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In an apartment complex, the number of people aged $51$ years and above is $30$ and there are at most $39$ people whose ages are below $51$ years. The average age of all the people in the apartment complex is $38$ years. What is the largest possible average age, in years, of the people whose ages are below $51$ years?

1. $27$
2. $28$
3. $26$
4. $25$

## 1 Answer

Let the number of people below $51 \; (<51) \; \text {years}$ be $x.$

The total number of people in an apartment complex $=$  The number of people whose ages $51 \; \text{years}$  and above $( \geq 51) +$ the number of people whose ages below $51 \; (<51) \; \text{years} = 30 + x$

The average age of all the people in the apartment complex $= 38 \; \text {years}.$

The total age of people in apartment complex $= (30+x) \times 38$

The smallest possible average age of people above $51 \; \text{years}$  is $51,$ and it gives the largest value for the other group.

The total age of people above $51 \; \text{years} = 30 \times 51 = 1530$

Now, the total age of people below $51 \; \text{years} = (30+x) \times 38 – 1530$

$\quad = 1140 + 38x – 1530 = 38x – 390$

The average age people below $51 \; \text{years} = \frac{38x – 390}{x} \quad \longrightarrow (1)$

The number of people whose ages are below $51 \; \text{years}$ is almost $39 \; (\leqslant 39).$

For getting the largest possible average age, the number of people should be $x = 39 .$

$\therefore$ The average age people below $51 \; \text{years} = \frac{(38 \times 39) – 390}{39} = \frac{39(38-10)}{39} = 28 \; \text{years}.$

Correct Answer $: \; \text{B}$
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