Installments MCQ Quiz - Objective Question with Answer for Installments - Download Free PDF

Given

Loan amount = ₹9,960

Interest rate = 7.5% per annum compounded annually

Repayment period = 2 years

Shortcut Trick 

Rate of interest = 7.5%

If Principal = 100

Amount = 107.5

So

Principal : Amount = 100 : 107.5 = 40 : 43

So  

                    Principal      Installment

First year          40               43

Second year     40²             43²

Both installments are equal so

                       Principal      Installment 

First year        40 × 43        43 × 43      

Second year     402               432

So

Total Principal = 40 × 43 + 40² = 3320

3320 units = Rs. 9960

1 unit = 3

so

Installment = 43² × 3 = Rs. 5547

Given:

Rate (r) = 20% per annum

Amount becomes double ⇒ A = 2P

Formula used:

A = P(1 + r/100)^t

Calculation:

⇒ 2P = P(1 + 20/100)^t

⇒ 2 = (1.2)^t

Now, take logarithm both sides:

⇒ log(2) = t × log(1.2)

⇒ t = log(2) ÷ log(1.2)

⇒ t = 0.3010 ÷ 0.0792

⇒ t ≈ 3.8 years

∴ The amount is doubled in approximately 4 years.

When a loan is repaid in equal yearly installments, the installment amount is calculated to ensure that the sum of the principal and the interest earned over a certain number of years is equal to the sum of the installments and their respective interests earned over the remaining repayment period.

Let each installment be denoted by \( x \).

The future value of Rs. 10000 at the end of 5 years is:

\( \text{Future Value} = 10000 + \left( 5 \times 10000 \times \frac{10}{100} \right) = 15000 \)

This value must be equal to the future value of all the installments paid at the end of 5 years:

\( 15000 = \left( x + 4x \times \frac{10}{100} \right) + \left( x + 3x \times \frac{10}{100} \right) + \left( x + 2x \times \frac{10}{100} \right) + \left( x + x \times \frac{10}{100} \right) + x \)

Solving this equation, we get \( x = 2500 \).

Thus, the answer is \( \boxed{2500} \).

Given:

Principal (P) = ₹15,400

Rate of interest (r) = 15% per annum

Annual installment = ₹2,500

Number of years = 3

Formula used:

Amount after 1 year = P(1 + r/100) - installment

Repeat the process for subsequent years.

Calculation:

After 1st year:

⇒ A = 15,400 × (1 + 15/100) - 2,500

⇒ = 15,400 × 1.15 - 2,500

⇒ = 17,710 - 2,500 = ₹15,210

After 2nd year:

⇒ A = 15,210 × 1.15 - 2,500

⇒ = 17,491.50 - 2,500 = ₹14,991.50

After 3rd year:

⇒ A = 14,991.50 × 1.15 - 2,500

⇒ = 17,240.225 - 2,500

⇒ = ₹14,740.225

∴ The man still owes ₹14,740.225 after three years.

Let the equal installment be x.

Given that the installment is paid at the beginning of the year.

Present value of the loan due at the end of two years = Sum of the present values of the installments paid

\(2310 \times(1.1)^{-2}=x+\frac{x}{1.1}\)

⇒ 2310 = (1.1)² x + (1.1)x

⇒ 2.31x = 2310

Therefore, x = ₹1,000

Calculation:

Total cost of the computer = Rs. 39000

Down payment = Rs. 19000

Balance = Rs. (39000 - 19000) = Rs. 20000.

Let the rate of interest be R% p.a.

Amount of Rs. 20000 for 5 months

=Rs.( 20000 + 20000 × 5/12 × R/100)=Rs.(20000+250R/3)

The customer pays the shopkeeper Rs. 4200 after 1 month,

Rs. 4200 after 2 months, ...... and Rs. 4200 after 5 months.

Thus, the shopkeeper keeps Rs. 4200 for 4 months, Rs. 4200 for 3 months, Rs. 4200 for 2 months, Rs. 4200 for 1 months and Rs. 4200 at the end.

