CMAT 2020

Number of ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women

= $$C^{10}_6\times C^8_5$$

= $$C^{10}_4\times C^8_3$$

= $$\frac{10\times9\times8\times7}{1\times2\times3\times4}\times\frac{8\times7\times6}{1\times2\times3}$$

= $$(10\times3\times7)\times(8\times7)$$

= $$210\times56=11760$$

=> Ans - (D)

Height of cylinder = 8 units and radius = 8 units

=> Farthest distance = hypotenuse = $$\sqrt{(8)^2+(16)^2}$$

= $$8\sqrt{5}$$ units

=> Ans - (A)

Prime factorization of :

$$2400=(2)^5\times(3)\times(5)^2$$

Now, to make it a perfect cube, we need to make the powers multiple of 3, and thus we need to divide it by = $$(2)^2\times3\times(5)^2=300$$

=> Ans - (D)

Let total number of people in the town = $$100x$$

People who have both brown hair and eyes = $$b=12x$$

=> People who have brown hair = $$a+b=40x$$

=> $$a=28x$$

Similarly, $$c=18x$$

$$\therefore$$ People who have neither brown hair nor brown eyes = $$100x-(12x+18x+28x)=42\%$$

=> Ans - (D)

If a sum of money becomes $$x$$ times in $$a$$ years, it will become $$y$$ times in $$b$$ years

=> $$(x)^{\frac{1}{a}}=(y)^{\frac{1}{b}}$$

Similarly, $$(8)^{\frac{1}{3}}=(16)^{\frac{1}{t}}$$

=> $$(2)^{\frac{3}{3}}=(2)^{\frac{4}{t}}$$

=> $$1=\frac{4}{t}$$

=> $$t=4$$ years

=> Ans - (A)

Let the two numbers be $$9x$$ and $$9y$$, where $$x$$ and $$y$$ are co prime numbers.

=> $$9x+9y=135$$

=> $$x+y=\frac{135}{9}=15$$

Now, pairs of natural numbers having sum 15 and co prime are = (1,14), (2,13), (4,11), (7,8)

Thus, 4 such pairs can be formed.

=> Ans - (C)

According to BODMAS,

first Multiply the given numbers

0.3 x0.3= 0.09

0.03 x0.03=0.0009

0.6 x 0.03=0.018

simplifying this according to the given question,

0.09-0.0009-0.018=0.0729

now $$\ \frac{\ 0.0729}{0.54}$$

=> 0.135

Therefore answer is option b 

Let the two numbers be $$x$$ and $$4x$$

=> $$x\times4x=84\times21$$

=> $$x^2=(21)^2$$

=> $$x=21$$

$$\therefore$$ Larger number = $$4\times21=84$$

=> Ans - (D)

Let $$x=2$$ and $$y=3$$

To find : $$\frac{3x+2y}{2x+5y}$$

= $$\frac{3(2)+2(3)}{2(2)+5(3)}$$

= $$\frac{12}{19}$$

=> Ans - (D)

Let the efficiency of a man, woman and child be $$3x,2x,x$$ espectively.

Thus, 1 day wage = $$(3x\times20)+(2x\times30)+(x\times36)=78$$

=> $$156x=78$$

=> $$x=\frac{1}{2}$$

$$\therefore$$ Wages of 15 men, 21 women and 30 children for 18 weeks = $$[(15\times3x)+(21\times2x)+(30\times x)]\times18\times7$$

= $$117\times\frac{1}{2}\times126$$

= $$117\times63=Rs.$$ $$7,371$$

=> Ans - (A)