Question

Damien played golf on each of the four days of his vacation. His scores on the first three days were $93,92$ and $89$, and his average for the four days was 90. What was his score on the fourth day?
${\text{(A) 84}}$
${\text{(B) 85}}$
${\text{(C) 86}}$
${\text{(D) 87}}$
${\text{(E) 88}}$

Answer 1

Hint: Here we have to find out the score on the fourth day. We will make use of the mean formula. We will substitute all the values and we will find the unknown value. The formula of mean is given as
${\text{Mean = }}\dfrac{{{\text{sum of terms}}}}{{{\text{number of terms}}}}$
Finally we get the required answer.

Complete step-by-step solution:
It is given that the question stated as Damien played golf for $4$days
Now we can write it as the number of terms in $4$.
Also, the mean of all the $4$ scores is $90$.
The score of day $1$, day $2$ and day $3$is $93,92$ and $89$ respectively.
Let the score of day $4$ be $x$
Now on using the formula of mean, the statement can be written as:
$ \Rightarrow 90 = \dfrac{{93 + 92 + 89 + x}}{4}$
On adding the numerator on the right-hand side we get:
$ \Rightarrow 90 = \dfrac{{274 + x}}{4}$
On cross multiplying we get:
$ \Rightarrow 90 \times 4 = 274 + x$
On multiply we get:
$ \Rightarrow 360 = 274 + x$
Now on taking like terms on the same side we get:
$ \Rightarrow 360 - 274 = x$
On subtracting we get:
$86 = x$
From the above equation we can see that $x = 86$ therefore, the score Damien got on day $4$ was $86$ which is the required answer.

Therefore, the correct option is ${\text{(C)}}$ which is $86$.

Note: Arithmetic mean should not be used when there are some extreme values in the distribution, since there are no extreme values over here, the formula of mean can be used to calculate the missing value.