LUANDATO wrote:Group A, which is a set of positive integers, has an average (arithmetic mean) of 8 while group B, a different set of positive integers, has an average of 10. In which group is the sum of the numbers greater?
(1) The average of all numbers in both groups combined is 8 2/3
(2) The sum of all numbers in both groups is 52
The OA is D.
Can any expert help me with this DS question please? I have some problems to understand it. Thanks!
Sum of all the numbers in a group = Average * number of integers
If we get to know the value of the number of integers in each set, we get the answer.
Question rephrased: Which group has a greater number of integers?
(1) The average of all numbers in both groups combined is 8 2/3.
We are given that the average of both the groups combined is 8 2/3 = 8.66.
Since the average of both the groups combined = 8.66 is closer to 8 than to 10, the number of integers in the first group must be greater than that of the second group. Sufficient.
(2) The sum of all numbers in both groups is 52.
Sum of all numbers in both groups = Sum of all numbers in the first group + Sum of all numbers in the second group
Say, the number of integers in the first group = x and the number of integers in the second group = y
52 = 8x + 10y
26 = 4x + 5y
y = (26 - 4x)/5
y = (25 + 1 - 4x)/5
y = 5 + (1 - 4x)/5
Since y is a positive integer, (1 - 4x) must be a multiple of 5.
Let's do some hit and trial. The first positive integer value of x that satisfies the condition is 4.
At x = 4, we have y = 5 + (1 - 4x)/5 = 5 + (1 - 4*4)/5 = 5 - 15/5 = 5 - 3 = 2
So, x = 4 and y = 2. The first group has a greater number of integers.
There is no need to find another possible positive integer value of x that satisfies the condition since y = 2 is the minimum possible value for the second group. At higher values of x, the value of y would go down further.
Thus, The first group has a greater number of integers. Sufficient.
The correct answer:
D
Hope this helps!
-Jay
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