# Compound Interest Multiple Choice Questions

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## Compound Interest MCQ

(01) An amount becomes Rs 24,000 in 4 years and Rs 27, 783 in 7 years at compound interest. Find the annual compound interest rate?

a. 10%
b. 5%
c. 15%
d. 20%

Let the Principal (P) =P
Rate (R) = R
Time (n1) = 4 years
Time (n2) = 7 years
Amount (A1) = Rs. 24,000
Amount (A2) = Rs. 27, 783

We have to find the Rate of interest

Hence the required rate of interest is 5%
Option (b) is the right answer

(02) The cost of machine decreases by 4% every year. If after three years its cost remains Rs 70778.8, Find its initial cost?

a. 80,000
b. 90,000
c. 85,000
d. 95,000

Let Amount = Rs 70778.8
Principal (P) = P
Rate (R) = 4
Time (n) = 3 years

A=P{ (1-\frac { R }{ 100 } ) }^{ n }\\\ \\ 70778.88=P{ (1-\frac { 4 }{ 100 } ) }^{ 3 }\\\ \\ 70778.88=P{ (\frac { 100-4 }{ 100 } ) }^{ 3 }\\\ \\ 70778.88=P{ (\frac { 96 }{ 100 } ) }^{ 3 }\\\ \\ 70778.88=P(\frac { 96 }{ 100 } \times \frac { 96 }{ 100 } \times \frac { 96 }{ 100 } )\\\ \\ P=\frac { 70778.88\times 100\times 100 }{ 96\times 96\times 96 } \\\ \\ P=8\times 100\times 100\\\ \\ P=₹80000

Hence the initial cost of machinery is Rs. 80,000
option (a) is the right answer

(03) Bhuvnesh lends Rs. 12,000 on the condition that he will charge 5% per annum for the first year, 6% per annum for next two years and10% per annum for the next three years. Find the amount he will receive as compound interest after the completion of time period?

a. Rs. 6500
b. Rs. 6232.32
c. Rs 6700.5
d. Rs 6843.4

Let the Principal (P) = Rs 12,000
Time (n1) = 1 years
Time (n2) = 2 years
Time (n3) = 3 years

Rate (R1)= 5%
Rate (R2) = 6%
Rate (R3) = 10%

\\ A=12000{ (1+\frac { 5 }{ 100 } ) }^{ 1 }{ (1+\frac { 6 }{ 100 } ) }^{ 2 }{ (1+\frac { 10 }{ 100 } ) }^{ 3 }\\\ \\ A=12000(\frac { 100+5 }{ 100 } ){ (\frac { 100+6 }{ 100 } ) }^{ 2 }{ (\frac { 100+10 }{ 100 } ) }^{ 3 }\\\ \\ A=12000(\frac { 105 }{ 100 } ){ (\frac { 106 }{ 100 } ) }^{ 2 }{ (\frac { 110 }{ 100 } ) }^{ 3 }\\\ \\ A=120(105){ (\frac { 53 }{ 50 } ) }^{ 2 }{ (\frac { 11 }{ 10 } ) }^{ 3 }\\\ \\ A=120\times 105\times \frac { 2809 }{ 2500 } \times \frac { 1331 }{ 1000 } \\\ \\ A=₹18843.44

The total amount to be paid back after 6 years => Rs 18843.44

Compound Interest = Amount – Principal
Compound Interest = Rs 1884.44 – 12,000 => Rs. 6843.4

Hence the total interest paid after 6 years => Rs. 6843.4
option (d) is the right answer

