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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-8 section-2

Each section contains 100 MCQs on Interest .
71
The simple interest on a certain sum is one-eighth of the sum when the number of years is equal to half of the rate percentage per annum. Find the simple interest (in Rs.) on Rs. 15,000 at the same rate of simple interest for 8 years.
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Let sum}} = P \cr & \frac{{P \times r \times t}}{{100}} = \frac{1}{8}P \cr & rt = \frac{{100}}{8} = \frac{{25}}{2}\,.....\,\left( {\text{i}} \right) \cr & {\text{Since }}t = \frac{r}{2} \cr & {\text{From equation}}\left( {\text{i}} \right) \cr & r \times \frac{r}{2} = \frac{{25}}{2} \cr & r = 5\% \cr & {\text{If }}P = 1500 \cr & t = 8{\text{ years}} \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{{P \times r \times t}}{{100}} \cr & = \frac{{15000 \times 8 \times 5}}{{100}} \cr & = {\text{Rs}}{\text{. }}6,000 \cr} $$
72
A person borrows Rs. 1,00,000 from a bank at 10% per annum simple interest and clears the debt in five years. If the instalment paid at the end of the first, second, third and fourth years to clear the debt are Rs. 10,000, Rs. 20,000, Rs. 30,000 and Rs. 40,000, respectively, what amount should be paid at the end of the fifth year to clear the debt?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Interest at the end of }}{{\text{1}}^{{\text{st}}}}{\text{ year}} \cr & = \frac{{1,00,000 \times 10}}{{100}} \cr & = 10,000 \cr & {\text{Paid amount}} = 10,000 \cr & {\text{Rest amount}} \cr & = 1,10,000 - 10,000 \cr & = 1,00,000 \cr & {\text{Interest for }}{{\text{2}}^{{\text{nd}}}}{\text{ year}} = 10,000 \cr & {\text{Amount paid}} = 20,000 \cr & {\text{Remaining amount}} \cr & = 1,10,000 - 20,000 \cr & = 90,000 \cr & {\text{Interest for }}{{\text{3}}^{{\text{rd}}}}{\text{ year}} = 9,000 \cr & {\text{Amount paid}} = 30,000 \cr & {\text{Remaining amount}} \cr & = 99,000 - 30,000 \cr & = 69,000 \cr & {\text{Interest for }}{{\text{4}}^{{\text{th}}}}{\text{ year}} = 6,900 \cr & {\text{Amount paid}} = 40,000 \cr & {\text{Remaining amount}} \cr & = 75,900 - 40,000 \cr & = 35,900 \cr & {\text{Interest for }}{{\text{5}}^{{\text{th}}}}{\text{ year}} \cr & = \frac{{35,900 \times 10}}{{100}} \cr & = 3,590 \cr & {\text{Amount paid at the end of }}{{\text{5}}^{{\text{th}}}}{\text{ year}} \cr & = 35,900 + 3,590 \cr & = 39,490 \cr} $$
73
A mobile phone is available for Rs. 25,000 or Rs. 5,200 down payment, followed by 4 equal monthly instalments. If the rate of interest is 25% p.a. simple interest, calculate the amount of each instalment.
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Cash payment}} = 25000 \cr & {\text{Down payment}} = 5200 \cr & {\text{Remaining}} = 19800 \cr & {\text{Installment}}\left( x \right) = 4 \cr} $$
\[\begin{array}{*{20}{c}} \Rightarrow &{19800}&{ - x} \\ {}&{19800}&{ - 2x} \\ {}&{\mathop {19800}\limits_{\_\_\_\_\_\_\_\_\_\_\_} }&{\mathop { - 3x}\limits_{\_\_\_\_\_\_\_\_\_} } \\ {}&{79200}&{ - 6x} \end{array}\]
$$\eqalign{ & \Rightarrow {\text{Interest}} = 4x - 19800 \cr & \Rightarrow \frac{{\left( {79200 - 6x} \right) \times 25}}{{12 \times 100}} = 4x - 19800 \cr & \Rightarrow 1920x - 50400 = 79200 - 60x \cr & \Rightarrow 1980x = 1029600 \cr & \Rightarrow x = 5200 \cr} $$
74
At which rate of simple interest does an amount become double in 12 years?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & 1 = \frac{{1 \times 12 \times r}}{{100}} \cr & r = 8\frac{1}{3}\% \cr} $$
75
On simple interest, a certain sum becomes Rs. 59,200 in 6 years and Rs. 72,000 in 10 years. If the rate of interest had been 2% more, then in how many years would the sum have become Rs. 76,000?
