Question

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for $20$ days she has to pay $\text{Rs.}1000$ as hostel charges whereas a student B, who takes food for $26$ days, pays $\text{Rs.}1180$ as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Find the fraction.

Answer 1

i) Let the fixed monthly hostel charges be $\text{Rs.}x$ and the cost of food per day taken from the mess be $y$.

As per the given condition,

$x+20y=1000$ ...(1)

$x+26y=1180$ ...(2)

Subtracting eq(1) from eq (2), we get

$\Rightarrow 6y=180$

$\Rightarrow y=30$

Substitute $y=30$ in eq (1), we get

$\Rightarrow x+20\left(30\right)=1000$

$\Rightarrow x+600=1000$

$\Rightarrow x=400$

So, the fixed monthly hostel charge is $\text{Rs}600$ and the cost of food per day is $\text{Rs}30$.

ii) Let the numerator be $x$ and the denominator be $y$, so the fraction will be $\frac{x}{y}$.

$\frac{x-1}{y}=\frac{1}{3}$

$\Rightarrow 3x-y=3\Rightarrow 3x-3=y$ ...(1)

$\frac{x}{y+8}=\frac{1}{4}$

$\Rightarrow 4x=y+8$ ...(2)

Substituting for $y$ in (2) we get,

$4x=3x-3+8$

$x=5$

Putting $x=5$ in eq(1), we get $y=12$
$\therefore$ fraction $=\frac{x}{y}=\frac{5}{12}$