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the answer is 6.

first let's sum your first three scores 75 + 91 + 95 =261

you know that you must average a 90 or better on the remaining tests. so let's say sum (X) / x = 90 where sum(x) is the sum of the remaining tests scores and x is the remaining number of tests.

you also know that you need a min of an 89 average or:

(261 + sum (X))/ (3+x) = 89

261 + sum (X) = 89(3+x)

261 + sum (X) = 267 + 89x

sum(x) = 6 + 89x

sum(x)/x = 6/x + 89

90 = 6/x + 89

1 = 6/x

x=6

Your equation would set up as follows.

(75+91+95) + (90X) / X +3 = 89

90X represents the minimum score times the number of tests left. Dividing by X (the number of tests left) + 3 (representing current number of tests taken) = 89 (desired average for an A)

Multiply both sides by X +3

261 + 90X = 89X + 267

Now separate your X

Subtract 261 from both sides

90X = 89X + 6

Subtract 89X from both sides

X = 6

You have 6 tests left to make an 89 average.

I hated those when I took algebra 2..

Lets say the next test's score = x,

90 = (75+91+95+x)/4

Solve for x...

And you get x=99

x is the 1st even type x + 2 is the 2nd even type and x + 4 would be the 0.33 even type all of those numbers upload as much as sixty six so x + (x + 2) + (x + 4) = sixty six 3x + 6 = sixty six 3x = sixty six - 6 3x = 60 x = 60/3 x = 20 So the three numbers are 20, 22, and 24 M = 4S M = 2S + 20 4S = 2S + 20 2S = 20 S = 10 Son = 10 years consequently guy = 40