Question

The List II gives the solution to the word problem in the List I, match them correctly.

Answer 1

Let the length of the park be $x$ and the width of the park be $y$
and let, $x=y+4$
Half the perimeter of the rectangular garden$\Rightarrow x+y=36$
$\Rightarrow y+4+y=36$
$\Rightarrow y=16$
$\therefore x=16+4\Rightarrow x=20$
$(x,y)\Rightarrow (20,16)$
Hence, A $\to$ 4.
Let the two numbers be $x$ and $y$
and let
$x=3y$
$\therefore x-y=26$
$\Rightarrow 3y-y=26$
$\Rightarrow y=13$
$\therefore x=3y\Rightarrow x=3\left(13\right)\Rightarrow x=39$
$\therefore (x,y)\Rightarrow (39,13)$
Hence, B$\to$1.
Let the age of jacob be $x$ and the age of his daughter be $y$
Then,
$x+5=3(y+5)$
Solving the above equation gives $x-3y=10......\left(1\right)$
and $x-5=7(y-5)$
Solving the above equation gives $x-7y=-30......\left(2\right)$
$\left(1\right)-\left(2\right)\Rightarrow y=10$
Substituting value of $y$ in $\left(1\right)\Rightarrow x=40$
$\therefore (x,y)\Rightarrow (40,10)$
Hence, C$\to$2.
Let the two numbers be $x$ and $y$
then,
$\frac{x+1}{y+1}=\frac{1}{2}$
$\Rightarrow 2x+2=y+1$
$\Rightarrow 2x-y=-1......\left(1\right)$
and,
$\frac{x-5}{y-5}=\frac{5}{11}$
$\Rightarrow 11x-55=5y-25$
$\Rightarrow 11x-5y=30......\left(2\right)$
$\left(1\right)\times 5+\left(2\right)\times 1\Rightarrow x=35$
Substituting value of $x$ in $\left(1\right)\Rightarrow y=71$
$\therefore (x,y)\Rightarrow (35,71)$
Hence, D $\to$3.