Question

A man invested 10000 rupees, split into the two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?

Answer 1

Answer:

2500 rupees is invested at rate of 8% & 7500 rupees is the amount invested at rate of 9%

Step-by-step explanation:

• Let x be the amount invested at rate of 8%.

• Hence 10000-x will be the amount invested at 9%.

We know the relation,

SI=(P×T×R) /100

here,

• SI = Simple Interest
• P = Principal Amount
• T = Time
• R = Rate of Interest

For the first one,

SI = (x × 1 × 8)/100

For the second one,

SI=[(10000-x) × 1 × 9]/100

Also given,

• He got 875 rupees as interest on both investments.

(8x/100) + ((10000-x) × 9/100) = 875

On solving this,

we get x = 2500 and 10000-x = 7500

∴ 2500 rupees is invested at rate of 8% & 7500 rupees is the amount invested at rate of 9%

Answer 2

Answer:

2500 at 8% and 7500 at 9%

Step-by-step explanation:

let us consider that man invested X rs at the raplte of 8% due to which remaining amount (10000-x) is invested at the rate of 9% for one year

Now

=> X×8/100+(10000-x)×9/100= 875

=> 8x/100+ 90000-9x/100= 875

=> -x = -2500

=> X = 2500

now

10000- 2500 = 7500