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Arithmetic: Compound Interest Test -4
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Question 1 |
Sudharshan invested Rs.15, 000 at interest @ 10 p.c. p.a. for one year. If the interest is compounded every six months what amount will Sudharshan get at the end of the year?
Rs.16, 537.50 | |
Rs.16, 500 | |
Rs.16, 525.50 | |
Rs.18, 150 |
Question 1 Explanation:
Let r be the rate therefore r = 10/2 = 5%
Since the interest is compounded every six months
So number of periods = 2
$ \displaystyle Amount=15000{{(1+10/200)}^{2}}=15000\,\text{ }\!\!\times\!\!\text{ }\,1.1025=16537.5$
Since the interest is compounded every six months
So number of periods = 2
$ \displaystyle Amount=15000{{(1+10/200)}^{2}}=15000\,\text{ }\!\!\times\!\!\text{ }\,1.1025=16537.5$
Question 2 |
The compound interest earned by Suresh on a certain amount at the end of two years at the rate of 8 p.c.p.a was Rs.1, 414.4. What was the total amount that Suresh got back at the end of two years in the form of principal plus interest earned?
Rs.9, 414.4 | |
Rs.9, 914.4 | |
Rs.9, 014.4 | |
Rs.8, 914.4 |
Question 2 Explanation:
Given that rate = 8%
And the compound interest = 1414.4
So we know that
$ \displaystyle \begin{array}{l}CI=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]\\\Rightarrow 1414.4=P\,\left[ {{\left( 1+\frac{8}{100} \right)}^{2}}-1 \right]\\\Rightarrow 1414.4=P\times 0.1664\\\Rightarrow P=\frac{1414.4}{0.1664}=Rs.8500\\\therefore \,\,\,Amount\,=Rs.\,\left( 8500+1414.4 \right)\\=Rs.\,9914.4s\end{array}$
And the compound interest = 1414.4
So we know that
$ \displaystyle \begin{array}{l}CI=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]\\\Rightarrow 1414.4=P\,\left[ {{\left( 1+\frac{8}{100} \right)}^{2}}-1 \right]\\\Rightarrow 1414.4=P\times 0.1664\\\Rightarrow P=\frac{1414.4}{0.1664}=Rs.8500\\\therefore \,\,\,Amount\,=Rs.\,\left( 8500+1414.4 \right)\\=Rs.\,9914.4s\end{array}$
Question 3 |
Mr. Rao invests a sum of Rs.41, 250 at the rate of 6 p.c.p.a. What approximate amount of compound interest will he obtain at the end of 3 years?
Rs.8, 100 | |
Rs.7, 425 | |
Rs.8, 210 | |
Rs.7, 879 |
Question 3 Explanation:
Principle value = Rs. 41,250
Rate = 6%
We also know that the
$ \displaystyle \begin{array}{l}C.I.=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]\\=41250\,\left[ {{\left( 1+\frac{6}{100} \right)}^{3}}-1 \right]\\=41250\,\left[ {{\left( 1.06 \right)}^{3}}-1 \right]\\=41250\,\left( 1.191-1 \right)\\=41250\times 0.191\\=Rs.\,7878.75\\=Rs.\,7879\end{array}$
Rate = 6%
We also know that the
$ \displaystyle \begin{array}{l}C.I.=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]\\=41250\,\left[ {{\left( 1+\frac{6}{100} \right)}^{3}}-1 \right]\\=41250\,\left[ {{\left( 1.06 \right)}^{3}}-1 \right]\\=41250\,\left( 1.191-1 \right)\\=41250\times 0.191\\=Rs.\,7878.75\\=Rs.\,7879\end{array}$
Question 4 |
What would be the compound interest obtained on an amount of Rs.7, 790 at the rate of 10 p.c.p.a. after two years?
Rs.1532.60 | |
Rs.1495.90 | |
Rs.1653.50 | |
Rs.1635.90 |
Question 4 Explanation:
This is very simple question of Compound Interest
Given principle value = Rs. 7,790
Rate = 10%
Time = 2 years
$ \displaystyle \begin{array}{l}C.I.=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]\\=7790\,\left[ {{\left( 1+\frac{10}{100} \right)}^{2}}-1 \right]\\=7790\,\left( \frac{121}{100}-1 \right)\\=\frac{7790\times 21}{100}=Rs.\,1635.9\end{array}$
Given principle value = Rs. 7,790
Rate = 10%
Time = 2 years
$ \displaystyle \begin{array}{l}C.I.=P\,\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]\\=7790\,\left[ {{\left( 1+\frac{10}{100} \right)}^{2}}-1 \right]\\=7790\,\left( \frac{121}{100}-1 \right)\\=\frac{7790\times 21}{100}=Rs.\,1635.9\end{array}$
Question 5 |
The simple interest accrued on an amount of Rs.14, 800 at the end of three years is Rs.6, 216. What would be the compound interest accrued on the same amount at the same rate in the same period?
Rs.6986.1142 | |
Rs.7042.2014 | |
Rs.7126.8512 | |
Rs.8321.4166 |
Question 5 Explanation:
Principle value = Rs. 14800
Simple Interest =Rs. 6216
Let r be the rate therefore
$ \displaystyle \begin{array}{l}r=\frac{6216\times 100}{14800\times 3}=14\\\therefore \,\,C.I.=P\,\left[ {{\left( 1+\frac{r}{100} \right)}^{3}}-1 \right]\\=Rs.\,14800\,\left[ {{\left( 1+\frac{14}{100} \right)}^{3}}-1 \right]\\=Rs.\,14800\,\left[ {{\left( 1.14 \right)}^{3}}-1 \right]\\=Rs.\,14800\,\left( 1.481544-1 \right)\\=Rs.\,14800\times 0.481544\\=Rs.\,7126.8512\end{array}$
Simple Interest =Rs. 6216
Let r be the rate therefore
$ \displaystyle \begin{array}{l}r=\frac{6216\times 100}{14800\times 3}=14\\\therefore \,\,C.I.=P\,\left[ {{\left( 1+\frac{r}{100} \right)}^{3}}-1 \right]\\=Rs.\,14800\,\left[ {{\left( 1+\frac{14}{100} \right)}^{3}}-1 \right]\\=Rs.\,14800\,\left[ {{\left( 1.14 \right)}^{3}}-1 \right]\\=Rs.\,14800\,\left( 1.481544-1 \right)\\=Rs.\,14800\times 0.481544\\=Rs.\,7126.8512\end{array}$
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