Simple interest
- A sum of Rs. 800 amounts to Rs. 920 in 3 years at the simple interest rate. If the rate is increased by 3% p.a., what will be the sum amount to in the same period ?
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Case I,
Here , Principal = ₹ 800 , Amount = ₹ 920
S.I. = Amount - Principal = 920 – 800 = ₹ 120Rate = Interest × 100 Principal × Time = 120 × 100 = 5% per annum 800 × 3
Case II,
Rate = 8% per annum
Correct Option: A
Case I,
Here , Principal = ₹ 800 , Amount = ₹ 920
S.I. = Amount - Principal = 920 – 800 = ₹ 120Rate = Interest × 100 Principal × Time = 120 × 100 = 5% per annum 800 × 3
Case II,
Rate = 8% per annumS.I. = 800 × 8 × 3 = ₹ 192 800 × 3
∴ Amount = Principal + S.I. = (800 + 192) = ₹ 992
- A person who pays income tax at the rate of 4 paise per rupee, find that a fall of interest rate from 4% to 3.75% diminishes his net yearly income by ₹48. What is his capital ?
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If the capital after tax deduction be p, then
p × (4 – 3.75) % = 48⇒ p × 0.25 = 48 100 ⇒ p × 25 = 48 10000 ⇒ p = 48 400
⇒ p = 48 × 400 = ₹ 19200
Correct Option: C
If the capital after tax deduction be p, then
p × (4 – 3.75) % = 48⇒ p × 0.25 = 48 100 ⇒ p × 25 = 48 10000 ⇒ p = 48 400
⇒ p = 48 × 400 = ₹ 19200∴ Required capital = 19200 × 100 = ₹ 20000 96
- A sum of money was lent at simple interest at a certain rate for 3 years. Had it been lent at 2.5% per annum higher rate, it would have fetched ₹540 more. The money lent was :
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If the sum lent be Rs. p, then
p × 2.5 × 3 = 540 100 p = 540 × 100 = ₹ 7200 2.5 × 3
Second method to solve this question :
P1 = P, R1 = R, T1 = 3
P2 = P, R2 = R + 2.5% , T2 = 3 , S.I. = Rs. 540∴ SI = P2 × R2 × T2 - P1 × R1 × T1 100
Correct Option: D
If the sum lent be Rs. p, then
p × 2.5 × 3 = 540 100 p = 540 × 100 = ₹ 7200 2.5 × 3
Second method to solve this question :
P1 = P, R1 = R, T1 = 3
P2 = P, R2 = R + 2.5% , T2 = 3 , S.I. = Rs. 540∴ SI = P2 × R2 × T2 - P1 × R1 × T1 100 540 = P × (R + 2.5%) × 3 - P × R × 3 100
54000 = 7.5PP = 540000 75
P = ₹ 7200
- A man loses ₹55.50 yearly when the annual rate of interest falls from 11.5% to 10%. His capital (in rupees) is
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Let his capital be p.
According to the question,p × 11.5 - p × 10 = 55.50 100 100
⇒ (11.5 – 10)p = 5550
⇒ 1.5p = 5550
Correct Option: A
Let his capital be p.
According to the question,p × 11.5 - p × 10 = 55.50 100 100
⇒ (11.5 – 10)p = 5550
⇒ 1.5p = 5550⇒ p = 5550 = ₹ 3700 1.5
- Mohan lent some amount of money at 9% simple interest and an equal amount of money at 10% simple interest each for two years. If his total interest was Rs. 760, what amount was lent in each case ?
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Let the sum lent in each case be p.
Then,
According to question,p × 9 × 2 + p × 10 × 2 = 760 100 100 p × 2 (9 + 10) = 760 100 ⇒ 2 × 19p = 760 100
Correct Option: D
Let the sum lent in each case be p.
Then,
According to question,p × 9 × 2 + p × 10 × 2 = 760 100 100 p × 2 (9 + 10) = 760 100 ⇒ 2 × 19p = 760 100 p = 760 × 100 = ₹ 2000 2 × 19
