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compound interest questions with solutions for bank exams

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A sum of money placed at compound interest double itself in 4 years. In how many years will it amount to four times itself?

(a) 2 years

(b) 4 years

(c) 8 years

(d) 10 years

(e) 12 years

Solution: (c)

Rs 1 becomes Rs 2 in 4 years

Rs 2 becomes Rs 4 in another 4 years

\(\displaystyle \Rightarrow \)Rs 1 becomes Rs 4 in (4 + 4) i.e., 8 years.

Alternate method

Let, Principal=Rs. 100%

Amount=Rs. 200

Rate=r%

Time=4 years

Now,

Amount= \(\displaystyle =P\times {{\left( {1+\frac{r}{{100}}} \right)}^{n}}\)

\(\displaystyle \begin{array}{l}200=100\times {{\left( {1+\frac{r}{{100}}} \right)}^{4}}\\2={{\left( {1+\frac{r}{{100}}} \right)}^{4}}——-(i)\end{array}\)

If sum become 4 times in the time ‘n’ years

Then,

\(\displaystyle \begin{array}{l}4={{\left( {1+\frac{r}{{100}}} \right)}^{n}}\\{{2}^{2}}={{\left( {1+\frac{r}{{100}}} \right)}^{n}}——(ii)\end{array}\)

Substituting  eqn (i) in( ii), we get

\(\displaystyle \begin{array}{l}{{\left[ {{{{\left( {1+\frac{r}{{100}}} \right)}}^{4}}} \right]}^{2}}={{\left( {1+\frac{r}{{100}}} \right)}^{n}}\\{{\left( {1+\frac{r}{{100}}} \right)}^{8}}={{\left( {1+\frac{r}{{100}}} \right)}^{n}}\end{array}\)

Therefore, n=8 years

The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:

(a) Rs. 400

(b) Rs. 500

(c) Rs. 600

(d) Rs. 700

(e) Rs. 800


Solution: (b)
Let the sum be Rs. P. Then
\(\displaystyle \begin{array}{l}P{{\left( {1+\frac{{10}}{{100}}} \right)}^{2}}-P=525\\\Rightarrow P{{\left( {\frac{{11}}{{10}}} \right)}^{2}}-P=525\\\Rightarrow P\left[ {{{{\left( {\frac{{11}}{{10}}} \right)}}^{2}}-1} \right]=525\\\Rightarrow P=\frac{{525\times 100}}{{21}}=2500\end{array}\)
Therefore Sum=Rs. 2500
So,
\(\displaystyle \begin{array}{l}SI=Rs.\left( {\frac{{2500\times 5\times 4}}{{100}}} \right)\\=Rs.500\end{array}\)

If the compound interest on a certain sum of money for 3 years at 10% p.a. be ₹ 993, what would be the simple interest?

(a) ₹ 800

(b) ₹ 950

(c) ₹ 900

(d) ₹ 1000

(e) None of these


Solution: (c)
Let Principal = ₹ P
\(\displaystyle p{{(1+\frac{{10}}{{100}})}^{3}}-p=993\)
\(\displaystyle \Rightarrow \)\(\displaystyle (\frac{{11}}{{10}}\times \frac{{11}}{{10}}\times \frac{{11}}{{10}}-1)p=993\)
\(\displaystyle \Rightarrow \)\(\displaystyle (\frac{{1331-1000}}{{1000}})p=993\)
P = \(\displaystyle \frac{{993\times 1000}}{{331}}=3000\)
Therefore, Simple interest = \(\displaystyle \frac{{3000\times 3\times 10}}{{100}}=900\)

The difference between compound interest and simple interest on a sum for 2 years at 10% per annum, when the interest is compounded annually is ₹ 16. If the interest were compounded half-yearly, the difference in two interests would be:

(a) ₹ 24.81

(b) ₹ 26.90

(c) ₹ 31.61

(d) ₹ 32.40

(e) None of these


Solution: (a)
For first year, S.I. = C.I.
Now, ₹ 10 is S.I. on ₹ 100.
Therefore, ₹16 is S.I. on \(\displaystyle \frac{{100}}{{10}}\times 16=160\)
So, S.I. on principal for 1 year at 10% is ₹160
Therefore, Principal = ₹ \(\displaystyle \frac{{100\times 160}}{{10\times 1}}=1600\)
Amount for 2 years compounded half yearly
\(\displaystyle [1600\times {{(1+\frac{5}{{100}})}^{4}}]=1944.81\)
Therefore, C.I. = ₹ (1944.81 – 1600) = ₹ 24.81.
S.I. = \(\displaystyle \frac{{1600\times 10\times 2}}{{100}}=320\)
Therefore, (C.I.) – (S.I.) = ₹ (344.81 – 320) = ₹ 24.81

A person lent out a certain sum on simple interest and the same sum on compound interest at certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and that of 2 years is 25 : 8. The rate of interest per annum is:

(a) 10%

(b) 11\(\displaystyle \frac{1}{2}\)%

(c) 12%

(d) 12 \(\displaystyle \frac{1}{2}\)%

(e) None of these


Solution: (d)
Let the principal be ₹ P and rate of interest be R% per annum.
Difference of C.I. and S.I. for 3 years
= \(\displaystyle [P\times {{(1+\frac{R}{{100}})}^{3}}-P]-(\frac{{P\times R\times 3}}{{100}})=\frac{{P{{R}^{2}}}}{{{{{10}}^{4}}}}(\frac{{300+R}}{{100}})\)
Difference of C.I. and S.I. for 2 years = \(\displaystyle P{{(\frac{R}{{100}})}^{3}}\)
\(\displaystyle \frac{{\frac{{P{{R}^{2}}}}{{{{{10}}^{4}}}}(\frac{{300+R}}{{100}})}}{{\frac{{P{{R}^{2}}}}{{{{{10}}^{4}}}}}}=\frac{{25}}{8}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{300+R}}{{100}}=\frac{{25}}{8}\)
\(\displaystyle \Rightarrow \)R = \(\displaystyle \frac{{100}}{8}=12\frac{1}{2}\%\)

A father left a will of Rs. 16400 for his two sons aged 17 and 18 years. They must get equal amount when they are 20 years, at 5% compound interest. Find the present share of the younger son?

(a) Rs. 8000

(b) Rs. 8200

(c) Rs. 8400

(d) Rs. 8600

(e) Rs. 8800


Solution: (a)
Let the share of the younger and elder sons be Rs. x and Rs. (16400 – x)
Then, amount of Rs. x after 3 years = Amount of Rs. (16400 – x) after 2 years
\(\displaystyle \begin{array}{l}\Rightarrow x{{\left( {1+\frac{5}{{100}}} \right)}^{3}}=\left( {16400-x} \right){{\left( {1+\frac{5}{{100}}} \right)}^{2}}\\\Rightarrow x\left( {1+\frac{5}{{100}}} \right)=\left( {16400-x} \right)\\\Rightarrow \frac{{21x}}{{20}}+x=16400\\\Rightarrow \frac{{41x}}{{20}}=16400\\\Rightarrow x=\frac{{16400\times 20}}{{41}}\\\Rightarrow x=8000\end{array}\)
Therefore, the amount the father deposits for his younger son is Rs. 8000