Math Question help!!!

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Four numbers are selected such that when each number is added to the average of the other three the following sums are obtained: 25, 37, 43, and 51. Determine the average of the four numbers.

Jun 11th, 2015

Let the 4 numbers be a,b,c and d.

When the first number is added to the average of the other three numbers the result is 25.

Therefore,

$a+\frac{b+c+d}{3}=25\\ \\ \frac{3a+b+c+d}{3}=25\\ \\ 3a+b+c+d=25\times3\\ \\ 3a+b+c+d=75---(1)\\$

When the second number is added to the average of the other three numbers the result is 37.

Therefore,

$b+\frac{a+c+d}{3}=37\\ \\ \frac{3b+a+c+d}{3}=37\\ \\ a+3b+c+d=37\times3\\ \\ a+3b+c+d=111---(2)\\$

When the third number is added to the average of the other three numbers the result is 43.

Therefore,

$c+\frac{a+b+d}{3}=43\\ \\ \frac{3c+a+b+d}{3}=43\\ \\ a+b+3c+d=43\times3\\ \\ a+b+3c+d=129---(3)\\$

When the fourth number is added to the average of the other three numbers the result is 51.

Therefore,

$d+\frac{a+b+c}{3}=51\\ \\ \frac{3d+a+b+c}{3}=51\\ \\ a+b+c+3d=51\times3\\ \\ a+b+c+3d=153---(4)\\$

Adding equations (1), (2), (3) and (4) we get

6a + 6b + 6c + 6d = 75 + 111 + 129 + 153

6(a + b + c + d) = 468

Divide both sides by 6

a + b + c + d = 468/6

a + b + c + d = 78

To get the average, divide both sides by 4

$\frac{a+b+c+d}{4}=\frac{78}{4}\\ \\ \frac{a+b+c+d}{4}=19.5\\$

So, average = 19.5

Jun 11th, 2015

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Jun 11th, 2015
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