Question

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

Answer 1

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Yearly interest  =130

⇒12x+10y=13000

⇒6x+5y=6500 .(i)

In the invested amounts are interchanged, then yearly interest increased by Rs 4.

⇒10x+12y=13400

⇒5x+6y=6700 ..(ii)

Subtracting equation (ii) from equation (i), we get

x−y=−200 .(iii)

Adding equation (ii) and (i), we get

11x+11y=13200

⇒x+y=1200 ..(iv)

Adding equations (iii) and (iv), we get

2x=1000⇒x=500

Putting x=500 in equation (iii), we get y=700

Thus, the person invested Rs 500 at the rate of 12% per year and Rs 700 at the rate of 10% per year.

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Answer 2

Answer:

ANSWER

Suppose the person invested Rs x at the rate of 12% simple interest and Rs y at the rate of 10% simple interest. Then,

Yearly interest =

100

12x

+

100

10y

∴

100

12x

+

100

10y

=130

⇒12x+10y=13000

⇒6x+5y=6500 .(i)

In the invested amounts are interchanged, then yearly interest increased by Rs 4.

∴

100

10x

+

100

12y

=134

⇒10x+12y=13400

⇒5x+6y=6700 ..(ii)

Subtracting equation (ii) from equation (i), we get

x−y=−200 .(iii)

Adding equation (ii) and (i), we get

11x+11y=13200

⇒x+y=1200 ..(iv)

Adding equations (iii) and (iv), we get

2x=1000⇒x=500

Putting x=500 in equation (iii), we get y=700

Thus, the person invested Rs 500 at the rate of 12% per year and Rs 700 at the rate of 10% per year.