∴ sum of the amounts of these installments

⇒  (Rs. 4200 + S.I. on Rs 4200 for 4 months) + (Rs. 4200 + S.I. on Rs. 4200 for 3 months) + ...... + (Rs. 4200 + S.I. on Rs. 4200 for 1 month) + Rs. 4200

⇒ Rs. (4200 × 5) + S.I. on Rs. 4200 for (4 + 3 + 2 + 1) months

⇒ Rs. 21000 + S.I. on Rs. 4200 for 10 months

⇒ Rs.(21000 + 4200 × R × 10/12×1/ 100)

⇒ (21000 + 35R)

(20000+250R/3) =  (21000 + 35R)

R = \(20\frac{20}{29}\)%
Alternate Method Total amount = 39000

Down payment = 19000

Remaining amount = 20000

Installment = 4200

So,

Principal 

At Starting         =   20000

after 1 month →  20000 - 4200 = 15800

After 2 months → 15800 - 4200 = 11600

After 3 months →  11600 - 4200 = 7400

After 4 months →  7400 - 4200 = 3200

So, 

total principal = 20000 + 15800 + 11600 + 7400 + 3200 = 58000

and

Interest = 4200 × 5 - 20000 = 1000

So,

rate of interest for 1 month = (1000/58000) × 100 = 100/58 = (50/29)%

rate of interest for 12 months or annual rate of interest = 12 ×  (50/29)% = (600/29)%

Given:

Rate of interest = 5% per annum

Two equal yearly instalments = Rs. 35,280

Concept:

For repayment in two equal annual instalments, each instalment consists of

principal as well as interest. For the first instalment, the interest is calculated

for one year, while of the second instalment, it is calculated for two years.

Calculation:

Let P be the initial borrowed sum.

⇒ According to the concept, we have: P = Rs. 35,280/(1 + 5/100) + Rs. 35,280/(1 + 5/100)²

⇒ Rs. 35,280/(1 + 1/20) + Rs. 35,280/(1 + 1/20)2

⇒ Rs. 35,280/(21/20) + Rs. 35,280/(21/20)2

⇒ Rs. 35,280 × 20/21 + Rs. 35,280 × 400/441

⇒ Rs. 35,280 × 20/21[1 + 20/21]

⇒ Rs. 35,280 × 20/21 × 41/21

⇒ P = Rs. 65600.

Therefore, the value of P is Rs. 65,600.

Given:

A = amount = ₹5,664

T = time = 4 years

R = rate of interest = 12%

Formula used:

Installment = (100 × A)/{100 × T + RT(T – 1)/2}

Calculation:

Installment = (100 × 5664)/(100 × 4 + 12 × 4 × 3/2)

⇒ (100 × 5664)/(400 +72)

⇒ 100 × (5664/472)

⇒ 100 × 12

⇒ 1200

∴ The annual instalment will be ₹1,200.

Mistake Points As per SSC, they assume the debt amount and solve the question.

The above is a Previous year's question, and SSC holds this solution correct.

Refer to the solution carefully.

Calculation:

Rate = 2% pa

As both installments are equal multiply I by 102, So

Total principle is 10200 + 10000 = 20200
According to the problem

20200 = 6000

1 = 6000/20200

102² = 6000/20200 × 102²

3090.30

Each installment equals to Rs. 3090.30.

∴ Option 2 is the correct answer. 

Given:

Principal = Rs.2000 

Rate = 5 % and time = 3 years

Formula used:

Installment = \(\frac{A\times 100}{N \times 100 + (N_{n-1} + N_{n-2} + ..+ 1)\times R}\)

Where A = Amount, R = rate, and N = number of years

Calculation:

S.I = (P × R × T)/100

⇒ (2000 × 5 × 3)/100 = 300

Amount (A) = 2000 + 300 = Rs.2300

Installment = \(\frac{A\times 100}{N \times 100 + (N_{n-1} + N_{n-2} + ..+ 1)\times R}\)

⇒ 2300 × 100/[3 × 100 + (2 + 1) × 5]

⇒ 230000/315 = 46000/63

⇒ 730\(\frac{10}{63}\)

∴ The correct answer is ₹730\(\frac{10}{63}\).

Mistake Points We calculate installment on Amount, 

Here, Principal is 2,000. 

Hence first we need to find the Amount, using the Simple Interest formula, 

and then solve the question. 

