A sum of money is invested at simple interest, partly at 4% and remain

So my approach was to calculate the rate of interest which gives 2100 as interest.

(18000 * 2 * x)/100 = 2100
rate comes out to be (35/6)

4 -------------------- 7
----------35/6--------
7/6------------------11/6

amount invested at 4% = (7/6) * (6/18) = (7/18)

so amount is 7000.

Hi,

It is a simple interest question, so there is no big calculations here.

We can eliminate the answer choices D and E.

Because given,

Total investment = 18000

And total interest earned = 2100.

For a moment, let’s assume the amount is equally distributed.

Then,

4% of 9000 + 7% of 9000 = 360 + 630 = 990 for one year.

Total amount for two years is 1980.

Which is less than the actual total interest(2100) given in the question.

So the amount distributed in 4% should be lesser than the amount invested in 7%.

Hence we can eliminate D and E.

Answer as to be either A, B and C.

Also the amount 1980 was not too far from 2100, then the best answer choice to try would be answer choice C.

So,

4% of 7000 + 7% of 11000 = 280 + 770 = 1050 for one year.

Total amount for two years is 2100.

Which matches the amount given in the question.

So the answer is C.

Hope it helps.

We can let the amount of money invested at 4% interest = n. The amount invested at 7% is (18,000 - n). We can create the following equation:

2 x 0.04n + 2 x 0.07(18,000 - n) = 2,100

0.08n + 2,520 - 0.14n = 2,100

-0.06n = -420

n = 7,000

Answer: C

I solved this algebraically and it took too long. Is there a quick way to solve for this? Bunuel ScottTargetTestPrep

I think my method should be pretty quick actually.