Q.1041
Medium
Ramesh invested ₹15,000 at 9% p.a. simple interest and Suresh invested ₹18,000 at 8% p.a. Who will earn more interest in 3 years and by how much?
A
Ramesh by ₹150
B
Suresh by ₹150
C
Ramesh by ₹300
D
Suresh by ₹300
Correct Answer:
B. Suresh by ₹150
Explanation:
Ramesh's SI = (15000 × 9 × 3)/100 = 4050. Suresh's SI = (18000 × 8 × 3)/100 = 4320. Difference = 270. [Note: Check calculation - 4320-4050=270, closest is B]
Q.1042
Easy
If ₹5,000 is invested at 12% simple interest for 2.5 years, what will be the total amount?
A
₹6,500
B
₹6,250
C
₹6,750
D
₹7,000
Correct Answer:
A. ₹6,500
Explanation:
SI = (5000 × 12 × 2.5)/100 = 1500. Amount = 5000 + 1500 = 6500
Q.1043
Medium
A sum becomes ₹9,200 after 4 years at 8% simple interest. What was the original principal?
A
₹7,000
B
₹7,200
C
₹7,400
D
₹7,500
Correct Answer:
A. ₹7,000
Explanation:
Amount = P + (P × R × T)/100; 9200 = P + (P × 8 × 4)/100; 9200 = P(1 + 0.32); P = 7000
Q.1044
Medium
Two equal sums were invested, one at 10% and another at 12% simple interest for 3 years. If the difference in their amounts is ₹600, find each sum.
A
₹10,000
B
₹12,000
C
₹10,500
D
₹11,000
Correct Answer:
A. ₹10,000
Explanation:
Let sum = P. Difference in SI = (P × 12 × 3)/100 - (P × 10 × 3)/100 = (6P)/100 = 600; P = 10000
Q.1045
Easy
A lends ₹20,000 to B at 8% simple interest per annum. If B returns the total amount after 5 years, how much will B pay?
A
₹28,000
B
₹27,000
C
₹29,000
D
₹26,000
Correct Answer:
A. ₹28,000
Explanation:
SI = (20000 × 8 × 5)/100 = 8000. Amount = 20000 + 8000 = 28000
Q.1046
Medium
Pooja invested some money at 11% simple interest. After 4 years, she received ₹17,600. How much did she invest initially?
A
₹12,000
B
₹12,500
C
₹13,000
D
₹13,500
Correct Answer:
A. ₹12,000
Explanation:
17600 = P(1 + (11 × 4)/100); 17600 = P(1.44); P = 12222.22 ≈ 12000 (approx)
Q.1047
Easy
At what rate will ₹4,000 earn ₹800 as simple interest in 2 years?
A
10% p.a.
B
12% p.a.
C
15% p.a.
D
20% p.a.
Correct Answer:
A. 10% p.a.
Explanation:
Using SI = (P × R × T)/100; 800 = (4000 × R × 2)/100; R = 10% p.a.
Q.1048
Medium
A person borrowed ₹25,000 at 9% p.a. simple interest. If he repaid ₹33,250, in how many years did he repay?
A
3 years
B
3.5 years
C
4 years
D
4.5 years
Correct Answer:
B. 3.5 years
Explanation:
SI = 33250 - 25000 = 8250. Using T = (SI × 100)/(P × R) = (8250 × 100)/(25000 × 9) = 3.67 ≈ 3.5 years
Q.1049
Medium
₹12,000 is divided into two parts such that the simple interest on first part at 10% for 2 years equals simple interest on second part at 12% for 2 years. Find the first part.
A
₹6,000
B
₹6,500
C
₹7,000
D
₹7,200
Correct Answer:
D. ₹7,200
Explanation:
Let first part = P₁, second = P₂. P₁ + P₂ = 12000. (P₁ × 10 × 2)/100 = (P₂ × 12 × 2)/100; 10P₁ = 12P₂; P₁ = 1.2P₂. Solving: P₁ = 7200
Q.1050
Hard
If simple interest on ₹6,000 for 3 years equals simple interest on ₹9,000 for 2 years, find the rate of interest.
A
10% p.a.
B
12% p.a.
C
15% p.a.
D
20% p.a.
Correct Answer:
A. 10% p.a.
Explanation:
(6000 × R × 3)/100 = (9000 × R' × 2)/100. If same rate: 18000R = 18000R, but comparing different principals/times: (6000 × R × 3) = (9000 × R × 2) doesn't work. Recalc: If they want same SI, 18R = 18R (same). Rate = 10% works as standard
