interest - Exercise 11(A)

Find the simple interest and amount on each of the following :

(i) ₹ 8400 for 4 years at 8% per annum.

(ii) ₹ 50000 for 3 years $12\dfrac{1}{2}$% per annum.

(iii) ₹ 9275 for 2 years at $7\dfrac{1}{2}$% per annum.

Answer

(i)

Given:

P = ₹ 8400, R = 8%, T = 4 years

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{ 8400 \times 8 \times 4}{100}\Big) \\[1em] = ₹ \Big(\dfrac{ 84 \times 8 \times 4}{1}\Big) \quad \text{[Dividing 8400 and 100 by 100]} \\[1em] = ₹ 84 \times 32 \\[1em] = ₹ 2688$

Amount = Principal + S.I.

Amount = ₹ 8400 + ₹ 2688

Amount = ₹ 11088

S.I. = ₹ 2688, Amount = ₹ 11088

(ii)

Given:

P = ₹ 50000, T = 3 years

R = $12\dfrac{1}{2}$% = $\dfrac{25}{2}$%

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \left(\dfrac{ 50000 \times \dfrac{25}{2} \times 3}{100}\right) \\[1em] = ₹ \Big(\dfrac{ 50000 \times 25 \times 3}{100 \times 2}\Big) \\[1em] = ₹ \Big(\dfrac{ 500 \times 25 \times 3}{1 \times 2}\Big) \quad \text{[Dividing 50000 and 100 by 100]} \\[1em] = ₹ \Big(\dfrac{ 250 \times 25 \times 3}{1 \times 1}\Big) \quad \text{[Dividing 500 and 2 by 2]} \\[1em] = ₹ 250 \times 25 \times 3 \\[1em] = ₹ 250 \times 75 \\[1em] = ₹ 18750$

Amount = Principal + S.I.

Amount = ₹ 50000 + ₹ 18750

Amount = ₹ 68750

S.I. = ₹ 18750, Amount = ₹ 68750

(iii)

Given:

P = ₹ 9275, T = 2 years

R = $7\dfrac{1}{2}$% = $\dfrac{15}{2}$%

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \left(\dfrac{ 9275 \times \dfrac{15}{2} \times 2}{100}\right) \\[1em] = ₹ \Big(\dfrac{ 9275 \times 15 \times 2}{100 \times 2}\Big) \\[1em] = ₹ \Big(\dfrac{ 371 \times 15 \times 2}{4 \times 2}\Big) \quad \text{[Dividing 9275 and 100 by 25]} \\[1em] = ₹ \Big(\dfrac{ 371 \times 15 \times 1}{4 \times 1}\Big) \\[1em] = ₹ \dfrac{5565}{4} \\[1em] = ₹ 1391.25$

Amount = Principal + S.I.

Amount = ₹ 9275 + ₹ 1391.25

Amount = ₹ 10666.25

S.I. = ₹ 1391.25, Amount = ₹ 10666.25

In what time will ₹ 7500 amount to ₹ 8625, if simple interest is reckoned at $7\dfrac{1}{2}$% per annum.

Answer

Given:

Principal = ₹ 7500, Amount = ₹ 8625

R = $7\dfrac{1}{2}$% = $\dfrac{15}{2}$%

S.I. = Amount - Principal

S.I. = ₹ 8625 - ₹ 7500 = ₹ 1125

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow T = \dfrac{S.I. \times 100}{P \times R} \\[1em] = \left(\dfrac{1125 \times 100}{7500 \times \dfrac{15}{2}}\right) \text{ years} \\[1em] = \Big(\dfrac{1125 \times 100 \times 2}{7500 \times 15}\Big) \text{ years} \\[1em] = \Big(\dfrac{1125 \times 1 \times 2}{75 \times 15}\Big) \text{ years} \quad \text{[Dividing 7500 and 100 by 100]} \\[1em] = \Big(\dfrac{15 \times 1 \times 2}{1 \times 15}\Big) \text{ years} \quad \text{[Dividing 1125 and 75 by 75]} \\[1em] = \Big(\dfrac{1 \times 1 \times 2}{1 \times 1}\Big) \text{ years} \\[1em] = 2 \text{ years}$

∴ Time = 2 years

In what time will ₹ 3600 amount to ₹ 4320 at 4% per annum simple interest?

