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# In a certain class, 1/5 of the boys are shorter than the shorter

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In a certain class, 1/5 of the boys are shorter than the shorter  [#permalink]

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08 May 2018, 00:42
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38% (02:59) correct 62% (03:17) wrong based on 103 sessions

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In a certain class, $$\frac{1}{5}$$ of the boys are shorter than the shortest girl in the class, and $$\frac{1}{3}$$ of the girls are taller than the tallest boy in the class. If there are 16 stu­dents in the class and no two people have the same height, what percent of the students are taller than the shortest girl and shorter than the tallest boy?

A. 25%

B. 50%

C. 62.5%

D. 66.7%

E. 75%

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Re: In a certain class, 1/5 of the boys are shorter than the shorter  [#permalink]

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08 May 2018, 01:45
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Total number of students is 16. Boys + Girls = 16.

1/5 of the boys are shorter than the shortest girl in the class.
This implies number of boys should be multiple of 5. It can be 5,10,15.

1/3 of the girls are taller than the tallest boy in the class.
This implies number of girls should be multiple of 3. It can be 3,6,9,12,15.

Let's find the number of boys and girls:
If number of boys is 5, then we can have 11 girls but 11 is not divisible by 3.
If number of boys is 15, then we can have 1 girls but 1 is not divisible by 3.
So we get No of boys is 10 and No. of Girls is 6.

Let's solve to find the answer:-

1/5 of the boys are shorter than the shortest girl in the class - No of boys = 1*10/5 = 2

And 1/3 of the girls are taller than the tallest boy in the class - No of Girls = 1*6/3 = 2.

what percent of the students are taller than the shortest girl and shorter than the tallest boy?

This can be found as given below.
16 - (2 + 2 ) - (1 + 1 ) = 10
(1 + 1 ) is for the shortest girl and the tallest boy.

Percentage = 10*100/16 = 62.5.

Ans - C
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Re: In a certain class, 1/5 of the boys are shorter than the shorter  [#permalink]

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08 May 2018, 01:39
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Bunuel wrote:
In a certain class, $$\frac{1}{5}$$ of the boys are shorter than the shortest girl in the class, and $$\frac{1}{3}$$ of the girls are taller than the tallest boy in the class. If there are 16 stu­dents in the class and no two people have the same height, what percent of the students are taller than the shortest girl and shorter than the tallest boy?

A. 25%

B. 50%

C. 62.5%

D. 66.7%

E. 75%

We'll draw out our data so it is easier to understand.
This is an Alternative approach.

1/5 of boys < shortest girl
tallest boy > 1/3 of girls

We need to find
shortest girl < ?? % students < tallest boy

all the girls except 1 are taller than the shortest girl and 1 -1/3 = 2/3 of them are shorter than the tallest boy
all the boys except 1 are shorter than the tallest boy and 1 -1/5 = 4/5 of them are taller than the shortest girl

so we need 2/3 of the girls and 4/5 of the boy - 2.
Let's find the number of girls and boys:
The number of boys is divisibly by 5 so it can be 5, 10 or 15.
If it is 5 there are 11 girls which is not divisible by 3. If it is 15 there is 1 girl which is not divisible by 3.
So there are 10 boys and 6 girls.

So, 2/3 of girls is 4 students and 4/5 of boys is 8 students.
Our percentage is (4+8-2)/16 = 10/16 = 5/8 = 0.625

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Re: In a certain class, 1/5 of the boys are shorter than the shorter  [#permalink]

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08 Jun 2019, 10:44
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This is a very good question which can be solved by simple calculations and pure logic. Beware of the trick element though, which can get you into trouble. Answer option E seems to be a very common trap answer, therefore.

The class has a total of 16 students, all of whom have distinct heights. The question mentions 1/5th of the boys and 1/3rd of the girls. This is a clear indication that the number of boys and girls should be multiples of 5 and 3 respectively. The only possible combination turns out to be 10 boys and 6 girls.

Therefore, as per the question, 2 boys are shorter than the shortest girl and 2 girls are taller than the tallest boy.

Let the shortest girl be Q and the tallest boy be M. Then, the entire question can be represented diagrammatically, like this:

Attachment:

08th June 2019 - Reply 1.JPG [ 21.85 KiB | Viewed 2712 times ]

The 2 boys who are shorter than the shortest girl will be shorter than the other girls too. Similarly, the 2 girls who are taller than the tallest boy will be taller than the other boys too.

How many boys are we left with who are taller than the shortest girl? 8, correct? But one of these 8 boys is M himself. So, there are 7 boys taller than the shortest girl, but shorter than the tallest boy.

How many girls are shorter than the tallest boy? 4, right? But, one of these 6 girls is Q herself. So, there are 3 girls who are shorter than the tallest boy but taller than the shortest girl.

In all, we have 10 students in between Q and M. 10 as a percentage of 16 is nothing but

$$\frac{10}{16}$$ * 100 = 62.5%

So, the correct answer option is C.

