Compound Interest Solved Problems

In this post we will discuss questions of compound interest. All the questions are fully solved with step by step process so that there will be no room of doubt for the students.

Remember that throughout the questions i have used different types of formulas to solve that questions quickly. Your job as a student to remember the formulas and understand the type of question where the formula is applicable.

All these compound interest questions are asked in competition exams like GRE, GMAT, CAT, CMAT, NMAT, SNAP, SSC, SSC-CGL, SSC-CHSL, RRB, Banking, SBI po, SBI clerk, NABARD, NDA, AFCAT etc.

Compound Interest Questions

(01) A sum of money amounts to Rs. 4840 in 2 years and to Rs. 5324 in 3 years at compound interest compounded annually. The rate of interest per annum is?

For this type of question we have shortcut formula which we have discussed below

b= 3 years
a= 2 years
B = rs 5324
A = rs 4840

We have to find rate of interest per annum

For this question, we will use following formula
R=(\frac { B }{ A } -1)\times 100\\\ \\ R=(\frac { 5324 }{ 4840 } -1)\times 100\\\ \\ R=(\frac { 5324-4840 }{ 4840 } )\times 100\\\ \\ R=\frac { 484 }{ 4840 } \times 100

R= 10 percent
Hence the required rate of interest is 10 percent

(02) A money lander borrows money at 4% per annum simple interest and pays the interest at the end of the year. He lends it at 6% per annum compound interest compounded half yearly and receive the interest at the end of year. In this way he gains rs 104.5 a year. The amount of money he borrows is?

Let the borrowed amount be x
Rate (r) = 4%
then interest paid by him = (P * R * T) /100
=> (x*4*1)/100 ==> 4x/100

Interest\quad received\quad by\quad the\quad money\quad lender=P[(1+\frac { r }{ 100 } )^{ n }-1]\\\ \\ where\quad r={ 6 }/{ 2 }=3percent\\ \\ and\quad n=2\\\ \\ =x[(1+\frac { 3 }{ 100 } )^{ 2 }-1]\\\ \\ =x[\frac { { (103) }^{ 2 } }{ { (100) }^{ 2 } } -1]\\\ \\ =x[\frac { { 103 }^{ 2 }-{ 100 }^{ 2 } }{ 10000 } ]\\\ \\ =x[\frac { (103+100)(103-100) }{ 10000 } ]\\\ \\=x[\frac { 203\times 3 }{ 10000 } ]=\frac { 609x }{ 10000 } \\\ \\ As\quad we\quad know,\ \frac { 609x }{ 10000 } -\frac { 4x }{ 100 } =104.50\\\ \\ =>\frac { 60900x-40000x }{ 10000 } =104.50\\\ \\ =>\frac { 209x }{ 10000 } =104.50\\\ \\ =>209x=1045000\\\ \\ =>x=\frac { 1045000 }{ 209 } =₹5000

Hence 5000rs is the amount the man borrowed

(03) A loan of rs. 12300 at 5% per annum compound interest is to be repaid in two equal annual installments at the end of every year. Find the amount of each installment

Sol:
Principal (P) to be repaid = Rs. 12300
Interest for first year at 5% = 12300 * 5/100 = Rs. 615

Let equal annual installment = S
After 1 year principal to be paid= 12300+615-S= Rs. 12915-S

Interest for second year= (12915-S) *5/100 =(12915 – S)/20

Total amount to be paid= Principal + First year interest + Second year interest
==> 12300 + 615 + (12915-S)/20 —(a)

Total Amount Paid= 2S — (b)

From equation (a) & (b)
2S=12915+\frac { (12915-S) }{ 20 } \\\ \\ 2S=\frac { (12915\times 20)+(12915-S) }{ 20 } \\\ \\ 40S=12915\times 21-S\\\ \\ 41S=12915\times 21\\\ \\ S=\frac { 12915\times 21 }{ 41 } \\\ \\ S=315\times 21\\\ \\ S=₹6615

Hence the amount of each installments is Rs. 6615

(04) At a certain rate per annum, the simple interest on a sum of money for 1 year is rs 260 and the compound interest on the same sum for two years is 540.80 rs. Find the rate of interest per annum?

Given,
Simple interest for 1 year= Rs. 260
Compound interest for 2 years = 540.80 Rs

Let
Principal = P
Rate of interest(R) =r

Simple interest = P*R*T/100
==> 260 = P* r * 1/100 —(a)

Compound\quad Interest=P[(1+\frac { R }{ 100 } )^{ n }-1]\\\ \\ 540.80=P[(1+\frac { r }{ 100 } )^{ 2 }-1]\\\ \\ 540.80=P[1+\frac { 2r }{ 100 } +\frac { { r }^{ 2 } }{ 10000 } -1]\\\ \\ 540.80=\frac { 2Pr }{ 100 } +\frac { { Pr }^{ 2 } }{ 10000 } \\\ \\ from\quad equation\quad (a)\\\ \\ 540.80=2\times 260+\frac { 260r }{ 100 } \\\ \\ 540.80=520+\frac { 260r }{ 100 } \\\ \\ 540.80=\frac { 52000+260r }{ 100 } \\\ \\ 54080=52000+260r

==>260r=54080-52000
==>260r=2080
==> r=8 percent

Hence 8% is the rate of interest per annum

(05) The compound interest on a certain sum of money at 5% per annum for 2 years is Rs. 246. The simple interest on the same sum for 3 years at 6% per annum

