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Algebra Difficulty 2.5 Junior Find the answer Canada

The ages of three students are consecutive integers. Their mean
(average) age is $13$ . A fourth
student joins the group and the mean of their four ages is $14$ . How old is the fourth student?

Pick one

Solution

Solution 1:

The mean age of the first three students is $13$ , and so the sum of their ages is $13\times3=39$ .

The mean age of the four students is $14$ , and so the sum of their ages is $14\times4=56$ .

The age of the fourth student is the difference between these two sums,
which is $56-39=17$ years old.

Solution 2:

In an ordered list of $3$
consecutive integers, the average of the list is always the middle
integer. Can you see why?

Therefore, three students whose ages are consecutive integers and whose
mean age is $13$ are $12$ , $13$ and $14$ years old.

Suppose the fourth student is $x$
years of age.

Since the mean age of the four students is $14$ , then $\dfrac{12+13+14+x}{4}=14$ or $39+x=14\times4$ , and so $x=56-39=17$ .

The fourth student is $17$ years
old.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.