Question

Susan invested certain amount of money in two schemes A and B , which offer interest at the rate of 8% per annum and 9% per annum , resp . She received Rs. 1860 as annum interest . However , had she interchanged the amount of investment in the two schemes , she would have received Rs. 20 more as annum interest . How much money did she invest in each scheme?​

Answer 1

$\huge {\red{\rm solution:-}}$

Let the money invested in scheme A be x

the money invested in scheme B be Y

_________________________________

interest rate of scheme A is 8%

interest rate of scheme B is 9%

_________________________________

using the formula of calculating interest,

$\boxed {\blue {\tt I = \dfrac {P.R.T}{100}}}$

$\tt I (a) = \dfrac {x×8×1}{100}$

$\tt 100×I(a)=8x$............. (1)

$\tt I (b)=\dfrac {y×9×1}{100}$

$\tt 100×I(b)=9y$............ (2)

add equations 1 and 2,

$\tt I (a)+I (b)=8x+9y/100$

$\tt 1860=8x+9y/100$

$\tt 186000=8x+9y$.......... (3)

_________________________________

case 2

let the amount invested in scheme A be y

let the amount invested in scheme B be x

calculating interest by the formula

I=P.R.t/100

$\tt I (a)=y×8×1/100$........... (4)

$\tt i (b)=x×9×1/100$......... (5)

$\tt i (a)+i(b)=8y+9x/100$

$\tt 1880=8y+9x/100$

$\tt 188000=8y+9x$......... (6)

_________________________________

solving equations 3 and 6,

$\tt (186000=8x+9y)×9$

$\tt (188000=9x+8y)×8$

$\tt 1674000=72x+81y$

$\tt 1504000=72x+64y$

$\tt 170000=17y$

$\tt y=10000$

$\tt 188000=80000+9x$

$\tt 188000-80000=9x$

$\tt 108000=9x$

$\tt 12000=x$

_________________________________

the amount of money invested in scheme A is 12000 rupees and in scheme B is 10000 rupees.

hope it helps you!

Answer 2

ANSWER

Let amount invested in A be Rs. x and in B be Rs. y.

As per the given statements, 0.08x+0.09y=1860.....(1)

And, 0.09x+0.08y=1880.....(2)

Multiplying equation (1) with 8 we get, 0.64x+0.72y=14880.....(3)

Multiplying equation (2) with 9 we get, 0.81x+0.72y=16920.....(4)

Subtracting equation (3) from (4), we get 0.17x=2040=>x=12000

Substituting x=12000 in the equation (1), we get 0.08(12000)+0.09y=1860⇒y=10000

Hence, amount invested A is Rs. 12000 and in B is Rs. 10000