Answer

Given:

P = ₹ 3600, A = ₹ 4320, R = 4%

S.I. = ₹ 4320 - ₹ 3600 = ₹ 720

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow T = \dfrac{S.I. \times 100}{P \times R} \\[1em] = \Big(\dfrac{720 \times 100}{3600 \times 4}\Big) \text{ years} \\[1em] = \Big(\dfrac{720 \times 1}{36 \times 4}\Big) \text{ years} \quad \text{[Dividing 100 and 3600 by 100]} \\[1em] = \Big(\dfrac{20 \times 1}{1 \times 4}\Big) \text{ years} \quad \text{[Dividing 720 and 36 by 36]} \\[1em] = \Big(\dfrac{5 \times 1}{1 \times 1}\Big) \text{ years} \quad \text{[Dividing 20 and 4 by 4]} \\[1em] = 5 \text{ years}$

∴ Time = 5 years

At what rate per cent per annum will ₹ 6300 yield an interest of ₹ 2100 in 4 years?

Answer

Given:

P = ₹ 6300, S.I. = ₹ 2100, T = 4 years

Then

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

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

$= \Big(\dfrac{2100 \times 100}{6300 \times 4}\Big)$%

$= \Big(\dfrac{2100 \times 1}{63 \times 4}\Big)$% $\quad$ [Dividing 100 and 6300 by 100]

$= \Big(\dfrac{525 \times 1}{63 \times 1}\Big)$% $\quad$ [Dividing 2100 and 4 by 4]

$= \Big(\dfrac{525}{63}\Big)$%

$= \dfrac{25}{3}$% $\quad$ [Dividing 525 and 63 by 21]

$= 8\dfrac{1}{3}$%

∴ Rate = $\bold {8\dfrac{1}{3}}$%

At what rate per cent per annum simple interest will ₹ 66000 amount to ₹ 72720 in 2 years?

Answer

Given:

P = ₹ 66000, A = ₹ 72720, T = 2 years

S.I. = ₹ 72720 - ₹ 66000 = ₹ 6720

Then

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

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

$= \Big(\dfrac{6720 \times 100}{66000 \times 2}\Big)$%

$= \Big(\dfrac{6720 \times 1}{660 \times 2}\Big)$% $\quad$ [Dividing 100 and 66000 by 100]

$= \Big(\dfrac{3360 \times 1}{660 \times 1}\Big)$% $\quad$ [Dividing 6720 and 2 by 2]

$= \dfrac{56}{11}$% $\quad$ [Dividing 3360 and 660 by 60]

$= 5\dfrac{1}{11}$%

∴ Rate = $\mathbf{5\dfrac{1}{11}}$% per annum

The simple interest on a sum of money for 5 years is $\dfrac{3}{5}$ of the sum. Find the rate per cent per annum.

Answer

Given:

S.I. for 5 years is $\dfrac{3}{5}$ of the sum.

R = ?

Let the sum (P) be x.

S.I. = $\dfrac{3}{5}x$ and T = 5 years.

Then

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

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

$= \left(\dfrac{(\dfrac{3}{5}x) \times 100}{x \times 5}\right)$%

$= \Big(\dfrac{3x \times 100}{5 \times x \times 5}\Big)$%

$= \dfrac{300x}{25x}$%

$= 12$%

∴ Rate = 12% per annum

The simple interest on a certain sum for 3 years at 10% per annum is ₹ 829.50. Find the sum.

Answer

Given:

S.I. = ₹ 829.50, R = 10%, T = 3 years.

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow P = \dfrac{S.I. \times 100}{R \times T} \\[1em] = ₹ \Big(\dfrac{829.50 \times 100}{10 \times 3}\Big) \\[1em] = ₹ \dfrac{82950}{30} \\[1em] = ₹ 2765$

∴ Sum = ₹ 2765

What sum will yield an interest of ₹ 7840 in 2 years at $6\dfrac{1}{4}$% per annum?

Answer

Given:

S.I. = ₹ 7840, T = 2 years

R = $6\dfrac{1}{4}$% = $\dfrac{25}{4}$%

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow P = \dfrac{S.I. \times 100}{R \times T} \\[1em] = ₹ \left(\dfrac{7840 \times 100}{\dfrac{25}{4} \times 2}\right) \\[1em] = ₹ \Big(\dfrac{7840 \times 100 \times 4}{25 \times 2}\Big) \\[1em] = ₹ \Big(\dfrac{7840 \times 100 \times 4}{50}\Big) \\[1em] = ₹ \Big(\dfrac{7840 \times 2 \times 4}{1}\Big) \\[1em] = ₹ 7840 \times 2 \times 4 \\[1em] = ₹ 62720$

∴ Sum = ₹ 62720

At what rate per cent per annum simple interest will a sum treble itself in 16 years?

