Question

Ages of five boys are ( 2a + 5 ), ( 3a – 4 ), ( a + 3 ), ( 2a - 5 ) and ( 2a + 6 ) respectively, if the average of ages is 21 find the value of ‘ a ’ .
( a ) 105
( b ) 100
( c ) 10
( d) 106

Answer 1

Hint: We can see that there is a factor ‘ a ‘ which is common in ages of all five boys so we have to find such ‘ a ‘ and we will find it with the concept of average value method. This method is used to find the average of any number of numerical values.

Complete step by step answer:
In question it is given that there are five boys and ages of these five boys are ( 2a + 5 ), ( 3a – 4 ), ( a + 3 ), ( 2a - 5 ) and ( 2a + 6 ) respectively. And, also it is given that the average age of these five boys is 21 years.
Now, we know that there is a concept known as Average Value. It is denoted by $\overline{n}$ and is defined as,
In general, if we have say total n numbers say ${{n}_{1}},{{n}_{2}},{{n}_{3}},........,{{n}_{n}}$ then, the average value of these total n objects will be equal to, Average value $\overline{n}=\dfrac{total\text{ }sum\text{ }of\text{ }all\text{ }n\text{ }values}{total\text{ }number\text{ }of\text{ }values}$ that is
$\overline{n}=\dfrac{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}+.......+{{n}_{n}}}{n}$ .
So here we have five values then n = 5 and let the values be say, ${{n}_{1}}=2a+5,{{n}_{2}}=3a-4,{{n}_{3}}=a+3,{{n}_{4}}=2a-5,{{n}_{5}}=2a+6$.
So, Average age value $\overline{n}=\dfrac{(2a+5)+(3a-4)+(a+3)+(2a-5)+(2a+6)}{5}$…… ( i )
It is given that the average value is equal to 21.
So , putting value of average in equation ( i ), we get
$21=\dfrac{(2a+5)+(3a-4)+(a+3)+(2a-5)+(2a+6)}{5}$
Taking 5 from denominator on right side to left side we get,
$21\times 5=(2a+5)+(3a-4)+(a+3)+(2a-5)+(2a+6)$
On solving we get,
$10a+5=105$,
On simplification, we get
$\begin{align}
  & 10a=105-5 \\
 & 10a=100 \\
 & a=10 \\
\end{align}$

So, the correct answer is “Option c”.

Note: Always remember the concept of Average value. It might happen that if we have more values to add we may make calculation mistakes so try to avoid the calculation mistake. In Cross multiplication, the number should be multiplied with each term when we open brackets.