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Ml aggarwal solutions

A person invested some money at 12% simple interest and some other amount at 10% simple

interest. He received a yearly interest of Rs, 1300. If he had interchanged the amounts, he would

have received Rs. 40 more as yearly interest. How much did he invest at different rates?

Let the amount invested at S.I be Rs x at rate = 12% p.a.

And another investment at S.I = Rs y at rate = 10% p.a.

Then according to the given conditions, we have

12x + 10y = 130000

6x + 5y = 65000 … (i)

And,

10x + 12y = 134000

5x + 6y = 67000 … (ii)

Multiplying (i) by 6 and (ii) by 5, we have

36x + 30y = 390000

25x + 30y = 335000

(-)—(-)-----(-)--------

11x = 55000

x = 55000/11

x = 5000

On substituting the value of x in (i), we get

6(5000) + 5y = 6500

30000 + 5y = 6500

5y = 65000 – 3000

y = 35000/5 = 7000

Therefore, the investment at 12% is Rs 5000 and the investment at 10% is Rs 7000.

A person invested some money at 12% simple interest and some other amount at 10% simple

interest. He received a yearly interest of Rs, 1300. If he had interchanged the amounts, he would

have received Rs. 40 more as yearly interest. How much did he invest at different rates?

Let the amount invested at S.I be Rs x at rate = 12% p.a.

And another investment at S.I = Rs y at rate = 10% p.a.

Then according to the given conditions, we have

12x + 10y = 130000

6x + 5y = 65000 … (i)

And,

10x + 12y = 134000

5x + 6y = 67000 … (ii)

Multiplying (i) by 6 and (ii) by 5, we have

36x + 30y = 390000

25x + 30y = 335000

(-)—(-)-----(-)--------

11x = 55000

x = 55000/11

x = 5000

On substituting the value of x in (i), we get

6(5000) + 5y = 6500

30000 + 5y = 6500

5y = 65000 – 3000

y = 35000/5 = 7000

Therefore, the investment at 12% is Rs 5000 and the investment at 10% is Rs 7000.

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