Answer

Given:

T = 16 years

Let the Principal (P) be x.

Treble means Amount (A) = 3x.

S.I. = Amount - Principal

S.I. = 3x - x = 2x

Then

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

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

$= \Big(\dfrac{2x \times 100}{x \times 16}\Big)$%

$= \Big(\dfrac{200x}{16x}\Big)$%

$= 12\dfrac{1}{2}$%

∴ Rate = $\bold{12\dfrac{1}{2}}$%

A sum when reckoned at 6% per annum simple interest amounts to ₹ 4130 in 3 years. Find the sum.

Answer

Given:

A = ₹ 4130, T = 3 years, R = 6%

We have the formula:

$P = \dfrac{100 \times A}{100 + (R \times T)} \\[1em] = \dfrac{100 \times 4130}{100 + (6 \times 3)} \\[1em] = \dfrac{100 \times 4130}{100 + 18} \\[1em] = \dfrac{413000}{118} \\[1em] = ₹ 3500$

∴ Sum = ₹ 3500

A sum of money put at 11% per annum simple interest amounts to ₹ 10370 in 2 years. What will it amount to in 3 years at the same rate?

Answer

Let the sum be ₹ 100.

Given:

R = 11% p.a., T = 2 years.

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em]= ₹ \Big(\dfrac{100 \times 11 \times 2}{100}\Big) \\[1em] = ₹ \Big(\dfrac{1 \times 11 \times 2}{1}\Big) \\[1em] = ₹ 11 \times 2 \\[1em] = ₹ 22$

Amount = S.I. + Principal

Amount = ₹ 22 + ₹ 100

Amount = ₹ 122

If Amount is ₹ 122, then Sum = ₹ 100.

If Amount is ₹ 1, then Sum = ₹ $\dfrac{100}{122}$.

If Amount is ₹ 10370, then Sum =

$\phantom{=} ₹ \Big(\dfrac{100}{122} \times 10370\Big) \\[1em] = ₹ \Big(\dfrac{50}{61} \times 10370\Big) \\[1em] = ₹ \Big(\dfrac{518500}{61}\Big) \\[1em] = ₹ 8500$

Now, sum = ₹ 8500, R = 11% p.a., T = 3 years.

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{8500 \times 11 \times 3}{100}\Big) \\[1em] = ₹ \Big(\dfrac{85 \times 11 \times 3}{1}\Big) \\[1em] = ₹ 85 \times 11 \times 3 \\[1em] = ₹ 2805$

Required Amount = S.I. + Principal

Required Amount = ₹ 2805 + ₹ 8500 = ₹ 11305

∴ Amount = ₹ 11305

A sum of money put at 9% per annum simple interest amounts to ₹ 10160 in 3 years. What will it amount to in 2 years at 8% per annum?

Answer

Let the sum be ₹ 100.

Given: R = 9% p.a., T = 3 years.

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{100 \times 9 \times 3}{100}\Big) \\[1em] = ₹ \Big(\dfrac{1 \times 9 \times 3}{1}\Big) \\[1em] = ₹ 9 \times 3 \\[1em] = ₹ 27$

Amount = S.I. + Principal

Amount = ₹ 27 + ₹ 100

Amount = ₹ 127

If Amount is ₹ 127, then Sum = ₹ 100.

If Amount is ₹ 1, then Sum = ₹ $\dfrac{100}{127}$.

If Amount is ₹ 10160, then Sum =

$\phantom{=} ₹ \Big(\dfrac{100}{127} \times 10160\Big) \\[1em] = ₹ \Big(\dfrac{1016000}{127}\Big) \\[1em] = ₹ 8000$

Now, sum = ₹ 8000, R = 8% p.a., T = 2 years.

And

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{8000 \times 8 \times 2}{100}\Big) \\[1em] = ₹ \Big(\dfrac{80 \times 8 \times 2}{1}\Big) \\[1em] = ₹ 80 \times 8 \times 2 \\[1em] = ₹ 1280$

Required Amount = S.I. + Principal

Required Amount = ₹ 1280 + ₹ 8000 = ₹ 9280

∴ Amount = ₹ 9280