Asifpirlo wrote:

A CEO is building an extra-wide garage in which to park his limousines. The garage is x feet wide, and at least 2 feet of space is required between each two cars and between the cars and the walls. Will all 9 limousines fit in the garage?

(1) The average width of the limousines is the square root of x.

(2) x = 100.

x = 9 (Average width of Limo) + Min [10 (2)] -----(1)

10 because 9 cars will have 8 gaps & 2 gaps for extreme ends (between wall & car)

Statement 1x = 9 \(\sqrt{x}\) +

Min [10 (2)]

Still we cant solve the equation as we don't know the value of x

Thus insufficient

Statement 2100 = 9 (Average width of Limo) +

Min [10 (2)]

Still we cant solve the equation as we don't know the value of "Average width of Limo"

Thus insufficient

Statement 1&2100 = 9 \(\sqrt{100}\) +

Min [10 (2)]

100 = 9 (10) +

Min [10 (2)]

100 =

Min 110

This clearly shows the garage won't be able to hold all the 9 limos

Thus Sufficient

Answer C

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