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09 Oct 2020, 03:00
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C
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The average (arithmetic mean) of three numbers is 100. If two of the numbers are 80 and 130, what is the third number?
A. 70
B. 80
C. 90
D. 100
E. 110
A. 70
B. 80
C. 90
D. 100
E. 110
09 Oct 2020, 03:11
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Bunuel
a+b+c = 100*3
80+130+c = 300
c = 90
ANswer: Option C
09 Oct 2020, 03:45
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Logic I:
Average: \(\frac{Sum }{ Total}\)
Sum: Average * Total
Sum: 100 * 3 = 300
=> 300 - (80 + 130) = 90
Answer C
Logic 2:
As the average is 100, let us suppose all three numbers were '100' each.
=> First number: 80 (-20 from 100)
=> Second number: 130 (+30 from 100)
Third number: (100 + 20 - 30 = 90) [As the first number is 20 less, then the third number should be '20' more to keep the average as 100. So, it becomes 120. Now the second number is already '30' more, so the third number will be 120 - 30 = 90]
Answer C
Average: \(\frac{Sum }{ Total}\)
Sum: Average * Total
Sum: 100 * 3 = 300
=> 300 - (80 + 130) = 90
Answer C
Logic 2:
As the average is 100, let us suppose all three numbers were '100' each.
=> First number: 80 (-20 from 100)
=> Second number: 130 (+30 from 100)
Third number: (100 + 20 - 30 = 90) [As the first number is 20 less, then the third number should be '20' more to keep the average as 100. So, it becomes 120. Now the second number is already '30' more, so the third number will be 120 - 30 = 90]
Answer C
