Simple Interest and Compound Interest — Exercise 8(A)

Find the simple interest and amount on :

₹ 4,500 for $2\dfrac{1}{2}$ years at $7\dfrac{2}{3}$% per annum

Answer

Here,

P = ₹ 4,500, T = $2\dfrac{1}{2}$ years = $\dfrac{5}{2}$ years and R = $7\dfrac{2}{3}$% p.a. = $\dfrac{23}{3}$% p.a.

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

$= ₹ \left(4500 \times \dfrac{23}{3} \times \dfrac{5}{2} \times \dfrac{1}{100}\right)$ = ₹ 862.50

Amount = Principal + S.I. = ₹ (4,500 + 862.50) = ₹ 5,362.50

Hence, simple interest = ₹ 862.50 and amount = ₹ 5,362.50.

Find the simple interest and amount on :

₹ 6,360 for 6 years 3 months at 8% per annum

Answer

Here,

P = ₹ 6,360, T = 6 years 3 months = $6\dfrac{3}{12}$ years = $\dfrac{25}{4}$ years and R = 8% p.a.

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

$= ₹ \left(6360 \times 8 \times \dfrac{25}{4} \times \dfrac{1}{100}\right)$ = ₹ 3,180

Amount = Principal + S.I. = ₹ (6,360 + 3,180) = ₹ 9,540

Hence, simple interest = ₹ 3,180 and amount = ₹ 9,540.

Find the simple interest and amount on :

₹ 19,200 for 11 months at $9\dfrac{3}{4}$% per annum

Answer

Here,

P = ₹ 19,200, T = 11 months = $\dfrac{11}{12}$ years and R = $9\dfrac{3}{4}$% p.a. = $\dfrac{39}{4}$% p.a.

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

$= ₹ \left(19200 \times \dfrac{39}{4} \times \dfrac{11}{12} \times \dfrac{1}{100}\right)$ = ₹ 1,716

Amount = Principal + S.I. = ₹ (19,200 + 1,716) = ₹ 20,916

Hence, simple interest = ₹ 1,716 and amount = ₹ 20,916.

Find the simple interest and amount on :

₹ 58,400 for 75 days at $6\dfrac{1}{2}$% per annum

Answer

Here,

P = ₹ 58,400, T = 75 days = $\dfrac{75}{365}$ years = $\dfrac{15}{73}$ years and R = $6\dfrac{1}{2}$% p.a. = $\dfrac{13}{2}$% p.a.

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

$= ₹ \left(58400 \times \dfrac{13}{2} \times \dfrac{15}{73} \times \dfrac{1}{100}\right)$ = ₹ 780

Amount = Principal + S.I. = ₹ (58,400 + 780) = ₹ 59,180

Hence, simple interest = ₹ 780 and amount = ₹ 59,180.

Find the simple interest on ₹ 8,600 from 18th October, 2006 to 13th March, 2007 at 8% per annum. Also find the amount.

Answer

The day on which the money is borrowed is not counted while the day on which it is returned is counted.

Time = (13 + 30 + 31 + 31 + 28 + 13) days = 146 days

= $\dfrac{146}{365}$ years = $\dfrac{2}{5}$ years

Also, P = ₹ 8,600 and R = 8% p.a.

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

$= ₹ \left(8600 \times 8 \times \dfrac{2}{5} \times \dfrac{1}{100}\right)$ = ₹ 275.20

Amount = Principal + S.I. = ₹ (8,600 + 275.20) = ₹ 8,875.20

Hence, simple interest = ₹ 275.20 and amount = ₹ 8,875.20.

Ashish lent ₹ 10,500 to Sunidhi at 7% per annum simple interest. After 5 years, Sunidhi discharged the debt by giving a watch and ₹ 13,000 in cash. What is the value of the watch?

Answer

Here,

P = ₹ 10,500, R = 7% p.a. and T = 5 years.

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

$= ₹ \left(\dfrac{10500 \times 7 \times 5}{100}\right)$ = ₹ 3,675

Amount = Principal + S.I. = ₹ (10,500 + 3,675) = ₹ 14,175

So, Sunidhi has to pay ₹ 14,175 in all.

Value of the watch = Amount − Cash paid = ₹ (14,175 − 13,000) = ₹ 1,175

Hence, the value of the watch is ₹ 1,175.

In what time will the simple interest on ₹ 7,560 be ₹ 1,102.50 at $6\dfrac{1}{4}$% per annum?

