Question

Mr. thomas invested an amount of rs. 13,900 divided in two different schemes a and b at the simple interest rate of 14% p.a. and 11% p.a. respectively. if the total amount of simple interest earned in 2 years be rs. 3508, what was the amount invested in scheme b?

Answer 1

$b=Rs 6400$

Explanation:  

given,Thomas amount invested$=Rs13900$

In two parts a,b      rate of interest of a is 14% p.a and rate of interest of b is 11%p.a and time is 2 years

if total amount of simple interest Rs 3508

$a+b=Rs13900$

$a=13900-b$

$S.I= P\times R\times T/100$

S.I of a$=a\times R\times T/100$

$=\left ( 13900-b \right )\times 14\times 2/100$

$=3892-28\times b/100$

S.I of b$=b\times R\times T/100$

$=b\times 11\times 2/100$

$=22\times b/100$

S.I of   $a+b=Rs3508$

$3892-28b/100+22b/100=3508$

$3892-6b/100=3508$

$-6b/100=3508-3892$

$-6b/100=-384$

$6b=38400$

$b=38400/6$

$b=Rs 6400$

$a+b=Rs13900$

$a=13900-6400$

$a=Rs 7500$

Answer 2

Explanation:

Given :  Invested Amount  Rs 13900 divided in 2 different schemes A and B at SI 14% pa and 11% pa

Total amount of SI earned in 2years = 3058

To Find : Amount invested in Scheme B

Solution:

Let say Amount invested in Scheme B  = 100P

Amount invested in Scheme A  = 13900 - 100P

= 100(139 - P)

SI = P x R x T /100

invested in Scheme A  =100(139 - P)

R = 14%

T = 2

SI = 100(139 - P) * 14 * 2 /100

= (139 - P)28

= 3892 - 28P

invested in Scheme B  =100P

R = 11%

T = 2

SI = 100P  * 11 * 2 /100

=22P

3892 - 28P + 22P =  3508

=> 3892 - 6P =  3508

=> 6P =  384

=> P = 64

=> 100P = 6400

Amount invested in Scheme B = Rs 6400