SOLUTION: (a). john invested 4000 for a year, part at 7% per annum and part at 9% per annum. the interest on the investment is 300. how much did john invest at each rate? (b). if 400 is d

Question 1163146: (a). john invested 4000 for a year, part at 7% per annum and part at 9% per annum. the interest on the investment is 300. how much did john invest at each rate?
(b). if 400 is deposited at a bank at 10% per annum compounded quarterly, how long will it take the deposit to double in value?
(c). a bowl collect 1.67cm height of water in its first week in a dewdrop. each week the height of the water increases by 4% more than it did the week before. by how much does it increase in nine weeks, including the first week?

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

One question per post -- it is the clear rule of the post. Re-post the other two problems separately.

For a standard algebraic solution to the problem, the setup would be something like this:

x amount at 7%, plus (4000-x) amount at 9%, equals 300

I leave it to you to finish solving the problem by that method.

Here is a quick and easy method for solving a "mixture" problem like this if a formal algebraic solution is not required.

(1) All $4000 invested at 7% would yield $280 interest; all at 9% would yield $360 interest.
(2) The actual interest amount, $300, is 20/80 = 1/4 of the way from $280 to $360; therefore, 1/4 of the total was invested at the higher rate.

ANSWER: 1/4 of $4000, or $1000, at 9%; the other $3000 at 7%.

CHECK: .09(1000)+.07(3000) = 90+210 = 300

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