Question

A person invested some money at 12% simple interest and some other amount at
10% simple interest. He received yearly interest of 1300. If he had interchanged the
amounts, he would have received 40 more as yearly interest. How much did he invest
at different rates?


GUYS PLZZ ANSWER FAST​

Answer 1

Answer:

Let, us suppose that :

The Person invested = Rs x at the rate of 12 percent simple interest

And

Rs y at the rate of 10 percent simple interest.

Then,

Yearly interest = 12x/100 + 10y/100

12x/100 + 10y/100 = 130

⇒ 12x + 10y = 13000

⇒ 6x + 5y = 6500 ----- ( 1 )

√ Invested amounts interchanged, then yearly interest increases.

Therefore, 10x/100 + 12y/100 = 134

10x + 12y = 13400

5x + 6y = 6700 ------- ( 2 )

Subtract eq ( 2 ) from ( 1 ) we get :-

x-y = - 200 ---- ( 3 )

Add eq (2) and ( 1 ) we get :-

11x + 11y = 13200

⇒ x+y = 1200 ---- ( 4 )

Add eq ( 3 ) and ( 4 ) we get :-

2x = 1000

x = 500

Put x = 500 in equation ( 3 ), then we get :-

y = 700

√ Atlast, I am going to conclude my whole answer :-

Hence, the person invested Rs 500 at the rate 12 percent per year.

And Rs700 at the rate of 10 percent per year.

mark me as brainlist...

Answer 2

$\sf\huge answer$

Let, us suppose that :

The Person invested = Rs x at the rate of 12 percent simple interest

And

Rs y at the rate of 10 percent simple interest.

Then,

Yearly interest = 12x/100 + 10y/100=1300

==>12x+10y=130000

==>6x+5y=65000-------(1)

√ Invested amounts interchanged, then yearly interest increases.

Therefore, 10x/100 + 12y/100 = 1340

10x + 12y = 134000

5x + 6y = 67000 ------- ( 2 )

eqn(1)×5==>

30x+25y=325000-------(3)

eqn(2)×6==>

30x+36y=402000-------(4)

(4)-(3)=>

11y=77,000

==>y=7000

value of y put in y ,we get

6x+5×7000=65000

==>6x=65000-35000

==>x=30000/6

==>x=5000

√ Atlast, I am going to conclude my whole answer :-

Hence, the person invested

• Rs 5000 at the rate 12 percent per year.

• And Rs7000 at the rate of 10 percent per year