Simple and Compound Interest Aptitude Solved Examples

Simple and Compound Interest Aptitude Questions and Answers with Detailed Solution:

Question No. 01
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9% p.a. in 5 years. What is the sum?
(A) Rs. 4462.50
(B) Rs. 8032.50
(C) Rs. 8900
(D) Rs. 8925
Answer: Option D
Explanation:
Principal = Rs. (100 × 4016.25)/(9 × 5)
                 = Rs. (401625/45) = Rs. 8925.

Question No. 02
Calculate the amount on Rs. 4480 at 8% per annum for 3 years.
(A) Rs. 5555.20
(B) Rs. 5545.20
(C) Rs. 5000
(D) Rs. 6555
Answer: Option A
Explanation:
S.I. = (P × N × R)/100
       = Rs. (4480 × 3 × 8)/100
       = Rs. 1075.20
Amount = Rs. (4480 + 1075.20)
                   = Rs. 5555.20

Question No. 03
A certain sum of money at simple interest amounts to Rs. 1260 in 2 years and to Rs. 1350 in 5 years. The rate percent per annum is?
(A) 35 %
(B) 25 %
(C) 50 %
(D) 45 %
Answer: Option B
Explanation:
S.I. for 3 years = Rs. (1350 - 1260) = Rs. 90
S.I. for 2 years = Rs. (90/3) × 2 = Rs. 60
Principal = Rs. (1260 - 60) = Rs. 1200
Rate, R = (100 × 60)/(1200 × 2) % = 25 %

Question No. 04
Mr. Roy 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?
(A) Rs. 6400
(B) Rs. 6500
(C) Rs. 7200
(D) Rs. 7500
Answer: Option A
Explanation:
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).
Then, [x × 14 × 2)/100] + [{(13900 - x) × 11 × 2}/100] = 3508
=> 28x - 22x = 350800 - (13900 × 22)
=> 6x = 45000
=> x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Question No. 05
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?
(A) 3.5 years
(B) 4 years
(C) 4.5 years
(D) 5 years
Answer: Option B
Explanation:
Time = [(100 × 81)/( 450 × 4.5)] years = 4 years.

Question No. 06
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) 3.6%
(B) 4.5%
(C) 5%
(D) None of these
Answer: Option D
Explanation:
Let the original rate be R%. Then, new rate = (2R)%.
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. for 1/3 year.
[(725 × R × 1)/100] + [362.50 × 2R × 1)/(100 × 3)] = 33.50
=> (2175 + 725) R = 33.50 × 100 × 3
=> (2175 + 725) R = 10050
=> (2900) R = 10050
=> R = (10050/2900) = 3.46
Original rate = 3.46%

Question No. 07
A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
(A) Rs. 35
(B) Rs. 245
(C) Rs. 350
(D) Cannot be determined
Answer: Option D
Explanation:
We need to know the S.I., principal and time to find the rate.
Since the principal is not given, so data is inadequate.

Question No. 08
Ravi 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?
(A) 3.6
(B) 6
(C) 18
(D) None of these
Answer: Option B
Explanation:
Let rate = R% and time = R years.
Then, (1200 × R × R)/100 = 432
=> 12R2 = 432
=> R2 = 36
=> R = 6.

Question No. 09
S.I. on Rs. 1500 at 7% per annum for a certain time is Rs. 210. Find the time;
(A) 3 years
(B) 5 years
(C) 2 years
(D) 1½ years
Answer: Option C
Explanation:
Time, N = (210 × 100)/(1500 × 7) = 2 years

Question No. 10
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
(A) Rs. 650
(B) Rs. 690
(C) Rs. 698
(D) Rs. 700
Answer: Option C
Explanation:
S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs. (39 × 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698.

Simple and Compound Interest:
Formula:                Simple and Compound Interest Formulas
Solved Examples:  Solved Examples: Set 01
Practice Test:         Practice Test: 01     Practice Test: 02     Practice Test: 03
                                 Practice Test: 04     Practice Test: 05     Practice Test: 06
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