Question

The mean of three positive numbers is 10 more than the smallest of the numbers and 10 less than the largest of the three. If twice the median of the three numbers is 80, then the mean of the given numbers is

• A. 30
• B. 40
• C. 50
• D. 60

Answer 1

The correct option is B 40

Let the three numbers be a, b and c such that a > b > c.
We are given that the mean of these numbers is 10 more than the smallest number and 10 less than the largest number.
∴ Mean of a, b, c = $\frac{a+b+c}{3}$
$\Rightarrow \frac{a+b+c}{3}=a-10=c+10$.....(i)
Now, the median of a, b, c is b. (a > b > c)
It is given that twice the median of the three numbers is 80.
$\Rightarrow 2\times \text{Median}=80$
$\Rightarrow 2\times b=80$
$\Rightarrow b=40$
Now, we have $\frac{a+b+c}{3}=a-10$ [From (i)]
$\Rightarrow a+40+c=3a-30$
$\Rightarrow 70+c=2a$
$\Rightarrow c=2a-70$ .........(ii)

Also,
$\Rightarrow a+40+2a–70=3(2a–70)+30$ [From (ii)]
$\Rightarrow 3a–30=6a–210+30$
$\Rightarrow 150=3a$
$\Rightarrow a=50$
$\Rightarrow c=2\times 50–70$ [From (ii)]
= 30
Thus, we have, a = 50, b = 40, c = 30.
$\Rightarrow$ Mean = $\frac{a+b+c}{3}$
$=\frac{50+40+30}{3}$
$=\frac{120}{3}$
$\Rightarrow \text{Mean}=40$
Hence, the correct answer is option (b).