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?

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 earned in first account for one year,

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get as,  

----- eqn 1

Step 2:

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

Step 3:

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

Step 4:

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

Hence we get,

By substituting eqn 1 , 2, 3 in eqn 4

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.

answer: ask

step-by-step explanation:

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm