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 $l = \frac{6800 - 64x}{50}$ , in the original equation, we get - $\frac{6800 - 64x}{50} \cdot (x+50) = 5000$ $(6800 - 64x)(x+50) = 250000$ $6800x + 340000 - 64x^2 - 3200x = 25000$ $6800x + 340000 - 64x^2 - 3200x = 250000$ $64x^2 - 3600x - 90000 = 0$ Solving we get: $x = \frac{225+375}{8} = \frac{600}{8}$ or $\frac{150}{8}$ 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 = $\frac{Option - 50}{2}$ LARGE = $Small + 50 = \frac{Option + 50}{2}$ 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 = $150 + \frac{50}{2} = 100$ Small = $150 - \frac{50}{2} = 50$ Now, total shirts $= \frac{5000}{100} + \frac{1800}{50} = 50 + 36 = 86$ Option C - Large =$200 + \frac{50}{2} = 125$ Small = $200 - \frac{50}{2} = 75$ Total shirts = $\frac{5000}{125} + \frac{1800}{75} = 40 + 24 = 64$