Simple Interest - Exercise 11(B)

Which of the following statements is incorrect?

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

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

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

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

Answer

The incorrect statement is R = $\dfrac{P \times T}{100 \times S.I.}$

The correct formula for Rate is R = $\dfrac{100 \times S.I.}{P \times T}$

Hence, option 2 is the correct option.

If the principal is ₹ 600, then the amount to be paid at the end of 3 years at $7\dfrac{1}{2}$% p.a. simple interest will be

1. ₹ 735
2. ₹ 750
3. ₹ 775
4. ₹ 785

Answer

Given:

P = ₹ 600, T = 3 years

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

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \left(\dfrac{600 \times \dfrac{15}{2} \times 3}{100}\right) \\[1em] = ₹ \Big(\dfrac{600 \times 15 \times 3}{100 \times 2}\Big) \\[1em] = ₹ \Big(\dfrac{6 \times 15 \times 3}{1 \times 2}\Big) \\[1em] = ₹ \Big(\dfrac{3 \times 15 \times 3}{1 \times 1}\Big) \\[1em] = ₹ 3 \times 15 \times 3 \\[1em] = ₹ 135$

Amount = S.I. + Principal

Amount = ₹ 135 + ₹ 600 = ₹ 735

Hence, option 1 is the correct option.

At what rate, ₹ 800 gives ₹ 208 as simple interest in 2 years?

1. 11%
2. 12%
3. 13%
4. 14%

Answer

Given:

Principal (P) = ₹ 800

Simple Interest (S.I.) = ₹ 208

Time (T) = 2 years

Then

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

$= \Big(\dfrac{208 \times 100}{800 \times 2}\Big)$%

$= \dfrac{20800}{1600}$%

$= \dfrac{208}{16}$%

$= 13$%

Hence, option 3 is the correct option.

In how many years will ₹ 900 give ₹ 351 as simple interest at 13% p.a.?

1. $1\dfrac{1}{2}$years

2. 2 years

3. $2\dfrac{1}{2}$

4. 3 years

Answer

Given:

Principal (P) = ₹ 900

Simple Interest (S.I.) = ₹ 351

Rate (R) = 13% p.a.

Then

$T = \dfrac{S.I. \times 100}{P \times R} \\[1em] = \Big(\dfrac{351 \times 100}{900 \times 13}\Big) \\[1em] = \dfrac{35100}{11700} \\[1em] = \dfrac{351}{117} \\[1em] = 3 \text{ years}$

Hence, option 4 is the correct option.

At what rate per cent per annum simple interest will a sum be double of itself in 8 years?

1. 10%

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

3. 15%

4. $17\dfrac{1}{2}$%

Answer

Given:

Let Principal (P) = x

Amount (A) = 2x

Time (T) = 8 years

S.I. = Amount - Principal

S.I. = 2x - x = x

Then

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

$= \Big(\dfrac{x \times 100}{x \times 8}\Big)$%

$= \dfrac{100}{8}$%

$= \dfrac{25}{2}$%

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

Hence, option 2 is the correct option.

At simple interest a sum becomes $\dfrac{7}{4}$ of itself in 5 years. The rate of interest is

1. 10% p.a.

2. 12% p.a.

3. $12\dfrac{1}{2}$% p.a.

4. 15% p.a.

Answer

Given:

Let the Principal (P) = ₹ 100

Amount (A): The problem says the sum becomes $\dfrac{7}{4}$ of itself.

Amount = ₹ $\dfrac{7}{4} \times 100$

Amount = ₹ 7 x 25 = ₹ 175

Time (T) = 5 years

S.I. = Amount - Principal

S.I. = ₹ 175 - ₹ 100 = ₹ 75

Then

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

$= \Big(\dfrac{75 \times 100}{100 \times 5}\Big)$%

$= \dfrac{75}{5}$%

$= 15$%

Hence, option 4 is the correct option.

Fill in the blanks :

(i) The money borrowed for a certain period is called ............... .

(ii) The rate per cent per annum is the interest on ............... for 1 year.

(iii) The simple interest on ₹ 800 invested at 13% per annum for 3 years is ............... .