Answer

Here,

P = ₹ 7,560, S.I. = ₹ 1,102.50 and R = $6\dfrac{1}{4}$% p.a. = $\dfrac{25}{4}$% p.a.

$\text{T} = \dfrac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \dfrac{100 \times 1102.50}{7560 \times \dfrac{25}{4}}\\[1em] = 100 \times 1102.50 \times \dfrac{1}{7560} \times \dfrac{4}{25}\\[1em] = \dfrac{441000}{189000}\\[1em] = \dfrac{7}{3}$

So, T = $\dfrac{7}{3}$ years = $2\dfrac{1}{3}$ years = 2 years 4 months.

Hence, the required time is 2 years 4 months.

In how much time will ₹ 25,600 amount to ₹ 35,664, when money is worth $9\dfrac{1}{4}$% per annum simple interest?

Answer

Here,

P = ₹ 25,600, A = ₹ 35,664 and R = $9\dfrac{1}{4}$% p.a. = $\dfrac{37}{4}$% p.a.

S.I. = A − P = ₹ (35,664 − 25,600) = ₹ 10,064

$\text{T} = \dfrac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \dfrac{100 \times 10064}{25600 \times \dfrac{37}{4}}\\[1em] = 100 \times 10064 \times \dfrac{1}{25600} \times \dfrac{4}{37}\\[1em] = \dfrac{4025600}{947200}\\[1em] = \dfrac{17}{4}$

So, T = $\dfrac{17}{4}$ years = $4\dfrac{1}{4}$ years = 4 years 3 months.

Hence, the required time is 4 years 3 months.

At what rate per cent per annum will ₹ 1,625 amount to ₹ 2,080 in $3\dfrac{1}{2}$ years?

Answer

Here,

P = ₹ 1,625, A = ₹ 2,080 and T = $3\dfrac{1}{2}$ years = $\dfrac{7}{2}$ years.

S.I. = A − P = ₹ (2,080 − 1,625) = ₹ 455

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

$= \dfrac{100 \times 455}{1625 \times \dfrac{7}{2}}$%

$= \left(100 \times 455 \times \dfrac{1}{1625} \times \dfrac{2}{7}\right)$%

$= \dfrac{91000}{11375}$%

= 8%

Hence, the required rate is 8% p.a.

At what rate per cent per annum will the simple interest on ₹ 6,720 be ₹ 1,911 in 3 years 3 months?

Answer

Here,

P = ₹ 6,720, S.I. = ₹ 1,911 and T = 3 years 3 months = $3\dfrac{3}{12}$ years = $\dfrac{13}{4}$ years.

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

$= \dfrac{100 \times 1911}{6720 \times \dfrac{13}{4}}$%

$= \left(100 \times 1911 \times \dfrac{1}{6720} \times \dfrac{4}{13}\right)$%

$= \dfrac{764400}{87360}$%

$= \dfrac{35}{4}$%

So, R = $\dfrac{35}{4}$% p.a. = $8\dfrac{3}{4}$% p.a.

Hence, the required rate is $\mathbf{8\dfrac{3}{4}}$% p.a.

At what rate per cent of simple interest will a sum of money double itself in 12 years?

Answer

Let the sum be ₹ $x$. Then,

P = ₹ $x$, A = ₹ $2x$, T = 12 years

S.I. = A − P = ₹ $(2x - x) = ₹ x$

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

$= \dfrac{100 \times x}{x \times 12}$%

$= \dfrac{100}{12}$%

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

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

Hence, the required rate is $\mathbf{8\dfrac{1}{3}}$% p.a.

Simple interest on a certain sum is $\dfrac{9}{16}$ of the sum. Find the rate per cent and the time if both are numerically equal.

Answer

Let the sum be ₹ $x$. Then, S.I. = ₹ $\left(\dfrac{9}{16}x\right)$.

Let the rate be R% per annum and the time be R years.

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow \dfrac{9x}{16} = \dfrac{x \times \text{R} \times \text{R}}{100}\\[1em] \Rightarrow \text{R}^2 = \dfrac{9}{16} \times 100\\[1em] \Rightarrow \text{R}^2 = \dfrac{900}{16}\\[1em] \Rightarrow \text{R} = \sqrt{\dfrac{900}{16}}\\[1em] = \dfrac{\sqrt{900}}{\sqrt{16}}\\[1em] = \dfrac{30}{4}\\[1em] = \dfrac{15}{2}$

So, R = $\dfrac{15}{2} = 7\dfrac{1}{2}$

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

Hence, rate = $\mathbf{7\dfrac{1}{2}}$% p.a. and time = $\mathbf{7\dfrac{1}{2}}$ years.

