The mean age is 34, so the sum of the four ages is $4 \times 34 = 136$. The median age is 33, which means the two middle ages must sum to $2 \times 33 = 66$.
The range is 32, so $d - a = 32$. Since $a < b < c < d$, we can also say that $d - c > 0$, $d - b > 0$, and $c - b > 0$. Therefore, \begin{align*}
d &> c > b > a. \\
d - c &> 0 \quad \Rightarrow \quad c < d. \\
d - b &> 0 \quad \Rightarrow \quad b < d. \\
c - b &> 0 \quad \Rightarrow \quad b < c.
\end{align*} Therefore, $d$ is the largest of the four ages, $c$ is the second largest, $b$ is the second smallest, and $a$ is the smallest.
Since $d$ is the largest, $d > a$, which means $d - a > 0$. Also, since $b$ is the second smallest, $b < c$.
We can already conclude that $c - a$ is at least $32 - 1 = 31$.
If we let $a = c - 31$, then $d = a + 32 = (c - 31) + 32 = c + 1$. Also, $b < c$ gives us $b < c - 31$, so $b - a < 0$. Therefore, $d - a = (c + 1) - (c - 31) = 32 > 0$. Hence, our values of $a$, $b$, $c$, and $d$ satisfy all the criteria.
The ages are $a = c - 31$, $b < c - 31$, $c$, and $d = c + 1$. The sum of these four ages is
\begin{align*}
(a + b + c + d) &= (c - 31) + b + c + (c + 1) \\
&= 3c - 30 + b + 1 \\
&= 3c + b - 19.
\end{align*}We are given that the sum of the ages is 136, so $$3c + b - 19 = 136,$$which means that $$3c + b = 155.$$Therefore, $3c + b$ is equal to 149, 152, 155, 158, 161, etc. In particular, $3c + b$ is a multiple of 3. However, the only value of $c$ for which $3c + b = 155$ is $c = 52$, which gives us a negative value of $b$. This means that there are no integers $a$, $b$, $c$, and $d$ that can satisfy the given conditions, so the answer is $\boxed{\text{none}}$.
The ages, in years, of four members of a family are represented
by a, b, c and d, where a < b < c < d. Their mean age is 34,
their median age is 33, and the range of their ages is 32.
What is the value of a?
1 answer
1 answer