(iv) If ₹ 2600 becomes ₹ 3900 in 5 years, then the rate of interest is ............... .

(v) If the principal is ₹ 3400, then the amount to be paid at the end of 5 years at 8% p.a. simple interest will be ............... .

Answer

(i) The money borrowed for a certain period is called principal.

(ii) The rate per cent per annum is the interest on ₹ 100 for 1 year.

(iii) The simple interest on ₹ 800 invested at 13% per annum for 3 years is ₹ 312.

(iv) If ₹ 2600 becomes ₹ 3900 in 5 years, then the rate of interest is 10%.

(v) If the principal is ₹ 3400, then the amount to be paid at the end of 5 years at 8% p.a. simple interest will be ₹ 4760.

Explanation

(i) In financial math, the initial sum of money you borrow or invest before interest is added is always called the Principal.

(ii) The word "percent" literally means "per hundred." So, a rate of 8% means you pay ₹ 8 for every ₹ 100 borrowed over one year.

(iii)

Given:

P = ₹ 800, R = 13%, T = 3 years

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{800 \times 13 \times 3}{100}\Big) \\[1em] = ₹ \Big(\dfrac{8 \times 13 \times 3}{1}\Big) \\[1em] = ₹ 8 \times 13 \times 3 \\[1em] = ₹ 312$

S.I. = ₹ 312

(iv)

Given:

P = ₹ 2600, A = ₹ 3900, T = 5 years

S.I. = Amount - Principal

S.I. = ₹ 3900 - ₹ 2600 = ₹ 1300

Then

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

$= \Big(\dfrac{1300 \times 100}{2600 \times 5}\Big)$%

$= \Big(\dfrac{1 \times 100}{2 \times 5}\Big)$%

$= \dfrac{100}{10}$%

$= 10$%

R = 10%

(v) Given:

P = ₹ 3400, R = 8%, T = 5 years

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{3400 \times 8 \times 5}{100}\Big) \\[1em] = ₹ \Big(\dfrac{34 \times 8 \times 5}{1}\Big) \\[1em] = ₹ 34 \times 40 \\[1em] = ₹ 1360$

Amount = S.I. + Principal

Amount = ₹ 1360 + ₹ 3400

Amount = ₹ 4760

Write true (T) or false (F) :

(i) The total money to be paid back to the lender is called interest.

(ii) Amount = Principal + Interest

(iii) The rate per cent per annum is the interest on ₹1 for 1 year.

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

(v) If a man borrows ₹ 5200 at 6% p.a. simple interest, the amount he has to return at the end of 5 years is ₹ 6760.

Answer

(i) False
Reason — The total money (Principal + Interest) is called the Amount. Interest is only the "extra" fee charged for borrowing the money.

(ii) True
Reason — This is the fundamental formula for calculating the total value of a loan or investment at the end of a time period.

(iii) False
Reason — The rate per cent is the interest on ₹ 100 for 1 year. (Per cent = per hundred).

(iv) False
Reason — This formula is inverted. The correct formula for Simple Interest is:

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

(v) True
Reason —

P = ₹ 5200, R = 6%, T = 5 years

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{5200 \times 6 \times 5}{100}\Big) \\[1em] = ₹ \Big(\dfrac{52 \times 6 \times 5}{1}\Big) \\[1em] = ₹ 52 \times 6 \times 5 \\[1em] = ₹ 1560$

Amount = S.I. + Principal

Amount = ₹ 1560 + ₹ 5200

Amount = ₹ 6760

Mr. Shah has two choices of investing his money. If he invests in a bank, he gets 8% simple interest. If he invests in his friend's company ADG, he gets 12% simple interest. Mr. Shah has ₹ 5,00,000 to invest.

(1) If Mr. Shah invests all the money with the bank, the interest received by him after 2 years will be:

1. ₹ 50,000
2. ₹ 60,000
3. ₹ 75,000
4. ₹ 80,000

(2) What sum invested in ADG company will amount to ₹ 4,80,000 in 5 years ?

1. ₹ 3,00,000
2. ₹ 3,50,000
3. ₹ 4,00,000
4. ₹ 4,20,000

(3) In what time will the money invested with the bank double itself ?

