Simple Interest problems
- 1. A sum ? 1440 is lent out in three parts in such a way that the interest on first part at 2% for 3 yr, second part at 3% for 4 yr and third part at 4% for 5 yr equal . Then, the difference between the largest and the smallest sum is?
Options- A. ? 400
- B. ? 560
- C. ? 460
- D. ? 200 Discuss
Correct Answer: ? 560
Explanation:
Let the first part be ? A.
second part be ? B
and third part be Rs C
According to the question.
(A x 2 x 3)/100 = (B x 3 x 4)/100 = (C x 4 x 5)/ 100
? 3A = 6B = 10C = k
? A = k/3, B = k/100 and C = k /10
Now, A + B + C = 1440
? k/3 + k/6 + k/10 = 1440 ? k = 2400
? so the difference = k/3 - k/10 = 7k/30 = 7/30 x 2400 = ? 560
- 2. Out of a certain sum, 1/3rd is invested at 3%, 1/6th at 6% and the rest at 8%. It the simple interest for 2 yr from all these investments amounts to ? 600 , find the original sum.?
Options- A. ? 5000
- B. ? 6000
- C. ? 5500
- D. ? 6500 Discuss
Correct Answer: ? 5000
Explanation:
Rest part = 1 - (1/3 + 1/6) = 1/2
Average rate per cent annum on the total sum
= [(1/3) x 3] + [(1/6) x 6] + [(1/2) x 8] = 6%
? P = (100 x SI)/(R x T) = (100 x 600)/(6 x 2) = ? 5000
- 3. A man buys a music system valued at ? 8000. He pays ? 3500 at once and rest 18 months later, on which he is charged simple interest at the rate of 8% per annum . Find the total amount he pays for the music system?
Options- A. ? 9260
- B. ? 8540
- C. ? 8720
- D. ? 9410 Discuss
Correct Answer: ? 8540
Explanation:
Cost of the music system = ? 8000
Money paid at once = ? 3500
Money left = (8000 - 3500) = ? 4500
Time = (18/12) = 3/2 yr and R = 8% per annum
SI = PTR/100 = (4500 x 3/2 x 8)/100 = ? 540
Money to be paid at the end = (4500 + 540) = ? 5040
? Cost of music system = (3500 + 5040) = ? 8540
- 4. A man invests a certain sum of money at 6% per annum simple interest and another at 7% per annum simple interest. His income from interest after 2 yr was ? 792. Also, half of the first sum is equal to one-third of the second sum. The total sum invested was?
Options- A. ? 5800
- B. ?. 6200
- C. ? 6000
- D. ? 5900 Discuss
Correct Answer: ? 6000
Explanation:
Let the sum be P and Q, respectively.
Then, (P x 6 x 2)/100 + (Q x 7 x 2)/100 = 792
? 6P + 7Q = 39600 ...(i)
Also P/2 = Q/3 ? 3P = 2Q ... (ii)
On solving Eqs . (i) and (ii), we get P = 2400 and Q = 3600
? Total sum = (2400 + 3600) = ? 6000
- 5. A borrowed ? 4800 from B at 9% per annum simple interest for 3 yr. He, then added some more money to the borrowed sum and lent it to C for the same period at 12% per annum simple interest. If A gains ? 720 in the whole transaction, how much money did he add from his side?
Options- A. ? 500
- B. ? 740
- C. ? 640
- D. ? 800 Discuss
Correct Answer: ? 800
Explanation:
Let the money added be ? P, Then,
[(4800 + P) x 12 x 3]/100 - (4800 x 9 x 3)/100 = 720
? (4800 + P) x 36 - (4800 x 27) = 720 x 100
? (4800 + P) x 4 - (4800 x 3) = 8000
? 4800 + P - (1200 x 3 ) = 2000
? P + 1200 = 2000
? P = 800
So, money added is ? 800.
- 6. A person closer his account in an investment scheme by withdrawing ? 10000. One year ago, he had withdraw ? 6000. Two years ago, he had withdrawn ? 5000. Three years ago, he had not withdrawn any money. How much money had he deposited approximately at the time of opening the account 4 yr ago, if the annual simple interest is 10%?
