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10 Jan 2025, 01:30
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C
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Difficulty:
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Question Stats:
81% (01:37) correct
19%
(02:22)
wrong
based on 159
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If x + y = 90 – z, what is the average (arithmetic mean) of (6x + 90), (6y – 20) and (6z – 100)?
A. 30
B. 90
C. 170
D. 180
E. It cannot be determined from the information given.
A. 30
B. 90
C. 170
D. 180
E. It cannot be determined from the information given.
10 Jan 2025, 07:59
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Given, \(x + y + z = 90\)
\(AM = \dfrac{(6x+90) + (6y-20) + (6z-100}{3} = \dfrac{6(x+y+z) + (90-20-100)}{3} = \dfrac{6(x+y+z) - 30}{3} = 2(x+y+z) - 10 = 2\times 90 - 10 = 170\)
ANS C
\(AM = \dfrac{(6x+90) + (6y-20) + (6z-100}{3} = \dfrac{6(x+y+z) + (90-20-100)}{3} = \dfrac{6(x+y+z) - 30}{3} = 2(x+y+z) - 10 = 2\times 90 - 10 = 170\)
ANS C
General Discussion
10 Jan 2025, 07:50
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x + y + z = 90
To find the average > (6(x+y+z - 5))/3 = 2(85) = 170
To find the average > (6(x+y+z - 5))/3 = 2(85) = 170
