# challenging ratio problem

• Mar 6th 2011, 07:37 PM
kingman
challenging ratio problem
Dear sir,
Would appreciate very much if someone can help me in the below question.
thanks
Kingman

The ratio of the number of red apples to the number of green apples in a box was 7:6. After twice as many red apples as green apples were sold, the ratio became 3: 4. There were 24 green apples in the box in the end. How many apples were in the box at first?
• Mar 7th 2011, 02:04 AM
HallsofIvy
Let the number of red apples in the box be x and the number of green apples be y. Then $\displaystyle \frac{x}{y}= \frac{7}{6}$ or $\displaystyle 6x= 7y$. Let the number of green apples sold be z so that the number of red apples sold is 2z. Then, at the end, there are x- 2z red apples and y- z green apples $\displaystyle \frac{x- 2z}{y- z}= \frac{3}{4}$ or $\displaystyle r4(x- 2z)= 3(y- z)$ which is the same as 4x= 3y+ z. Of course, we also have (x- 2z)+ (y- z)= x+ y- 3z= 24.

Solve the three equations 6x= 7y, 4x= 3y+ z, and x+ y- 3z= 24 for x, y, and z. The answer to the question "How many apples were in the box at first?" is x+ y.

Are you sure of your numbers? I do NOT get an integer number of apples.
• Mar 7th 2011, 02:58 AM
kingman
Thanks for the reply; the question if correct and the answer if 78 apples.
• Mar 7th 2011, 03:18 AM
Wilmer
Quote:

Originally Posted by kingman
Thanks for the reply; the question if correct and the answer if 78 apples.

Red: 42 ; -24 ; 18
Grn: 36 ; -12 ; 24
42 + 36 = 78
• Mar 7th 2011, 03:27 AM
Soroban
Hello, kingman!

Another approach . . .

Quote:

The ratio of Red apples to Green apples in a box was 7:6.
After twice as many Reds as Greens were sold, the ratio became 3:4.
There were 24 Greens in the box in the end.
How many apples were in the box at first?

At first, there were $\displaystyle 7a$ Reds and $\displaystyle 6a$ Greens, for some integer $\displaystyle \,a.$

Then $\displaystyle \,x$ Greens were sold, and $\displaystyle \,2x$ Reds were sold.

This left: $\displaystyle (7a - 2x)$ Reds and $\displaystyle (6a-x)$ Greens.

The ratio became 3:4. . . $\displaystyle \dfrac{7a-2x}{6a-x} \:=\:\dfrac{3}{4} \quad\Rightarrow\quad x \,=\,2a$ .[1]

There were 24 Greens left: .$\displaystyle 6a-x \:=\:24$ .[2]

Substitute [1] into [2]: . $\displaystyle 6a - 2a \,=\,24 \quad\Rightarrow\quad a \,=\,6$

Therefore, there were 42 Reds and 36 Greens . . . a total of 78 apples.

• Mar 7th 2011, 04:10 AM
kingman
Dear sir,
thanks for the answer but Can you please explain how you arrive at the statement 'Grn: 36 ; -12 ; 24'
but not Grn: 18 ; 6 ; 12
thanks
• Mar 7th 2011, 07:26 AM
Wilmer
Quote:

Originally Posted by kingman
Dear sir,
thanks for the answer but Can you please explain how you arrive at the statement
'Grn: 36 ; -12 ; 24' but not Grn: 18 ; -6 ; 12

That leaves 12 green apples; but "There were 24 green apples in the box in the end."
I edited my post to remove the "2 boxes" I wrongly assumed.