Compound Interest = rs. 246
Rate (R)=5 %
Time (n)= 2 years

C.I=P(1+\frac { R }{ 100 } )^{ n }-1\\\ \\ 246=P(1+\frac { 5 }{ 100 } )^{ 2 }-1\\\\ \\ 246=P(\frac { 100+5 }{ 100 } )^{ 2 }-1\\\ \\ 246=P{ (\frac { 105 }{ 100 } ) }^{ 2 }-1\\\ \\ 246=P{ (\frac { 21 }{ 20 } ) }^{ 2 }-1\\\ \\ 246=P(\frac { 441 }{ 400 } -1)\\\ \\ 246=P(\frac { 441-400 }{ 400 } )\\\ \\ 246=P(\frac { 41 }{ 400 } )\\\ \\ P=\frac { 246\times 400 }{ 41 } \\\ \\ P=₹2400

Now we have to calculate simple interest
P= Rs 2400
R= 6% per annum
T = 3 years

Simple interest => (P * R * T)/100
==> S.I = 2400 *6*3/100
==> S.I= rs 432

Hence the simple interest is 432 rs

(06) The compound interest on a certain sum for two successive years are rs 225 and 238.5 rs. Find the rate of interest per annum

S.I on Rs. 225 for 1 year =Difference of C.I for two years
S.I on Rs. 225 for 1 year = Rs. 238.5 – Rs 225
S.I on Rs. 225 for 1 year = Rs 13.5

We know that:
Rate=\frac { S.I\times 100 }{ Principal\times Time } \\\ \\ R=\frac { 13.50\times 100 }{ 225\times 1 } \\\ \\ R=\frac { 1350 }{ 225 } \\\ \\ R=6\quad percent.\


Hence the required rate is 6% per annum

(07) A sum of Rs 12,000 deposited at compound interest becomes double after 5 years. How much it will be after 20 years?

Principal = rs 12,000
Time (n) = 5 years
Amount = 24000 rs

A=P{ (1+\frac { R }{ 100 } ) }^{ n }\\\ \\ 24000=12000{ (1+\frac { R }{ 100 } ) }^{ 5 }\\\ \\ \frac { 24000 }{ 12000 } ={ (1+\frac { R }{ 100 } ) }^{ 5 }\\\ \\ 2={ (1+\frac { R }{ 100 } ) }^{ 5 }\\\ \\ 2^{ 4 }=[{ (1+\frac { R }{ 100 } ) }^{ 5 }]^{ 4 }…………..(multiplying\quad both\quad the\quad sides\quad by\quad power\quad of\quad 4)\\\ \\ 16={ (1+\frac { R }{ 100 } ) }^{ 20 }

The sum amount to be 16 times of principal after 20 years
12000*16 =192000 rs

Hence after 20 years the amount will be Rs. 192000

(08) On a certain sum of money, the difference between compound interest for a year, payable half yearly, and simple interest for a year is Rs. 56. If the rate of interest for bot cases is 16%, then the sum is?

Given
Compound Interest – Simple Interest = Rs. 56
Interest is compounded half yearly then
Rate ( R ) = 16/2 = 8 percent
Time (n) = 2 years

For this type of question, we will use the special formula
Remember it is important to remember these formulas as it would help us to solve questions fast in the examination hall

C.I-S.I=P{ (\frac { R }{ 100 } ) }^{ n }\\\ \\ 56=P{ (\frac { 8 }{ 100 } ) }^{ 2 }\\\ \\ 56=P{ (\frac { 2 }{ 25 } ) }^{ 2 }\\\ \\ 56=P(\frac { 4 }{ 625 } )\\\ \\ P=\frac { 56\times 625 }{ 4 } \\\ \\ P=₹8750

Hence Rs. 8750 is the initial principal for this question

(09) A sum of money at compound interest amounts to thrice itself in three years. In how any years it will be 9 times itself?

Let the sum be x
x becomes 3x in three years then,

Principal (P)=x
Amount (A) =3x
Time (n)= 3 years

P{ (1+\frac { R }{ 100 } ) }^{ n }=A\\\ \\ =>(1+\frac { R }{ 100 } ​)^{ 3 }=3…….(a)

Let x becomes 9x in n years, then;
Principal (P)=x
Amount (A)=9x
Time (n)=n

=>(1+\frac { R }{ 100 } ​)^{ n }=9x\\\ \\ =>(1+\frac { R }{ 100 } ​)^{ n }=9…….(b)\\\ \\ =>[(1+\frac { R }{ 100 } ​)^{ 3 }]^{ 2 }={ 3 }^{ 2 }=9\quad \quad \quad \quad From(a)\\\ \\ =>(1+\frac { R }{ 100 } ​)^{ 6 }=9\quad ……(c)\\\ \\ =>From(b)and(c)\\\ \\ =>n=6

Required number of years is 6

(10) An amount of money at compound interest grows upto rs. 3840 in 4 years and up to rs. 3936 in 5 years. Find the rate of interest?

Given,
b = 5 years
a = 4 years and

B= rs 3936
A = rs. 3840

For this type of question we have to use following formula

R=(\frac { B }{ A } -1)\times 100\\\ \\ R=(\frac { 3936 }{ 3840 } -1)\times 100\\\ \\ R=(\frac { 3936-3840 }{ 3840 } )\times 100\\\ \\ R=(\frac { 96 }{ 3840 } )\times 100\\\ \\ R=(\frac { 1 }{ 40 } )\times 100\\\ \\ R=\frac { 100 }{ 40 } =2.5\quad percent.\

Hence the rate of interest is 2.5%