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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-5 section-2

Each section contains 100 MCQs on Interest .
41
Tushar borrowed a sum of Rs. 12000 at 15% per annum from a money - lender on 13th January, 1987 and return the amount on 8th June, 1987 to clear his debt. Then the amount paid by Tushar to the money - lender to clear his debt was = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Time = 18 + 28 + 31 + 30 + 31 + 8 = 146 days
$$\eqalign{ & {\text{Simple Interest}} \cr & {\text{ = }}\frac{{12000 \times 15 \times 146}}{{365 \times 100}} \cr & {\text{ = Rs}}{\text{. 720}} \cr & \therefore {\text{Amount will be }} \cr & {\text{ = Rs}}{\text{.}}\left( {12000 + 720} \right){\text{ }} \cr & {\text{ = Rs}}{\text{. 12720}} \cr} $$
42
Vishwas borrowed a total amount of Rs. 30000, part of it on simple interest rate of 12 p.c.p.a. and remaining on simple interest rate of 10 p.c.p.a. If at the end of 2 year she paid in all Rs. 36480 to settle the loan amount, what was the amount borrowed at 12 p.c.p.a ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the sum borrowed at 12% p.a. be Rs. x
and that borrowed at 10% p.a. be Rs. (30000 - x)
S.I. at the end of 2 years
= Rs. (36480 - 30000)
= Rs. 6480
$$\therefore \left( {\frac{{x \times 12 \times 2}}{{100}}} \right) + $$    $$\left[ {\frac{{\left( {30000 - x} \right) \times 10 \times 2}}{{100}}} \right]$$     $$ = 6480$$
$$\eqalign{ & \Leftrightarrow 24x + 600000 - 20x = 648000 \cr & \Leftrightarrow 4x = 48000 \cr & \Leftrightarrow x = 12000 \cr} $$
43
A sum of Rs. 18750 is left by a will by a father to be divided between the two sons, 12 and 14 years of age, so that when they attain maturity at 18, the amount (principal + interest) received by each at 5 percent simple interest will be the same. Find the sum alloted at present to each son.
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the two sums be Rs. x and Rs. (18750 - x).
Then,
  $$ = x + \frac{{x \times 5 \times 6}}{{100}} = \left( {{\text{18750}} - x} \right) + $$       $$\frac{{\left( {{\text{18750}} - x} \right) \times 5 \times 4}}{{100}}$$
$$\eqalign{ & \Leftrightarrow x + \frac{{30x}}{{100}} = \left( {{\text{18750}} - x} \right) + 3750 - \frac{{20x}}{{100}} \cr & \Leftrightarrow 2x + \frac{x}{2} = 22500 \cr & \Leftrightarrow \frac{{5x}}{2} = 22500 \cr & \Leftrightarrow x = \left( {\frac{{22500 \times 2}}{5}} \right) \cr & \Leftrightarrow x = 9000 \cr} $$
So the other sum will be
= ( 18750 - 9000)
= 9750
Hence,
The two sums are Rs. 9000, Rs. 9750
44
I had Rs. 10000 with me. Out of this money I lent some money to A for 2 years @ 15% simple interest. I lent the remaining money to B for an equal number of years @18% simple interest. After 2 years, I found that A had given me Rs. 360 more as interest as compared to B. The amount of money which I had lent to B must have been.
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the sum lent to A be Rs. x. and that lent to B be Rs. (10000 - x)
Then,
$$\eqalign{ & \Rightarrow \frac{{x \times 15 \times 2}}{{100}} - \frac{{\left( {10000 - x} \right) \times 18 \times 12}}{{100}} = 360 \cr & \Rightarrow 30x - 360000 + 36x = 36000 \cr & \Rightarrow 66x = 396000 \cr & \Rightarrow x = 6000 \cr} $$
Hence,
Sum lent to B
= Rs. (10000 - 6000)
= Rs. 4000
45
A certain sum of money amount to Rs 2200 at 5% interest Rs 2320 at 8% interest in the same period of time. The period of time is = ?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{P}} + {\text{SI = }}\frac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}} + {\text{P}} \cr & \Rightarrow 2200{\text{ = }}\frac{{{\text{P}} \times 5 \times {\text{T}}}}{{100}} + {\text{P}} \cr} $$
⇒ 2200 × 100 = 5PT + 100P ......... (i)
$$ \Rightarrow 2320 = \frac{{{\text{P}} \times 8 \times {\text{T}}}}{{100}} + {\text{P}}$$
⇒ 2320 × 100 = 8PT + 100P
⇒ 2320 × 100 = 3PT + 5PT + 100P ............(ii)
Value of equation (i) put equation (ii)
⇒ 2320 × 100 = 3PT + 2200 × 100
⇒ 3PT = 120 × 100
⇒ PT = 4000
Value of PT in equation (i)
⇒ 2200 × 100 = 5 × 4000 + 100P
⇒ 220000 - 20000 = 100P
