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

Question

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.

DS61602.01
OG2020 NEW QUESTION

ScottTargetTestPrep · Accepted Answer

We are given that the number of green large = red large and green small = red small.

We need to determine the value of (large green)/(total green).

Statement One Alone:

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

Only knowing that we have 400 green small toys is not enough information to determine large green/green.

Statement one alone is not sufficient to answer the question.

Statement Two Alone:

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

Since 2/3 of the total are small, 1/3 are large. Thus, (2/3)/2 = 1/3 are small green and (1/3)/2 = 1/6 are large green. Thus, the large green toys are (1/6)/(1/3 + 1/6) = 1/(2 + 1) = 1/3 of the total number of green toys.

Answer: B

firas92 · Answer

Use a 2x2 table to solve

Let small red toys be  x  which makes small green toys also  x  (they are equal)
Similarly, large red toys and large green toys are each  y

So the total number of green toys is  x+y

We are asked to find  \frac{y}{x+y}

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

x=400

No info about  y  or how  x  is related to  y . Not sufficient

(2) In the production lot,  \frac{2}{3}  of the toys produced are small.

So  \frac{1}{3}  of the total toys produced are large

This means  \frac{2y}{2(x+y)}=\frac{1}{3}

Therefore,  \frac{y}{x+y}  is  \frac{1}{3}

Sufficient

Answer is  (B)

DavidTutorexamPAL · Answer

The Logical approach to this question will focus on the fact that we are asked to find the ratio, so we don't necessarily need exact numbers.
Statement (1) doesn't give us any information regarding the ratio between the groups of small vs. large toys, and is thus insufficient.
Statement (2) tells us that 2/3 are small, meaning 1/3 are large, and since the number of red and green toys is equal for each size group, then 1/3 of the green toys are large and the other 2/3 are small (as they make half of these groups). 
The correct answer is (B).

Posted from my mobile device

ZoltanBP · Answer

For the red category, let  s  and  l  be the numbers of small and large toys produced, respectively. For the green category, the production volumes are the same as those for the red category according to the original information. Globally,  2s  and  2l  are the numbers of small and large toys produced, respectively.

The original question: For the green category,  \frac{l}{s+l}=?

1) We know that  s=400 , but no information is given about  l . Thus, we can't get a unique value to answer the original question.  \implies   Insufficient

2) We know that  \frac{2s}{2s+2l}=\frac{s}{s+l}=\frac{2}{3}

1-\frac{s}{s+l}=1-\frac{2}{3}

\frac{l}{s+l}=\frac{1}{3}

Thus, the answer to the original question is a unique value.  \implies   Sufficient

Answer: B

rnn · Answer

Hi  ScottTargetTestPrep

I did not understand this sentence "Thus, (2/3)/2 = 1/3 are small green and (1/3)/2 = 1/6 are large green. Thus, the large green toys are (1/6)/(1/3 + 1/6) = 1/(2 + 1) = 1/3 of the total number of green toys"

Could you kindly elaborate here more please?

EMPOWERgmatRichC · Answer

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

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

That is because in the problem stem, it says “If, for each size, there are equal numbers of red and green toys in a certain production lot.” That means, for the small size, half are green and the other half is red. Since 2/3 of the toys are small, (2/3)/2 = 1/3 of all toys are small green toys. Similarly, since 1/3 of the toys are lare, (1/3)/2 = 1/6 of all toys are large green toys. The question asks what fraction of green toys are large. So we have 1/6 of the toys are large green and (1/3 + 1/6) of the toys are green, so the fraction is (1/6)/(1/3 + 1/6) = 1/(2 + 1) = 1/3.

Basshead · Answer

The wording of this question makes it challenging. We're told for each size, there are equal numbers of red and green toys.

For example, for small toys, red = green = 10. Large toys: red = green = 20.

What fraction of the total numbers of green toys is large? Notice the question is asking for a ratio -- not a specific amount.

(1) If 400 of the small toys are green, then 400 of the small toys are red. However, we have no information about the large toys; INSUFFICIENT.

(2) If 2/3 of the toys produced are small, then 1/3 of the toys produced are big.

Of the 1/3 toys, HALF will be green and HALF will be red. Therefore, we can conclude 1/6 of the green toys is large. SUFFICIENT.

Answer is B.

GMATGuruNY · Answer

For each size, the number of red = the number of green.
One more approach for Statement 2:

Let the total number of toys = 6x.
Small toys  = \frac{2}{3}*6x = 4x  --> 2x small red, 2x small green
Large toys = 6x-4x = 2x --> x large red, x large green

Resulting fraction:
 \frac{large-green}{total-green} = \frac{x}{2x+x} = \frac{x}{3x} = \frac{1}{3} 
SUFFICIENT.

