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!

_________________

Sandy

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