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

Last updated on Jun 3, 2025

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Latest Installments MCQ Objective Questions

Installments Question 1:

A sum of ₹9,960 was borrowed at 7.5% per annum compound interest and paid back in two equal annual instalments. What was the amount of each instalment?

  1. 5,475
  2. 5,547
  3. 5,745
  4. 5,457
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 5,547

Installments Question 1 Detailed Solution

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

Installments Question 2:

At 20% per annum rate, an amount is doubled in approximately in ______ years at compound interest.

  1. 2
  2. 5
  3. 4
  4. 3

Answer (Detailed Solution Below)

Option 3 : 4

Installments Question 2 Detailed Solution

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.

Installments Question 3:

Nikita took a loan of Rs. 10000 from a bank at a simple interest rate of 10% per annum. She decided to repay the loan in five equal installments, paying each installment at the end of each year. How much did Nikita pay annually?

  1. 3000
  2. 2500
  3. 2100
  4. 2750

Answer (Detailed Solution Below)

Option 2 : 2500

Installments Question 3 Detailed Solution

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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 \( \boxed{2500} \).

Installments Question 4:

A man borrowed ₹15,400 at 15% per annum compound interest. At the end of every year, he pays ₹2,500 as part repayment. How much (in ₹) does he still owe after three such instalments?

  1. 14,740.225
  2. 14,870.725
  3. 14,650.525
  4. 14,900.252

Answer (Detailed Solution Below)

Option 1 : 14,740.225

Installments Question 4 Detailed Solution

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.

Installments Question 5:

A sum of ₹2310 is due to be repaid at the end of two years. If it has to be repaid in two equal annual instalments (the instalments being paid at the beginning of the year) at 10% p.a. compounded annually, find the value of each instalment.

  1. ₹1210
  2. ₹1000
  3. ₹1100
  4. ₹1331

Answer (Detailed Solution Below)

Option 2 : ₹1000

Installments Question 5 Detailed Solution

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

Top Installments MCQ Objective Questions

A computer is available for Rs. 39,000 on cash payment or Rs. 19,000 as cash payment followed by five monthly installments of Rs. 4,200 each. What is the rate of interest per annum under the instalment plan?

  1. \(20\frac{19}{29}\) %
  2. \(20\frac{17}{29}\) %
  3. \(20\frac{20}{29}\) %
  4. \(20\frac{18}{29}\) %

Answer (Detailed Solution Below)

Option 3 : \(20\frac{20}{29}\) %

Installments Question 6 Detailed Solution

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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)%

A sum of Rs. P was borrowed and paid back in two equal yearly instalments, each of Rs. 35,280. If the rate of interest was 5% per annum and interest is compounding annually, then the value of P is ________.

  1. Rs. 64,400
  2. Rs. 65,600
  3. Rs. 65,400
  4. Rs. 64,800

Answer (Detailed Solution Below)

Option 2 : Rs. 65,600

Installments Question 7 Detailed Solution

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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.

What annual instalment will discharge a debt of ₹5,664 in 4 years at 12% simple interest?

  1. ₹1,230
  2. ₹1,210
  3. ₹1,200
  4. ₹1,220

Answer (Detailed Solution Below)

Option 3 : ₹1,200

Installments Question 8 Detailed Solution

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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.

A sum of Rs. 6,000 is to be paid back in two equal annual installments; each installment is to be paid at the end of every year. How much is each installment if the interest is compounded annually at 2% p.a.? (Rounded off up to two decimal places)

  1. Rs. 2,092.29
  2. Rs. 3,090.30
  3. Rs. 2,291.29
  4. Rs. 3,589.30

Answer (Detailed Solution Below)

Option 2 : Rs. 3,090.30

Installments Question 9 Detailed Solution

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Calculation:

Rate = 2% pa

Year    
I 100 102
II 1002

1022


As both installments are equal multiply I by 102, So

Year    
I 100 × 102 1022
II 1002

1022

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. 

A person borrowed ₹2,000 at 5% annual simple interest repayable in 3 equal annual installments. What will be the annual installment?

  1. ₹730\(\frac{10}{63}\)
  2. ₹840\(\frac{9}{61}\)
  3. ₹640\(\frac{11}{63}\)
  4. ₹250\(\frac{10}{63}\)

Answer (Detailed Solution Below)

Option 1 : ₹730\(\frac{10}{63}\)

Installments Question 10 Detailed Solution

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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. 

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. How much is the down payment?

  1. ₹1,55,945
  2. ₹1,95,455
  3. ₹1,94,555
  4. ₹1,45,955

Answer (Detailed Solution Below)

Option 2 : ₹1,95,455

Installments Question 11 Detailed Solution

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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.

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. The rate of interest is: 

  1. \(11{ 9 \over 11}\)%
  2. \(21{ 9 \over 11}\)%
  3. \(10{ 2 \over 11}\)%
  4. \(9{ 1 \over 11}\)%

Answer (Detailed Solution Below)

Option 2 : \(21{ 9 \over 11}\)%

Installments Question 12 Detailed Solution

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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 \(21{ 9 \over 11}\)%.

A person borrows Rs. 1,00,000 from a bank at 10% per annum simple interest and clears the debt in five years. If 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, respectively, what amount should be paid at the end of the fifth year to clear the debt?

  1. Rs. 38,250
  2. Rs. 30,000
  3. Rs. 40,450
  4. Rs. 36,450

Answer (Detailed Solution Below)

Option 2 : Rs. 30,000

Installments Question 13 Detailed Solution

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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.

Damani purchased an item costing ₹7,500 and paid ₹3,500 as a down payment for the same. If the simple interest charged for the remaining amount is 9% per annum and Damani cleared all dues after 4 months of the purchase, how much did Damani pay after 4 months as interest?

  1. ₹120
  2. ₹100
  3. ₹132
  4. ₹125

Answer (Detailed Solution Below)

Option 1 : ₹120

Installments Question 14 Detailed Solution

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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

A man borrows a certain sum of money and pays it back in 2 years in two equal instalments. If Compound Interest is reckoned at 5% per annum in case of annual compounding and he pays back annually Rs. 882, what sum did he borrow?

  1. Rs. 1,600
  2. Rs. 1,640
  3. Rs. 1,682
  4. Rs. 1,650

Answer (Detailed Solution Below)

Option 2 : Rs. 1,640

Installments Question 15 Detailed Solution

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Given Data:

Pays back annually Rs. 882 in 2 equal instalments.

Compound Interest is reckoned at 5% per annum.

Concept:

Compound Interest formula and the concept of annuity.

Calculation:

The man paid rupees 882 as the amount at the end of the first year and another rupees 882, as the amount at the end of the second year.

Principal for the first year = 882/(1 + (5/100)) = 840

Principal for the second year = 882/(1 + (5/100))2 = 800

Total principal = 840 + 800 = 1640

Therefore, sum did he borrow is Rs. 1640.

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