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The average (arithmetic mean) of 6, 21, x, and y is 13, where x and y

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Bunuel
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The average (arithmetic mean) of 6, 21, x, and y is 13, where x and y [#permalink] New post  08 Sep 2021, 04:53
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The average (arithmetic mean) of 6, 21, x, and y is 13, where x and y are integers with a product of 100. Which of the following could be x?

A. 50
B. 20
C. 10
D. 4
E. 1
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Re: The average (arithmetic mean) of 6, 21, x, and y is 13, where x and y [#permalink] New post  08 Sep 2021, 05:20
Bunuel wrote:
The average (arithmetic mean) of 6, 21, x, and y is 13, where x and y are integers with a product of 100. Which of the following could be x?

A. 50
B. 20
C. 10
D. 4
E. 1


\(\frac{6+21+x+y}{ 4}\) =13
=> 27+x+y = 52
=> x+y = 25
Also given in question x*y =100

Method 1:
by values we can directly get 20 and 5
Only option B does satisfy

Method 2:
x*y =100 =>x = \(\frac{ 100}{y}\)
\(\frac{10}{y }\)+ y =25
solve the quadratic equation to get values .........
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Re: The average (arithmetic mean) of 6, 21, x, and y is 13, where x and y [#permalink] New post  03 Aug 2022, 21:45
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Theory

    ➡ Average or Arithmetic Mean = Sum of all the Values / Total Number of Values
    ➡ Sum of All the values = Average * Total Number of Values

The average (arithmetic mean) of 6, 21, x, and y is 13

Sum = Average * 4 = 13 * 4 = 52
=> 6 + 21 + x + y = 52
=> x + y = 52 - 27 = 25
=> y = 25 - x ...(1)

x and y are integers with a product of 100

=>x * y = 100

Substituting value of y from (1)
=> x * (25 - x) = 100
=> 25x - \(x^2\) = 100
=> \(x^2\) - 25x +100 = 0
=> \(x^2\) - 5x - 20x + 100 = 0
=> x*(x-5) - 20*(x-5) = 0
=> (x-5) * (x-20) = 0
=> x = 5 or 20

So, Answer will be B.
Hope it helps!

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Re: The average (arithmetic mean) of 6, 21, x, and y is 13, where x and y [#permalink]
03 Aug 2022, 21:45
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