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17 Aug 2017, 00:52
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If the average (arithmetic mean) of 7 numbers is 21, what is the standard deviation of the numbers?
1) The smallest number of them is 21
2) The greatest number of them is 21
=>Condition 1)
Since the minimum number and the average are same, all numbers should be 21.
Condition 2)
Since the maximum number and the average are same, all numbers should be 21.
Ans: D
1) The smallest number of them is 21
2) The greatest number of them is 21
=>Condition 1)
Since the minimum number and the average are same, all numbers should be 21.
Condition 2)
Since the maximum number and the average are same, all numbers should be 21.
Ans: D
18 Aug 2017, 01:50
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Is the total profit from the sales of 3 products greater than $6?
1) The least profit from the sales of the 3 products is at least $2.1
2) The 2nd largest profit from the sales of the 3 products is at least $3.1.
=> Condition 1)
Since the minimum profit of 3 products is $2.1, the total profit of them is greater than or equal to $2.1 x 3 = $6.3, which is greater than $6.
Thus this is sufficient.
Condition 2)
Since the 3nd largest profit is at least $3.1, the sum of the first and the second products is at least $6.2, which is greater than $6.
Thus this is sufficient too.
Ans: D
1) The least profit from the sales of the 3 products is at least $2.1
2) The 2nd largest profit from the sales of the 3 products is at least $3.1.
=> Condition 1)
Since the minimum profit of 3 products is $2.1, the total profit of them is greater than or equal to $2.1 x 3 = $6.3, which is greater than $6.
Thus this is sufficient.
Condition 2)
Since the 3nd largest profit is at least $3.1, the sum of the first and the second products is at least $6.2, which is greater than $6.
Thus this is sufficient too.
Ans: D
20 Aug 2017, 19:08
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If xy≠0, is |x|>|y|？
1) x=-2y
2) x=y^4
=>Condition 1)
|x| = |-2y| = 2|y| > |y|.
Thus this is sufficient.
Condition 2)
If y = 2, then x = 16 and so |x| > |y|.
If y = ½, then x = 1/16 and so |x| < |y|.
This is not sufficient.
Ans: A
1) x=-2y
2) x=y^4
=>Condition 1)
|x| = |-2y| = 2|y| > |y|.
Thus this is sufficient.
Condition 2)
If y = 2, then x = 16 and so |x| > |y|.
If y = ½, then x = 1/16 and so |x| < |y|.
This is not sufficient.
Ans: A
