What you need is a way to organize the information so your brain keeps cool. What we have here is a sets table question, solvable by a 3*3 grid: put like/dislike lima in the columns, like/dislike Brussels sprouts in the rows, and don't forget a total column and row. Then organize the information in the question stem and statements, clearly mark which box you need to find, and see if the information you have allows you to reach that box.
The attached file shows the tables for stat. (1) and (2).
Stat. (1) is sufficient because you can reach the like BS / dislike Lima box:
120 total / total students.
2/3 of those dislike Lima = 80 (goes in Dislike Lima / Total)
3/5 of the 80 also dislike BS = 48 (goes in dislike Lima / dislike BS)
The remaining 32 are the required dislike Lima / like BS, so stat. (1) is sufficient.
Stat. (2) is also sufficient: if 40 of the students like Lima, and we know that 2/3 of the students dislike Lima, then the 40 lima likers constitute the remaining 1/3 of the students. if 40 is 1/3 of the total/total (marked as X), then the total students is 3*40 = 120 - and the rest is the same as stat. (1). The two statements basically say the same thing.
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Quant #5
Source: Beat The GMAT — Quantitative Reasoning |
Here is how I approached the problem.(solved in under 2 minutes)
I assumed let no of students be =30 [divisible by both 3 and 5]
Like Dislike
Sprouts - -
Lima - 20 (2/3*10)
Like Dislike
Sprouts - 12 (3/5*20)
Lima - 20
Like Dislike
Sprouts - 12
Lima 10 20 [ total no of students =30 and there is no intersection of Like and dislike Lima beans]
Like Dislike
Sprouts 18 12 [same reason -total students =30]
Lima 10 20
Now question was Like sprout + dislike lima = 18 [like spout]-10[like Lima]=8 Dislike Lima
SO we can infer,if we know the number of students,we can find the answer
1) sufficient
2) Lets x= no of students
2/3x=dislike lima beans
then 1/3 like lima beans
1/3x=40 or x=120 [same as 1].sufficient
My answer D
I assumed let no of students be =30 [divisible by both 3 and 5]
Like Dislike
Sprouts - -
Lima - 20 (2/3*10)
Like Dislike
Sprouts - 12 (3/5*20)
Lima - 20
Like Dislike
Sprouts - 12
Lima 10 20 [ total no of students =30 and there is no intersection of Like and dislike Lima beans]
Like Dislike
Sprouts 18 12 [same reason -total students =30]
Lima 10 20
Now question was Like sprout + dislike lima = 18 [like spout]-10[like Lima]=8 Dislike Lima
SO we can infer,if we know the number of students,we can find the answer
1) sufficient
2) Lets x= no of students
2/3x=dislike lima beans
then 1/3 like lima beans
1/3x=40 or x=120 [same as 1].sufficient
My answer D
Geva@MasterGMAT wrote:What you need is a way to organize the information so your brain keeps cool. What we have here is a sets table question, solvable by a 3*3 grid: put like/dislike lima in the columns, like/dislike Brussels sprouts in the rows, and don't forget a total column and row. Then organize the information in the question stem and statements, clearly mark which box you need to find, and see if the information you have allows you to reach that box.
The attached file shows the tables for stat. (1) and (2).
Stat. (1) is sufficient because you can reach the like BS / dislike Lima box:
120 total / total students.
2/3 of those dislike Lima = 80 (goes in Dislike Lima / Total)
3/5 of the 80 also dislike BS = 48 (goes in dislike Lima / dislike BS)
The remaining 32 are the required dislike Lima / like BS, so stat. (1) is sufficient.
Stat. (2) is also sufficient: if 40 of the students like Lima, and we know that 2/3 of the students dislike Lima, then the 40 lima likers constitute the remaining 1/3 of the students. if 40 is 1/3 of the total/total (marked as X), then the total students is 3*40 = 120 - and the rest is the same as stat. (1). The two statements basically say the same thing.
Geva - Nice! Was totally able to get the concept. Can we say that all like/dislikes types of question can be solved using the set table way?
Rishab1988 - Your way even though it works, seems a bit complicated to me compared to Geva's solution!
Thanks to both of you for chipping in!

















