CAT 2017 Question Paper Slot 2

Let the total number of shirts be x. Hence number of non defective shirts = x - 15% of x = 0.85x
Number of shirts left for export = No of non defective shirts - number of shirts sold in domestic market
= No of non defective shirts - 20% of No of non defective shirts
= 80% of No of non defective shirts
Hence 8840 = 0.8 * (0.85x) . Solving for x we get, x = 13000

Let the average height of 22 toddlers be 3x.
Sum of the height of 22 toddlers = 66x
Hence average height of the two toddlers who left the group = x
Sum of the height of the remaining 20 toddlers = 66x - 2x = 64x
Average height of the remaining 20 toddlers = 64x/20 = 3.2x
Difference = 0.2x = 2 inches => x = 10 inches
Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

Let the manufacturing price of the table = $$x$$
Hence the price at which the wholesaler bought from the manufacturer = $$1.1 \times x$$
The price at which the retailer bought from the wholesaler = $$1.3 \times 1.1 \times x$$
The price at which the customer bought from the retailer  = $$1.5 \times 1.3 \times 1.1 \times x$$
$$1.5 \times 1.3 \times 1.1 \times x = 4290$$
=> x = 2000

Let the time taken by the outlet pipe to empty = x hours
Then, $$\frac{1}{8} - \frac{1}{x} = \frac{1}{10}$$
=> $$x = 40$$
Hence time taken by the outlet pipe to make the tank half-full = 40/2 = 20 hour

Let the number of dozens of candies he bought of each variety be x
Hence total cost = 12x + 15x = 27x
Total selling price = 16.50*2x = 33x
Profit = 33x - 27x = 6x
Given 6x = 150 => x = 25
Hence he bought 50 dozens of candies in total

Let initial population and production be x,y and final population be z
Final production = 1.4y, final percapita = 1.27 times initial percapita
=> $$\frac{1.4y}{z} $$ = $$ 1.27 \times \frac{y}{x}$$
=> $$\frac{z}{x} = \frac{1.4}{1.27} \approx 1.10$$
Hence the percentage increase in population = 10%

a : b = 3:4 and b : c = 2:1 => a:b:c = 3:4:2
=> a = 3x, b = 4x , c = 2x
=> a + b + c = 9x
=> a + b + c is a multiple of 9.
From the given options only, option C is a multiple of 9

Let the distance traveled by the car be x KMs
Distance traveled by the bike = 168 KMs
Speed of car is double the speed of bike
=> $$\frac{x}{3:40 - 2:00}$$ = $$2 \times \frac{168}{3:40 - 1:00}$$
=> $$\frac{x}{100}$$ = $$2 \times \frac{168}{160}$$
=> x = 210
Hence the distance between A and B is x + 168 = 378 KMs

Let the time take by kamal to complete the task be x days.
Hence we have $$\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$$
=> x = 40 days.
Ratio of the work done by them = $$\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$$ = 4 : 5 : 1
Hence the wage earned by Kamal = 1/10 * 1000 = 100

The proportion of water in the first mixture is $$\frac{1}{3}$$
The proportion of Liquid A in the first mixture is $$\frac{2}{3}$$

The proportion of water in the second mixture is $$\frac{1}{4}$$
The proportion of Liquid B in the second mixture is $$\frac{3}{4}$$

The proportion of water in the third mixture is $$\frac{1}{5}$$
The proportion of Liquid C in the third mixture is $$\frac{4}{5}$$

As they are mixed in the ratio 4:3:2, the final amount of water is $$4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$$
The final amount of Liquid A in the mixture is $$4\times\frac{2}{3} = \frac{8}{3}$$
The final amount of Liquid B in the mixture is $$3\times\frac{3}{4} = \frac{9}{4}$$
The final amount of Liquid C in the mixture is $$2\times\frac{4}{5} = \frac{8}{5}$$

Hence, the ratio of Water : A : B : C in the final mixture is $$\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$$

From the given choices, only option C  is correct.