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# Find the least values of $x$ and $y$ which satisfy the equations: $77y - 30x = 295.$${\text{A}}{\text{. }}x = 32,y = 27 \\ {\text{B}}{\text{. }}x = 21,y = 34 \\ {\text{C}}{\text{. }}x = 47,y = 35 \\ {\text{D}}{\text{. }}x = 46,y = 32 \\$

Last updated date: 18th Sep 2024
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Hint: In order to solve this problem you need to make the equation such that you can break some of the terms into the mixed fraction and then you need to assume the integer as variable assuming its minimum value as 1 and then with reference to that you need to solve for x and y solving the equation algebraically to get the right answer.

According to the question, we have to find the least values of $x$ and $y$. So, we solve for positive integers.
As given $77y - 30x = 295$ …….(1)
By Rearranging the above equation, we get
$\Rightarrow \dfrac{{77y}}{{30}} - x = \dfrac{{295}}{{30}}$
In the above equation break down the $y$ term as follows
$\Rightarrow \dfrac{{17y}}{{30}} + 2y - x = \dfrac{{295}}{{30}}$
And if we notice carefully $\dfrac{{295}}{{30}}$ can further be broken down as $9 + \dfrac{{25}}{{30}}$
So Now $\dfrac{{17y}}{{30}} + 2y - x = 9 + \dfrac{{25}}{{30}}$ or we can write $\dfrac{{17y - 25}}{{30}} + 2y - x = 9$
And as in the beginning it is stated that we are solving for positive integers.
Therefore, $\dfrac{{17y - 25}}{{30}}$ must be an integer value.
Now, we multiply by $23$ as when it is multiplied by $23$ then $23 \times 17 = 391$ and when $391$ divided by $30$ leaves remainder $1$. i.e.,
$\Rightarrow \dfrac{{391y - 575}}{{30}} =$ integer. This equation when broken down gives
$\Rightarrow 13y + \dfrac{y}{{30}} - 19 - \dfrac{5}{{30}}$
$\Rightarrow 13y - 19 + \dfrac{{y - 5}}{{30}}$
$\Rightarrow \dfrac{{y - 5}}{{30}} =$ Integer
Let the integer be p
$\dfrac{{y - 5}}{{30}} =$p
By cross multiplication, we get
$\Rightarrow y = 30{\text{p}} + 5$……………..(2)
By putting the value of $y$ in equation (1)
$\Rightarrow 77\left( {30{\text{p}} + 5} \right) - 30x = 295$
$\Rightarrow 2310{\text{p}} + 385 - 30x = 295$
$\Rightarrow 30x = 2210{\text{p}} - 90$
$\Rightarrow x = 77{\text{p}} - 30$ …………….(3)
From the equation (2) and (3) we can see that the value of $x$ and $y$ is negative for integer p $\leqslant 1$
which is not possible as we are solving for positive integers.
So, the least value of p is $1$.
Substituting p$= 1$ in (2) and (3)
$\Rightarrow y = 35, x = 47$

So, the correct answer is “Option C”.

Note: When you get to solve such problems you need to know that an equation can be modified by dividing or multiplying something on both sides. You also need to know that $a\dfrac{b}{c} = \dfrac{{ac + b}}{c}$. Knowing this will solve your problem and will give you the right answer.