Question

Rs. 4000 lend at Simple Interest in two different parts. First Part at 8% p.a. and Second Part at 10% p.a. for 1 Year. Simple Interest Collected after 1Yr is Rs. 352. Then Find the Amount Given at 8% p.a.​

Answer 1

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Explanation:

Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x).

Then,() + []

= (100000 * )

<=> = = 9750

<=> 2x = (1100000 - 975000) = 125000

<=> x = 62500.

Sum invested at 9% = Rs. 62500.

Sum invested at 11% = Rs. (100000 - 62500) = Rs. 37500.

Answer 2

AnswEr :

Let the Amount given at 8% be Rs.x and, given at 10% is Rs.(4000 - x)

$\bold{First \: Part} \begin{cases} \sf{Principal=Rs. x} \\ \sf{Rate=8\% \: p.a.} \\ \sf{Time=1 \: Yr.}\end{cases}$

$\bold{Second \: Part} \begin{cases} \sf{Principal=Rs.(4000 - x)} \\ \sf{Rate=10\% \: p.a.} \\ \sf{Time=1 \: Yr.}\end{cases}$

⠀

• According to Question Now ;

$\leadsto \sf{SI_1 + SI_2 = Rs. 352}$

$\leadsto\sf{ \dfrac{PRT_1}{100} + \dfrac{PRT_2}{100} = 352 }$

$\leadsto\sf{ \dfrac{x \times 8 \times 1}{100} + \dfrac{(4000 - x) \times 10 \times 1}{100} = 352 }$

$\leadsto\sf{8x + 40000 - 10x = 352 \times100 }$

$\leadsto\sf{40000 - 2x = 35200 }$

$\leadsto \sf{40000 - 35200 = 2x}$

$\leadsto\sf{4800 = 2x}$

$\leadsto\sf{ \cancel\dfrac{4800}{2} = x }$

$\leadsto\boxed{\sf{x = Rs.\:2400}}$

⠀

$\therefore$ Rs. 2400 is given at 8%p.a. for 1 Year.