Question

The average of the first three numbers is double of the fourth number. If the average of all the four numbers is 12. Find the fourth number.
A) 16
B) \[\dfrac{{48}}{7}\]
C) 20
D) 18

Answer 1

Hint:
Let’s assume the four numbers. We have two different conditions given in the question. We will use them to find our answer so let’s start!

Complete step by step solution:
Let the four numbers be \[{n_1},{n_2},{n_3},{n_4}\].
Now average is the sum of the items divided by total number of items.
So average of first three numbers is
\[ \Rightarrow \dfrac{{{n_1} + {n_2} + {n_3}}}{3}\]
But it is given that the average of the first three numbers is double of the fourth number.
So, the equation becomes
\[ \Rightarrow \dfrac{{{n_1} + {n_2} + {n_3}}}{3} = 2{n_4}\]
Now the second condition is given that the average of all the four numbers is 12.
So it can be written as,
\[ \Rightarrow \dfrac{{{n_1} + {n_2} + {n_3} + {n_4}}}{4} = 12\]
On cross multiplying we get,
\[ \Rightarrow {n_1} + {n_2} + {n_3} + {n_4} = 48\]
But from equation of first condition we get
\[
   \Rightarrow {n_1} + {n_2} + {n_3} = 3 \times 2{n_4} \\
   \Rightarrow {n_1} + {n_2} + {n_3} = 6{n_4} \\
 \]
Now we can replace the sum of first three numbers by \[6{n_4}\]
So the equation becomes,
\[
   \Rightarrow 6{n_4} + {n_4} = 48 \\
   \Rightarrow 7{n_4} = 48 \\
   \Rightarrow {n_4} = \dfrac{{48}}{7} \\
 \]
So the value of fourth number is \[\dfrac{{48}}{7}\]

So, the correct option is B.

Note:
Students note here that they have not mentioned here any particular kind of numbers like they are consecutive even numbers, odd numbers; first 4 prime numbers etc. Nothing is given. So in such cases consider them as consecutive natural numbers only.