A man has rs.1900. He lent some of it at 5% pa and rest at 9% pa. The total amount of interest he got after 4 yrs was rs.580.What was the sum?

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Answer 1 · 2018-07-31T13:07:30

Amount lent @ 9% Interest = Rs 1250 and amount lent at 5% Interest = Rs 650.

Taking it as simple Interest:

$I = \frac{P N R}{100}$ $I =( P N R) / 100$

Let x be the amount lent @ 9% Interest.

Then 1900 - x is the amount lent at 5% Interest.

$I = R s 580, N = 4 y e a r s$ $I = Rs 580, N = 4 years$

$\frac{x \cdot 4 \cdot 9}{100} + \frac{(1900 - x) \cdot 4 \cdot 5}{100} = 580$ $(x * 4 * 9)/ 100 + ((1900-x) * 4 * 5)/100 = 580$

$36 x + 38000 - 20 x = 58000$ $36 x + 38000 - 20x = 58000$

$16 x = 58000 - 38000 = 20000$ $16 x = 58000 - 38000 = 20000$

$x = \frac{20000}{16} = R s 1250$ $x = 20000/16 = Rs 1250$

$1900 - x = 1900 - 1250 = R s 650$ $1900 - x = 1900 - 1250 = Rs 650$

Answer 2 · 2018-07-31T14:51:33

Amount lent @ 9% Interest = Rs 869.71 and amount lent @ 5% is = Rs 1030.29

Taking it as compound interest.

$A = P {(1 + \frac{R}{100})}$ $A = P (1 + R/100)^$

Let x be the amount lent at 9% Interest.

Then 1900 - x is the amount lent @ 5% Interest.

$P = 1900, I = 580, N = 4 y e a r s$ $P = 1900, I = 580, N = 4 years$

$1900 + 580 = x {(1 + \frac{9}{100})}^{4} + (1900 - x) {(1 + \frac{5}{100})}^{4}$ $1900 + 580 = x (1 + 9/100)^4 + (1900-x) (1 + 5/100)^4$

$x {1.09}^{4} + 1900 \cdot {1.05}^{4} - x {1.05}^{4} = 2480$ $x 1.09^4 + 1900 * 1.05^4 - x 1.05^4 = 2480$

$1.4116 x - 1.2155 x = 2480 - 1900 \cdot 1.2155 = 170.55$ $1.4116 x - 1.2155x = 2480 - 1900* 1.2155 = 170.55$

$x = \frac{170.55}{1.4116 - 1.2155} = 869.71$ $x = 170.55 / (1.4116 - 1.2155) = 869.71$

$1900 - x = 1900 - 869.71 = 1030.29$ $1900 - x = 1900 - 869.71 = 1030.29$