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 = 172.85Hence 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!

_________________

Sandy

If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test