Question

If two third, one half and one seventh of a number is added to itself the result is $37$.
${\text{A}}.14\dfrac{2}{{97}}$
${\text{B}}.16\dfrac{2}{{97}}$
${\text{C}}.18\dfrac{2}{{97}}$
\[{\text{D}}{\text{.}}\]${\text{None}}$

Answer 1

Hint: From the question, we have to find the number by using the number system concept. First we have to find the number one third, one half and one seventh of a number and after some addition of the numbers to get the result $37$.
The number system includes the unit's digit, reminders, digital root, divisibility rules, integers and number of zeros. Besides these topics, questions are also based on real numbers, prime numbers, natural numbers and whole numbers.

Formula used: Let the unknown number be ${\text{x}}$.
Let the two-third of a number be $\left( {\dfrac{2}{3}} \right) \times {\text{x}} = \dfrac{{2{\text{x}}}}{3}$.
Let the one-half of a number be $\left( {\dfrac{1}{2}} \right) \times {\text{x}} = \dfrac{{\text{x}}}{2}$ .
Let the one-seventh of a number be $\left( {\dfrac{1}{7}} \right) \times {\text{x}} = \dfrac{{\text{x}}}{7}$ .

Complete step-by-step answer:
From the given condition two third, one half and one seventh of a number is added to itself to get the result $37$.
 First, we are going to add the two third of a number, one half of a number and one seventh of a number is added to itself. We get,
$ \Rightarrow \left( {\dfrac{{2{\text{x}}}}{3}} \right) + \left( {\dfrac{{\text{x}}}{2}} \right) + \left( {\dfrac{{\text{x}}}{7}} \right) + {\text{x}}$
Second, we make the above term as an equation from the given information. That is, from the given result, the above form is equal to $37$. Then we get the equation as
$\left( {\dfrac{{2{\text{x}}}}{3}} \right) + \left( {\dfrac{{\text{x}}}{2}} \right) + \left( {\dfrac{{\text{x}}}{7}} \right) + {\text{x}} = 37$
Third, we have to find the number ${\text{x}}$ by using Least Common Multiple (LCM). Here, the Least Common Multiple of $3,2,7$ is $42$. Then we get,
$ \Rightarrow \left( {\dfrac{{2{\text{x}} \times 14}}{{3 \times 14}}} \right) + \left( {\dfrac{{{\text{x}} \times {\text{21}}}}{{2 \times 21}}} \right) + \left( {\dfrac{{{\text{x}} \times 6}}{{7 \times 6}}} \right) + \left( {\dfrac{{{\text{x}} \times 42}}{{1 \times 42}}} \right) = 37$
\[ \Rightarrow \left( {\dfrac{{28{\text{x}}}}{{42}}} \right) + \left( {\dfrac{{{\text{21x}}}}{{42}}} \right) + \left( {\dfrac{{{\text{6x}}}}{{42}}} \right) + \left( {\dfrac{{42{\text{x}}}}{{42}}} \right) = 37\]
In the above term, all the denominators are the same on the left hand side (LHS) and we are going to add the numerator on the left hand side (LHS). We get,
$ \Rightarrow \left( {\dfrac{{28{\text{x}} + 21{\text{x}} + 6{\text{x}} + 42{\text{x}}}}{{42}}} \right) = 37$
$ \Rightarrow \left( {\dfrac{{97{\text{x}}}}{{42}}} \right) = 37$
Now, cross multiply the $42$ into the right hand side (RHS). We get,
$ \Rightarrow 97{\text{x}} = 37 \times 42$
$ \Rightarrow 97{\text{x}} = 1554$
$ \Rightarrow {\text{x}} = \left( {\dfrac{{1554}}{{97}}} \right)$
Now, we have to make the improper fraction into the mixed fraction form.
$ \Rightarrow {\text{x}} = 16\left( {\dfrac{2}{{97}}} \right)$
Hence, the unknown number ${\text{x}}$ is $16\left( {\dfrac{2}{{97}}} \right)$.

$\therefore $ The option B is the correct answer.

Note: In the above, the problem may go wrong on taking Least Common Multiple (LCM) and on converting the solution (improper fraction into mixed fraction). Linear equations in one variable are used when we have one unknown quantity. Linear equations in two variables are used when we have two unknown quantities.