Question

5
4
An honest person invested some amount at the rate of 12% simple interest and some other at the rate
of 10% simple interest. He received yearly interest of Rs 130. But if he had interchanged amount
invested, he would have received Rs 4 more as interest.
How much amount did he invest at different rates?
Which mathematics concepts is used in this problem? which mathematics concept is used in this problem ​

Answer 1

Answer:

Step-by-step explanation:

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.