CMAT 2018 Slot 1 Question Paper

Given monthly savings of Amit and Bharat are in the ration 37:18. Let the savings of Amit and Bharat be 37k and 18k, where 'k' is a constant.

$$\Rightarrow$$ Total  monthly  savings  of  Amit  and  Bharat = 55k.

Given Total annual savings of Amit and Bharat = 1,32,000.

$$\Rightarrow$$ Total  monthly  savings = 11,000.

$$\Rightarrow$$ 55k = 11000

$$\Rightarrow$$ k = 200.

Hence monthly savings of Amit and Bharat are 7400 and 3600 respectively.

From the given data,

Let the monthly incomes of Amit and Bharat be 5x and 4x respectively, where 'x' is a constant.

Similalrly, let the monthly expenditures of Amit and Bharat be 19y and 21y respectively, where 'y' is a constant.

Savings =  Income - Expenditure

$$\Rightarrow$$ 5x - 19y = 7400..........................(1)

and  4x - 21y = 3600...............................(2)

Solving both the equations we get x = 3000 and y = 400.

Therefore, the monthly income of Amit is 5x i.e., 15000.

In $$\triangle$$AOB, AO = OB (Equal radii)

∴ $$\triangle$$AOB is isosceles.

∴ ∠ABO = ∠BAO = (180°-114°)/2 = 33°

Hence, option D.

As can be seen from the above venn diagram distribution

Number of households reading only one newspaper = 22 + 26 + 16 = 64.

Hence, option C.

60 minutes = 360 degrees or 2$$\pi$$ radians.

$$\Rightarrow$$ 1 minute = 60  degrees.

$$\Rightarrow$$ 30 minutes = 180  degrees  or  $$\pi$$ radians.

Therefore area (A) swept by minute hand of length 42 mm in 30 minutes =  area of semi-circle of radius (r = 42 mm) .

$$\Rightarrow A = \pi r^2\div 2 = 3.14\times 42^2\div 2 = 2769.5 mm^2$$.

Let say f(x) = $$x^3 + 2x^2 + bx + 68$$ and given (x+4) is a factor of f(x).

$$\Rightarrow f(x) = (x+4)\times k$$.................(1)

where 'k' is the quotient when f(x) is divided by (x+4).

Substituting 'x= -4' in the equation (1), we get

f(-4) = 0

$$\Rightarrow -4^3 + 2 (-4)^2 + b (-4) + 68 = 0$$

$$\Rightarrow -64 + 32 - 4b + 68 = 0$$

$$\Rightarrow 4b = 36$$

$$\Rightarrow b = 9$$.

$$Average  speed = Total  distance  covered\div Total  time  taken$$.

Total distance covered = 12 + 36 + 32 = 80 km.

Total time taken = $$12\div 6 + 36\div 9 + 32\div 4 = 2 + 4 + 8 = 14 hrs$$.

$$\Rightarrow Average  speed = 80\div 14 = 5.71 km/hr$$.

Let the capacity of the cistern be 600 units.

From the given data, the efficiencies of pipes A and B are 5 units/ min and 4 units/min respectively.

Let the efficiency of outlet pipe C be 'k' units/min.

Given the time taken to fill the cistern when all the three pipes are open = 100 minutes

$$\Rightarrow Efficiency  of  pipes\times time  taken = Capacity  of  cistern$$.

$$\Rightarrow (5+4-k)\times 100 = 600$$

$$\Rightarrow 9-k = 6$$

$$\Rightarrow k = 3$$

Therefore the time taken(t)  by pipe C to empty the cistern = $$Capacity  of  the  cistern\div efficiency  of  pipe  C$$

$$\Rightarrow t = 600\div 3 = 200 minutes = 3 hrs 20 min$$.

As per the given conditions, weeks 1 and 3 will have similar pay structure and weeks 2 and 4 will have similar pay structure.

For weeks 1 and 3

wage/hr for working = 60

wage/hr during rest = 30

Total payment per day = (60*6 )+ (30*3) = 450

Total payment for weeks 1 and 3 = 450 * 12 = 5400

For weeks 2 and 4

wage/hr for working = 20

wage/hr during rest = 10

Total payment per day = (20*3 )+ (10*6) = 120

Total payment for weeks 2 and 4 = 120 * 12 = 1440

$$\therefore $$ Total monthly salary = 5400 + 1440 = 6840

Hence, option A.

Probability = $$Expected  number  of  outcomes\div Total  number  of  outcomes$$.

Total number of outcomes = Total number of ways to draw randomly 3 balls out of 12 balls = 12C3 ways.

Expected number of outcomes = Number of ways to draw 2 yellow balls out of 8 and 1 black ball out of 4 = $$8C2\times 4C1$$

$$ Hence Probability = 8C2\times 4C1\div 12C3 = 28\times 4\div 220 = \frac{28}{55}$$

$$cotA = \frac{1}{tanA}$$

let $$tanA = x, then  cotA = \frac{1}{x}$$

Given, x + $$\frac{1}{x} = \sqrt 5$$.............................(1)

Cubing on both sides, we get

$$(x + \frac{1}{x})^3 = \sqrt 5^3$$

$$\Rightarrow x^3 + \frac{1}{x}^3 + 3\times x\times \frac{1}{x}\times (x + \frac{1}{x}) = 5\sqrt 5$$

$$\Rightarrow x^3 + \frac{1}{x}^3 + 3\times \sqrt 5 = 5\sqrt 5$$

$$\Rightarrow x^3 + \frac{1}{x}^3 = 2\sqrt 5$$.

$$\Rightarrow tanA^3 + cotA^3 = 2\sqrt 5$$.