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(a) 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.
OR
(b) Find the value of $k$ for which the following pair of linear equations will have infinitely many solutions :
$kx + 3y - (k - 3) = 0$ and $12x + ky - k = 0$
Hence, find any two solutions of the given pair of equations.
OR
(b) Find the value of $k$ for which the following pair of linear equations will have infinitely many solutions :
$kx + 3y - (k - 3) = 0$ and $12x + ky - k = 0$
Hence, find any two solutions of the given pair of equations.
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(a) Let the fraction be $\frac{x}{y}$ [$\frac{1}{2}$ mark]
$\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3x - y = 3 \dots (i)$ [$\frac{1}{2}$ mark]
$\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4x - y = 8 \dots (ii)$ [$\frac{1}{2}$ mark]
On solving the equations $(i)$ and $(ii)$, we get $x = 5, y = 12$ [$1$ mark]
Required fraction is $\frac{5}{12}$ [$\frac{1}{2}$ mark]
OR
(b) For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$ [$1$ mark]
$k^2 = 36$ and $k^2 - 3k = 3k$
$(k = \pm 6)$ and $(k = 6, 0)$
$\therefore k = 6$ [$1$ mark]
For $k = 6$, equations are $6x + 3y = 3$ and $12x + 6y = 6$
any two correct solutions [$\frac{1}{2} + \frac{1}{2}$ mark]
$\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3x - y = 3 \dots (i)$ [$\frac{1}{2}$ mark]
$\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4x - y = 8 \dots (ii)$ [$\frac{1}{2}$ mark]
On solving the equations $(i)$ and $(ii)$, we get $x = 5, y = 12$ [$1$ mark]
Required fraction is $\frac{5}{12}$ [$\frac{1}{2}$ mark]
OR
(b) For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$ [$1$ mark]
$k^2 = 36$ and $k^2 - 3k = 3k$
$(k = \pm 6)$ and $(k = 6, 0)$
$\therefore k = 6$ [$1$ mark]
For $k = 6$, equations are $6x + 3y = 3$ and $12x + 6y = 6$
any two correct solutions [$\frac{1}{2} + \frac{1}{2}$ mark]