CAT Exam 2021 Slot 3 Solved Paper

Let the number of large shirts be $l$ and the number of small shirts be $s$
Let the price of a small shirt be $x$ and that of a large shirt be $x + 50 $
Now, $s + l = 64 $
$\l (x + 50) = 5000 $
$s x = 1800 $
Adding them, we get:
$l x + s x + 50l = 6800$
$64x +50l = 6800$
Substituting $ l = {6800 - 64 x} / 50$ , in the original equation, we get -
${(6800 - 64 x)}/50 . {(x + 50)} = 5000$
$ (6800 - 64 x) (x + 50) = 250000$
$6800 x + 340000 − 64 x^2 -3200 x = 25000$
$6800 x +340000 - 64 x^2 - 3200 x = 250000 $
$64 x^2 - 3600 x - 90000 = 0$
Solving we get:
$ x = {225 + 375}/8 = 600/8 $ or $150/8$
SO, $x = 75$
$x + 50 = 125$
Answer - $ 75 + 125 = 200$
Alternate approach: By options.
Hint: Each option gives the sum of the costs of one large and one small shirt. We know that $\large = \small + 50$
Hence, $\small + \small + 50 = \option$
SMALL = ${\Option - 50}/2$
LARGE = $\Small + 50 = {Option + 50}/2$
Option A and Option D gives us decimal values for SMALL and LARGE.Hence we will consider them later.
Lets start with Option B -
Large = $150 + 50 / 2 = 100$
Small = $150 - 50 / 2 = 50$
Now, total shirts $= 5000/100 + 1800/50 = 50 + 36 = 86$
Option C -
Large =$ 200 + 50 / 2 = 125$
Small = $200 - 50 / 2 = 75$
Total shirts = $5000/125 + 1800/75 = 40 + 24 = 64$