1. $7\dfrac{1}{2}$years

2. 9 years

3. $12\dfrac{1}{2}$years

4. $11\dfrac{3}{4}$years

(4) If Mr. Shah wishes to invest partly in the bank and partly in ADG company such that after 2 years he receives the same interest from both, then find the sum that he would invest in the bank.

1. ₹ 2,00,000
2. ₹ 2,50,000
3. ₹ 3,00,000
4. ₹ 3,50,000

Answer

(1)

Given:

P = ₹ 5,00,000, R = 8%, T = 2 years.

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{5,00,000 \times 8 \times 2}{100}\Big) \\[1em] = ₹ 5,000 \times 8 \times 2 \\[1em] = ₹ 80,000$

Hence, option 4 is the correct option.

(2)

Given:

A = ₹ 4,80,000, R = 12%, T = 5 years.

Then

$P = \dfrac{100 \times A}{100 + (R \times T)} \\[1em] = ₹ \Big(\dfrac{100 \times 4,80,000}{100 + (12 \times 5)}\Big) \\[1em] = ₹ \Big(\dfrac{100 \times 4,80,000}{100 + 60}\Big) \\[1em] = ₹ \Big(\dfrac{4,80,00,000}{160}\Big) \\[1em] = ₹ 3,00,000$

Hence, option 1 is the correct option.

(3)

Given:

R = 8%

Let P = x, Amount = 2x

S.I. = Amount - P

S.I. = 2x - x = x

Then

$T = \dfrac{100 \times \text{S.I.}}{P \times R} \\[1em] = \Big(\dfrac{100 \times x}{x \times 8}\Big) \text{ years} \\[1em] = \dfrac{100}{8} \text{ years} \\[1em] = \dfrac{25}{2} \text{ years} \\[1em] = 12\dfrac{1}{2} \text{ years}$

Hence, option 3 is the correct option.

(4)

Given:

R (Bank) = 8%

R (ADG) = 12%

T = 2 years

Total Sum = ₹ 5,00,000.

Let Bank Investment = x, then ADG Investment = (5,00,000 - x)

Then

$S.I. \text {(Bank)} = S.I. \text {(ADG)} \\[1em] \Rightarrow \dfrac{x \times 8 \times 2}{100} = ₹ \dfrac{(5,00,000 - x) \times 12 \times 2}{100} \quad \text{[Cancel 100 and 2 from both sides]} \\[1em] \Rightarrow 8x = ₹ (5,00,000 - x) \times 12 \\[1em] \Rightarrow \dfrac{8x}{12} = ₹ (5,00,000 - x) \\[1em] \Rightarrow \dfrac{2x}{3} = ₹ (5,00,000 - x) \\[1em] \Rightarrow \dfrac{2x}{3} + x = ₹ 5,00,000 \\[1em] \Rightarrow \dfrac{5x}{3} = ₹ 5,00,000 \\[1em] \Rightarrow x = ₹ \dfrac{5,00,000 \times 3}{5} \\[1em] \Rightarrow x = ₹ \dfrac{1,00,000 \times 3}{1} \\[1em] \Rightarrow x = ₹ 3,00,000$

Hence, option 3 is the correct option.

Assertion: The interest on ₹ 700 at 5% p.a. for 12 months is ₹ 35.

Reason: S.I. = $\dfrac{\text{PRT}}{100}$, where P = principal, R = rate per cent per annum and T = time in months.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is true but Reason (R) is false.

Explanation

Given:

P = ₹ 700, R = 5%, T = 12 months = 1 year

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{700 \times 5 \times 1}{100}\Big) \\[1em] = ₹ \Big(\dfrac{7 \times 5 \times 1}{1}\Big) \\[1em] = ₹ 7 \times 5 \\[1em] = ₹ 35$

The Assertion is True.

Reason:

S.I. = $\dfrac{\text{PRT}}{100}$, where P = principal, R = rate per cent per annum and T = time in months.

This is False.

In the standard formula, T must always be in years. If time is given in months, it must be converted (divided by 12) before using this specific formula.