If you forget to exclude M and Q from your calculation, you will end up marking 75% as the answer, which as we mentioned, could be a very common trap answer in this question.

Hope this helps!
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Re: In a certain class, 1/5 of the boys are shorter than the shorter  [#permalink]

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04 Dec 2019, 12:35
GMATPrepNow better if u explain it, and advance thanks
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Re: In a certain class, 1/5 of the boys are shorter than the shorter  [#permalink]

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04 Dec 2019, 12:58
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Top Contributor
Bunuel wrote:
In a certain class, $$\frac{1}{5}$$ of the boys are shorter than the shortest girl in the class, and $$\frac{1}{3}$$ of the girls are taller than the tallest boy in the class. If there are 16 stu­dents in the class and no two people have the same height, what percent of the students are taller than the shortest girl and shorter than the tallest boy?

A. 25%

B. 50%

C. 62.5%

D. 66.7%

E. 75%

GIVEN: There are 16 students and 1/5 of them are boys.
This means the number of boys must be a multiple of 5.
There are 3 possible cases:
i) 5 boys and 11 girls
ii) 10 boys and 6 girls
iii) 15 boys and 1 girl

GIVEN: There are 16 students and 1/3 of them are girls.
This means the number of girls must be a multiple of 3.
When we check the three possible cases above, we see that only one case (case ii) is such that the number of girls is divisible by 3.

So we now know that there are 10 boys and 6 girls

Let A, B, C, D, E, F represent the heights of the 6 girls arranged in ASCENDING order
Let Q, R, S, T, U, V, W, X, Y, Z represent the heights of the 10 boys arranged in ASCENDING order

1/5 of the boys are shorter than the shortest girl in the class
1/5 of 10 = 2
So, 2 boys are shorter than the shortest girl in the class
We have: Q, R, A [ these 3 heights must be arranged in ascending order]

1/3 of the girls are taller than the tallest boy in the class
1/3 of 2 = 2
So, 2 girls are taller than the tallest boy in the class
We have: Z, E, F [ these 3 heights must be arranged in ascending order]

NOTE: The remaining students must lie BETWEEN A and Z, however there is no way to determine the relationships between each boy and each girl within this range.

So one possible configuration is as follows: Q, R, A, S, T, U, V, W, X, Y, B, C, D, Z, E, F

What percent of the students are taller than the shortest girl and shorter than the tallest boy?
Shortest girl is A and the tallest boy is Z
As we can see from the above diagram, there are 10 such students

10/16 = 5/8 = 62.5%

Cheers,
Brent
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Re: In a certain class, 1/5 of the boys are shorter than the shorter  [#permalink]

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04 Dec 2019, 13:36
GMATPrepNow wrote:
Bunuel wrote:
In a certain class, $$\frac{1}{5}$$ of the boys are shorter than the shortest girl in the class, and $$\frac{1}{3}$$ of the girls are taller than the tallest boy in the class. If there are 16 stu­dents in the class and no two people have the same height, what percent of the students are taller than the shortest girl and shorter than the tallest boy?

A. 25%

B. 50%

C. 62.5%

D. 66.7%

E. 75%

GIVEN: There are 16 students and 1/5 of them are boys.
This means the number of boys must be a multiple of 5.
There are 3 possible cases:
i) 5 boys and 11 girls
ii) 10 boys and 6 girls
iii) 15 boys and 1 girl

GIVEN: There are 16 students and 1/3 of them are girls.
This means the number of girls must be a multiple of 3.
When we check the three possible cases above, we see that only one case (case ii) is such that the number of girls is divisible by 3.

So we now know that there are 10 boys and 6 girls

Let A, B, C, D, E, F represent the heights of the 6 girls arranged in ASCENDING order
Let Q, R, S, T, U, V, W, X, Y, Z represent the heights of the 10 boys arranged in ASCENDING order

1/5 of the boys are shorter than the shortest girl in the class
1/5 of 10 = 2
So, 2 boys are shorter than the shortest girl in the class
We have: Q, R, A [ these 3 heights must be arranged in ascending order]

1/3 of the girls are taller than the tallest boy in the class
1/3 of 2 = 2
So, 2 girls are taller than the tallest boy in the class
We have: Z, E, F [ these 3 heights must be arranged in ascending order]

NOTE: The remaining students must lie BETWEEN A and Z, however there is no way to determine the relationships between each boy and each girl within this range.

So one possible configuration is as follows: Q, R, A, S, T, U, V, W, X, Y, B, C, D, Z, E, F

What percent of the students are taller than the shortest girl and shorter than the tallest boy?
Shortest girl is A and the tallest boy is Z
As we can see from the above diagram, there are 10 such students

10/16 = 5/8 = 62.5%

Cheers,
Brent

thank u so much.
Re: In a certain class, 1/5 of the boys are shorter than the shorter   [#permalink] 04 Dec 2019, 13:36
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