Simple interest
- In simple interest rate per annum a certain sum amounts to Rs. 5,182 in 2 years and Rs. 5,832 in 3 years. The principal in rupees is
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Let the principal be Rs. P.
∴ Principle + Simple Interest = Amount
According to the question,
P + S.I. for 2 years = Rs. 5182 ...(i)
P + S.I. for 3 years = Rs. 5832 ...(ii)
By equation (ii) – (i),
S.I. for 1 year = Rs. (5832 – 5182) = Rs. 650
∴ S.I. for 2 years = Rs. (2 × 650) = Rs. 1300Correct Option: C
Let the principal be Rs. P.
∴ Principle + Simple Interest = Amount
According to the question,
P + S.I. for 2 years = Rs. 5182 ...(i)
P + S.I. for 3 years = Rs. 5832 ...(ii)
By equation (ii) – (i),
S.I. for 1 year = Rs. (5832 – 5182) = Rs. 650
∴ S.I. for 2 years = Rs. (2 × 650) = Rs. 1300
∴ Principal = Amount - SI = Rs. (5182 – 1300) = Rs. 3882
- ₹6,000 becomes ₹7,200 in 4 years at a certain rate of simple interest. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be
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SI = Amount - Principle = (7200 – 6000) = 1200
∴ SI = Principal × Time × Rate 100 ⇒ 1200 = 6000 × 4 × R 100 ⇒ R = 1200 × 100 = 5% 6000 × 4
New rate of R = 5 × 1.5 = 7.5%
Correct Option: B
SI = Amount - Principle = (7200 – 6000) = 1200
∴ SI = Principal × Time × Rate 100 ⇒ 1200 = 6000 × 4 × R 100 ⇒ R = 1200 × 100 = 5% 6000 × 4
New rate of R = 5 × 1.5 = 7.5%Then, SI = 6000 × 7.5 × 5 100
SI = ₹2250
∴ Amount = Principle + SI = (6000 + 2250) = ₹8250
- A sum of money at simple interest trebles itself in 15 years. It will become 5 times of itself in
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Let the principal be P.
Case-I2P = P × R × 15 100 ⇒ 2P = 2 × 100 = 40 % 15 3
Case-II
SI = 4P∴ 4P = P × 40 × T 100 ⇒ T= 4 × 300 = 30 yaers 40
Second method to solve this question :R = (3 - 1) × 100% 15 R = 2 × 100% = 2 × 20% 15 15
Correct Option: C
Let the principal be P.
Case-I2P = P × R × 15 100 ⇒ 2P = 2 × 100 = 40 % 15 3
Case-II
SI = 4P∴ 4P = P × 40 × T 100 ⇒ T= 4 × 300 = 30 yaers 40
Second method to solve this question :R = (3 - 1) × 100% 15 R = 2 × 100% = 2 × 20% 15 15 R = 40 % 3 T = (n - 1) Years R T = (5 - 1) × 100 (40 ÷ 3)
T = 30 years.
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then the required time will beIn certain years a sum of money is doubled to itself at 6 1 % simple interest per annum, 4
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According to the question,
If principal be Rs. p, then
S.I. = Rs. p∴ Time = SI × 100 Principal × Rate = p × 100 p × 25 4 = 400 = 16 years 25
Second method to solve this question :
Here , n = 2T = (n - 1) × 100% R
Correct Option: A
According to the question,
If principal be Rs. p, then
S.I. = Rs. p∴ Time = SI × 100 Principal × Rate = p × 100 p × 25 4 = 400 = 16 years 25
Second method to solve this question :
Here , n = 2T = (n - 1) × 100% R T = 2 - 1 25 × 100 4 T = 400 = 16 years. 25
- The simple interest on a sum of money is 8/25 of the sum. If the number of years is numerically half the rate percent per annum, then the rate percent per annumis
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Let Rate = R% per annum
∴ Time = R years. 2 ∴ Rate = SI × 100 Principal × Time ⇒ R = 8 × 100 25 ( R/2 )
Correct Option: B
Let Rate = R% per annum
∴ Time = R years. 2 ∴ Rate = SI × 100 Principal × Time ⇒ R = 8 × 100 25 ( R/2 ) ⇒ R² = 8 × 200 = 64 25
⇒ R = √64 = 8% per annum
