Question

Q3. Mr. Govind invested an account of rs. 13900 divided in two different schemes s1 and s2 at the simple interest rate of 14% p.A. And 11% p.A. Respectively. If the total amount of simple interest earned in two years was rs. 3508, what was the amount invested in scheme s2?

Answer 1

Answer:

Step-by-step explanation:

Solution :-

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x)

$\sf\implies (\frac{x\times14\times2}{100})+(\frac{(13900-x)\times11\times2}{100})=Rs. \: 3508$

$\sf\implies 28x - 22x = 350800 - (13900\times22)$

$\sf\implies 6x = 45000$

$\sf\implies x=\frac{45000}{6}$

$\sf\implies x=7500$

$\sf\implies Sum \: invested \: in \: Scheme \: B= Rs. \: (13900 - 7500)$

$\bf\implies Sum \: invested \: in \: Scheme \: B=Rs. \: 6400$

Hence, the amount invested in scheme S2 is Rs. 6400

Answer 2

AnswEr :

Let's Divide Rs. 13900 in two different parts.

$\bold{Scheme \: 1} \begin{cases} \sf{Principal=Rs. (13900 - x)} \\ \sf{Rate=14\% \: p.a.} \\ \sf{Time=2\: Yr. }\end{cases}$

$\bold{Scheme \: 2} \begin{cases} \sf{Principal=Rs. x} \\ \sf{Rate=11\% \: p.a.} \\ \sf{Time=2\: Yr. }\end{cases}$

• According to the Question Now :

$\leadsto \sf SI_1 + SI_2 = Rs.3508$

$\leadsto \sf \dfrac{prt_1}{100} + \dfrac{prt_2}{100} = Rs.3508$

$\leadsto \sf ((13900 - x) \times 14 \times 2) + (x \times11 \times 2 ) = 3508 \times 100$

$\leadsto \sf 389200 - 28x + 22x = 350800$

$\leadsto \sf 389200 - 350800 = 28x - 22x$

$\leadsto \sf38400 = 6x$

$\leadsto \sf x = \cancel\dfrac{38400}{6}$

$\leadsto\large \boxed{ \sf x = Rs. 6400}$

⠀

∴ Money Invested on S₂ is Rs. 6400