Simple interest
- 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.
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According to the question,
[P + (P x R x T)/100] - [P + (P x 4 x T)/100] = 400 - 200Correct Option: A
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
- The simple interest on a certain sum of money for 21/2 yr at 12% per annum is ₹ 20 less than the simple interest on the same sum for 31/2 yr at 10% per annum. Find the sum.
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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)] = 20Correct Option: D
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
- 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.
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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 = ₹ 120Correct Option: A
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
- 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 Two years ago, he had withdrawn ₹ 5000. Three years ago, he had not withdrawn any money.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% ?
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∴ 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After withdrawn ₹ 5000 from ₹ 121P/100, the balance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)/100Correct Option: D
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
- 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 ?
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Let the money added be ₹ P, Then,
[(4800 + P) x 12 x 3]/100 - (4800 x 9 x 3)/100 = 720Correct Option: D
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.
