A company produces a certain toy in only 2 sizes, small or large, and in only 2 colors, red or green. If, for each size,

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 amFollowed by: 2 members

by Gmat_mission » Fri Jan 08, 2021 4:29 am

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A company produces a certain toy in only 2 sizes, small or large, and in only 2 colors, red or green. If, for each size, there are equal numbers of red and green toys in a certain production lot, what fraction of the total number of green toys is large?

(1) In the production lot, 400 of the small toys are green.
(2) In the production lot, 2/3 of the toys produced are small.

Answer: B

Source: Official Guide

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Source: Beat The GMAT — Data Sufficiency | Fri Jan 08, 2021 4:29 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

Re: A company produces a certain toy in only 2 sizes, small or large, and in only 2 colors, red or green. If, for each s

by deloitte247 » Sun Jan 17, 2021 7:32 am

Let the small red toys = x, automatically, small green toys = x
Let the large red toys = y, automatically, large green toys = y
Total number of red toys = x + y, and the total number of green toys = x + y

Target question => what fraction of the total number of green toys is large?
$$i.e\ eveluate\ \frac{y}{x+y}$$

Statement 1: In the production lot, 400 of the small toys are green.
x = 400
But there is no information regarding the value of y. Therefore, statement 1 is NOT SUFFICIENT.

Statement 2: In the production lot, 2/3 of the toys produced are small.
$$1-\frac{2}{3}=\frac{1}{3}$$
1/3 of total toys are large
Large green toys = y / (x+y); large red toys will also be y / (x+y).
$$So,\ all\ large\ toys\ \frac{2y}{2\left(x+y\right)}\ and\ \frac{2y}{2\left(x+y\right)}=\frac{1}{3}$$ $$\frac{y}{x+y}=\frac{1}{3};\ fraction\ of\ large\ green\ toys\ =\ \frac{1}{3}$$
Statement 2 is SUFFICIENT.

Answer = B

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