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25 Oct 2018, 02:26
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:53) correct
32%
(01:40)
wrong
based on 101
sessions
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If the average (arithmetic mean) of four unequal numbers is 40, how many of the numbers are greater than 40?
(1) No number is greater than 70.
(2) Two of the four numbers are 19 and 20.
(1) No number is greater than 70.
(2) Two of the four numbers are 19 and 20.
25 Oct 2018, 03:30
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The Ans is (C). Considering only 1st statement: we might get 1, 2 or 3 as an answer. Hence not sufficient
Considering only 2nd statement, we might get 1 or 2 as an answer. Not sufficient.
Combining 1st and 2nd, if the two nos. are 19 and 20, the remaining 121 has to be contributed by another 2 nos. Both of these have to be more than 40 as one number alone being restricted by 70 would not be sufficient and hence, 2 nos. will be more than 40.
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Considering only 2nd statement, we might get 1 or 2 as an answer. Not sufficient.
Combining 1st and 2nd, if the two nos. are 19 and 20, the remaining 121 has to be contributed by another 2 nos. Both of these have to be more than 40 as one number alone being restricted by 70 would not be sufficient and hence, 2 nos. will be more than 40.
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25 Oct 2018, 04:40
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The answer is C.
1) the answer could be 0 (they are all 40!), 1 (39, 40, 40, 41), 2 (39,39,41,41) or 3 (37, 41,41,41). Insufficient!
2) If the sum of the 1st 2 number is 39 (19+20), the sum of the next 2 must be 121 (160-39). This can be accomplished by 2 answers larger than 40 (60,61), or one (30, 81). Insufficient!
Combined: 2 numbers cannot be sum to 121, with a number smaller than 41 and a number smaller than 71 (40+70=110, the max) - BOTH number have to be larger than 40 (60+61, 70+51, etc). Sufficient! Answer C.
1) the answer could be 0 (they are all 40!), 1 (39, 40, 40, 41), 2 (39,39,41,41) or 3 (37, 41,41,41). Insufficient!
2) If the sum of the 1st 2 number is 39 (19+20), the sum of the next 2 must be 121 (160-39). This can be accomplished by 2 answers larger than 40 (60,61), or one (30, 81). Insufficient!
Combined: 2 numbers cannot be sum to 121, with a number smaller than 41 and a number smaller than 71 (40+70=110, the max) - BOTH number have to be larger than 40 (60+61, 70+51, etc). Sufficient! Answer C.
