Question

Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on the total investment. How much money did he put in each account?

Answer 1

Mathew invested $600 and $2400 in each account.

Solution:

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest
`(I_1)` earned in first account for one year,

`\text {simple interest}=\frac{\text {pnr}}{100}`

Where

p = amount invested in first account

n = number of years

r = rate of interest

hence, by using above equation we get
`(I_1)` as,

`I_(1)=(P * 1 * 3)/(100)` ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest
`(I_2)` earned in second account,

`I_(2) = ((3000-P) * 1 * 8)/(100) \text { ------ eqn } 2`

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

`I = (3000 * 1 * 4)/(100) ----- eqn 3`

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

`I = I_1+ I_2 ---- eqn 4`

By substituting eqn 1 , 2, 3 in eqn 4

`(3000 * 1 * 4)/(100) = (P * 1 * 3)/(100) + ((3000-P) * 1 * 8)/(100)`

`(12000)/(100) = (3 P)/(100) + ((24000-8 P))/(100)`

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600

Hence, Mathew invested $600 and $2400 in each account.