Simple interest
- A man borrowed ₹ 12000 for 4 yr at 73/4 per annum and a year later he again borrowed another ₹ 12000 for 3 yr at the same rate. How much should he pay at the end to settle the loans ?
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In the first case
P = Rs. 12000, T = 4 yr and R = 31/4 % per annum
SI = PTR/100 = (12000 x 4 x 31/4)/100 = ₹ 3720
A = (P + SI) = (12000 + 3720) = ₹ 15720
In the second case
P = ₹ 12000, T = 3 yr and R = 31/4 % per annum
SI = PTR/100 = (12000 x 3 x 31/4)/100 = ₹ 2790
A = (P + SI ) = (12000 + 2790) = ₹ 14790Correct Option: A
In the first case
P = Rs. 12000, T = 4 yr and R = 31/4 % per annum
SI = PTR/100 = (12000 x 4 x 31/4)/100 = ₹ 3720
A = (P + SI) = (12000 + 3720) = ₹ 15720
In the second case
P = ₹ 12000, T = 3 yr and R = 31/4 % per annum
SI = PTR/100 = (12000 x 3 x 31/4)/100 = ₹ 2790
A = (P + SI ) = (12000 + 2790) = ₹ 14790
∴ Total memory to be paid at the end = (15720 + 14790) = ₹ 30510
- A man took a loan at simple interest at 7% in the first year with an increase of 0.5% in each subsequent year. He pays ₹ 3690 as interest after 5 yr . How much loan did he take ?
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Let the loan taken be ₹ P Then,
[(P x 7 x 1)/100] + [(P x 7.5 x 1)/100] + [(P x 8 x 1)/100] + [(P x 8.5 x 1)/100] + [(P x 9 x 1)/100] = ₹ 3690Correct Option: C
Let the loan taken be ₹ P Then,
[(P x 7 x 1)/100] + [(P x 7.5 x 1)/100] + [(P x 8 x 1)/100] + [(P x 8.5 x 1)/100] + [(P x 9 x 1)/100] = ₹ 3690
⇒ P x (7 + 7.5 + 8 + 8.5 + 9)/100 = 3690
⇒ (P x 40)/100 = 3690
∴ P = (3690 x 100)/40 = ₹ 9225
Hence, the loan taken was ₹ 9225.
- What will be the difference in the compound interest on ₹ 50000 at 12% for one year, when the interest is paid yearly and half- yearly ?
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CI on ₹ 50000 at the rate of 12% for one year, when the interest is paid half-yearly
= 50000(1 + 6/100)2 - 50000
CI when the interest is paid yearly
= 50000(1 + 12/100)1 - 50000Correct Option: C
CI on ₹ 50000 at the rate of 12% for one year, when the interest is paid half-yearly
= 50000(1 + 6/100)2 - 50000
= [50000 x (53/50) x (53/50)] - 50000 = ₹ 6180
CI when the interest is paid yearly
= 50000(1 + 12/100)1 - 50000
= 50000(28/25 - 1) = 50000(3/25) = ₹ 6000
∴ Required difference = CI - SI = 6180 - 6000 = ₹ 180
- Vikram borrowed ₹. 6450 at 5% simple interest repayable in 4 equal instalments. What will be the annual instalment payable by him ?
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Let the annual instalment be Rs. P.
Amount of ₹ 100 after 4 yr = 100 + (100 x 5 x 4)/100 = ₹ 120
∴ Present Value (PV) of ₹ 120 due after 4 yr = ₹ 100
Present Value (PV) of ₹ due after 4 yr = 100P/120 = 5P/6
Similary, PV of ₹ P due after 3 yr = 20P/23
PV of ₹ P due after 2 yr = 10P/11
PV of ₹ P due after 1 yr = 20P/21Correct Option: B
Let the annual instalment be Rs. P.
Amount of ₹ 100 after 4 yr = 100 + (100 x 5 x 4)/100 = ₹ 120
∴ Present Value (PV) of ₹ 120 due after 4 yr = ₹ 100
Present Value (PV) of ₹ due after 4 yr = 100P/120 = 5P/6
Similary, PV of ₹ P due after 3 yr = 20P/23
PV of ₹ P due after 2 yr = 10P/11
PV of ₹ P due after 1 yr = 20P/21
∴ 5P/6 + 20P/23 + 10P/11 + 20P/11 = 6450
⇒ P = ₹ 1810
- Subbarao was approached by two neighbour for loan. He had ₹ 2540, a part of which he lent to one person at 12% interest per annum and the other part was lent to the second person at 12.5%. At the end of a year, Subbarao received ₹ 311.60 as interest on the total loan . Calculate the amount of money of money lent by him at 12% interest . ?
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Let the amount of money lent at 12% interest be ₹ P.
∴ Amount of money lent at 12.5% interest = ₹ (2540 - P)
According to the question;
(P x 12 x 1)/100 + [(2540 - P) x 12.5 x 1]/100 = 311.60Correct Option: D
Let the amount of money lent at 12% interest be ₹ P.
∴ Amount of money lent at 12.5% interest = ₹ (2540 - P)
According to the question;
(P x 12 x 1)/100 + [(2540 - P) x 12.5 x 1]/100 = 311.60
⇒ -0.5P/100 + 31750/100 = 311.60
⇒ 0.5P = 31750 - 31160 = 590
∴ P = ₹ 1180
