Interest aptitude mcq

 

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Interest

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Question 1
Rs. 2100 is lent at compound interest of 5% per annum for 2 years. Find the amount after two years.
A
Rs. 2320
B
Rs. 2315.25
C
Rs. 2300
D
Rs. 2310
Civics Test Questions answers

Question 1 Explanation: 
\begin{align} & {\text{We}}\, {\text{can}}\, {\text{use}}\, {\text{formula}}\, {\text{of}}\, {\text{compound}}\, {\text{interest}} \cr & A = P \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^n} \cr & A = 2100 \times {\left[ {1 + \left( {\frac{5}{{100}}} \right)} \right]^2} \cr & A = 2100 \times {\left[ {\frac{{105}}{{100}}} \right]^2} \cr & A = \frac{{\left( {2100 \times 11025} \right)}}{{10000}} \cr & {\text{Hence, }}\, {\text{Amount}}\, A = Rs.\, 2315.25 \cr\end{align}
Question 2
If the rate increases by 2%, the simple interest received on a sum of money increases by Rs. 108. If the time period is increased by 2 years, the simple interest on the same sum increases by Rs. 180. The sum is:
A
Rs. 1800
B
Rs. 5400
C
Rs. 3600
D
Data inadequate
Aptitude test Questions answers

Question 2 Explanation: 
\begin{align} & {\text{Let the sum be Rs}}{\text{. }}x \cr & {\text{Rate be R}}\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{Time be T years}}{\text{.}} \cr & {\text{Then, }} \cr & \left[ {\frac{{x \times \left( {{\text{R}} \times 2} \right) \times {\text{T}}}}{{100}}} \right] - \left( {\frac{{x \times {\text{R}} \times {\text{T}}}}{{100}}} \right) = 108 \cr & \Leftrightarrow 2x{\text{T}} = 10800\, .....(i) \cr & And, \cr & \left[ {\frac{{x \times {\text{R}} \times \left( {{\text{T}} + 2} \right)}}{{100}}} \right] - \left( {\frac{{x \times {\text{R}} \times {\text{T}}}}{{100}}} \right) = 108 \cr & \Leftrightarrow 2x{\text{R}} = 18000\, .....(ii) \cr\end{align} Clearly, from (i) and (ii), we cannot the find the value of x.

So, the data is inadequate.

Question 3
A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
A
None of these
B
5%
C
3.60%
D
4.50%
English grammar Questions answers

Question 3 Explanation: 
Let the original rate be R%. Then, new rate = (2R)%.

Note:

Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s).

\begin{align} & \therefore \left( {\frac{{725 \times R \times 1}}{{100}}} \right) + \left( {\frac{{362.50 \times 2R \times 1}}{{100 \times 3}}} \right) \cr & = 33.50 \cr & \Rightarrow \left( {2175 + 725} \right)R = 33.50 \times 100 \times 3 \cr & \Rightarrow \left( {2175 + 725} \right)R = 10050 \cr & \Rightarrow \left( {2900} \right)R = 10050 \cr & \Rightarrow R = \frac{{10050}}{{2900}} = 3.46 \cr & \therefore Original\, rate = 3.46\% \cr\end{align}

Question 4
If the simple interest on Rs. 1 for 1 month is 1 paisa, then the rate percent per annum will be =?
A
12%
B
10%
C
8%
D
6%
Puzzles Questions answers

Question 4 Explanation: 
\begin{align} & {\text{t}} = {\text{1 month = }}\frac{1}{{12}}{\text{year}} \cr & {\text{SI = 1 paisa = Rs}}{\text{. }}\frac{1}{{100}} \cr & {\text{r}}\% = \frac{{{\text{SI}} \times {\text{100}}}}{{{\text{P}} \times {\text{T}}}} = \frac{{1 \times 100 \times 12}}{{100 \times 1 \times 1}} \cr & {\text{r}}\% = 12\% \cr\end{align}
Question 5
Rs. 1000 is invested at 5% per annum simple interest. If the interest is added to the principal after every years, the amount will become Rs. 2000 after =?
A
20 years
B
15 years
C
18 years
D
162/3 years
Aptitude test Questions answers

Question 5 Explanation: 
\begin{align} & {\text{Principal = Rs}}{\text{. 1000 }} \cr & {\text{Rate = 5}}\% \cr & {\text{Interest for first 10 years}} \cr & = \frac{{1000 \times 5 \times 10}}{{100}} \cr & = {\text{Rs}}{\text{. 500}} \cr & {\text{After 10 years principal}} \cr & = {\text{(1000}} + {\text{500)}} \cr & {\text{ = Rs}}{\text{. 1500}} \cr & {\text{Remaining interest}} \cr & {\text{ = Rs}}{\text{. (2000}} - {\text{1500)}} \cr & {\text{ = Rs}}{\text{. 500}} \cr & {\text{Required time }} \cr & {\text{ = }}\frac{{500}}{{1500}} \times \frac{{100}}{5} \cr & \Rightarrow \frac{{100}}{5} = \frac{{20}}{3} \cr & \Rightarrow 6\frac{2}{3}{\text{years}} \cr & {\text{Total time}} \cr & = \left( {10 + 6\frac{2}{3}} \right){\text{years}} \cr & {\text{ = 16}}\frac{2}{3}{\text{years}} \cr\end{align}
There are 5 questions to complete.

 

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