What sum will yield ₹ 406 as simple interest in 1 year 2 months at $6\dfrac{1}{4}$% per annum?

Answer

Here,

S.I. = ₹ 406, T = 1 year 2 months = $1\dfrac{2}{12}$ years = $\dfrac{7}{6}$ years and R = $6\dfrac{1}{4}$% p.a. = $\dfrac{25}{4}$% p.a.

$\text{P} = \dfrac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \dfrac{100 \times 406}{\dfrac{25}{4} \times \dfrac{7}{6}}\\[1em] = 100 \times 406 \times \dfrac{4}{25} \times \dfrac{6}{7}\\[1em] = \dfrac{974400}{175}\\[1em] = 5568$

Hence, the required sum is ₹ 5,568.

What sum will amount to ₹ 1,748 in $2\dfrac{1}{2}$ years at $7\dfrac{1}{2}$% per annum simple interest?

Answer

Let the sum be ₹ $x$.

Then, P = ₹ $x$, T = $2\dfrac{1}{2}$ years = $\dfrac{5}{2}$ years and R = $7\dfrac{1}{2}$% p.a. = $\dfrac{15}{2}$% p.a.

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

$= ₹ \left(x \times \dfrac{15}{2} \times \dfrac{5}{2} \times \dfrac{1}{100}\right) = ₹ \left(\dfrac{3x}{16}\right)$

Amount = P + S.I. = ₹ $\left(x + \dfrac{3x}{16}\right) = ₹ \left(\dfrac{19x}{16}\right)$

But, amount = ₹ 1,748.

$\Rightarrow \dfrac{19x}{16} = 1748\\[1em] \Rightarrow x = 1748 \times \dfrac{16}{19}\\[1em] \Rightarrow x = 1472$

Hence, the required sum is ₹ 1,472.

A sum of money becomes $\dfrac{8}{5}$ of itself in 5 years at a certain rate of simple interest. Find the rate of interest.

Answer

Let the sum be ₹ P. Then, A = ₹ $\left(\dfrac{8}{5}\text{P}\right)$ and T = 5 years.

S.I. = A − P = ₹ $\left(\dfrac{8\text{P}}{5} - \text{P}\right) = ₹ \left(\dfrac{3\text{P}}{5}\right)$

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow \dfrac{3\text{P}}{5} = \dfrac{\text{P} \times \text{R} \times 5}{100}\\[1em] \Rightarrow \dfrac{3\text{P}}{5} = \dfrac{\text{PR}}{20}\\[1em] \Rightarrow \text{R} = \dfrac{3 \times 20}{5}\\[1em] \Rightarrow \text{R} = 12$

Hence, the rate of interest is 12% p.a.

What sum of money lent at $12\dfrac{1}{2}$% per annum will produce the same interest in 4 years as ₹ 8,560 produces in 5 years at 12% per annum?

Answer

For ₹ 8,560, we have P = ₹ 8,560, R = 12% p.a. and T = 5 years.

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

$= ₹ \left(\dfrac{8560 \times 12 \times 5}{100}\right)$ = ₹ 5,136

Now, S.I. = ₹ 5,136, R = $12\dfrac{1}{2}$% p.a. = $\dfrac{25}{2}$% p.a. and T = 4 years.

$\text{P} = \dfrac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \dfrac{100 \times 5136}{\dfrac{25}{2} \times 4}\\[1em] = 100 \times 5136 \times \dfrac{2}{25} \times \dfrac{1}{4}\\[1em] = \dfrac{1027200}{100}\\[1em] = 10272$

Hence, the required sum is ₹ 10,272.

If ₹ 1,250 amount to ₹ 1,550 in 3 years at simple interest, what will ₹ 3,200 amount to in 4 years at the same rate?

Answer

Here,

P = ₹ 1,250, A = ₹ 1,550 and T = 3 years.

S.I. = A − P = ₹ (1,550 − 1,250) = ₹ 300

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

$= \dfrac{100 \times 300}{1250 \times 3}$%

$= \dfrac{30000}{3750}$%

= 8%

So, the rate of interest is 8% p.a.

