CMAT 2019 Question Paper

In a triangle, the angular bisector divides the side which it intersects in the ratio of other two sides.

So, In the triangle ABC, the angular bisector AD divides the side BC in the ration of AB and AC

$$\Rightarrow$$ BD : DC = AB : AC

Given BC = 10 cm and DC = 6 cm 

$$\Rightarrow$$ BD = BC - DC = 10 - 6 = 4 cm

Then $$AB =\frac{BD}{DC}\times AC = \frac{4}{6}\times 4.2 = 2\times 1.4 = 2.8$$ cm 

Based on the given constraints, there will be two cases in which task 1 is either assigned to one among person 3 or 4 or to one among person 5 or 6

Case 1:

Assigning Task 1 to either Person 3 or 4.

No. of ways of assigning task 1 = 2 ways

No. of ways of assigning task 2 = 1 way 

Remaining 4 tasks can be assigned among the remaining 4 persons in 4! ways.

Therefore, Total number of ways = 2 x 1 x 4! = 48

Case 2:

Assigning Task 2 to either Person 5 or 6.

No. of ways of assigning task 1 = 2 ways

No. of ways of assigning task 2 = 2 ways 

Remaining 4 tasks can be assigned among the remaining 4 persons in 4! ways.

Therefore, Total number of ways = 2 x 2 x 4! = 96

Hence, total number of possible arrangements= 48 + 96 = 144

Let us consider the fucntion f(x) = $$a\cos x + b\sin x + c$$ 

The range of f(x) if given as $$c - \sqrt{a^2 + b^2} \leq f(x) \leq c + \sqrt{a^2 + b^2}$$

Then the maximum value of  3 $$cosx+4 sinx+8$$  is = 8 + $$\sqrt{3^2 + 4^2}$$ 

= 8 + 5 = 13

Let the number be 'x'

one-fourth of the number = $$\frac{1}{4}\times x = \frac{x}{4}$$

one-third of one-fourth of the number = $$\frac{1}{3}\times \frac{x}{4} = \frac{x}{12}$$

Given, one-third of one-fourth of the number is = 15

$$\Rightarrow \frac{x}{12} = 15$$

$$\Rightarrow x = 180$$

Hence the number is 180.

Then three-tenth of the number = $$\frac{3}{10}\times x = \frac{3}{10}\times 180 = 54$$

Let the number of men be 'm' and number of women be 'w' in the class.

Given Average score of men = 70

$$\Rightarrow$$ Total score of men$$\div$$number of men = 70

$$\Rightarrow$$ Total score of men = 70$$\times$$m

Also, Average score of women = 83

$$\Rightarrow$$ Total score of women$$\div$$number of women = 83

$$\Rightarrow$$ Total score of women = 83$$\times$$w

Given the class average = 76

$$\Rightarrow$$ (Total score of men + Total score of women)$$\div$$(number of men + number of women) = 76

$$\Rightarrow \frac{70\times m + 83\times w}{m + w} = 76 $$

Let the ratio of number of men to women $$\frac{m}{w} = k$$

$$\Rightarrow$$ 70k + 83 = 76k + 76

$$\Rightarrow$$ 6k = 7 

$$\Rightarrow$$ k = 7 : 6.

Hence the ratio of men to women in the class is 7 : 6.

Probability of A or B = Probability of A + Probability of B - Probability of A and B.

$$\Rightarrow$$ Probability of drawing king or heart = Probability of drawing king + Probability of drawing heart - Probability of drawing kings which are hearts from the deck of cards.

A deck of cards contain 52 cards, of which there are 4 kings and 13 hearts and 1 king which is heart also. 

Probability of drawing a king = $$\frac{4}{52}$$

Probability of drawing a heart = $$\frac{13}{52}$$

Probability of drawing a king which are heart also = $$\frac{1}{52}$$

$$\Rightarrow$$ Probability of drawing king or heart = $$\frac{4}{52} + \frac{13}{52} - \frac{1}{52}$$

= $$\frac{16}{52}$$

= $$\frac{4}{13}$$

Let the three consecutive odd numbers be 'x-2', 'x' and 'x+2'

Given the sum of three consecutive odd numbers is 20 more than the first number

$$\Rightarrow$$ x-2 + x + x+2 = 20 + x-2

$$\Rightarrow$$ 2x = 18

$$\Rightarrow$$ x = 9

Therefore the three consecutive odd numbers are 7,9 and 11.

The middle number is 9.

When a pair of dice are thrown simultaneously the maximum sum possible is 12.

The possible cases of getting a sum of 9 are:

                    dice 1     dice 2

                    four         five.....................(1)

                    five          four.....................(2)

                     six          three....................(3)

                   three         six.......................(4)  

So a total of 4 cases are possible

The possible cases of getting a sum of 10 are:

                    dice 1     dice 2

                     four        six......................(1)

                      six         four.....................(2)

                     five        five.....................(3)

So a total of 3 cases are possible

The possible cases of getting a sum of 11 are:

                   dice 1       dice 2

                    five           six...................(1)

                     six           five..................(2)

So a total of 2 cases are possible

The possible cases of getting a sum of 12 are:

                  dice 1      dice 2

                   six           six.....................(1)

So a total of 1 case is possible.

Total number ways of getting a sum of 9 or more = 4 + 3 + 2 + 1 = 10 ways

Total outcomes when two dice are rolled simultaneously = 6 * 6 = 36 ways

Probability of getting a sum of 9 or more = Expected number of outcomes/ Total outcomes

= $$\frac{10}{36}$$

= $$\frac{5}{18}$$

Given the length, breadth and height of a rectangular cuboid are in the ratio 1: 2 : 3.

Let the length, breadth and height be k, 2k, and 3k units where 'k' is a constant.

Volume of the cuboid(V') = $$l\times b\times h$$ units = $$k\times 2k\times 3k$$  = $$6k^3$$ units

Given the length, breadth and height are increased by 100% each .

Then the new length, breadth and height will be doubled.

So new length, breadth and height of the rectangular cuboid are 2k, 4k, and 6k units.

New Volume of the cuboid (V') = $$2k\times 4k\times 6k = 48k^3$$ units.

Increase in the volume of the cuboid = V' - V = $$48k^3 - 6k^3 = 42k^3 = 7 [6k^3] = 7V$$

Therefore the volume has increased 7 times the initial volume.

let the length of the halls be 'a' and 'a+1' m respectively and height of the halls be 'h' m.

Area of 4 walls of first square hall (A1) = $$(4\times a\times h) m^2$$

Given, cost of painting is Rs. 5 per sq.m

$$\Rightarrow$$  cost of painting 4 walls of the first square hall = A1 * 5 = $$4\times a\times h\times 5 = Rs. 20*a*h$$.

Given that this cost of painting = Rs. 6000

$$\Rightarrow$$ 20*a*h = 6000...........(1)

Similarly,

Area of 4 walls of second square hall (A2) = $$(4\times a+1\times h) m^2$$

$$\Rightarrow$$  cost of painting 4 walls of the second square hall = A2 * 5 = $$4\times a+1\times h\times 5 = Rs. 20*(a+1)*h$$.

Given that this cost of painting = Rs. 6100

$$\Rightarrow$$ 20*(a+1)*h = 6100............(2)

Dividing (2) with (1), we get

$$\frac{a+1}{a} = \frac{6100}{6000}$$

$$\Rightarrow 1 + \frac{1}{a} = \frac{61}{60}$$

$$\Rightarrow a = 60$$

Substituting this value in equation (1), we get

20*60*h = 6000

$$\Rightarrow$$ h = 5 m.

So, the height of the two walls is 5 m.