(04) Find the amount at which the difference between the simple interest and compound interest at 5% per annum for 3 years is Rs. 91.5

a. Rs 12,000
b. Rs 11,000
c. Rs. 13,000
d. Rs 14,000

Rate of interest (R) = 5%
Time (n) = 3 years

Amount(A)=P{ (1+\frac { R }{ 100 } ) }^{ n }\\\ \\ A=P{ (1+\frac { 5 }{ 100 } ) }^{ 3 }\\\ \\ \ A=P{ (\frac { 21 }{ 20 } ) }^{ 3 }\\\ \\ \\\ \\ \ C.I=A-P\\\ \\ C.I=P{ (\frac { 21 }{ 20 } ) }^{ 3 }-P\\\ \\ \\\ \\ \ S.I=\frac { P\times R\times T }{ 100 } \\\ \\ S.I=\frac { P\times 5\times 3 }{ 100 } \\\ \\ S.I=\frac { 15P }{ 100\ } \\\ \\ \\\ \\ Given\quad that;\quad C.I-S.I=₹91.5\\\ \\ [P{ (\frac { 21 }{ 20 } ) }^{ 3 }-P]-\frac { 15P }{ 100\ } =91.5\\\ \\ [P(\frac { 9261 }{ 8000 } )-P]-\frac { 15P }{ 100\ } =91.5\\\ \\ (\frac { 9261P }{ 8000 } -P)-\frac { 15P }{ 100\ } =91.5\\\ \\ (\frac { 9261P-8000P }{ 8000 } )-\frac { 15P }{ 100\ } =91.5\\\ \\ \frac { 1261P }{ 8000 } -\frac { 15P }{ 100\ } =91.5\\\ \\ \frac { 1261P-15P\times 80 }{ 8000 } =91.5\\\ \\ \frac { 1261P-1200P }{ 8000 } =91.5\\\ \\ \frac { 61P }{ 8000 } =91.5\\\ \\ \ P=₹12000\<strong> </strong>

Hence Rs. 12,000 is the required amount.
Option (a) is the right answer

(05) Rakesh borrows a sum at 3% per annum compounded yearly and lends it at 5% per annum compounded half yearly. In this way he earns a profit of Rs 1320 at the end of the year. Find the amount he initially borrows?

a. Rs. 61,000
b. Rs. 62,000
c. Rs. 64,000
d. Rs. 65,000

When Rakesh lends money
Time (n1) = 2 half years
Rate (R1) = 5% per annum = 5/2 => 2.5 => 1/40 compounded half yearly

Rate 1/40 means when principal is 40 then amount is one more than principal (amount 41)

40 (Principal) ———-> 41 (Amount)
40*40 (Principal) ——> 41 * 41 (Amount)
1600 (Principal) ——— 1681 (Amount)
So let us assume that Rakesh lends 1600 units and get back 1681 unit back

Now, when Rakesh borrows money
Time (n2) = 1 year
Rate (r2) = 3/100

Rate 3/100 means that if principal is Rs. 100 then the amount will be 103 rs
Rs. 100 (Principal) ————-> Rs. 103 (Amount)
100 * 16 (Principal) ———–> 103 * 16 (Amount)
1600 (principal) ————–>1648 (Amount)
Here we assumed that Rakesh borrowed 1600 units and returned 1648 units

Its given in the question that
Lending unit – Borrowing unit = Rs. 1320
1681 – 1648 unit = Rs. 1320
33 unit = Rs. 1320
1 unit =1320/33

if 1 unit =1320/33, then 1600 unit (sum he borrowed) is
==> 1600 unit = (1320/33) * 1600
==> 1600 unit = Rs. 64,000

Hence rakesh borrowed Rs. 64,000
option (c) is the right answer

(06) Rakesh took a loan of Rs. 6300 from a bank and decided to repay it back in three equal annual installments at 10% compound interest per annum. Find the amount of each installment

a. Rs 2631.16
b. Rs 2533.32
c. Rs 2532.10
d. Rs 2530.5

Let the value of each installment be P
Rate (R) = 10% per annum

Its been said that Rakesh will pay back the loan in three future installments.
So the value of loan is equal to the present value of future installments.