Now, P = ₹ 3,200, R = 8% p.a. and T = 4 years.

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

$= ₹ \left(\dfrac{3200 \times 8 \times 4}{100}\right)$ = ₹ 1,024

Amount = Principal + S.I. = ₹ (3,200 + 1,024) = ₹ 4,224

Hence, ₹ 3,200 will amount to ₹ 4,224 in 4 years.

A sum of money lent at simple interest amounts to ₹ 3,224 in 2 years and ₹ 4,160 in 5 years. Find the sum and the rate of interest.

Answer

S.I. for 3 years = (Amount after 5 years) − (Amount after 2 years)

= ₹ (4,160 − 3,224)

= ₹ 936

S.I. for 1 year = ₹ $\left(\dfrac{936}{3}\right)$ = ₹ 312

S.I. for 2 years = ₹ (312 × 2) = ₹ 624

∴ Sum = (Amount after 2 years) − (S.I. for 2 years)

= ₹ (3,224 − 624)

= ₹ 2,600

Now, P = ₹ 2,600, S.I. = ₹ 624 and T = 2 years.

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

= $\left(\dfrac{100 \times 624}{2600 \times 2}\right)$% p.a.

= 12% p.a.

Hence, the required sum is ₹ 2,600 and the rate of interest is 12% p.a.

A lends ₹ 2,500 to B and a certain sum to C at the same time at 7% per annum simple interest. If, after 4 years, A altogether receives ₹ 1,120 as interest from B and C, find the sum lent to C.

Answer

Interest earned from B = ₹ $\left(\dfrac{2500 \times 7 \times 4}{100}\right)$ = ₹ 700

∴ Interest earned from C = ₹ (1,120 − 700) = ₹ 420

Now, for C, we have S.I. = ₹ 420, R = 7% p.a. and T = 4 years.

$\text{P} = \dfrac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \dfrac{100 \times 420}{7 \times 4}\\[1em] = \dfrac{42000}{28}\\[1em] = 1500$

Hence, the sum lent to C is ₹ 1,500.

The simple interest on a certain sum for 3 years at 8% per annum is ₹ 96 more than the simple interest on the same sum for 2 years at 9% per annum. Find the sum.

Answer

Let the sum be ₹ $x$. Then,

S.I. on ₹ $x$ for 3 years at 8% p.a. = ₹ $\left(\dfrac{x \times 8 \times 3}{100}\right) = ₹ \left(\dfrac{24x}{100}\right)$

S.I. on ₹ $x$ for 2 years at 9% p.a. = ₹ $\left(\dfrac{x \times 9 \times 2}{100}\right) = ₹ \left(\dfrac{18x}{100}\right)$

Since the simple interest for 3 years at 8% is ₹ 96 more than the simple interest for 2 years at 9%,

$\Rightarrow \dfrac{24x}{100} - \dfrac{18x}{100} = 96\\[1em] \Rightarrow \dfrac{6x}{100} = 96\\[1em] \Rightarrow x = \dfrac{96 \times 100}{6}\\[1em] \Rightarrow x = 1600$

Hence, the required sum is ₹ 1,600.

Two equal sums of money were lent at simple interest at 11% p.a. for $3\dfrac{1}{2}$ years and $4\dfrac{1}{2}$ years respectively. If the difference in interests for two periods was ₹ 412.50, find each sum.

Answer

Let each sum be ₹ P.

S.I. for $4\dfrac{1}{2}$ years = ₹ $\left(\dfrac{\text{P} \times 11 \times 9}{2 \times 100}\right)$

S.I. for $3\dfrac{1}{2}$ years = ₹ $\left(\dfrac{\text{P} \times 11 \times 7}{2 \times 100}\right)$

Since the difference between the two interests is ₹ 412.50,

$\Rightarrow \dfrac{\text{P} \times 11 \times 9}{2 \times 100} - \dfrac{\text{P} \times 11 \times 7}{2 \times 100} = 412.50\\[1em] \Rightarrow \dfrac{11\text{P}}{200} \times (9 - 7) = 412.50\\[1em] \Rightarrow \dfrac{11\text{P}}{200} \times 2 = 412.50\\[1em] \Rightarrow \dfrac{11\text{P}}{100} = 412.50\\[1em] \Rightarrow \text{P} = \dfrac{412.50 \times 100}{11}\\[1em] \Rightarrow \text{P} = 3750$

Hence, each sum is ₹ 3,750.

