A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year he earned $112 in interest. How much money was invested in each account?
Which of the following equations could be used to solve the problem?
0.02x + 0.04(800) = 112
0.02x + 0.04(x + 800) = 112
0.04x + 0.02(x + 800) = 112
Answers
b is the correct answer for this question
Given :-
- A man invests a certain amount of money at 2% interest .
- And $800 more than that amount in another account at 4% interest.
- At the end of one year he earned $112 in interest.
To Find :-
How much money was invested in each account?
Which of the following equations could be used to solve the problem ?
A) 0.02x + 0.04(800) = 112
B) 0.02x + 0.04(x + 800) = 112
C) 0.04x + 0.02(x + 800) = 112.
Solution :-
Let us assume that, the man invests $ x at 2 % interest for 1 year.
So,
→ Principal = $ x
→ Rate = 2 % per annum.
→ Time = 1 year.
we know that,
→ Simple interest = (Principal * Rate * Time) / 100
Putting all values we get,
→ SI = (x * 2 * 1)/100
→ SI = $ (0.02x) .
__________
Now, given that, he adds $800 more in the amount and invest at 4% interest.
So,
→ Principal = $ (x + 800).
→ Rate = 4 % per annum.
→ Time = 1 year.
we know that,
→ Simple interest = (Principal * Rate * Time) / 100
Putting all values we get,
→ SI = {(x + 800)* 4 * 1)/100
→ SI = $ 0.04(x + 800) .
__________
Now, we have given that, At the end of one year he earned $112 in interest from both investment.
Therefore,
→ SI on x + SI on (x + 800) = 112
→ 0.02x + 0.04(x + 800) = 112. (Option C) (Ans.)
if we want to solve , how much money was invested in each account .
Than,
→ 0.02x + 0.04(x + 800) = 112
→ 0.02x + 0.04x + 32 = 112
→ 0.06x = 112 - 32
→ 0.06x = 80
→ x = (80/0.06)
→ x = $1333.33 .(Ans.)
Therefore,
→ (x + 800) = 1333.33 + 800 = $2133.33 (Ans.)
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