6300=(\frac { P }{ { (1+\frac { R }{ 100 } ) }^{ 1 }+{ (1+\frac { R }{ 100 } ) }^{ 2 }+{ (1+\frac { R }{ 100 } ) }^{ 3 } } )\\\ \\ 6300=(\frac { P }{ { (1+\frac { 10 }{ 100 } ) }^{ 1 }+{ (1+\frac { 10 }{ 100 } ) }^{ 2 }+{ (1+\frac { 10 }{ 100 } ) }^{ 3 } } )\\\ \\ 6300=(\frac { P }{ { (\frac { 100+10 }{ 100 } ) }+{ (\frac { 100+10 }{ 100 } ) }^{ 2 }+{ (\frac { 100+10 }{ 100 } ) }^{ 3 } } )\\\ \\ 6300=(\frac { P }{ { (\frac { 110 }{ 100 } ) }+{ (\frac { 110 }{ 100 } ) }^{ 2 }+{ (\frac { 110 }{ 100 } ) }^{ 3 } } )\\\ \\ 6300=(\frac { P }{ { (\frac { 11 }{ 10 } ) }+{ (\frac { 11 }{ 10 } ) }^{ 2 }+{ (\frac { 11 }{ 10 } ) }^{ 3 } } )\\\ \\ \ 6300=(\frac { P }{ { (\frac { 11 }{ 10 } ) }+{ (\frac { 121 }{ 100 } ) }+{ (\frac { 1331 }{ 1000 } ) } } )\\\ \\ 6300=P[\frac { 10 }{ 11 } +\frac { 100 }{ 121 } +\frac { 1000 }{ 1331 } ]\\\ \\ 6300=P[\frac { 1210+1100+1000 }{ 1331 } ]\\\ \\ 6300\times 1331=3310P\\\ \\ P=\frac { 6300\times 1331 }{ 3310 } \\\ \\ P=₹2533.32

Hence the value of each installment is Rs. 2533.32
option (b) is the right answer

(07) The cost of TV is Rs. 12,000. A customer bought it after paying Rs. 4000 as down payment and he promises to pay the rest in three equal installments at 5% rate of interest. Find the value of each installment paid by the customer

a. Rs 2937.66
b. Rs 2941.50
c. Rs. 2800
d. Rs 2950.6

Total T.V cost = Rs. 12,000
Down Payment = Rs. 4000
Balance = 12,000-4000 = rs. 8000

So this Rs. 8000 is paid back in three equal installments at 5% rate of interest.
The loan amount Rs. 8000 is equal to present value of all the future installents

Hence the value of installments is Rs 2937.66

option (a) is the right answer

(08) I lent Rs 2500 to Rakesh at 24 percent per annum compound interest for 15 months. If the interest is compounded after every 5 months, than find the amount that Rakesh will pay back to me?

a. Rs 3320.4
b. Rs 3325.5
c. Rs 3327.5
d. Rs 3333.5

Interest is payable after every 5 onths then

Rate = 24 * (5/12) = 10%
Time (n) = 15 months ==> 3 (five months)
Principal = Rs 2500

Amount(A)=P{ (1+\frac { R }{ 100 } ) }^{ n }\\\ \\ A=2500{ (1+\frac { 10 }{ 100 } ) }^{ 3 }\\\ \\ A=2500{ (\frac { 110 }{ 100 } ) }^{ 3 }\\\ \\ A=2500\times \frac { 1331 }{ 1000 } \\\ \\ A=₹3327.50

Hence, Rakesh will pay back Rs. 3327.50

option (c) is the right answer

(09) A merchant gives a loan at different rates for different years. The rate of compound interest for first two years is 10% per annum for the next three years is 5% per annum, for the last year is 6.25% per annum respectively. A poor man took a loan of Rs. 6400 from him and paid after the completion of time period. How much poor man paid?

a. Rs. 9500
b. Rs. 9525
c. Rs. 9550
d. Rs. 9575

Rate of interest for first 2 years (R1) = 10%, Time (n1) =2
Rate of interest for next 3 years (R2)=5% , Time (n2)=3
rate of interest for last year (R30 = 6.25% , Time (n3)=1