Divide ₹ 6,000 into two parts so that the simple interest on the first part for 9 months at 12% per annum is equal to the simple interest on the second part for $1\dfrac{1}{2}$ years at 10% per annum.

Answer

Let the first part be ₹ $x$.

Then, second part = ₹ $(6000 - x)$.

S.I. on ₹ $x$ at 12% p.a. for 9 months, i.e. $\dfrac{3}{4}$ years

= ₹ $\left(x \times 12 \times \dfrac{3}{4} \times \dfrac{1}{100}\right) = ₹ \left(\dfrac{9x}{100}\right)$

S.I. on ₹ $(6000 - x)$ at 10% p.a. for $1\dfrac{1}{2}$ years, i.e. $\dfrac{3}{2}$ years

= ₹ $\bigg\lbrace (6000 - x) \times 10 \times \dfrac{3}{2} \times \dfrac{1}{100} \bigg\rbrace = ₹ \bigg\lbrace \dfrac{15(6000 - x)}{100}\bigg\rbrace$

Since both the interests are equal,

$\Rightarrow \dfrac{9x}{100} = \dfrac{15(6000 - x)}{100}\\[1em] \Rightarrow 9x = 90000 - 15x\\[1em] \Rightarrow 24x = 90000\\[1em] \Rightarrow x = 3750$

Second part = ₹ (6,000 − 3,750) = ₹ 2,250

Hence, the two parts are ₹ 3,750 and ₹ 2,250.

Divide a sum of ₹ 13,500 into two parts such that if one part be lent at $8\dfrac{1}{3}$% per annum for 2 years 9 months and the other at $7\dfrac{1}{2}$% per annum for 1 year 8 months, the total interest received is ₹ 2,375.

Answer

Let the first part be ₹ $x$.

Then, second part = ₹ $(13500 - x)$.

2 years 9 months = $2\dfrac{9}{12}$ years = $\dfrac{11}{4}$ years

1 year 8 months = $1\dfrac{8}{12}$ years = $\dfrac{5}{3}$ years

S.I. on ₹ $x$ at $\dfrac{25}{3}$% p.a. for $\dfrac{11}{4}$ years

S.I. = ₹ $\left(x \times \dfrac{25}{3} \times \dfrac{11}{4} \times \dfrac{1}{100}\right) = ₹ \left(\dfrac{11x}{48}\right)$

S.I. on ₹ $(13500 - x)$ at $\dfrac{15}{2}$% p.a. for $\dfrac{5}{3}$ years

S.I. = ₹ $\bigg\lbrace (13500 - x) \times \dfrac{15}{2} \times \dfrac{5}{3} \times \dfrac{1}{100}\bigg\rbrace = ₹ \bigg\lbrace \dfrac{13500 - x}{8}\bigg\rbrace$

Since the total interest received is ₹ 2,375,

$\Rightarrow \dfrac{11x}{48} + \dfrac{13500 - x}{8} = 2375\\[1em] \Rightarrow 11x + 6(13500 - x) = 2375 \times 48\\[1em] \Rightarrow 11x + 81000 - 6x = 114000\\[1em] \Rightarrow 5x = 33000\\[1em] \Rightarrow x = 6600$

Second part = ₹ (13,500 − 6,600) = ₹ 6,900

Hence, the two parts are ₹ 6,600 and ₹ 6,900.

In what time will a sum of money lent at $8\dfrac{1}{3}$% simple interest become 4 times of itself.

Answer

Let the sum be ₹ $x$. Then,

P = ₹ $x$, A = ₹ $4x$, R = $8\dfrac{1}{3}$% p.a. = $\dfrac{25}{3}$% p.a.

S.I. = A − P = ₹ $(4x - x) = ₹ 3x$

$\text{T} = \dfrac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \dfrac{100 \times 3x}{x \times \dfrac{25}{3}}\\[1em] = 100 \times 3x \times \dfrac{1}{x} \times \dfrac{3}{25}\\[1em] = \dfrac{900}{25}\\[1em] = 36$

Hence, the sum will become 4 times of itself in 36 years.

A certain sum of money lent out at $6\dfrac{2}{3}$% p.a. produces the same simple interest in 6 years as ₹ 3,200 lent out at $8\dfrac{2}{5}$% p.a. for 7 years. Find the sum.

Answer

Let the required sum be ₹ P.

