Simple interest
- The amount 2,100 became ₹2,352 in 2 years at simple interest. If the interest rate is decreased by 1%, what is the new interest ?
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As per the given in question ,
S.I. = Amount - Principal = 2352 – 2100 = ₹ 252Rate = Interest × 100 Principal × Time Rate = 252 × 100 = 6% per annum 2100 × 2
New rate = 5%
Correct Option: A
As per the given in question ,
S.I. = Amount - Principal = 2352 – 2100 = ₹ 252Rate = Interest × 100 Principal × Time Rate = 252 × 100 = 6% per annum 2100 × 2
New rate = 5%∴ S.I. = 252 × 5 = ₹ 210 6
- ₹800 amounts to ₹920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to
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Given that , Amount = ₹ 920 and Principal = ₹ 800
∴ S.I. = Amount - Principal = ₹ (920 – 800) = ₹ 120∴ Rate = Interest × 100 Principal × Time Rate = 120 × 100 = 5% per annum 800 × 3
New rate = 8% per annum
Correct Option: D
Given that , Amount = ₹ 920 and Principal = ₹ 800
∴ S.I. = Amount - Principal = ₹ (920 – 800) = ₹ 120∴ Rate = Interest × 100 Principal × Time Rate = 120 × 100 = 5% per annum 800 × 3
New rate = 8% per annum∴ SI = 800 × 8 × 3 = ₹ 192 100
∴ Amount = Principal + SI = (800 + 192) = ₹ 992
- A sum of 2,400 amounts to ₹3,264 in 4 years at a certain rate of simple interest. If the rate of interest is increased by 1%, the same sum in the same time would amount to
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Here , Principal = 2,400 and Amount = ₹3,264
S.I. = 3264 – 2400 = ₹ 864Rate = S.I. × 100 Principal × Time Rate = 864 × 100 = 9% per annum 2400 × 4
New rate = 10% per annum
Correct Option: D
Here , Principal = 2,400 and Amount = ₹3,264
S.I. = 3264 – 2400 = ₹ 864Rate = S.I. × 100 Principal × Time Rate = 864 × 100 = 9% per annum 2400 × 4
New rate = 10% per annum∴ SI = 2400 × 10 × 4 = ₹ 960 100
∴ Amount = Principal + SI = 2400 + 960 = 3360
- A sum was lent at simple interest at a certain rate for 2 years. Had it been lent at 3% higher rate, it would have fetched ₹300 more. The original sum of money was :
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If the principal be p, then
We can find the required answer with the help of given formula ,∴ SI = P × R × T 100 p × 3 × 2 = 300 100 ⇒ p = 300 × 100 = ₹ 5000 3 ⇒ 2
Second method to solve this question :
P1 = P, R1 = R, T1 = 2.
P2 = P, R2 = R + 3, T2 = 2 , S.I.= ₹ 300∴ SI = P2 × R2 × T2 - P1 × R1 × T1 100 300 = P × (R + 3) × 2 - PR2 100
Correct Option: A
If the principal be p, then
We can find the required answer with the help of given formula ,∴ SI = P × R × T 100 p × 3 × 2 = 300 100 ⇒ p = 300 × 100 = ₹ 5000 3 ⇒ 2
Second method to solve this question :
P1 = P, R1 = R, T1 = 2.
P2 = P, R2 = R + 3, T2 = 2 , S.I.= ₹ 300∴ SI = P2 × R2 × T2 - P1 × R1 × T1 100 300 = P × (R + 3) × 2 - PR2 100 300 = 6P 100
P = ₹ 5000
- A sum of money was invested at a certain rate of simple interest for 2 years . Had it been invested at 1% higher rate, it would have fetched ₹24 more inter-est. The sum of money is :
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Given that , SI = ₹ 24 , R = 1% , T = 2 years
Let Principal = P
We know that ,SI = P × R × T 100 P × 1 × 2 = 24 100 P = 2400 = ₹ 1200 2
Second method to solve this question :
P1 = P, R1 = R, T1 = 2.
P2 = P, R2 = R + 1, T2 = 2 , S.I.= Rs. 24
Correct Option: A
Given that , SI = ₹ 24 , R = 1% , T = 2 years
Let Principal = P
We know that ,SI = P × R × T 100 P × 1 × 2 = 24 100 P = 2400 = ₹ 1200 2
Second method to solve this question :
P1 = P, R1 = R, T1 = 2.
P2 = P, R2 = R + 1, T2 = 2 , S.I.= Rs. 2424 = P(R + 1) 2 - PR2 100
2400 = 2PR + 2P – 2PR
P = ₹ 1200
