Compound Interest MCQ Quiz - Objective Question with Answer for Compound Interest - Download Free PDF

Given:

Principal (P) = ₹50,000

Rate of interest (r) = 12% per annum (compounded half-yearly, so r/2 = 6%)

Time (t) = 1 year (compounded half-yearly, so n = 2)

Formula used:

A = P\((1+\frac{r}{2×100})^{2t}\)

Where,

A = Amount, P = Principal, r = Rate of interest, t = Time

Calculations:

A = 50,000 × \((1+\frac{12}{2×100})^{2×1}\)

⇒ A = 50,000 × \((1+\frac{6}{100})^{2}\)

⇒ A = 50,000 × \((1+0.06)^{2}\)

⇒ A = 50,000 × \((1.06)^{2}\)

⇒ A = 50,000 × 1.1236

⇒ A = ₹56,180

∴ The correct answer is option 2.

Given:

Principal (P) = ₹10,000

Rate (r) = 10% per annum

Time (t) = 2 years

Formula used:

A = P\((1+\frac{r}{100})^t\)

Where, A = Amount

Calculation:

A = 10,000\((1+\frac{10}{100})^2\)

⇒ A = 10,000\((1+\frac{1}{10})^2\)

⇒ A = 10,000\((1.1)^2\)

⇒ A = 10,000 × 1.21

⇒ A = ₹12,100

∴ The correct answer is option 2.

Given:

Total will amount = ₹52500

Ages of the two sons = 18 years and 19 years

They will receive equal amounts at the age of 21 years.

Rate of compound interest (R) = 10% per annum.

Formula Used:

Amount (A) = Principal (P) × (1 + R/100)^T

Where P is the present share, R is the rate of interest, and T is the time period.

Calculation:

Let the present share of the younger son (age 18) be P₁.

Let the present share of the elder son (age 19) be P₂.

We know that P₁ + P₂ = 52500 (Equation 1)

The younger son will receive the amount after (21 - 18) = 3 years.

Amount for the younger son (A₁) = P₁ × (1 + 10/100)³ = P₁ × (1.1)³ = P₁ × 1.331

The elder son will receive the amount after (21 - 19) = 2 years.

Amount for the elder son (A₂) = P₂ × (1 + 10/100)² = P₂ × (1.1)² = P₂ × 1.21

According to the will, they must get equal amounts when they are 21 years old.

A₁ = A₂

P₁ × 1.331 = P₂ × 1.21

P₂ = P₁ × (1.331 / 1.21) = P₁ × (1331 / 1210)

Substitute the value of P₂ in Equation 1:

P₁ + P₁ × (1331 / 1210) = 52500

P₁ × (1 + 1331 / 1210) = 52500

P₁ × (1210 + 1331) / 1210 = 52500

P₁ × (2541 / 1210) = 52500

P₁ = 52500 × (1210 / 2541)

P₁ = 63525000 / 2541

P₁ = 25000

∴ The present share of the younger son is ₹25000.

Given:

Total will amount = Rs. 44100

Ages of the two sons = 14 years and 15 years

Age at which they get equal amounts = 17 years

Rate of compound interest (R) = 10% per annum

Formula Used:

Amount at compound interest, A = P × (1 + R/100)ⁿ, where P is the principal and n is the number of years.

Calculation:

Let the present share of the younger son (age 14) be x.

Then the present share of the elder son (age 15) will be 44100 - x.

For the younger son, the amount after (17 - 14) = 3 years will be:

Amount_younger = x × (1 + 10/100)³ = x × (1 + 0.1)³ = x × (1.1)³ = 1.331x

For the elder son, the amount after (17 - 15) = 2 years will be:

Amount_elder = (44100 - x) × (1 + 10/100)² = (44100 - x) × (1.1)² = (44100 - x) × 1.21

According to the will, they must get equal amounts when they are 17 years old:

Amount_younger = Amount_elder

1.331x = 1.21 × (44100 - x)

1.331x = 1.21 × 44100 - 1.21x

1.331x = 53361 - 1.21x

1.331x + 1.21x = 53361

2.541x = 53361

x = 53361 / 2.541

x = 21000

The present share of the younger son is Rs. 21000.

Given:

P = 7000, t = 1 year, r = 6%

Formula Used:

CI_yearly = P(1 + r/100) - P

CI_half-yearly = P(1 + r/200)² - P

Difference = |CI_half-yearly - CI_yearly|

Calculations:

CI_yearly = 7000(1 + 0.06) - 7000 = 7420 - 7000 = 420

CI_half-yearly = 7000(1 + 0.03)² - 7000 = 7000(1.0609) - 7000 = 7426.3 - 7000 = 426.3

Difference = |426.3 - 420| = 6.3

∴ Difference = ₹ 6.3

Gi​ven:

Amount = 27 P in 3 years

Concept:

In compound interest, the ratio of the amount and the principal is given by:

\(\frac{A}{P} = (1 + \frac{R}{100})^n\)

Calculation:

We know that,

\(\frac{A}{P} = (1 + \frac{R}{100})^n\)

\(⇒ \frac{27}{1} = (1 + \frac{R}{100})^3 \)

\(⇒ 3^3 = (1 + \frac{R}{100})^3 \)

\(⇒ 3 = (1 + \frac{R}{100}) \)

⇒ R/100 = 3 - 1 = 2

⇒ R = 200%

Hence, the annual interest rate is 200%.