Discuss
Answer & Solution
Answer: Option B
Solution:
\[\begin{gathered} P\xrightarrow{{{\text{6 years}}}}59,200 \hfill \\ P\xrightarrow{{{\text{10 years}}}}72,000 \hfill \\ \end{gathered} \]
$$\eqalign{ & {\text{Difference 4 years}} \cr & = \left( {72,000 - 59,200} \right) \cr & = 12,800 \cr & {\text{1 year}} = \frac{{12,800}}{4} = 3,200 \cr & \therefore {\text{6 years S}}{\text{.I}}{\text{.}} \cr & = 3,200 \times 6 = 19,200 \cr & P = 59,200 - 19,200 = 40,000 \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{{P \times r \times t}}{{100}} \cr & 19,200 = \frac{{40,000 \times r \times 6}}{{100}} \cr & r = 8\% \cr & {\text{If }}r = 8\% + 2\% = 10\% \cr & P = 40,000 \cr & t = ? \cr & A = 76,000 \cr & \therefore {\text{S}}{\text{.I}}{\text{.}} = 36,000 \cr & 36,000 = \frac{{40,000 \times 10 \times t}}{{100}} \cr & t = 9{\text{ years}} \cr} $$
76
A sum is deposited in a bank which gives simple interest. The sum becomes 1.25 times in 3 years. If there is a requirement of Rs. 7,60,000 after seven years, how much amount (in Rs.) should one deposit to fulfil the requirement?
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the sum = 100
Interest mcq question image
$$\eqalign{ & {\text{Rate}} = \frac{{25}}{3}\% \cr & \Rightarrow {\text{Total amount in}} \cr & = 100 + \frac{{25}}{3} \times 7 \cr & = 100 + \frac{{175}}{3} \cr & = \frac{{475}}{3}\mu \to 76000 \cr & 1\mu \to 4800 \cr & 100\mu \to 480000 \cr} $$
77
A certain sum is lent at 4% per annum for 3 years 8% per annum for next 4 years and 12% per annum beyond 7 years. If for a period of 11 years, the simple interest obtained is Rs. 27,600, then the sum is (in Rs.):
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & 27600 = \frac{{P\left( {3 \times 4 + 4 \times 8 + 4 \times 12} \right)}}{{100}} \cr & \frac{{27600 \times 100}}{{12 + 32 + 48}} = P \cr & P = \frac{{27600 \times 100}}{{92}} \cr & P = {\text{Rs}}{\text{. }}30000 \cr} $$
78
If the amount obtained by A by investing Rs. 9,100 for three years at a rate of 10% p.a. on simple interest is equal to the amount obtained by B by investing a certain sum of money for five year at a rate of 8% p.a. on simple interest, then 90% of the sum invested by B (in Rs.) is:
Discuss
Answer & Solution
Answer: Option D
Solution:
Amount obtained by A is equal to amount obtained by B
Amount = Principal + Simple Interest
$$\eqalign{ & 9100 + \frac{{9100 \times 10 \times 3}}{{100}} = B + \frac{{B \times 8 \times 5}}{{100}} \cr & 9100 + 2730 = \frac{{140}}{{100}}B \cr & B = 11830 \times \frac{{100}}{{140}} \cr & B = 8450 \cr} $$
90% of invested amount by B
$$\eqalign{ & = 8450 \times \frac{{90}}{{100}} \cr & = 7605 \cr} $$
79
A person deposited Rs. 15,600 in a fixed deposit at 10% per annum simple interest. After every second year he adds his interest earned to the principal. The interest at the end of 4 years is:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{S}}{\text{.I}}{\text{. for 2 years}} \cr & = \frac{{15600 \times 2 \times 10}}{{100}} \cr & = 3120 \cr & {\text{Principal for next 2 years}} \cr & = 15600 + 3120 \cr & = 18720 \cr & {\text{S}}{\text{.I}}{\text{. for next 2 years}} \cr & = \frac{{18720 \times 10 \times 2}}{{100}} \cr & = 3744 \cr & {\text{Total S}}{\text{.I}}{\text{.}} \cr & = 3120 + 3744 \cr & = 6864 \cr} $$
80
A person borrowed 1,200 at 8% per annual and Rs. 1,800 at 10% per annum, as simple interest for the same period. He had to pay Rs. 1380 in all as interest. Find the time? (in year)
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \frac{{1200 \times t \times 8}}{{100}} + \frac{{1800 \times t \times 10}}{{100}} = 1380 \cr & 96t + 180t = 1380 \cr & 276t = 1380 \cr & t = 5 \cr} $$