Question

Divide $1162$ into three parts such that $4$ times the first part, $5$ times the second part and $7$ times the third part are equal.
(A) $490,392,280$
(B) $492,392,380$
(C) $493,329,389$
(D) $393,290,360$

Answer 1

Hint:This is a general question of linear equation in one variable so for that we make a general equation by using the given condition.Whenever the question say \[a\] time of $x$ and $b$ times of $y$ and $c$ time of $z$ we can written it as $a \times x = b \times y = c \times z$. And we can put this equal to any constant like $K$ and from that we find the value of $x,y,z$.

Complete step-by-step answer:
Given : we have to divide $1162$ into three parts such that $4$ times the first part, $5$times the second part and $7$ times the third part are equal.
So from this we can write
$4 \times $first part = $5 \times $second part = $7 \times $ third part
And we can assume it is equal to $x$
$4 \times $first part = $5 \times $second part = $7 \times $ third part = $x$
Now equating the terms
$4 \times $first part = $x$
So from here we get first part = $\dfrac{x}{4}$
$5 \times $second part = $x$
So from here we get second part = $\dfrac{x}{5}$
$7 \times $ third part = $x$
So from here we get third part = $\dfrac{x}{7}$
Now we know that
$1162 = $ first part + second part + third part ( Because we divide 1162 in three part )
So $\dfrac{x}{4} + \dfrac{x}{5} + \dfrac{x}{7} = 1162$
Now we take LCM of $\left( {4,5,7} \right)$ is $4 \times 5 \times 7 = 140$
So $\dfrac{{35x + 28x + 20x}}{{140}} = 1162$
Now by solving this we get
$\dfrac{{83x}}{{140}} = 1162$
From this we can say
$83x = 1162 \times 140$
And $x = \dfrac{{1162 \times 140}}{{83}}$
So this become $x = 1960$
Now
$\therefore $ First number = $\dfrac{x}{4} = \dfrac{{1960}}{4} = 490$
$\therefore $ Second number = $\dfrac{x}{5} = \dfrac{{1960}}{5} = 392$
$\therefore $ Third number = $\dfrac{x}{7} = \dfrac{{1960}}{7} = 280$

So, the correct answer is “Option A”.

Note:Whenever if amount or any other thing divided to $A$ and $B$ in a ratio $m:n$ Then value of $A$ is given by $A = \dfrac{m}{{m + n}} \times $Total value and value of $B$ is given by by $B = \dfrac{n}{{m + n}} \times $Total value.