Math, asked by upnext2024, 9 months ago

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

Answered by parthmankar21
5

b is the correct answer for this question

Answered by RvChaudharY50
2

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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