Question

The average of $6$ numbers is $30$. If the average of the first four is $25$ and that of the last three is \[35\], the fourth number is
$\left( {\text{A}} \right)25$
\[({\text{B) }}30\]
\[(C{\text{) }}35\]
\[(D{\text{) 4}}0\]

Answer 1

Hint: Here, we have to find the fourth number.
First, we assume a variable for all the $6$ numbers.
Then, we have to form equations.
By substituting one equation to the other, we will get the fourth number.

Formula used: \[\dfrac{{{\text{Sum of the total of all observations}}}}{{{\text{Total no}}{\text{.of observation}}}} = {\text{Average}} {\text{value}}\]

Complete step-by-step solution:
Let us assume that the $6$ numbers be ${\text{a, b, c, d, e and f}}$.
So we need to find the fourth number that is ${\text{d}}$
It is given that the average of $6$ numbers is $30$.
By using the formula and we can write it as,
$\dfrac{{{\text{a + b + c + d + e + f}}}}{{\text{6}}}{\text{ = }}30$ ,
By taking cross multiplication we get,
\[ \Rightarrow {\text{a + b + c + d + e + f}} = 30 \times 6\]
Let us multiply we get,
\[ \Rightarrow {\text{a + b + c + d + e + f}} = 180........(1)\]
Again it is stated in the question, that the average of the first four numbers is $25$.
By using the formula and we can write it as
\[\dfrac{{a + b + c + d}}{4} = 25\]
By taking cross multiplication we get,
\[ \Rightarrow {{a + b + c + d = 25 \times 4}}\]
On multiplying the RHS we get,
\[ \Rightarrow {\text{a + b + c + d = 100}}...........{\text{(2)}}\]
Also it is stated in the question that the average of last three numbers is $35$,
By using the formula and we can write it as
\[\dfrac{{{\text{d + e + f}}}}{3} = 35\]
By taking cross multiplication we get,
\[{{d + e + f = 35 \times 3}}\]
Let us multiply the terms and we get
\[ \Rightarrow {\text{d + e + f}} = 105...........(3)\]
Now, we need to find the fourth number that is the value of \[{\text{d}}\]
Putting \[(2)\] and \[(1)\] we get
\[1{\text{00 + e + f }} = 180\]
On subtracting \[100\] on both sides we get,
\[{\text{e + f}} = 180 - 100\]
On subtracting the values we get
\[ \Rightarrow {\text{e + f}} = 80\]
From\[(3)\],
\[{\text{d + e + f}} = 105\]
We can substitute \[{\text{e + f}} = 80\] in \[{\text{d + e + f = 105}}\] to get \[{\text{d}}\].
\[{\text{d}} + 80 = 105\]
On subtracting \[80\] on both sides we get,
\[{\text{d}} = 105 - 80\]
Let us subtract the values and we get,
\[{\text{d}} = 25.\]
$\therefore $ The fourth number is $25$.

Hence the correct option is A

Note: This type of question is simply a kind of logical reasoning and such questions always contain certain clues.
Here, the clues are, ‘there are 6 numbers’ then ‘average of first four numbers’ and ‘average of last three numbers’. and when we add up 4 and 3, it gives 7.
This indicates that there is a number which is common and by this approach we need to form and substitute the equations.