Compound Interest Multiple Choice Questions

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

\ 24000=\quad P{ (1+\frac { R }{ 100 } ) }^{ 4 }\\\ \\ 24000=\quad P(\frac { 100+R }{ 100 } )^{ 4 }\quad …….(a)\\\ \\ \\\ \\ 27783=\quad P{ (1+\frac { R }{ 100 } ) }^{ 7 }\\\ \\ 27783=\quad P(\frac { 100+R }{ 100 } )^{ 7 }\quad …….(b)\\\ \\ Dividing\quad equation(a)\quad from\quad equation(b),\quad we\quad get;\\\ \\ \\\ \\ \frac { 27783 }{ 24000 } ={ (\frac { 100+R }{ 100 } ) }^{ 3 }\\\ \\ \frac { 9261 }{ 8000 } ={ (\frac { 100+R }{ 100 } ) }^{ 3 }\\\ \\ \sqrt [ 3 ]{ \frac { 9261 }{ 8000 } } =\frac { 100+R }{ 100 } \\\ \\ \frac { 21 }{ 20 } =\frac { 100+R }{ 100 } \\\ \\ \frac { 21 }{ 20 } \times 100=100+R\\\ \\ 21\times 5=100+R\\\ \\ R=105-100\\\ \\ R=5\quad percent.

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

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

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

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

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

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

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

Each\quad Installment=(\frac { P }{ \frac { 1 }{ { (1+\frac { R }{ 100 } ) }^{ 1 } } +\frac { 1 }{ { (1+\frac { R }{ 100 } ) }^{ 2 } } +\frac { 1 }{ { (1+\frac { R }{ 100 } ) }^{ 3 } } } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 1 }{ { (1+\frac { 5 }{ 100 } ) }^{ 1 } } +\frac { 1 }{ { (1+\frac { 5 }{ 100 } ) }^{ 2 } } +\frac { 1 }{ { (1+\frac { 5 }{ 100 } ) }^{ 3 } } } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 1 }{ { (\frac { 100+5 }{ 100 } ) } } +\frac { 1 }{ { (\frac { 100+5 }{ 100 } ) }^{ 2 } } +\frac { 1 }{ { (\frac { 100+5 }{ 100 } ) }^{ 3 } } } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 1 }{ { (\frac { 105 }{ 100 } ) } } +\frac { 1 }{ { (\frac { 105 }{ 100 } ) }^{ 2 } } +\frac { 1 }{ { (\frac { 105 }{ 100 } ) }^{ 3 } } } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 1 }{ { (\frac { 21 }{ 20 } ) } } +\frac { 1 }{ { (\frac { 21 }{ 20 } ) }^{ 2 } } +\frac { 1 }{ { (\frac { 21 }{ 20 } ) }^{ 3 } } } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 20 }{ { 21 } } +{ (\frac { 20 }{ 21 } ) }^{ 2 }+{ (\frac { 20 }{ 21 } ) }^{ 3 } } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 20 }{ { 21 } } [1+{ \frac { 20 }{ 21 } }+{ (\frac { 20 }{ 21 } ) }^{ 2 }] } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 20 }{ { 21 } } [1+{ \frac { 20 }{ 21 } }+{ \frac { 400 }{ 441 } }] } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 20 }{ { 21 } } [\frac { 441+420+400 }{ 441 } ] } )\\\ \\ Each\quad Installment=(\frac { 8000 }{ \frac { 20 }{ { 21 } } [\frac { 1261 }{ 441 } ] } )\\\ \\ Each\quad Installment=8000\times \frac { 21 }{ 20 } \times \frac { 441 }{ 1261 } \\\ \\ Each\quad Installment=₹2937.66

Hence the value of installments is Rs 2937.66

option (a) is the right answer

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

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

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

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

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

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

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