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by bubbliiiiiiii » Tue Mar 29, 2011 1:58 am
If the average (arithmetic mean) of four different positive integers is greater than 3 and less than 4, what is the range of the four numbers?

(1) One number is greater than 7

(2) The median of the four numbers is 2.5


A drawer contains 8 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

(1) The probability is less than 0.2 that the first sock is black.

(2) The probability is more than 0.8 that the first sock is white.

OA D



If the average (arithmetic mean) of four different positive integers is greater than 3 and less than 4, what is the range of the four numbers?

(1) One number is greater than 7

(2) The median of the four numbers is 2.5

OA E
[spoiler][/spoiler]
Regards,

Pranay
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Source: — Data Sufficiency |

by srcc25anu » Tue Mar 29, 2011 3:29 am
Q2. from stat 1: we know P(black on first draw) < 0.2 i.e. there is less than 1.6 black socks out of 8. since socks cant be decimal, there is one black sock out of 8. hence P(both are black) = 0 since there is only 1 black sock
hence SUFFICIENT

From stat 2: P(white on first draw) > 0.8 so there must be greater than 6.4 white socks out of 8. again socks cant be decimals so minimum number of white socks = 7 out of 8. hence P(both are black) = 0
SUFFICIENT

Ans D
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by srcc25anu » Tue Mar 29, 2011 3:35 am
Q3. sum of 4 integers must be between 12 and 16 exclusive i.e.sum can be (13, 14, 15)

from stat 1: numbers can be 1 2 3 8 (sum = 14) or 1 2 3 9 (sum = 15)
hence not sufficient

stat 2: median = 2.5
hence numbers can be 1 2 3 8 (median = 2.5) or 1 2 3 9 (median = 2.5)
hence Not sufficient

even together, the two statements are not sufficient to answer the question
hence E
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by Geva@EconomistGMAT » Tue Mar 29, 2011 4:33 am
Q1 the answer is E: if 3<average<4, then the total is 3*4 < total < 4*4, or the total is between 12 and 16.

The twin sets of 1, 2, 3, 8 and 1, 2, 3, 9 satisfy both statements and the question stem, and give a different range for either one.
Geva
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