Question

Question 13
The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what could be the highest possible number of remaining two numbers?
(a) 81
(b) 40
(c) 100
(d) 71

Answer 1

Mean of three numbers = 40 and lowest number= 19 [given]
Let the three observations be 19, x and y, respectively.
$\therefore Mean=\frac{Sumofallobservations}{Totalnumberofobservations}\Rightarrow 40=\frac{19+x+y}{3}[\because mean=40,given]\Rightarrow 3\times 40=19+x+y$
$\Rightarrow$ 120 = 19 + x + y
$\Rightarrow$ x + y = 120 - 19
$\Rightarrow$ x + y = 101
Since, 19 is the lowest observation.
Hence, for the highest possible value of remaining two numbers, one must be 20.
Let x = 20
From Eq. (i), we get
20 + y = 101
$\Rightarrow$ y = 101 - 20
$\Rightarrow$ y = 81