100+ Simple Interest Questions and Answers for Competitive Exams - 1
Question: 1
Rs.6000 becomes Rs.7200 in 4 years at a certain rate of simple interest. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be
(A) Rs. 8000
(B) Rs. 8250
(C) Rs. 9000
(D) Rs. 9250
Ans: B
P = Rs. 6000, S.I. = Rs. (7200 – 6000) = Rs. 1200, T = 4 yrs.
∴ Rate = $$({100 × 1200} / {6000 × 4})$$% = 5%.
New rate = (1.5 × 5)% = 7.5%.
New S.I. = Rs. $$({6000 × 7.5 × 5} / {100})$$ = Rs. 2250.
∴ New amount = Rs. (6000 + 2250) = Rs. 8250.
Question: 2
The simple interest on a certain sum of money at the rate of 5% p.a. for 8 years is Rs.840. At what rate of interest the same amount of interest can be received on the same sum after 5 years?
(A) 6%
(B) 7%
(C) 8%
(D) 9%
Ans: C
S.I. = Rs. 840, R = 5%, T = 8 years.
Principal = Rs. $$({100 × 840} / {5 × 8})$$ = Rs. 2100.
Now, P = Rs. 2100, S.I. = Rs. 840, T = 5 years.
∴ Rate = $$({100 × 840} / {2100 × 5})$$% = 8%.
Question: 3
Suganya invested Rs.7500 at simple interest @ 11 p.c.p.a. She further invested some amount at simple interest @ 15 p.c.p.a. Total interest earned at the end of the year became 12 p.c.p.a. Find the amount invested at the rate of 15 p.c.p.a.
(A) Rs. 2000
(B) Rs. 2500
(C) Rs. 3000
(D) Rs. 3500
Ans: B
Let the required sum be x.
Then, 11% of 7500 + 15% of x = 12% of (7500 + x)
⇒ $$({11}/{100} × 7500)$$ + $$({15}/{100}x)$$ = $${12}/{100}(7500 + x)$$
⇒ 82500 + 15x = 90000 + 12x ⇒ 3x = 7500 x = 2500
Hence, required sum = Rs. 2500.
Question: 4
What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?
(A) 1 : 3
(B) 2 : 3
(C) 2 : 4
(D) 4 : 1
Ans: B
Let the principal be P and rate of interest be R%.
Required ratio = $$[({P × R × 6}/{100})/({P × R × 9}/{100})]$$ = $${6PR}/{9PR}$$ = $${6}/{9}$$ = 2 : 3.
Question: 5
A person borrowed Rs.500 @ 3% per annum S.I. and Rs.600 @ $$44{1}/{2}$$% per annum on the agreement that the whole sum will be returned only when the total interest becomes Rs.126. The number of years, after which the borrowed sum is to be returned, is
(A) 2 years
(B) 3 years
(C) 4 years
(D) 5 years
Ans: B
Let the time be x years.
Then, $$({500 × 3 × x} / {100})$$ + $$({600 × 9 × x} / {100 × 2})$$ = 126.
⇔ 15x + 27x = 126 ⇔ 42x = 126 ⇔ x =3.
∴ Required time = 3 years.
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