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Intern Joined: 17 Apr 2017
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Need help with these questions. [#permalink] 00:00

Question Stats: 50% (00:11) correct 50% (00:00) wrong based on 2 sessions
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. GRE Prep Club Legend  Joined: 07 Jun 2014 Posts: 4857 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) Followers: 112 Kudos [?]: 1865 , given: 397 Re: Need help with these questions. [#permalink] 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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Sandy
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Intern Joined: 17 Apr 2017
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Re: Need help with these questions. [#permalink]
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
GRE Prep Club Legend  Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: Need help with these questions. [#permalink]
Expert's post
Please share the solution for question 2

Regards,
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Sandy
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Intern Joined: 18 May 2016
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Re: Need help with these questions. [#permalink]
I also get B as an answer for question 2.
Otherwise, D would also make sense if it is assumed that in either one class there could be NO boys at all.

Hope anyone can post a solution for 2.

Thanks!
Intern Joined: 04 Mar 2018
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Re: Need help with these questions. [#permalink]
For the 2nd question, I think, since the average decreases as the sample set increases with smaller numbers. Similarly, the no. of females in French were more and in Spanish were less. And school is a meld of these two, therefore, the ratio of female to male will decrease as the Spanish class had more males then the females, the ratio will surely be smaller than the French class alone.
Intern Joined: 23 Jan 2017
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Re: Need help with these questions. [#permalink]
sandy wrote:

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!

Your solution actually revealed the answer of the question 2 is A (Quantity A > QuantityB).
Because Quantity B is always less than x => B < A => The answer is A

My Solution is:

Female in French and Spanish: x, y
Male in French and Spanish: a, b
Hence, x/a > y/b => y <x*b/a (because b>0, we can multiple both sides for b)

Quantity A: x/a
Quantity B: (x+y)/(a+b)
Because y<bx/a as mentioned => Quantity B< (x + bx/a)/(a+b) = N
Multiple both numerator and denominator of N for a => Quantity B < [x(a+b)]/[a*(a+b)] = x/a = Quantity A

So the answer is A: Quantity A is greater
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Re: Need help with these questions. [#permalink]
Expert's post
Please follow the rules for posting on the board.

Every question must go under the right sub-forum and one question at a time, not 2 or more questions in a single spot.

Thank you.

Regards
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Intern Joined: 04 Apr 2018
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Re: Need help with these questions. [#permalink]
If you face problems like this you can alsi check on dum dum.co and connect with some prep buddy nearby this will help you get instant solutions to your prob Re: Need help with these questions.   [#permalink] 26 Apr 2018, 12:53
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