Hence, option 3 is the correct option.

Assertion: Amount received after depositing ₹ 800 for a period of 3 years at the rate of 12% p.a. simple interest is ₹ 1096.

Reason: Amount = Principal + Interest.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is false but Reason (R) is true.

Explanation

Given:

P = ₹ 800, T = 3 years, R = 12%

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{800 \times 12 \times 3}{100}\Big) \\[1em] = ₹ \Big(\dfrac{8 \times 12 \times 3}{1}\Big) \\[1em] = ₹ 8 \times 12 \times 3 \\[1em] = ₹ 288$

Amount = S.I. + Principal

Amount = ₹ 288 + ₹ 800

Amount = ₹ 1088

Since the Assertion claims the amount is ₹ 1096, the Assertion is False.

Reason:

Amount = Principal + Interest.

This is True. This is the correct mathematical definition of Amount.

Hence, option 4 is the correct option.

The simple interest on a sum of money is $\dfrac{4}{9}$ of the principal. Find the rate per cent, if time and rate per cent are numerically equal.

1. $5\dfrac{2}{3}$%

2. $7\dfrac{2}{3}$%

3. $6\dfrac{2}{3}$%

4. $6\dfrac{2}{3}$%

Answer

Given:

S.I. = $\dfrac{4}{9}$ of the principal

Time and Rate are numerically equal.

Let the Principal (P) be x.

S.I. = $\dfrac{4}{9}x$

Let R = T = y

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow \dfrac{4}{9}x = \dfrac{x \times y \times y}{100} \\[1em] \Rightarrow \dfrac{4}{9} = \dfrac{y^2}{100} \\[1em] \Rightarrow y^2 = \dfrac{4 \times 100}{9} \\[1em] \Rightarrow y^2 = \dfrac{400}{9} \\[1em] \Rightarrow y = \sqrt{\dfrac{400}{9}} \\[1em] \Rightarrow y = \dfrac{20}{3} \\[1em] \Rightarrow y = 6\dfrac{2}{3}$

∴ Rate = $6\dfrac{2}{3}$%

Hence, option 3 is the correct option.

A farmer borrowed ₹ 5500 at 8% per annum. After 5 years, he cleared the account by giving ₹ 6000 and a cow. The cost of the cow is:

1. ₹ 2100
2. ₹ 1900
3. ₹ 1700
4. ₹ 1500

Answer

Given:

P = ₹ 5500, R = 8%, T = 5 years

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \Big(\dfrac{5500 \times 8 \times 5}{100}\Big) \\[1em] = ₹ \Big(\dfrac{55 \times 8 \times 5}{1}\Big) \\[1em] = ₹ 55 \times 40 \\[1em] = ₹ 2200$

Amount = Principal + S.I.

Amount = ₹ 5500 + ₹ 2200

Amount = ₹ 7700

The farmer cleared the account by giving ₹ 6000 and a cow.

Cost of cow = Amount - ₹ 6000

Cost of cow = ₹ 7700 - ₹ 6000 = ₹ 1700

Hence, option 3 is the correct option.

Ameesha took a loan of ₹ 1200 with simple interest for as many years as the rate of interest. If she paid ₹ 432 as interest at the end of the loan period, what was the time?

1. 3.6 years
2. 6 years
3. 18 years
4. Cannot be determined

Answer

Given:

P = ₹ 1200, S.I. = ₹ 432

Time and Rate are numerically equal.

Let R = T = y

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 432 = \dfrac{1200 \times y \times y}{100} \\[1em] \Rightarrow 432 = \dfrac{1200 \times y^2}{100} \\[1em] \Rightarrow 432 = 12 \times y^2 \\[1em] \Rightarrow y^2 = \dfrac{432}{12} \\[1em] \Rightarrow y^2 = 36 \\[1em] \Rightarrow y = \sqrt{36} \\[1em] \Rightarrow y = ±6$

Since time cannot be negative,

∴ Time = 6 years

Hence, option 2 is the correct option.

