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 $$=\dfrac{x}{y}=\dfrac{5}{12}$$MathematicsRS AgarwalStandard X

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