Shortcut Trick

A sum becomes 27 times in 3 years

3^x = 27

⇒ 3^x = 3³

⇒ x = 3

Rate = (x - 1) × 100%

⇒ (3 - 1) × 100% = 200%

∴ The annual interest rate is 200%.

Given:

Principal = Rs. 15,000

Amount = Rs. 19,965

Time = 15 months

Condition = on every 5 months

Concept used:

Condition = on every 5 months

New rate = Rate × 5/12

New time = Time × 12/5

Calculations:

Let the new rate be R%

According to the question,

New time = Time × 12/5

⇒ 15 × 12/5 = 36 months = 3 years

Simplifying the values by dividing it by 15 to its lowest possible values, we get Principal = 1000 and Amount = 1331

Now, new time period is 3 years, hence taking the cube roots of Principal and Amount,

⇒ R = 10%

New rate = Rate × 5/12

⇒ 10 = Rate × 5/12

⇒ Rate = (10 × 12)/5

⇒ Rate = 24%

∴ Rate is 24% per annum.

Alternate MethodGiven:

Principal = Rs. 15,000

Amount = Rs. 19,965

Time = 15 months

Condition = on every 5 months

Concept used:

Condition = on every 5 months

New rate = Rate × 5/12

New time = Time × 12/5

Formulae used:

(1) Effective rate for 3 years = 3R + 3R²/100 + R³/10000

(2) A = P(1 + R/100)^T

Where, A → Amount

P → Principal

R → Rate of interest

T → Time

Calculations:

According to the question,

Let the new rate be R%

New time = Time × 12/5

⇒ 15 × 12/5 = 36 months = 3 years

Amount = P(1 + R/100)^T

⇒ 19,965 = 15,000(1 + R/100)³

⇒ 19,965/15,000 = (1 + R/100)³

⇒ 1331/1000 = (1 + R/100)³

⇒ (11/10)³ = (1 + R/100)³

⇒ 11/10 = 1 + R/100

⇒ (11/10) – 1 = R/100

⇒ 1/10 = R/100

⇒ R = 10%

New rate = Rate × 5/12

⇒ 10 = Rate × 5/12

⇒ Rate = (10 × 12)/5

⇒ Rate = 24%

∴ Rate is 24% per annum

Additional InformationCompound Interest means interest earned on interest. Simple interest always occurs on only principal but compound interest also occurs on simple interest. So, if time period is 2 years, compound interest will also apply on simple interest of first year.

Given: 

Hari invested Rs.100 for three years at a simple interest rate of 11.03%.

Tipu invested a sum for three years at 10%.

Concept used:

Simple Interest, SI = (P × R × T)/100

Compound interest, CI = P(1 + R/100)ⁿ - P

Calculation:

Let the principal amount that Tipu invested be Rs. P.

After three years,

Hari gets simple interest on the sum he invested,

⇒ (100 × 11.03 × 3)/100

⇒ Rs. 33.09

Tipu gets compound interest on the sum he invested,

⇒ [P × (1 + 10/100)³] - P

⇒ P × 0.331

According to the question,

P × 0.331 = 33.09

⇒ P = 99.969..

⇒ P ≈ 100

∴ Tipu should invest Rs. 100 to get the same amount after three years but at 10% compound interest.

Shortcut Trick S.I = (P × R × t)/100

⇒ \((100×11.03×3)\over100\) = 33.09

Amount = Principal + S.I

⇒ 100 + 33.09 = 133.09

Successive % = a + b + c + \((ab+bc+ca)\over100\) + \(abc\over100^2\)

Here, a = b = c = 10%

Successive % = 10 + 10 + 10 + (300/100) + 1000/10000

Successive % =  33.1%

Compound interest 10% in 3 years

⇒ \(133.09\over133.1\) × 100 = Rs.100

Given:

Principal = Rs.13000 

Rate of interest = 15%

Concept used:

Rate of interest for 12 months = 15%

Rate of interest for 8 months = 15 × (8/12) = 10%

And 2 years = 24 months

Total 8-monthly time = 24/8 = 3

Formula:

Let P = Principal, R = rate of interest and n = time period

Compound interest = P(1 + R/100)ⁿ  - P

Calculation:

∴ Compound interest = 13000(1 + 10/100)³  - 13000

⇒ 13000 × (1331/1000)

⇒ 17303 - 13000

= Rs.4303

Given:

The Sum becomes Rs. 7800 in 3 years and Rs. 11232 in 5 years

Formula used:

At compound interest, the final amount = \(P\left(1+\frac{r}{100} \right)^{n}\)

Where, P = The sum of the amount

r = Rate of interest

n = Time (years)

Calculation:

Here, Rs. 7800 becomes Rs. 11232 at compound interest in two years.

