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Intern
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Need help with these questions. [#permalink] New post 17 Apr 2017, 12:28
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1. Mary sold 15 widgets last week. The median sale price of all widgets was $130 and the average sale price of all widgets was $150.
Quantity A: $165
Quantity B: The least possible sale price of the most expensive widget.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


2. Each student in a school takes either the French class or Spanish class, but not both. Each class has some male and some female students. The ratio of the number of female students to the number of male students is less for Spanish class than for French class.
Quantity A: The ratio of the number of female students to the number of male students in French class.
Quantity B: The ratio of the number of female students to the number of male students in the school.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


3. Function f(100)= the product of all the even integers from 2 to 100, inclusive.
Quantity A: 40
Quantity A: The largest prime factor of f(100)
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
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Re: Need help with these questions. [#permalink] New post 19 Apr 2017, 03:21
Expert's post
Hey!


PLEASE share the source of these questions and also the official answers!


1. We have total number of widgets = 15.

Median value = $130 which is 8th value on the list when arranged in ascending order.

Let x be the maximum price of widget or the value in the 15th position.

The prices ........ 130,........ x.

For minimum x we need to put as high as possible values at every other position. Thus only possible set under this circumstance is:

130, 130, 130, 130, 130, 130, 130, 130, x, x, x, x, x, x, x.

Now we also know that the average is 150. So the sum of the above prices = 150 \times 15 = 2250.

Further the sum of prices from the series we constructed = 130 \times 8 + 7 \times x.

Equating both prices and solving for x we get x = 172.85

Hence Quantity B is greater


2. Let N1 be number of boys in French Class and N2 be number of boys in Spanish and let x be girl:boy in French class and y be girl:boy in Spanish class.

Now given that y < x

Quantity A: x

Quantity B: \frac{x*N1 + y*N2}{N1+N2}

Now Quantity B is a weighted average of x and y and hence its value will lie between x and y. Since x is the upper bound Quantity B is always less than x. Unless N2 = 0 in which case both are equal. But since it is mentioned both classes have some male and female students hence Quantity B is greater!

3. Given f(100)= the product of all the even integers from 2 to 100, inclusive

Largest prime factor of f(100) is the largest prime number n which has 2 \times n < 100. We are looking for largest prime number less than 50. In this case it is 47.

2 \times 47 = 84 is a term included in f(100).

Quantity B is 47. Hence B is correct option!
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Re: Need help with these questions. [#permalink] New post 19 Apr 2017, 04:41
sandy wrote:
Hey!


PLEASE share the source of these questions and also the official answers!


1. We have total number of widgets = 15.

Median value = $130 which is 8th value on the list when arranged in ascending order.

Let x be the maximum price of widget or the value in the 15th position.

The prices ........ 130,........ x.

For minimum x we need to put as high as possible values at every other position. Thus only possible set under this circumstance is:

130, 130, 130, 130, 130, 130, 130, 130, x, x, x, x, x, x, x.

Now we also know that the average is 150. So the sum of the above prices = 150 \times 15 = 2250.

Further the sum of prices from the series we constructed = 130 \times 8 + 7 \times x.

Equating both prices and solving for x we get x = 172.85

Hence Quantity B is greater


2. Let N1 be number of boys in French Class and N2 be number of boys in Spanish and let x be girl:boy in French class and y be girl:boy in Spanish class.

Now given that y < x

Quantity A: x

Quantity B: \frac{x*N1 + y*N2}{N1+N2}

Now Quantity B is a weighted average of x and y and hence its value will lie between x and y. Since x is the upper bound Quantity B is always less than x. Unless N2 = 0 in which case both are equal. But since it is mentioned both classes have some male and female students hence Quantity B is greater!

3. Given f(100)= the product of all the even integers from 2 to 100, inclusive

Largest prime factor of f(100) is the largest prime number n which has 2 \times n < 100. We are looking for largest prime number less than 50. In this case it is 47.

2 \times 47 = 84 is a term included in f(100).

Quantity B is 47. Hence B is correct option!




Hi Sandy.

This is the source: brightlinkprep(dot)com/3-hardest-gre-questions-from-april-2017/ [new member so no URLS for me :I]

Personally I trust this site as I have met the teacher a couple of times, great guy.


Answers
1. B
2. A
3. B
CEO
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Joined: 07 Jun 2014
Posts: 2841
GRE 1: 323 Q167 V156
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Re: Need help with these questions. [#permalink] New post 19 Apr 2017, 05:20
Expert's post
Please share the solution for question 2

Regards,
_________________

Sandy
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Re: Need help with these questions.   [#permalink] 19 Apr 2017, 05:20
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