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Interest - Aptitude MCQ Questions and Solutions with Explanations

Interest MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Interest with easy and logical explanations. Interest MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-4 section-1

Each section contains 100 MCQs on Interest .
31
A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments. If the rate of the interest be 10% compounded annually, then the value of each installment is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Let X = equal installment at the end of one year( rate% annually) .
Now 1st year,
P =210,
Interest = $$\frac{{{\text{PTR}}}}{{100}}$$  = 210 * 0.1 = 21.
Let X is to be paid as an equal installment.
Now, at the beginning of 2nd year,
P = 210 + 21 - X,
Interest at the end of 2nd year,
= (231 - X) * 0.1 = 23.1 - 0.1X.
Hence,total installment,
2X = 210 + 21 + 23.1 - 0.1X,
X = $$\frac{{{\text{254}}{\text{.1}}}}{{2.1}}$$  = 121.
32
There is a decrease of 10% yearly on an article. If this article was bought 3 years ago and present cost is Rs. 5,832 then what was the cost of article at buying time?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$A = P {\left( {1 - \frac{R}{{100}}} \right)^n}$$
Where A = Value of goods after n years
P = Initial Price
R = Rate of depriciation
$$\eqalign{ & \therefore P = \frac{{5832}}{{{{\left( {1 - \frac{{10}}{{100}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {1 - \frac{1}{{10}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {\frac{9}{{10}}} \right)}^3}}} \cr & \Rightarrow P = 5832 \times \frac{{10}}{9} \times \frac{{10}}{9} \times \frac{{10}}{9} \cr & \Rightarrow P = 8000 \cr} $$
33
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
Discuss
Answer & Solution
Answer: Option C
Solution:

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Therefore Principal = Rs. (815 - 117) = Rs. 698.

34
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x)
Then,
$$ {\frac{{x \times 14 \times 2}}{{100}}} + $$   $$ {\frac{{\left( {13900 - x} \right) \times 11 \times 2}}{{100}}} $$     $$ = 3508$$
⇒ 28x - 22x = 350800 - (13900 x 22)
⇒ 6x = 45000
⇒ x = 7500
So, sum invested in Scheme B
= Rs. (13900 - 7500)
= Rs. 6400
35
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Principal}} \cr & = Rs.\,\left( {\frac{{100 \times 4016.25}}{{9 \times 5}}} \right) \cr & = Rs.\,\left( {\frac{{401625}}{{45}}} \right) \cr & = Rs.\,8925 \cr} $$
36
How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Time}} = \left( {\frac{{100 \times 81}}{{450 \times 4.5}}} \right){\text{years}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\,{\text{years}} \cr} $$
37
Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{rate}}\, = R\% \,{\text{and}} \cr & {\text{Time}}\,{\text{ = }}\,{\text{R}}\,{\text{years}}{\text{}} \cr & {\text{Then,}}\, {\frac{{1200 \times R \times R}}{{100}}} = 432 \cr & \Rightarrow 12{R^2} = 432 \cr & \Rightarrow {R^2} = 36 \cr & \Rightarrow R = 6 \cr} $$
38
A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{S}}{\text{.I}}{\text{.}} = Rs.\,\left( {15500 - 12500} \right) \cr & \,\,\,\,\,\,\,\,\,\, = Rs.\,3000 \cr & {\text{Rate}} = \left( {\frac{{100 \times 3000}}{{12500 \times 4}}} \right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\% \cr} $$
39
An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{sum}}\,{\text{be}}\,{\text{Rs}}{\text{.}}\,{\text{100}}{\text{.}}\,{\text{Then,}} \cr & {\text{S}}{\text{.I}}{\text{.}}\,{\text{for}}\,{\text{first}}\,{\text{6}}\,{\text{months}} \cr & = Rs.\,\left( {\frac{{100 \times 10 \times 1}}{{100 \times 2}}} \right) \cr & = Rs.\,5 \cr & {\text{S}}{\text{.I}}{\text{.}}\,{\text{for}}\,{\text{last}}\,{\text{6}}\,{\text{months}} \cr & = Rs.\,\left( {\frac{{105 \times 10 \times 1}}{{100 \times 2}}} \right) \cr & = Rs.\,5.25 \cr} $$
So, amount at the end of 1 year
= Rs. (100 + 5 + 5.25)
= Rs. 110.25
∴ Effective rate = (110.25 - 100) = 10.25%
40
A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{rate}}\,{\text{be}}\,R\% \,{\text{p}}{\text{.a}}{\text{.}}\,{\text{Then,}} \cr & {\frac{{5000 \times R \times 2}}{{100}}} + {\frac{{3000 \times R \times 4}}{{100}}} \cr & = 2200. \cr & \Rightarrow 100R + 120R = 2200 \cr & \Rightarrow R = {\frac{{2200}}{{220}}} = 10 \cr & \therefore {\text{Rate}} = 10\% \cr} $$