Let, the rate of interest = R

So, 11232 = \(7800\left(1+\frac{R}{100} \right)^2\)

⇒ [(100 + R)/100]² = 11232/7800

⇒ [(100 + R)/100]2 = 144/100

⇒ [(100 + R)/100]2 = (12/10)²

⇒ [(100 + R)/100] = (12/10)

⇒ 100 + R = 1200/10 = 120

⇒ R = 120 - 100 = 20

∴ The rate per cent is 20%

Given:

Principal = Rs.12000 

Time = 5 years 

Formulas used:

Amount = Principal × (1 + r/100)ⁿ

Calculation:

Amount = Principal × (1 + r/100)⁵

⇒ 24000 = 12000 × (1 + r/100)5

⇒ 24000/12000 = (1 + r/100)5

⇒ 2 = (1 + r/100)5          (1) 

⇒ At the end of 15 years, 

⇒ Amount = 12000 × (1 + r/100)¹⁵

⇒ Amount = 12000 × [(1 + r/100)⁵ ]³       (From 1) 

⇒12000 × 2³

⇒12000 × 8 

⇒ 96000 

∴ The amount at the end of 15 years will be Rs.96000

Shortcut Trick 

∴ The amount at the end of 15 years will be 8 times of 12000 = Rs.96000

Given:

Certain sum amounts to Rs. 1758 in two years and to Rs. 2,021.70 in 3 years at compound interest when compounded annually.

Concept used:

When compounded annually, the amount received at the end of the period is

Amount = P[1 + r/100]t

Where, P = Principal, r = Rate of interest p.a., t = Time period

Calculation:

Let the rate be R%

P(1 + R/100)2 = 1758  ....(i)

P(1 +R/100)3 = 2021.7 ....(ii)

Dividing equation (ii) by (i)

⇒ 1 + R/100 = 2021.7/1758

⇒ R/100 = (2021.7 – 1758)/1758

⇒ R = (263.7 × 100)/1758 = 15%

∴ The rate of interest p.a. is 15%.

Shortcut TrickDifference between the amount of 2 yr and 3 yr = 2021.7 - 1758 = 263.7

Now, this sum of Rs. 263.70 is earned as interest on Rs. 1758 (2 years' SI) taken as principal.

Therefore, the reqd rate % = (263.70/1758) × 100 = 15%.

Given:

Lend amount = Rupees 72,000

Rate = 12% per annum

Time = 3 years

Compounded annually

Concept used:

CI = Total amount - Principal

P(1 + R/100)^N - P

Where, P = Principal, R = Rate of interest, N = Time (in years)

Calculation:

Amount at the end of 1st year

⇒ 72000 × (1 + 12/100)

⇒ 72000 × (112/100)

⇒ Rs. 80640 

Amount at the end of 2nd year

⇒ 80640 × (1 + 12/100) 

⇒ 80640 × (112/100) 

⇒ 90316.8 ≈ Rs. 90317

Interest at the end of 3rd year

⇒ 90317 × (1 + 12/100) - 90317

⇒ 90317 × (112/100) - 90317

⇒ 101155 - 90317

⇒ Rs. 10838

∴ The interest for the 3rd year is Rs. 10838.

Shortcut Trick 

Given:

Principal = Rs. 60,000

Rate = 9%

Compound Interest = Rs. 11,286

Amount = Principal + Compound Interest

Formula used:

Amount = P(1 + Rate/100)^Time

Amount = Principal + Compound Interest

Calculation:

Amount = 60,000 + 11,286 = 71,286

Amount = P(1 + Rate/100)Time

⇒ 71,286 = 60,000(1 + 9/100)^Time

 ⇒ 71,286 = 60,000[(100 + 9)/100]^Time

⇒ 71,286/60,000 = (109/100)^Time

⇒ (11,881/10,000) = (109/100)^Time

⇒ (109/100)2 = (109/100)Time

⇒ Time = 2

∴ The time period is 2 years.

Given:

Each installment = ₹ 980

Rate of interest = 4%

Number of years = 2

Formula used:

In case of compound interest, borrowed money = x/(1 + r/100) + x/(1 + r/100)² + ..... + x/(1 + r/100)ⁿ

Calculation:

The borrowed money = \(\frac{980}{(1+\frac{4}{100})}+\frac{980}{(1+\frac{4}{100})^{2}}\)

= \(\frac{980}{\frac{26}{25}}+\frac{980}{\frac{26}{25}×\frac{26}{25}}\)

= [980 × (25/26)] + [980 × (25/26) × (25/26)]

= 980 [(25/26) + (25/26) × (25/26)]

= \(980 \left[\frac{(25×26)+(25×25)}{26×26} \right]\)

= 980 × [(650 + 625)/676]

= 980 × (1275/676)

= 1848.372...

≈ 1850 (in ₹, to the nearest tens)

∴ The sum (in ₹, to the nearest tens) borrowed was ₹ 1850