Question

Keith had 694 green, yellow and blue marbles altogether. he had 3 times as many blue marbles as yellow marbles. there were 90 fewer green than blue marbles. how many blue marbles did Keith have?

Answer 1

Let $g$ the number of green, $y$ the number of yellow and $b$ the number of blue marbles respectively.

We have

$g + y + b = 694$ (we will call this equation 1)

Also, $b = 3*y$ (yellow marbles are three times as many as blue marbles), so equation 1 becomes

$g + y + 3y = 694$

But, $g = b - 90$ (green marbles are as many as blue marbles - 90)
and because $b = 3*y$, we have that $g = 3*y - 90$

So, equation 1 is now

$3y - 90 + y + 3y = 694$
$7y - 90 = 694$

We add 90 to both sides so

$7y = 694 + 90 = 784$

We divide each side by 7, so finally

$y = 112$

and we have

$b = 3*y = 3*112 = 336$
and
$g = b - 90 = 336 - 90 = 246$

To check if our results are correct, we can see that

$g + b + y = 246 + 336 + 112 = 694$,

so we are indeed correct.

Answer 2

Let the number of blue marbles be x

number of yellow marbles = x/3

number of green = x- 90

ATQ

x + x/3 + x - 90 = 694

2x + x/3 = 784

(6x + x)/3 = 784

7x/3 = 784

x = 784 x 3/7

x = 112 x 3

x = 336

number of blue marbles = 336