A certain organization sells boxes of 32 oranges for $8.00 per box, bo

A certain organization sells boxes of 32 oranges for $8.00 per box, boxes of 20 grapefruit for $7.00 per box
Price PER ORANGE = $8.00/32 = $1/4 = $0.25
Price PER GRAPEFRUIT = $7.00/20 = $0.35

Each combination box contains TWICE as many oranges as grapefruit
Let G = the NUMBER of grapefruit in the box
So, 2G = the NUMBER of oranges in the box

Combination boxes of oranges and grapefruit for $8.50 per box
We can write: (G)(0.35) + (2G)($0.25) = 8.50
Simplify: 0.35G + 0.5G = 8.50
Simplify: 0.85G = 8.50
Solve: G = 8.50/0.85 = 10

So, there are 10 grapefruit in the combination box

Answer: D

Cheers,
Brent

"If the unit price of each kind of fruit is independent of the kind of box in which it is packed" ---- this is to create unwanted confusion.
Given:
I. boxes of 32 oranges for $8.00 per box
price per orange="a"= 8$/32= (1/4)$ ----- (i)

II. boxes of 20 grapefruit for $7.00 per box
price per grapefruit="b"= 7$/20 ----- (ii)

III. combination boxes of oranges and grapefruit for $8.50 per box
let x be the number of oranges and y be the number of grapefruits in combination box
==> 8.5$=(x*a+y*b) ------ (iii)

IV. each combination box contains twice as many oranges as grapefruit
x=2y ----- (iv)

substituting (i), (ii), and (iv) in (iii) we get (or values of a, b and value of x in terms of y)
8.5$= (1/4)$*2y + (7/20)$*y
or
8.5= (5/20)*2y + (7/20)$*y
or
8.5= (17/20)*y
or y=10
or
number of grapefruits in combination box = 10 = option D

rate per orange ; 8/32 ; 1/4 and rate per gf = 7/20
so for a box of both price ; 8.5 $
which contains ; 1/4 * oranges + 7/20 * gf
and given o=2*gf
so
8.5 = 1/4*2*gf + 7/20 * gf
solve for gf = 10
OPTION D

D

32 oranges price --- $8
1 orange price ------1/4

20 grapefruit price---- $7
1 grapefruit price---------7/20

Combination Box:
orange = 2 graprefruit
o =2g;

Also; No of oranges * price of one orange + No of grapefruit * price for one grapefruit = $8.5

o*(1/4) + g*(7/20) = 85/10
2g(1/4) + g*(7/20) = 85/10

2g(5/20) + g*(7/20) = 85/10
17g/20 = 85/10
g= 10
_________________

1 orange costs = $8/32 = $0.25
1 grapefruit costs = $7/20 = $0.35
Let number of grapefruit in a combination box = g
Therefore, (0.25*2g) + (0.35*g) = 8.50
Solving for g,
g=10

Answer: D

Solution

Given
In this question, we are given that

• A certain organization sells

o boxes of 32 oranges for $8.00 per box
o boxes of 20 grapefruit for $7.00 per box
o combination boxes of oranges and grapefruit for $8.50 per box


• Each combination box contains twice as many oranges as grapefruit

To find
We need to determine

• The number of grapefruits in a combination box

Approach and Working out

• Cost of 1 orange = 8/32 = $0.25
• Cost of 1 grapefruit = $7/20
• Thus, $8.5 = 0.25(2x) + 7x/20 (where x represents the number of grapefruits in a combination box)

o Implies, x = 10

Therefore, Option D is the correct answer.

Correct Answer: Option D
_________________

We see that oranges cost 8/32 = $0.25 each and a grapefruit costs 7/20 = $0.35. If we let x = the number of grapefruit in a combination box, then 2x = the number of oranges in a combination box, and we can create the equation:

0.25(2x) + 0.35x = 8.5

0.5x + 0.35x = 8.5

0.85x = 8.5

x = 10

Answer: D
_________________

\(\frac{8}{32} = 0.25\) per orange
\(\frac{7}{20} = 0.35\) per grapefruit

A combination box = $8.50

\(0.25(2x) + 0.35(x) = 8.50\)

\(0.5x + 0.35x = 8.50\)

\(0.85x = 8.50\)

\(x = 10\)

Answer is D.
_________________

Oranges: O
Grapefruit: G

Combination box: G + O = $8.50

=> G * \(\frac{7}{20}\) + O * \(\frac{8}{32}\) = 8.50

=> Since, O = 2G

=> G * \(\frac{7}{20}\) + 2G * \(\frac{8}{32}\) = 8.50

=> \(\frac{(7G)}{(20)}\) + \(\frac{(8G)}{(16)}\) = 8.50

=> \(\frac{[28G + 40G] }{ 80}\) = 8.50

=> 68G = 8.50 * 80 = 680

=> G = \(\frac{680}{68}\) = 10

Answer D

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