Given:

A car with a price of ₹6,50,000 is bought by making some down payment. On the balance, a simple interest of 10% is charged in lump sum and the money is to be paid in 20 equal annual instalments of ₹25,000. 

Concept used:

Amount becomes in lump sum = P(1 + R%)

P = Principal

R = Rate of interest

Calculation:

Let the downpayment be of Rs. X.

​According to the question,

(650000 - X) × (1 + 10%) = 25000 × 20

⇒ (650000 - X) × (1 + 10%) = 500000

⇒ (650000 - X) = 500000 ÷ (1 + 10%)

⇒ (650000 - X) ≈ 454545.45

⇒ X ≈ 650000 - 454545

⇒ X ≈ 195455

∴ The downpayment was Rs. 195455.

Given:

A sum of Rs. 10 is lent by a child to his friend to be returned in 11 monthly instalments of Rs. 1 each, the interest being simple. 

Concept used:

Simple Interest, SI = (P × R × T)/100

Amount = P + SI

where

P = Principal amount

R = Rate of interest per year

T = Time in years

Calculation:

Let the rate of interest be R% p.a.

Total interest paid = 11 - 10  = Rs. 1

Since the installments are being paid monthly, the first, second, ...., and second-last EMI will incur interest for the next 10, 9, ...., and 1 month respectively. Since the last EMI completes the repayment, it will not incur any interest.

According to the concept,

(1 × R/100 × 10/12) + (1 × R/100 × 9/12) + (1 × R/100 × 8/12) + .... + (1 × R/100 × 1/12) = 1

⇒ \(\frac {R}{1200}\) (10 + 9 + ... + 1) = 1

⇒ \(\frac {R}{1200} \times \frac {11 \times 10}{2}\) = 1

⇒ R = 1200/55

⇒ R = \(21{ 9 \over 11}\)%

∴ The rate of interest is \(21{ 9 \over 11}\)%.

Given:

A person borrows Rs. 1,00,000 from a bank at 10% per annum simple interest and clears the debt in five years.

The installment paid at the end of the first, second, third, and fourth years to clear the debt are Rs. 10,000, Rs. 20,000, Rs. 30,000, and Rs. 40,000

Concept used:

S.I = (P × T × R)/100

Here,

P = Principle

T = Time

R = Rate

Calculation:

1st year interest = 100000 × 10% = 10000

Principal amount after 1st year = 100000 - 10000 = 90000

2nd year interest = 90000 × 10% = 9000

Principal amount after 2nd year = 90000 - 20000 = 70000

3rd year interest = 70000 × 10% = 7000

Principal amount after 3rd year = 70000 - 30000 = 40000

4th year interest = 40000 × 10% = 4000

Principal amount after 4th year = 40000 - 40000 = 0

At the end of the fifth year, the person has to pay the total interest remaining:

Total Interest = 10000 + 9000 + 7000 + 4000 = 30000

So, the person should pay Rs. 30,000 at the end of the fifth year to clear the debt.

Given:

Purchased an item at Rs. 7500

Down-payment paid = Rs. 3500

Rate = 9%

Time = 4 months

Formula used:

SI = \(\dfrac{principle × rate × time}{100}\)

1 year = 12 months

4 month = \(\dfrac{4}{12}\)

Calculations:

Amount left to be paid = Rs. (7500 - 3500) = Rs. 4000

SI = \(\dfrac{4000 × 9 × 4}{12 × 100}\)

= 40 × 3 = 120

∴ The answer is Rs. 120

Given Data:

Interest rate = 20%

Number of years = 3

Each installment amount = ₹1,250

Concept:

This problem involves the concept of compound interest and annuity. The present value of the loan can be found by summing the present value of each installment.

Calculation:

The present value of each installment can be calculated using the formula PV = P / (1 + r)ⁿ, where PV is the present value, P is the installment, r is the interest rate, and n is the

number of years.

⇒ So, the present value of the loan = ₹1,250 / (1 + 20/100)¹ + ₹1,250 / (1 + 20/100)² + ₹1,250 / (1 + 20/100)³ ≈ ₹2,633.

Therefore, the money borrowed by Sohan was approximately ₹2,633.