Average

[MoniMini]

Hi
The Average of 5 numbers is 51. If the average of the first 3 numbers is 42 and the average of the last 3 numbers is 59. Find the 3rd number.

Okay. So...
(a + b + c + d + e) / 5 = 51
and
(a + b + c) / 3 = 42
and
(c + d + e) / 3 = 59
so
a + b + c = 42 into 3 i.e. 126
c + d + e = 59 into 3 i.e. 177
So, to find the value of 'c' we
should subtact 126 from 177 (hope i'm right)
I checked the answer behind the book, and
answer i.e. 51 was correct but I can't find
any reason why we subtracted 126 from 177.
Yeah, I know it is a bit silly to ask why something
is done when you know how to do it.
But somehow I'm just not being able to figure
out why that step is done.

Any help will be appreciated.

Thanks,
~MoniMini

 

[soroban]

Hello, MoniMini!

The average of 5 numbers is 51.
The average of the first 3 numbers is 42; the average of the last 3 numbers is 59.
Find the 3rd number.

We have: .\(\displaystyle \begin{Bmatrix} \dfrac{a+b+c+d+e}{5} \:=\:51 & \Rightarrow & a+b+c+d+e \:=\:255 & [1] \\ \\ \dfrac{a+b+c}{3} \:=\:42 & \Rightarrow & a + b + c \:=\:126 & [2] \\ \\ \dfrac{c+d+e}{3} \:=\:59 & \Rightarrow & c+d+e \:=\:177 & [3] \end{Bmatrix}\)


\(\displaystyle \begin{array}{cccccccc}\text{Add [2] and [3]:} & a + b + 2c + d + e &=& 303 \\ \text{Subtract [1]:} & a + b + c + d + e &=& 255 \\ \text{Therefore: } & c\;=\;48\end{array}\)

 

[MoniMini]

Thanks

Aah! I see. I guess the book had the answer misprinted.
Thanks a Lot @soroban