CMAT 2018 Slot 2 Question Paper

Average Salary Expenditure (in Lakh Rupees) per annum = Total salary expenditure in all these years$$\div$$Total number of years

= $$\frac{576+682+648+672+740}{5}$$

= 663.6

Option A:

In 2011, number of Refrigerators sold =60

        number of Cell phones sold = 336

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{60}{336}\times 100 = 17.85$$   

Option B:

In 2012, number of Refrigerators sold = 79

             number of Cell phones sold = 404

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{79}{404}\times 100$$ = 19.5

Option C:

In 2013,number of Refrigerators sold = 93

             number of Cell phones sold = 411

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{93}{411}\times 100$$ = 22.6

Option D:

In 2014, number of Refrigerators sold = 112

              number of Cell phones sold = 442

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{112}{442}\times 100$$ = 25.33

Hence Option D is the correct answer.

SInce PQ is the diameter, the angle subtended by it at R is 90 deg. i.e., $$\angle$$ PRQ = 90 deg.

Let $$\angle$$ RPQ = $$\theta$$, then $$\angle$$RQP = 90 - $$\theta$$

As the angles subtended by a chord in same segment are equal, $$\angle$$RPQ = $$\angle$$RSQ = $$\theta$$

In triangle RSQ, $$\angle$$QRS + $$\angle$$RSQ + $$\angle$$RQS = 180 

$$\Rightarrow$$ $$\angle$$QRS + $$\theta$$ + 35 + 90 - $$\theta$$ = 180

$$\Rightarrow$$ $$\angle$$QRS = 180 - 125 = 55 deg.

Hence $$\angle$$QRS = 55 deg.

Given  x = $$\sqrt[6]{5}$$ and y = $$\sqrt[5]{4}$$

which can also be written as x = $$5^{\frac{5}{30}}$$ and y = $$4^{\frac{6}{30}}$$    

which can be further written as x = $$\sqrt[30]{5^5}$$ and y = $$\sqrt[30]{4^6}$$

As we know $$4^6 > 5^5$$

$$\Rightarrow \sqrt[30]{4^6}  > \sqrt[30]{5^5} $$

$$\Rightarrow$$ y > x                     

Given, If a and b are positive real numbers then a * b denotes √ab

Consider 4 * 16 = $$\sqrt{4\times 16} = \sqrt64 = 8$$

then 8 * 8 = $$\sqrt{8\times 8} = \sqrt64 = 8$$

Hence the value of  8 * (4 * 16) = 8

Given the proportion of ages of three men are 3:5:7.

Let their ages be 3k, 5k, 7k ,where k is any constant.

Given average of ages of three men = 50 

$$\Rightarrow \frac{3k+5k+7k}{3} = 50$$

$$\Rightarrow \frac{15k}{3} = 50$$

$$\Rightarrow 5k = 50$$

$$\Rightarrow k = 10$$

Therefore the ages of three men are 30, 50, and 70 years.

The age of the youngest men is 30 years.

If 64 mangoes are sold at Rs.2000, each mango will be sold at Rs. $$\frac{2000}{64}$$

Hence Selling price (S.P) of each mango = Rs. 31.25

Given loss percentage of vendor at this S.P = 40%

Loss percentage = $$\frac{C.P - S.P}{C.P}\times 100$$ 

$$\Rightarrow \frac{40}{100} = \frac{C.P - S.P}{C.P}$$

$$\Rightarrow S.P = 0.6\times C.P$$

$$\Rightarrow C.P = \frac{31.25}{0.6} = 52$$

Therefore Cost Price of 1 mango (C.P) = Rs. 52

Let us calculate the S.P of each mango in order to get a 20% gain.

Gain percentage = $$\frac{S.P - C.P}{C.P}\times 100$$

$$\Rightarrow \frac{20}{100} = \frac{S.P - C.P}{C.P}$$

$$\Rightarrow S.P = 1.2\times C.P$$

$$\Rightarrow S.P = 62.5$$

So, to get a gain of 20% we need to sell each mango at Rs. 62.5 

Let say we sold 'x' number of mangoes.

Selling price of these 'x' number of mangoes (S.P) = Rs. 62.5x

But given that this S.P = Rs. 1000

$$\Rightarrow 62.5x = 1000$$

$$\Rightarrow x = \frac{1000}{62.5} = 16$$.

Therefore a total of 16 mangoes are to be sold for Rs. 1000 to get a gain of 20%.

Area of triangular metal plate with base(b) = 88 cm and altitude(h) = 64 cm is given as A = $$\frac{1}{2} b\times h = \frac{1}{2} 88\times 64 = 2816 cm^2$$

Given this area is to be reduced to one-fourth by making a hole in the shape of circle

$$\Rightarrow$$ Reduction in the area of the triangle = Area of the circular hole

$$\Rightarrow\frac{3}{4}\times A=\pi r^2$$

$$\Rightarrow \frac{\frac{3}{4}\times 2816}{\pi} = r^2$$

$$\Rightarrow r = \sqrt672 = 4\sqrt42 $$

So, the radius of the circular hole = $$4\sqrt42$$ cm.

Let us consider $$\frac{2 + \sqrt3}{2 - \sqrt3}$$ 

Rationalising the denominator by multiplying and diving with 2+$$\sqrt3$$ we get,

$$\frac{(2 + \sqrt3)\times (2 + \sqrt3) }{(2 - \sqrt3)\times (2 + \sqrt3) } = \frac {(2 + \sqrt3)^2}{4 - 3} = (2 + \sqrt3)^2$$

Now,

 $$\sqrt{\frac{2 + \sqrt3}{2 - \sqrt3}} = \sqrt{(2 + \sqrt3)^2} = 2 + \sqrt3 = 2 + 1.732 = 3.732$$

Initial amount of Kerosene(I) = 5% of 5 kg mixture of petrol and kerosene = $$\frac{5}{100}\times 5000 = 250$$ grams 

Let say 'x' gm of Kerosene is added to the mixture.

The final amount of Kerosene in the mixture after the addition is given as 10%.

$$\Rightarrow \frac{250 + x}{5000 + x} = \frac{10}{100}$$

$$\Rightarrow 2500 + 10x = 5000 + x$$

$$\Rightarrow 9x = 2500$$

$$\Rightarrow x = 275$$ gm approximately.

Therefore additionally 275 gm of kerosene is to be added to the mixture to make it 10%.