Simple interest
- A certain sum at simple interest amount to ₹ 1350 in 5 yr and to ₹ 1620 in 8 yr what is the sum ?
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Given, A1 = ₹ 1350, A2 =₹ 1620
T1 = 5 yr and T2 = 8 yr
Let principal amount be ₹ P.
∴ In time = 8 - 5 = 3 yr
Simple interest will be
1620 - 1350 = ₹ 270
∴ R = [(A2 - A1 ) x 100] / [A1 T2 - A2 T1]
= [(1620 - 1350) x 100] / [(1350 x 8) - (1620 X 5)]Correct Option: C
Given, A1 = ₹ 1350, A2 =₹ 1620
T1 = 5 yr and T2 = 8 yr
Let principal amount be ₹ P.
∴ In time = 8 - 5 = 3 yr
Simple interest will be
1620 - 1350 = ₹ 270
∴ R = [(A2 - A1 ) x 100] / [A1 T2 - A2 T1]
= [(1620 - 1350) x 100] / [(1350 x 8) - (1620 X 5)]
= (270 x 100)/[10800 - 8100]
= 27000/2700
⇒ R = 10%
∴ P = (SI x 100) / (R x T)
= (270 x 100) / (10 x 3)
= ₹ 900
- In 4 yr. ₹ 6000 amounts to ₹ 8000. In what time at the same rate, will ₹ 525 amount to ₹ 700 ?
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Amount = ₹ 8000
Time (T) = 4 yr; Principal (P) = ₹ 6000
Simple interest (SI) = (A - P)
= Amount - Principal
= 8000 - 6000 = ₹ 2000
Rate (R) = ?
According to the formula. SI = (P x R x T)/100
⇒ 2000 = (6000 x R x 4)
⇒ R = (6000 x 100)/(6000 x 4) = (25/3) %
Now, again Amount (A) = ₹ 700
Principle (P) = ₹ 525, Time (T) = ?
Rate (R) = 25/3%
Simple interest = A - P
⇒ 700 - 525 = ₹ 175
Using formula, SI = (P x R x T) / 100Correct Option: C
Amount = ₹ 8000
Time (T) = 4 yr; Principal (P) = ₹ 6000
Simple interest (SI) = (A - P)
= Amount - Principal
= 8000 - 6000 = ₹ 2000
Rate (R) = ?
According to the formula. SI = (P x R x T)/100
⇒ 2000 = (6000 x R x 4)
⇒ R = (6000 x 100)/(6000 x 4) = (25/3) %
Now, again Amount (A) = ₹ 700
Principle (P) = ₹ 525, Time (T) = ?
Rate (R) = 25/3%
Simple interest = A - P
⇒ 700 - 525 = ₹ 175
Using formula, SI = (P x R x T) / 100
⇒ 175 = [525 x (25/3) x T] / 100
⇒ T = (175 x 100 x 3) / (525 x 25)
= 4 yr
- A Principal amounts to ₹ 944 in 3 yr and to ₹ 1040 in 5 yr, each sum being invested at the same simple interest. The principal was
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By formula, P = (A2T1 - A1T2) / (T1 - T2)
Here, A1 = ₹ 944, T1 = 3
A2 = ₹ 1040, T2 = 5Correct Option: A
By formula, P = (A2T1 - A1T2) / (T1 - T2)
Here, A1 = ₹ 944, T1 = 3
A2 = ₹ 1040, T2 = 5
∴ P = [(1040 x 3) - (944 x 5)] / (3 - 5)
= (3120 - 4720) / (-2)
= (-1600) / (-2)
= ₹ 800
∴ Principal = ₹ 800
- A sum of ₹ 1550 was lent partly at 5% and partly at 8% per annum simple interest. The total interest received after 4 yr was ₹ 400, The ratio of the money lent at 5% to that lent at 8% is
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Let the sum lent at 5% = P
∴ Sum lent at 8% = (1550 - P)
Then, [(P x 5 x 4)/100] + [{(1550 - P) x 8 x 4}/100] = 400Correct Option: A
Let the sum lent at 5% = P
∴ Sum lent at 8% = (1550 - P)
Then, [(P x 5 x 4)/100] + [{(1550 - P) x 8 x 4}/100] = 400
⇒ 20P - 32P + 1550 x 32 = 40000
⇒ - 12P + 49600 = 40000
⇒ - 12P = - 9600
∴ p = ₹ 800
Sum lent at 8% = 1550 - 800 = ₹ 750
∴ Required ratio = 800 : 750 = 16 : 15
- Neeta borrowed some money at the rate of 6% per annum for the first 3 yr. at the rate of 9% per annum for the next 5 yr and a the rate of 13% per annum for the period beyond 8 yr. If she pays a total interest of ₹ 8160 at the and of 11 yr how much money did she borrow ?
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Let the sum borrowed = P
Then, according to the question,
[(P x 6 x 3)/100] + [(P x 9 x 5)/100] + [(P x 13 x 3)/100] = 8160Correct Option: C
Let the sum borrowed = P
Then, according to the question,
[(P x 6 x 3)/100] + [(P x 9 x 5)/100] + [(P x 13 x 3)/100] = 8160
⇒ (18P + 45P +39P) / 100 = 8160
⇒ 102P/100 = 8160
⇒ P = (8160 x 100)/102
∴ P = ₹ 8000
