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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 $$\dfrac{1}{3}$$ when $$1$$ is subtracted from the numerator and it becomes $$\dfrac14$$ when $$8$$ is added to its denominator. Find the fraction.


Solution

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
$$\Rightarrow x+20(30) = 1000$$
$$\Rightarrow x+600 = 1000$$
$$\Rightarrow x=400$$
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 $$\dfrac { x }{ y }$$.
$$\dfrac{ x-1}{ y}=\dfrac { 1 }{ 3 } $$   

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

$$\dfrac { x }{ y+8}=\dfrac14 $$       

$$\Rightarrow 4x=y+8$$       ...(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 $$=\dfrac{x}{y}=\dfrac{5}{12}$$



Mathematics
RS Agarwal
Standard X

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