Principal = Rs. 6400

Amount(A)=P{ (1+\frac { { R }<em>{ 1 } }{ 100 } ) }^{ { n }</em>{ 1 } }{ (1+\frac { { R }<em>{ 2 } }{ 100 } ) }^{ { n }</em>{ 2 } }{ (1+\frac { { R }<em>{ 3 } }{ 100 } ) }^{ { n }</em>{ 3 } }\\\ \\ A=6400{ (1+\frac { 10 }{ 100 } ) }^{ 2 }{ (1+\frac { 5 }{ 100 } ) }^{ 3 }{ (1+\frac { 6.25 }{ 100 } ) }^{ 1 }\\\ \\ \ A=6400{ (\frac { 11 }{ 10 } ) }^{ 2 }{ (\frac { 21 }{ 20 } ) }^{ 3 }{ (\frac { 106.25 }{ 100 } ) }\\\ \\ A=64{ (\frac { 121 }{ 100 } ) }{ (\frac { 9261 }{ 8000 } ) }{ (106.25) }\\\ \\ A=₹9524.93\

Hence the poor man paid Rs 9525
option (b) is the right answer

(10) A man borrows Rs. 9000 at 10% compound interest per annum. After the end of each year he returns Rs. 3000. At the end of third year how much money should he return to settle all his debt

a. Rs. 5045
b. Rs. 5049
c. Rs . 5050
d. Rs. 5051

Calculation for 1 year
Principal = Rs 9000
Rate = 10%
Time = 1 year

Interest = (P * R * T)/100 => 9000*10/100 => Rs. 900

Rs. 900 is the interest for 1 year.

Total Amount becomes => 9000 + 900 => Rs. 9900
Amount repaid => Rs. 3000
Remaining Amount ==> 9900 – 3000 => Rs. 6900

Calculation for second year
Principal = Rs. 6900
Rate = 10%
Time = 1 year

Interest => (P * R * T)/100 ==> 6900 * 10/100 => Rs 690

Amount to be paid=> 6900 + 690 ==> Rs. 7590
Amount paid => Rs 3000
Remaining amount ==> 7590 – 3000 => Rs. 4590

Calculation for third year
Principal => Rs. 4590
Rate => 10% per annum
time = 1 year

Interest => P * R * T/100 => 4590 *10/100 => Rs. 459

Total amount => 4590 + 459 => Rs. 5049
hence Rs 5049 is to be paid back in the third year to settle all the debt

option (b) is the right answer

(11) Rs 22830 was divided into two parts and given to A & B and were told that they should spent at 5% per annum. A used it for 9 years and B used it for 11 years and then they left with equal amount. Find the money given to each person initially?

a. 10830, 12000
b. 12000, 13000
c. 11540 , 12000
d. 12000, 14000

Case 01: For person ” A”
Rate (R) = 5 , Time (n1) = 9

As the money is decreasing we will use following formula

{ A }<em>{ 1 }=P{ (1-\frac { R }{ 100 } ) }^{ { n }</em>{ 1 } }\\\ \\ { A }<em>{ 1 }=P{ (1-\frac { 5 }{ 100 } ) }^{ 9 }\\\ \\ { A }</em>{ 1 }=P{ (\frac { 19 }{ 20 } ) }^{ 9 }………(a)

Case 02 : For person “B”
Rate (R) = 5% , Time (n2)= 11 years

{ A }<em>{ 1 }=P{ (1-\frac { R }{ 100 } ) }^{ { n }</em>{ 2 } }\\\ \\ { A }<em>{ 1 }=P{ (1-\frac { 5 }{ 100 } ) }^{ 11 }\\\ \\ \ { A }</em>{ 1 }=P{ (\frac { 19 }{ 20 } ) }^{ 11 }………(b)\\\ \\ \ Subtracting\quad equation\quad 'a'\quad and\quad equation\quad 'b'\\\ \\ P{ (\frac { 19 }{ 20 } ) }^{ 11 }-P{ (\frac { 19 }{ 20 } ) }^{ 9 }\\\ \\ P{ (\frac { 19 }{ 20 } ) }^{ 2 }\\\ \\ P(\frac { 361 }{ 400 } )……..(c)

from equation ‘c’ we get A & B share, that is;
A’s share = 361x
B’s share = 400x

A share + B share = 22830
361x + 400x =22830
761x = 22830
x = 22830/761
x = Rs. 30

