Simple interest
- 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
- A and B borrowed Rs. 3000 and Rs. 3200 respectively at the same rate of
rate of interest.interest for 2 1 years. If B paid Rs. 40 more interest than A, find the 2
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Rate of interest = r % per annum
SI = Principal × Time × Rate 100
According to the question,3200 × 5 × r - 3000 × 5 × r = 40 100 × 2 200
⇒ 80r – 75r = 40
⇒ 5r = 40 ⇒ r = 40 ÷ 5 = 8% per annum
Second method to solve this question :Here, P1 = Rs. 3000, R1 = R, T1 = 5 years 2
Correct Option: C
Rate of interest = r % per annum
SI = Principal × Time × Rate 100
According to the question,3200 × 5 × r - 3000 × 5 × r = 40 100 × 2 200
⇒ 80r – 75r = 40
⇒ 5r = 40 ⇒ r = 40 ÷ 5 = 8% per annum
Second method to solve this question :Here, P1 = Rs. 3000, R1 = R, T1 = 5 years 2 P2 = Rs. 3200, R2 = R, T2 = 5 years 2
Difference S.I. = Rs. 40⇒ 40 = 3200 × R × (5 / 2) - 3000 × R × (5 / 2) 100
⇒ 4000 = 8000R – 7500R
⇒ R = 8%
- In what time will the simple interest be 2/5 of the principal at 8 percent per annum?
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Let the principal be p.
∴ Simple Interest = 2 p 5
Rate = 8% per annum∴ Time = Interest × 100 Principal × Rate Time = (2 / 5)p × 100 = 40 = 5 years p × 8 8
Second method to solve this question :Here, n = 2 and R = 8% 5
⇒ RT = (n × 100)T = n × 100 R
Correct Option: C
Let the principal be p.
∴ Simple Interest = 2 p 5
Rate = 8% per annum∴ Time = Interest × 100 Principal × Rate Time = (2 / 5)p × 100 = 40 = 5 years p × 8 8
Second method to solve this question :Here, n = 2 and R = 8% 5
⇒ RT = (n × 100)T = n × 100 R T = 2 × 100 5 8
T = 5 years
- The rate of simple interest for which a sum of money becomes 5 times of itself in 8 years is :
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Let Principal = Rs. p
According to question ,
∴ Amount = Rs. 5p
Simple Interest = Amount - Principal = Rs. (5p – p) = Rs. 4p∴ Rate = SI × 100 Principal × Time
Correct Option: C
Let Principal = Rs. p
According to question ,
∴ Amount = Rs. 5p
Simple Interest = Amount - Principal = Rs. (5p – p) = Rs. 4p∴ Rate = SI × 100 Principal × Time Rate = 4p × 100 = 50% per annum p × 8
