Simple interest
- ₹800 becomes ₹956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will ₹800 become in 3 years ?
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We know that ,
S.I. = Amount - Principle = 956 – 800 = Rs. 156∴ Rate = S.I. × 100 Principal × Time R = 156 × 100 = 6.5% per annum 800 × 3
∴ New rate = 6.5 + 4 = 10.5%
Correct Option: C
We know that ,
S.I. = Amount - Principle = 956 – 800 = Rs. 156∴ Rate = S.I. × 100 Principal × Time R = 156 × 100 = 6.5% per annum 800 × 3
∴ New rate = 6.5 + 4 = 10.5%∴ SI = Principal × Time × Rate 100 SI = 800 × 3 × 10.5 = ₹ 252 100
∴ Amount = Principle + SI = 800 + 252 = ₹ 1052
- A person borrows ₹5,000 for 2 years at 4% per annum simple interest. He immediately
His gain in the transaction islends it to another person at 6 1 % per annum simple interest for 2 years. 4
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Here , P = ₹ 5000 , R = 4% , T = 2 years
Case I :SI = 5000 × 2 × 4 = ₹ 400 100
Case II :Here , P = ₹ 5000 , T = 2 years , R = 25 % 4 SI = 5000 × 25 × 2 = ₹ 625 100 × 4
Correct Option: C
Here , P = ₹ 5000 , R = 4% , T = 2 years
Case I :SI = 5000 × 2 × 4 = ₹ 400 100
Case II :Here , P = ₹ 5000 , T = 2 years , R = 25 % 4 SI = 5000 × 25 × 2 = ₹ 625 100 × 4
∴ Gain = ₹ (625 – 400) = ₹ 225
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In what time will 72 become ₹81 at 6 1 % per annum simple interest ? 4
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Let the time be t years.
Simple Interest = Amount - Principle = ₹(81 – 72) = ₹9
We know that ,SI = P × R × t 100 9 = 72 × 25 × t 4 × 100
Correct Option: A
Let the time be t years.
Simple Interest = Amount - Principle = ₹(81 – 72) = ₹9
We know that ,SI = P × R × t 100 9 = 72 × 25 × t 4 × 100 ⇒ t = 9 × 400 = 2yrs 72 × 25
- What annual instalment will discharge a debt of ₹6450 due in 4 years at 5% simple interest ?
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Let each instalment be y Then,
⇒ y + y × 5 × 1 + y + y × 5 × 2 + y + y × 5 × 3 + y = 6450 100 100 100 ⇒ y + y + y + y + y + 3y + y = 6450 20 10 20 ⇒ 21y + 11y + 23y + y = 6450 20 10 20 ⇒ 21y + 22y + 23y + 20y = 6450 20 ⇒ 86y = 6450 20 ⇒ y = 6450 × 20 = ₹ 1500 86
Second method to solve this question :
Here , A = ₹ 6450 , T = 4 years , R = 5%Equal instalment = A × 200 T[200 + (T - 1) × R]
Correct Option: A
Let each instalment be y Then,
⇒ y + y × 5 × 1 + y + y × 5 × 2 + y + y × 5 × 3 + y = 6450 100 100 100 ⇒ y + y + y + y + y + 3y + y = 6450 20 10 20 ⇒ 21y + 11y + 23y + y = 6450 20 10 20 ⇒ 21y + 22y + 23y + 20y = 6450 20 ⇒ 86y = 6450 20 ⇒ y = 6450 × 20 = ₹ 1500 86
Second method to solve this question :
Here , A = ₹ 6450 , T = 4 years , R = 5%Equal instalment = A × 200 T[200 + (T - 1) × R] Equal instalment = 6450 × 200 4[200 + (4 - 1) × 5] Equal instalment = 6450 × 200 4 × 215 Equal instalment = 6450 × 200 4 × 215 Equal instalment = 6450 × 50 215
Equal instalment = ₹ 1500
- A sum of money lent out at simple interest amounts to 720 after 2 years and to 1020 after a further period of 5 years. The sum is :
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As per the given in question ,
Principal + SI for 2 years = ₹720 .... (i)
Principal + SI for 7 years = ₹1020 .....(ii)
Subtracting equation (i) from (ii) get,
SI for 5 years = (1020 – 720) = ₹300
∴ SI for 2 years = 300 × ( 2 ÷ 5 ) = ₹120
∴ Principal = Amount - SI = (720 – 120) = ₹600
Second method to solve this question :
Here , A2 = ₹ 1020 , T1 = 2 years , A1 = ₹ 720 , T2 = 7 yearsP = A2T1 - A1T2 T1 - T2
Correct Option: B
As per the given in question ,
Principal + SI for 2 years = ₹720 .... (i)
Principal + SI for 7 years = ₹1020 .....(ii)
Subtracting equation (i) from (ii) get,
SI for 5 years = (1020 – 720) = ₹300
∴ SI for 2 years = 300 × ( 2 ÷ 5 ) = ₹120
∴ Principal = Amount - SI = (720 – 120) = ₹600
Second method to solve this question :
Here , A2 = ₹ 1020 , T1 = 2 years , A1 = ₹ 720 , T2 = 7 yearsP = A2T1 - A1T2 T1 - T2 P = 1020 × 2 - 720 × 7 2 - 7 P = 2040 - 5040 - 5 P = - 3000 - 5
P = ₹ 600