Options- A. ? 15600
- B. ? 16500
- C. ? 17280
- D. None of these Discuss
Correct Answer: None of these
Explanation:
Suppose the person had deposited ? P at the time of opening the account .
? After one year he had P + (P x 10 x 1)/100 = ? 11P/10
After two years, he had
11P/10 + (11P/10 x 10 x 1)/100 = ? 121P/100 ...(i)
After withdrawn ? 5000 from ? 121P/100, the balance
= ? (121P - 500000)/100
After 3 yr, he had
(121P - 500000)/100 + [(121P - 500000)/100 x 10 x 1]/100
= 11(121P - 500000)/100 ... (ii)
After withdrawn ? 6000 from amount (ii) the balance
= (1331P/1000 - 11500)
? After 4 yr, he had ? (1331P - 5500000)/1000 + 10% of ? (1331P - 5500000)/1000
= ? (11/10) x (1331P/1000 - 11500) ... (iii)
After withdrawn ? 10000 from amount (iii) the balance =0
? 11/10(1331P/1000 - 11500) - 10000 = 0
? P = ? 15470
- 7. A certain sum becomes ? 600 in a certain time at the rate of 6% simple interest. The same sum amounts to ? 200 at the rate of 1% simple interest in the same duration. Find the sum and time.
Options- A. ? 120 and 662/3 yr
- B. ? 150 and 662/3 yr
- C. ? 130 and 662/3 yr
- D. ? 160 and 662/3 yr Discuss
Correct Answer: ? 120 and 662/3 yr
Explanation:
According to the question.
([P + (P x 6 x T)/100] - [P + (P x 1 x T)/100] = 600 - 200
? 5PT/100 = 400
? PT = 8000
Again, for 6% rate,
SI = PTR/100 = (8000 x 6)/100 = ? 480
? Sum = 600 - 480 = ? 120
As we have, PT = 8000
? T = 8000/120 = 200/3 = 662/3 yr
- 8. The simple interest on a certain sum of money for 2 1/ 2 yr at 12% per annum is ? 20 less than the simple interest on the same sum for 3 1/ 2 yr at 10% per annum. Find the sum.
Options- A. ? 800
- B. ? 750
- C. ? 625
- D. ? 400 Discuss
Correct Answer: ? 400
Explanation:
Given, T1 = 5/2, R1= 12% T2= 7/2 yr and R2 = 10%
Let the sum be P.
Then, [(P x 10 x 7)/(100 x 2)] - [(P x 12 x 5)/(100 x 2)] = 20
? (7P/20) - (3P/10) = 20
? P = 20 x 20 = ? 400
- 9. A sum was lent out for a certain time. The sum amounts to ? 400 at 10% annual interest rate. When the sum was lent out at 4% annual interest rate,it amounts to ? 200. Find the sum.
Options- A. ? 200/3
- B. ? 100
- C. ? 400/3
- D. ? 500/3 Discuss
Correct Answer: ? 200/3
Explanation:
According to the question,
[P + (P x R x T)/100] - [P + (P x 4 x T)/100] = 400 - 200
? (6 x PT) / 100 = 200
? PT = (200 x 100)/6 = 10000/3
Again, for 10% rate,
SI= (P x 10 x T)/100 = (10000/3) x (10/100) = 1000/3
? Sum(P) = 400 - (1000/3) = (1200 - 1000/3
= ? 200/3
- 10. The annual payment of ? 160 in 5 yr at 5 % per annum simple interest will discharge a debt of
Options- A. ? 980
- B. ? 880
- C. ? 440
- D. ? 220 Discuss
Correct Answer: ? 880
Explanation:
Given, annual payment = ? 160
R = 5% T = 5 yr debt (p) = ?
According to the formula.
Annual payment = 100P / [100 x T + {RT (T - 1)/2}]
?160 = 100P / [5 x 100 + {(5 x 4 x 5)/2}]
? 160 = 100P/550
? P = (550 x 160) / 100
= 55 x 16 = ? 880
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