Question

A man invested an amount at 12% per annum simple interest and another amount at 10% per annum
simple interest. He received an annual interest of R. 2600. But, if he had interchanged the amounts
invested, he would have received *. 140 less. Find the amount invested at each rate. ​

Answer 1

Answer:

The amount invested at each rate is ₹371.5. (approx).

Step-by-step explanation:

Let the amount be x

Therefore, S.I. = p*r*t/100

= x*12*25/100

= 3x

New Amount = old amount+interest

= x + 3x = 4x

Therefore, S.I. = 4x*10*10/100

= 4x

Therefore, 3x + 4x = 2600 - 140

Or, 7x = 2460

Or, x = 2460/7 = 351.5(approx).

Answer 2

Answer:

Let the amounts invested at 10% and 8% be Rs.x and Rs.y respectively.

Then as per the question

10x + 8y = 135000 …………..(i)

After the amounts interchanged but the rate being the same, we have

8x +10y = 130500 …………….(ii)

Adding (i) and (ii) and dividing by 9, we get

2x + 2y = 29500 ……………(iii)

Subtracting (ii) from (i), we get

2x – 2y = 4500

Now, adding (iii) and (iv), we have

4x = 34000

x = 34000/4 = 8500

x=8500

Putting x = 8500 in (iii), we get

2 × 8500 + 2y = 29500

2y = 29500 – 17000 = 12500

y = 12500/2 = 6250

y=6250

Hence, the amounts invested are Rs. 8,500 at 10% and Rs. 6,250 at 8%.