A sum of money is lent at simple interest. If the money gets doubled in 5 years, then the rate of interest is:

1. 20% pa
2. 25% pa
3. 15% pa
4. 22% pa

Answer

Given:

T = 5 years

Let the Principal (P) be x.

Money is doubled, so Amount (A) = 2x

S.I. = Amount - Principal

S.I. = 2x - x = x

Then

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

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

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

$= \dfrac{100}{5}$%

$= 20$%

∴ Rate = 20% per annum

Hence, option 1 is the correct option.

A certain sum of money at simple interest doubles in 10 years. In how many years, at the same simple interest, will it be tripled?

1. 15 years
2. 18 years
3. 20 years
4. 24 years

Answer

Given:

The sum doubles in 10 years.

Let the Principal (P) be x.

When sum is doubled, Amount = 2x

S.I. = Amount - Principal

S.I. = 2x - x = x and T = 10 years

Then

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

$= \Big(\dfrac{x \times 100}{x \times 10}\Big)$%

$= \dfrac{100}{10}$%

$= 10$%

Now, when sum is tripled, Amount = 3x

S.I. = 3x - x = 2x and R = 10%

Then

$T = \dfrac{S.I. \times 100}{P \times R} \\[1em] = \Big(\dfrac{2x \times 100}{x \times 10}\Big) \text{ years} \\[1em] = \dfrac{200}{10} \text{ years} \\[1em] = 20 \text{ years}$

∴ Time = 20 years

Hence, option 3 is the correct option.

The simple interest on a sum for 6 years is ₹ 29,250. The rate of interest for the first 2 years is 7% per annum and for the next 4 years is 16% per annum. The sum is:

1. ₹ 32,210
2. ₹ 36,815
3. ₹ 37,500
4. ₹ 38,200

Answer

Given:

Total S.I. for 6 years = ₹ 29,250

For first 2 years, R = 7% p.a., T = 2 years

For next 4 years, R = 16% p.a., T = 4 years

Let the sum (P) be x.

S.I. for first 2 years :

$S.I._1 = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{x \times 7 \times 2}{100} \\[1em] = \dfrac{14x}{100}$

S.I. for next 4 years :

$S.I._2 = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{x \times 16 \times 4}{100} \\[1em] = \dfrac{64x}{100}$

Total S.I. = S.I.₁ + S.I.₂

$\Rightarrow ₹ 29250 = \dfrac{14x}{100} + \dfrac{64x}{100} \\[1em] \Rightarrow ₹ 29250 = \dfrac{14x + 64x}{100} \\[1em] \Rightarrow ₹ 29250 = \dfrac{78x}{100} \\[1em] \Rightarrow x = ₹ \dfrac{29250 \times 100}{78} \\[1em] \Rightarrow x = ₹ \dfrac{2925000}{78} \\[1em] \Rightarrow x = ₹ 37500$

∴ Sum = ₹ 37,500

Hence, option 3 is the correct option.

The simple interest on ₹ 32,000 at 8.5% per annum for the period from 10th February, 2019, to 24th April, 2019 is:

1. ₹ 544
2. ₹ 604
3. ₹ 615
4. ₹ 644

Answer

Given:

P = ₹ 32,000, R = 8.5% p.a.

Time period is from 10th February, 2019 to 24th April, 2019.

Number of days (excluding the day of borrowing and including the day of repayment) :

February (from 11th to 28th) = 18 days

March = 31 days

April (from 1st to 24th) = 24 days

Total number of days = 18 + 31 + 24 = 73 days

T = $\dfrac{73}{365}$ year = $\dfrac{1}{5}$ year

Then

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = ₹ \left(\dfrac{32000 \times 8.5 \times \dfrac{1}{5}}{100}\right) \\[1em] = ₹ \Big(\dfrac{32000 \times 8.5}{100 \times 5}\Big) \\[1em] = ₹ \Big(\dfrac{32000 \times 8.5}{500}\Big) \\[1em] = ₹ \Big(\dfrac{320 \times 8.5}{5}\Big) \\[1em] = ₹ \Big(\dfrac{2720}{5}\Big) \\[1em] = ₹ 544$

∴ S.I. = ₹ 544

Hence, option 1 is the correct option.