CAT 2021 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.
Question 9 : Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is
Let’s consider the total number of pens to be
(100 +x)
Let's consider the fixed-wage of the labour to be ‘w’
case (i): the net profit is 300
CP = 8(100 +x) + w
SP = 12 x 100 + 11 x = 1200 + 11x
Profit = SP - CP
300 = 1200 + 11x - 8(100 +x) - w
w - 3x = 100 → eq(1)
Case (ii): the net loss is 300
CP = 8(100 +x) + w
SP = 12 x 100 + 9 x = 1200 + 9x
Loss = CP - SP
300 = 8(100 +x) + w - 1200 - 9x
w - x = 700 → eq(2)
By solving equation (1) & (2) we get w = 1000.
Therefore, the wage of the employee is 1000.
Alternate solution:
We can also do this with a little intuition. The Rs 2 decrease per pen results in 300 loss from the case of 300 profit.
The net value of 600 resulted by selling the remaining pens at Rs 2 lesser.
If the number of remaining pens is x, then 2x = 600
So, x = 300.
Total pens = 100 + 300 = 400.
We can get the wage of an employee(w) by considering profit/loss
Profit = 300
100(12) + 300(11) - 400(8) -w = 300
w = 1000
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