GMATinsight · Answer

Answer: Option B

Video solution by  GMATinsight

https://www.youtube.com/watch?v=-xQELz6HV9I

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=be4gWF5IRG4

Engineer1 · Answer

How is everyone getting 1/3?

Moreover, it took me 2 mins 35 ish seconds. I did not want to take chance with no calculation. Any tips to reduce time here? MartyMurray Thank you.

MartyMurray · Answer

We can answer this one in under a minute using mental math.

The key to quickly answering the question is seeing the implications of the following in the passage:

for each size, there are equal numbers of red and green toys in a certain production lot

That information indicates two things:

- In the entire production lot, there are equal numbers of red and green toys. After all, if there are equal numbers of red and green toys for each size, then the numbers of red and green toys in the two sizes must add up to the same total for red and green. So, red toys and green toys each make up half the toys.

- The fraction of toys in a particular size is also the fraction of toys in a color that are in that size. After all, if there are equal numbers of red and green toys for each size, then the number of red or green toys for a size will simply be half the number of toys for that size, and red toys and green toys each make up half of the toys. So, the fraction of toys that are of a size is (number in size)/(total), and the fraction of toys in a color that are in a size is (1/2 number in size)/(1/2 total). We see that (number in size)/(total) = (1/2 number in size)/(1/2 total). So, fraction of total in a size = fraction of a color in a size.

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

This information about a single absolute number does not indicate anything about the fraction of toys that are small or the fraction of toys that are large or the fraction of green toys that are small or the fraction of green toys that are large.

Insufficient.

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

This information indicates that 1/3 of the toys produced are large.

That information, in turn, indicates that 1/3 of the green toys are large, given what we have discussed above.

Sufficient.

The correct answer is (B).

sahilsehdev97 · Answer

Question - 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, of the toys produced are small.

ANSWER: We need to find  \frac{Total number of Large}{Total number of green}

Use a Table to ease things: 
1) 400 are small green, so input the same:
    
  SMALL  
  LARGE  
  TOTAL  
 
   RED

GREEN  
  400

TOTAL

not enough information to answer the rest. INSUFFICIENT

2) 2/3 of toys produced are small.
From this statement, we can express the total number of Large toys and the overall total number of toys through algebraic expressions. Let total number of toys be N so:

SMALL  
  LARGE  
  TOTAL  
 
   RED

GREEN

TOTAL  
 2N/3 
  N/3  
  N

Since the question says that for each size, there are equal numbers of red and green toys, we can determine the expressions for the other spaces in the table, so it will be as follows:
    
  SMALL  
  LARGE  
  TOTAL  
 
   RED  
  N/3  
  0.5N/3  
  1.5N/3  
 
   GREEN  
  N/3  
  0.5N/3  
  1.5N/3  
 
   TOTAL  
 2N/3 
  N/3  
  N

So to answer the original question, regarding the fraction:
 \frac{Total number of large}{total number of green}  = N/3 divided by 1.5N/3

Using basic arithmetic, the N will be cancelled out, leaving a fraction in its place. Thus it is SUFFICIENT

ANSWER: B 
 Note: I realize that it seems like a lot of steps to do in the test environment. The only reason I put a long explanation here is to help people understand the logic of how to approach this question. Some replies I have seen just skip directly to the answer with a very rushed explanation on how to do it. I wanted to highlight all the steps here. Thanks.

steffe440 · Answer

1) 400 are small toys are green, that means we have 400 small toys that are read, so we know a total of 800 small toys, but large toys could be anything, there is no connection.

2) However here we know that 1/3 is large toys, and we know that 1/6 of all toys will be large green toys. We also know that 1/3 so of small toys will be green, hence the total number of greens will be 1/3 + 1/6 which gives us 1/2 of all toys will be green where as 1/6 of all toys are large and green, hence (1/6)/(1/2) would yield as the right answer to the question, 1/3.

Adit_ · Answer

Statement 1:
We only know about small toys but not the large toys.
INSUFFICIENT.

Statement 2:

When the required contingency table is drawn we see that:
For R: a+b
For G: a+b 
With a and b for small and large respectively.

Here:
2a = 2/3x.
2b = x.

We need: 
a/(a+b)

Since we have a and b in terms of x, they get cancelled.
SUFFICIENT.

Answer:  Option B

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

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	SMALL	LARGE	TOTAL
RED	N/3	0.5N/3	1.5N/3
GREEN	N/3	0.5N/3	1.5N/3
TOTAL	2N/3	N/3	N

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

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