Initial amount given to A ==> 361 * 30 ==> Rs. 10830
Initial amount given to B ==> 400 * 30 ==> Rs. 12,000

Option (a) is the right answer

(12) Rakesh deposited certain sum at the beginning of every year and bank gives 10% compound interest on the sum. At the and of third year the amount in his account is Rs. 7282. Find how much amount he deposited each year?

a. Rs. 1200
b. Rs. 1500
c. Rs 3000
d. Rs 2000

Rate = 10%
Amount = Rs 7282

Let each installment be P

According to question
A=\frac { P }{ \frac { 1 }{ (1+\frac { R }{ 100 } ) } +\frac { 1 }{ { (1+\frac { R }{ 100 } ) }^{ 2 } } +\frac { 1 }{ { (1+\frac { R }{ 100 } ) }^{ 3 } } } \\\ \\ 7282=\frac { P }{ \frac { 1 }{ (1+\frac { 10 }{ 100 } ) } +\frac { 1 }{ { (1+\frac { 10 }{ 100 } ) }^{ 2 } } +\frac { 1 }{ { (1+\frac { 10 }{ 100 } ) }^{ 3 } } } \\\ \\ \ 7282=\frac { P }{ \frac { 10 }{ 11 } +{ (\frac { 10 }{ 11 } ) }^{ 2 }+{ (\frac { 10 }{ 11 } ) }^{ 3 } } \\\ \\ 7282=P[\frac { 11 }{ 10 } +{ (\frac { 11 }{ 10 } ) }^{ 2 }+{ (\frac { 11 }{ 10 } ) }^{ 3 }]\\\ \\ \ 7282=P(\frac { 3641 }{ 1000 } )\\\ \\ \ P=₹2000

Hence each installment is Rs.2000

Option (d) is the right answer

(13) Bhuvnesh bought a scooter on the condition that he would pay rs. 12,000 instantly and rs. 1680 after 1 year and Rs. 5292 after two years. If the payment is at 5% compound interest, then what was the cost price of scooter?

a. Rs 18200
b. Rs 18100
c. Rs 18400
d. Rs 18000

Case 01:
Amount (A) = Rs. 1680
Rate (R) = 5%
time (n) = 1 year

A=P{ (1+\frac { R }{ 100 } ) }^{ n }\\\ \\ 1680=P{ (1+\frac { 5 }{ 100 } ) }^{ 1 }\\\ \\ \ 1680=P(\frac { 21 }{ 20 } )\\\ \\ P=\frac { 1680\times 20 }{ 21 } \\\ \\ P=₹1600 </p> <p></p> <p>Case 02: Amount (A1) = Rs 5292 Rate (r) = 5% Time = 2</p> [latex] 5292=p{ (1+\frac { 5 }{ 100 } ) }^{ 2 }\\\ \\ 5292=p{ (\frac { 100+5 }{ 100 } ) }^{ 2 }\\\ \\ \ 5292=p(\frac { 441 }{ 400 } )\\\ \\ p=₹4800

Cost Price = 12000 + 1600 + 4800 = Rs. 18400

Hence the cost price is Rs. 18400

Option (c) is the right answer

(14) The PNB lends Rs. 1331 to Rakesh at compound interest and got Rs. 1728 after three years. What is the rate of interest if the interest is compounded annually

a. 9.09%
b. 10%
c. 6.06%
d. 5.05%

Principal = Rs. 1331
Amount = Rs. 1728
Time = 3 years

1728=1331{ (1+\frac { R }{ 100 } ) }^{ 3 }\\\ \\ \frac { 1728 }{ 1331 } ={ (1+\frac { R }{ 100 } ) }^{ n }\\\ \\ \sqrt [ 3 ]{ \frac { 1728 }{ 1331 } } =1+\frac { R }{ 100 } \\\ \\ \frac { 12 }{ 11 } =\frac { 100+R }{ 100 } \\\ \\ \ R=\frac { 100 }{ 11 } =9.09

hence the rate of interest is 9.09%

option (a) is the correct answer