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A company produces a certain toy in only 2 sizes, small or

Expert replies
by AbeNeedsAnswers » Wed May 01, 2019 12:50 pm

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B

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E

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Difficulty

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.

B

Source: Official Guide 2020
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Source: — Data Sufficiency |

by Jay@ManhattanReview » Thu May 02, 2019 1:59 am
AbeNeedsAnswers wrote: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.

B

Source: Official Guide 2020
Say the company produces x numbers of small and y numbers of large toys. Thus,

Number of red toys = Number of green toys =x/2 + y/2 = (x + y)/2

We have to get the value of y / [(x + y)/2] = 2y / (x + y) = 2 / (x/y + 1)

If we get the value of x/y, we get the answer.

Let's take each statement one by one.

(1) In the production lot, 400 of the small toys are green.

=> x/2 = 400 => x = 800. Can't get the value fo x/y. Insufficient.

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

=> 2/3(x + y) = x
2x/3 + 2y/3 = x
2y/3 = x/3

x/y = 2. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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by [email protected] » Wed May 15, 2019 6:36 pm
Hi All,

We're told that a company produces a certain toy in only 2 sizes, small or large, and in only 2 colors, red or green - and that for each size, there are EQUAL numbers of red and green toys in a certain production lot. We're asked for the fraction of the total number of GREEN toys that are LARGE. This question can be approached with a mix of logic and TESTing VALUES.

(1) In the production lot, 400 of the small toys are green.

With the information in Fact 1, we know that there are 400 small RED toys (since there are EQUAL numbers of red and green toys in each size), but we don't know how many LARGE GREEN toys there are, so the answer to the question would change depending on number.
Fact 1 is INSUFFICIENT

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

With Fact 2, we know that 2/3 of the toys are SMALL, so the remaining 1/3 of the toys are LARGE. The prompt tells us that there are EQUAL numbers of red and green toys in each size). These ratios are enough to answer the question; you can prove it with Algebra or by TESTing VALUES.

IF....
TOTAL toys = 6
Total Small = 4 (2 red and 2 green)
Total Large = 2 (1 red and 1 green)
Total fraction of GREEN toys that are LARGE = 1/3

IF....
TOTAL toys = 12
Total Small = 8 (4 red and 4 green)
Total Large = 4 (2 red and 2 green)
Total fraction of GREEN toys that are LARGE = 2/6 = 1/3
Etc.
The answer will ALWAYS be 1/3.
Fact 2 is SUFFICIENT

Final Answer: B

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Rich
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matrix representation

by SampathKp » Wed Dec 18, 2019 9:28 am
AbeNeedsAnswers wrote: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.

B

Source: Official Guide 2020
Given information can be represented in the following matrix

S . L . T
R . X . Y . X+Y
G . X . Y . X+Y
T . 2X . 2Y . 2(X+Y)

Question asked is 2Y/X+Y = Y/X+Y. we need to see from which options we get answer Y/X+Y

From (1) , X=400 , we have no information on Y, so this is NOT Sufficient to get value of Y/X+Y
From (2) , 2x = 2/3 , X= 1/3 also if 2/3 of Total no: of toys is small, then 1/3 of the toys will be large. so 2Y = 1/3 , Y= 1/6
so we have value of X and Y , hence we can compute Y/X+Y. Hence (2) is sufficient to answer the question.

Answer is B
PS Y/X+Y = 1/6/1/3+1/6 . = 1/3. so 1/3 of total numbers are toys are green colour.
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