Question

The mean of the three numbers is $40$. All three numbers are different natural numbers. If the lowest is $19$, what could be the highest possible number of the remaining two numbers?

• A. $81$
• B. $40$
• C. $100$
• D. $71$

Answer 1

The correct option is A

$81$

The explanation for the correct option.

Let us consider that the remaining two numbers are $a\mathrm{and}b$.

Since the mean of the three numbers is $40$ and the lowest number is $19$.

$$\begin{gathered} \therefore \frac{a+b+19}{3}=40 \\ \Rightarrow a+b+19=120 \\ \Rightarrow a+b=120-19 \\ \Rightarrow a+b=101 \end{gathered}$$

Let one number $b$ is $20$ because the lowest number is $19$.

$$\begin{gathered} \therefore a+20=101 \\ \Rightarrow b=101-20 \\ \Rightarrow b=81 \end{gathered}$$

Thus, the highest possible number is $81$.

Hence, the option $A$is correct.