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03 Jan 2018, 21:13
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C
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Difficulty:
45%
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Question Stats:
68% (01:59) correct
32%
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wrong
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If the average (arithmetic mean) of four positive numbers is 40, how many of the numbers are less than 40?
(1) Two of the numbers are greater than 50.
(2) The smallest number is greater than 20.
(1) Two of the numbers are greater than 50.
(2) The smallest number is greater than 20.
03 Jan 2018, 22:20
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Sum of 4 nbr is 160. Statement 1 says 2 of nbr is greater than 50. So sum of other 2 nbr should be less than 60. Either both can be less than 40 or one can be less than 40. So statement 1 is not sufficient. Statement 2 alone is not sufficient. Combined statements says sum of remaining 2 is less than 60 and one is greater than 20 so other should be less than 40. So both statements are required. Answer choice C
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04 Jan 2018, 02:11
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Bunuel
Lets say the four numbers are a, b, c, d in ascending order. Given that a+b+c+d = 40*4 = 160
(1) Say c and d are greater than 50. This means sum c+d > 100. So a+b < 60.
Now we dont know how less than 60 is the sum a+b. If a+b = 55 (say), then still we could have both numbers less than 40 (a=25, b=30) or we could have only one number less than 40 (a=10, b=45). So cant say. Not Sufficient.
(2) Say the smallest number is a, which is 21. This means b+c+d = 160-21 = 139. Here we could have all three numbers > 40 (46, 46, 47) Or only two > 40 (30, 50, 50). So we cant say. Not sufficient.
Combining the two statements, a > 20, and c+d > 100. Since c+d > 100, a+b < 60. Now if a=20, then also b cannot be 40 (because then sum will touch 60). And since we are given that a > 20, then b definitely has to be < 40 (then only sum will be less than 60). So, these two numbers, a & b, must be less than 40. Sufficient.
Hence C answer
