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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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= (12x/100) + (10y/100)

Therefore, (12x/100)+(10y/100)= 1300

= 12x + 10y = 1300×100

= 6x + 5y = 130000 ÷ 2

= 6x + 5y = 65000............{i}

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In the invested amounts are interchanged, then yearly interest increased by Rs 40.

therefore, (12x/100)+(10y/100)= 134000

6x + 5y = 67000..............{ii}

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

x-y = -2000......................{iii}

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

x+y = 12000.....................{iv}

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

2x = 10000

x= 5000

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Putting x in equation (iii)

we get y = 7000.

Thus, the person invested Rs 5000 at the rate of 12% per year and Rs 7000 at the rate of 10% per year.

Hope it helps ✨