Question

If y varies directly with x, and if $y=12$ when $x=15$, how do you find x when $y=21$?

Answer 1

In the above problem, it is said that y varies directly with x so the relation between y and x will look as follows: $y=kx$. Here, “k” is the constant. Now, we have given when $x=15$ then $y=12$ so we are going to substitute the value of x and y and then we will get the value of k. Also, we have asked to find the value of x when $y=21$ so we are going to substitute this value of y and the value of k which we have calculated above in $y=kx$ and hence, will find the value of x.

Complete step by step solution:
It is given that y varies directly with x then we can write the relation between y and x as follows:
$y=kx$………… (1)
In the above equation, “k” is a constant. And we are going to find the value of k by substituting $x=15$ and $y=12$ in the above equation and we get,
$\Rightarrow 12=k\left( 15 \right)$
Dividing 15 on both the sides we get,
$\Rightarrow \dfrac{12}{15}=k$
In the above equation, numerator and denominator will be divided by 3 and we get,
$\Rightarrow \dfrac{4}{5}=k$
From the above, we have calculated the value of k as $\dfrac{4}{5}$. Now, substituting this value of k in eq. (1) we get,
$y=\dfrac{4}{5}x$
In the above question, we are asked to find the value of x when $y=21$ so substituting this value of y in the above equation and we get,
$21=\dfrac{4}{5}x$
Multiplying 5 on both the sides we get,
$\begin{aligned} & \Rightarrow 21\times 5=4x \\\ & \Rightarrow 105=4x \\\ \end{aligned}$
Dividing 4 on both the sides we get,
$\Rightarrow \dfrac{105}{4}=x$
Hence, we have found the value of x when $y=21$.

Note: The mistake that could be possible in the above problem is in writing the relation between x and y in regard to the relation which says that y varies directly with x. You may write the relation between x and y as:
$y=x$
The above relation is showing y varies directly with x but this is the special case when the constant which we have taken above $\left( k=1 \right)$ but after making this mistake when you substitute the value of x and y which is given as $y=12$ when $x=15$ then the above equation will look like:
$12=15$
As L.H.S is not equal to R.H.S, this means that the relation which we have just written between x and y is incorrect and then multiply the constant “k” with x.