Simple interest
- An old article is available for ₹ 12,000 at cash payment or is available for ₹ 7,000 cash payment and a monthly instalment of ₹630 for 8 months. The rate per cent per annum is
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As per the given in question , we have
Simple Interest = (7000 + 630 × 8) – 12000
Simple Interest = (7000 + 5040) – 12000
Simple Interest = 12040 – 12000 = 40
Total Principal = 5000 + 4370 + 3740 + 3110 + 2480 + 1850 + 1220 + 590 = ₹ 22360Correct Option: A
As per the given in question , we have
Simple Interest = (7000 + 630 × 8) – 12000
Simple Interest = (7000 + 5040) – 12000
Simple Interest = 12040 – 12000 = 40
Total Principal = 5000 + 4370 + 3740 + 3110 + 2480 + 1850 + 1220 + 590 = ₹ 22360
Rate = 40 × 100 ×12 ≈ 2.1% 22360 × 1
- 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 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 sum of ₹ 10,000 is lent partly at 8% and remaining at 10% per annum. If the yearly interest on the average is 9.2%, the two parts are :
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Let p be lent at 8%, then (10000 – p) is lent at 10%.
According to question ,∴ 10000 × 9.2 × t = p × 8 × t + (10000 - p) × 10 × t 100 100 100 ⇒ 920000 = 8pt + (10000 - p)10t 100 100 100
Correct Option: A
Let p be lent at 8%, then (10000 – p) is lent at 10%.
According to question ,∴ 10000 × 9.2 × t = p × 8 × t + (10000 - p) × 10 × t 100 100 100 ⇒ 920000 = 8pt + (10000 - p)10t 100 100 100
⇒ 92000t = 8pt + (10000 – p) 10t
⇒ 92000t = 8pt + (10000 – p) 10t
⇒ 92000 = 8p + 100000 – 10p
⇒ 2p = 8000
⇒ p = 4000
∴ First part = ₹ 4000
Second part = 10000 - 4000 = ₹ 6000
- 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