S.I. on ₹ P at $\dfrac{20}{3}$% p.a. for 6 years = ₹ $\left(\dfrac{\text{P} \times 20 \times 6}{3 \times 100}\right)$

S.I. on ₹ 3,200 at $\dfrac{42}{5}$% p.a. for 7 years = ₹ $\left(\dfrac{3200 \times 42 \times 7}{5 \times 100}\right)$

Since the two simple interests are equal,

$\Rightarrow \dfrac{\text{P} \times 20 \times 6}{3 \times 100} = \dfrac{3200 \times 42 \times 7}{5 \times 100}\\[1em] \Rightarrow \dfrac{2\text{P}}{5} = \dfrac{940800}{500}\\[1em] \Rightarrow \dfrac{2\text{P}}{5} = \dfrac{9408}{5}\\[1em] \Rightarrow 2\text{P} = 9408\\[1em] \Rightarrow \text{P} = 4704$

Hence, the required sum is ₹ 4,704.

Naveen and Praveen borrowed ₹ 42,000 and ₹ 55,000 respectively for $3\dfrac{1}{2}$ years at the same rate of interest. If Praveen has to pay ₹ 3,640 more than Naveen, find the rate of interest.

Answer

Let the rate of interest be R% p.a. and T = $3\dfrac{1}{2}$ years = $\dfrac{7}{2}$ years.

Interest paid by Naveen = ₹ $\left(42000 \times \text{R} \times \dfrac{7}{2} \times \dfrac{1}{100}\right)$ = ₹ 1,470R

Interest paid by Praveen = ₹ $\left(55000 \times \text{R} \times \dfrac{7}{2} \times \dfrac{1}{100}\right)$ = ₹ 1,925R

Since Praveen has to pay ₹ 3,640 more interest than Naveen,

$\Rightarrow 1925\text{R} - 1470\text{R} = 3640\\[1em] \Rightarrow 455\text{R} = 3640\\[1em] \Rightarrow \text{R} = \dfrac{3640}{455}\\[1em] \Rightarrow \text{R} = 8$

Hence, the rate of interest is 8% p.a.

A sum of money was put at simple interest at a certain rate for 2 years. If this sum had been put at 3% higher rate, it would have earned ₹ 720 more as interest. Find the sum.

Answer

Let sum of money be ₹ P and rate of interest be R%.

Case I : T = 2 years

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

$= \dfrac{\text{P} \times \text{R} \times 2}{100}$

Case II : T = 2 years, new rate = (R + 3)%

∴ S.I. = $\dfrac{\text{P} \times (\text{R} + 3) \times 2}{100}$

But the difference in these two S.I.'s is ₹ 720.

$\therefore \dfrac{\text{P} \times (\text{R} + 3) \times 2}{100} - \dfrac{\text{P} \times \text{R} \times 2}{100} = 720\\[1em] \Rightarrow 2\text{PR} + 6\text{P} - 2\text{PR} = 720 \times 100\\[1em] \Rightarrow 6\text{P} = 72000\\[1em] \Rightarrow \text{P} = \dfrac{72000}{6}\\[1em] \Rightarrow \text{P} = 12000$

Hence, the required sum is ₹ 12,000.

A sum of money invested at 6% p.a. simple interest for a certain period of time yields ₹ 960 as interest. If this sum had been invested for 5 years more, it would have yielded ₹ 2,160 as interest. Find the sum.

Answer

Let the sum be ₹ P and the time be T years.

$\dfrac{\text{P} \times 6 \times \text{T}}{100} = 960\\[1em] \Rightarrow 6\text{PT} = 96000\\[1em] \Rightarrow \text{PT} = 16000 \qquad \text{.....(i)}$

Also, when the time is (T + 5) years,

$\dfrac{\text{P} \times 6 \times (\text{T} + 5)}{100} = 2160\\[1em] \Rightarrow 6\text{PT} + 30\text{P} = 216000\\[1em] \Rightarrow \text{PT} + 5\text{P} = 36000 \qquad \text{.....(ii)}$

Substituting PT = 16000 from equation (i) in equation (ii),

$16000 + 5\text{P} = 36000\\[1em] \Rightarrow 5\text{P} = 36000 - 16000\\[1em] \Rightarrow 5\text{P} = 20000\\[1em] \Rightarrow \text{P} = 4000$

Hence, the required sum is ₹ 4,000.