$$\eqalign{ & \Rightarrow {\text{P = }}\frac{{200000}}{{100}} \cr & \Rightarrow {\text{P = Rs 2000}} \cr & {\text{Using this formula}} \cr & \left( {\because {\text{SI = }}\frac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}} \right) \cr & \therefore {\text{200}} = \frac{{2000 \times 5 \times {\text{T}}}}{{100}} \cr & \Rightarrow {\text{T}} = \frac{{200}}{{100}}{\text{ = 2 years}} \cr & \cr & {\text{ }}{\bf{Alternate:}} \cr & \left( {8 - 5} \right)\% = 2320 - 2200 \cr & \Rightarrow 3\% = 120 \cr & \Rightarrow 1\% = 40 \cr & \Rightarrow 5\% = 200 \cr & {\text{Principal = 2200 - 200}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = Rs}}{\text{. 2000}} \cr & {\text{SI = }}\frac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}} \cr & \Rightarrow {\text{200}} = \frac{{2000 \times 5 \times {\text{T}}}}{{100}} \cr & \Rightarrow {\text{T}} = \frac{{200}}{{100}}{\text{ = 2 years}} \cr} $$
46
For 2 years, a sum was put at SI at a certain rate. If the rate was 3% higher, it would have fetched Rs. 300 more. What will be the sum ?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let principal is P}} \cr & {\text{then,}} \cr & {\text{300 = }}\frac{{{\text{P}} \times 3 \times 2}}{{100}} \cr & {\text{P = 5000}} \cr} $$
47
A money lender claims to lend money at the rate of 10% per annum simple interest. However, he takes the interest in advance when he lends a sum for one year. At what interest rate does he lend the money actually ?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Amount}} \to {\text{10}} \cr & \,\,\,\,\, \downarrow \cr & \,\,\,\,\,90 \to \frac{{10}}{{90}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\, = 11\frac{1}{9}\% \cr} $$
48
A sum of Rs. 1550 was lent partly at 5% and partly at 8% p.a. simple interest. The total interest received after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the sum lent at 5% be Rs. x and that lent 8% be Rs. (1550 - x).
Then,
$$\eqalign{ & \left( {\frac{{x \times 5 \times 3}}{{100}}} \right) + \left[ {\frac{{\left( {1550 - x} \right) \times 8 \times 3}}{{100}}} \right] = 300 \cr & \Leftrightarrow 15x - 24x + \left( {1550 \times 24} \right) = 30000 \cr & \Leftrightarrow 9x = 7200 \cr & \Leftrightarrow x = 800. \cr & \therefore {\text{Required ratio}} = 800:750 \cr & = 16:15 \cr} $$
49
An amount of Rs. 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and second, 11% p.a. If the total interest at the end of one year is $$9\frac{3}{4}$$ %, then the amount invested in each share was -
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x).
Then,
$$\eqalign{ & = \left( {\frac{{x \times 9 \times 1}}{{100}}} \right) + \left[ {\frac{{\left( {100000 - x} \right) \times 11 \times 1}}{{100}}} \right] \cr & = \left( {100000 \times \frac{{39}}{4} \times \frac{1}{{100}}} \right) \cr & \Leftrightarrow \frac{{9x + 1100000 - 11x}}{{100}} = \frac{{39000}}{4} = 9750 \cr & \Leftrightarrow 2x = \left( {1100000 - 975000} \right) = 125000 \cr & \Leftrightarrow x = 62500 \cr & \therefore {\text{Sum invested at 9}}\% \cr & = {\text{Rs}}{\text{. }}62500 \cr & {\text{Sum invested at 11% }} \cr & {\text{ = Rs}}{\text{.}}\left( {100000 - 62500} \right) \cr & = {\text{Rs}}{\text{. }}37500 \cr} $$
50
A sum of Rs. 1440 is lent out in three parts in such away that the interests on first part at 2% for 3 years, second part at 3% for 4 years and third part at 4% for 5 years are equal. Then the difference between the largest and the smallest sum is -
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the parts be Rs. x, Rs. y and Rs. [1440 - (x + y)].
Then,
$$\eqalign{ & = \frac{{x \times 2 \times 3}}{{100}} = \frac{{y \times 3 \times 4}}{{100}} \cr & = \frac{{\left[ {1440 - \left( {x + y} \right)} \right] \times 4 \times 5}}{{100}} \cr & \therefore 6x = 12y\,or\,x = 2y. \cr & So,\frac{{x \times 2 \times 3}}{{100}} = \frac{{\left[ {1440 - \left( {x + y} \right)} \right] \times 4 \times 5}}{{100}} \cr & \Leftrightarrow 12y = \left( {1440 - 3y} \right) \times 20 \cr & \Rightarrow 72y = 28800 \cr & \Rightarrow y = 400 \cr & {\text{First part}} = x = 2y = {\text{Rs}}{\text{. }}800, \cr & {\text{Second part}} = {\text{Rs}}{\text{. }}400 \cr & {\text{Third part}} \cr & = {\text{Rs}}.\left[ {1440 - \left( {800 + 400} \right)} \right] \cr & = {\text{Rs}}{\text{. }}240. \cr & \therefore {\text{Required difference}} \cr & {\text{ = Rs}}{\text{.}}\left( {800 - 240} \right) \cr & = {\text{Rs}}.560 \cr} $$