CAT 2019 Slot 2 Question Paper

It is given that AD and BE are medians which are perpendicular to each other.

The lengths of AD and BE are 12cm and 9cm respectively.

It is known that the centroid G divides the median in the ratio of 2:1

Area of $$\triangle$$ ABC = 2* Area of the triangle ABD

Area of $$\triangle\ $$ABD = Area of $$\triangle\ $$ AGB + Area of $$\triangle\ $$ BGD

Since $$\angle\ AGB\ =\ \angle\ BGD\ =\ 90$$ (Given)

Area of $$\triangle\ $$ AGB = $$\ \frac{\ 1}{2}\times\ 8\times\ 6$$ = 24

Area of $$\triangle\ $$ BGD = $$\ \frac{\ 1}{2}\times\ 6\times\ 4$$ = 12

 Area of $$\triangle\ $$ABD = 24+12=36

Area of $$\triangle\ ABC\ =\ 2\times\ 36=72$$

A: 15, 19, 23, 27, . . . . , 415

B: 14, 19, 24, 29, . . . , 464

Here the first common term = 19

Common difference = LCM of 5, 4=20

19+(n-1)20 $$\le\ $$ 415

(n-1)20 $$\le\ $$ 396

(n-1) $$\le\ $$ 19.8

n=20

$$\left(ax+by\right)^2=65^2$$

$$a^2x^2\ +\ b^2y^2+\ 2abxy\ =\ 65^2$$
$$k = ay - bx$$

$$k^2\ =\ a^2y^2+b^2x^2-2abxy$$

$$(a^2 + b^2)(x^2 + y^2 )= 25* 169$$

$$a^2x^2+a^2y^2+b^2x^2+b^2y^2=\ 25\times\ 169$$

$$k^2=\ 65^2\ -\ \left(25\times\ 169\right)$$

k = 0

D is the correct answer.

Let the cost of each bicycle= 100b

CP of 10 bicycles = 1000b

It is given that he sold six of these at a profit of 25% and the remaining four at a loss of 25%

SP of 10 bicycles = 125b*6+75b*4 

=1050b

Profit = 1050b-1000b =50b

50b=2000

CP = 100b = 4000

Let the score of D = 100d

The score of C = 20% less than that of D = 80d

The score of B = 25% more than C = 100d

The score of A = 10% less than B =90d

90d=72

100d= 72*100/90

= 80

Let the salaries of Ramesh, Ganesh and Rajesh in 2010 be 6x, 5x, 7x respectively

Let the salaries of Ramesh, Ganesh and Rajesh in 2015 be 3y, 4y, 3y respectively

It is given that Ramesh’s salary increased by 25% during 2010-2015,3y = 1.25*6x

y=2.5x

Percentage increase in Rajesh's salary = 7.5-7/7=0.07

=7%

Each interior angle in an n-sided polygon = $$\frac{\left(n-2\right)180\ }{n}$$

It is given that each interior angle of B is $$\frac{3}{2}$$ times each interior angle of A and b = 2a

$$\frac{\left(b-2\right)180\ }{b}$$ = $$\ \frac{\ 3}{2}\times\ $$ $$\frac{\left(a-2\right)180\ }{a}$$

$$2\times\ \left(b-2\right)\times\ a\ =\ 3\times\ \left(a-2\right)\times\ b$$

2(ab-2a) = 3(ab-2b)

ab-6b+4a=0

a*2a-12a+4a=0

$$2a^2-8a=0$$

a(2a-8) = 0

a cannot be zero so 2a=8

a=4, b = 4*2=8

a+b = 12

Each interior angle of a regular polygon with 12 sides = $$\ \frac{\ \left(12-2\right)\times\ 180}{12}$$

=150

Given, f(mn) = f(m)f(n)
when m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1

when m=1,  n= 2, f(2) = f(1)*f(2) ==> f(1) = 1

when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $$[f(2)]^2$$

Similarly f(8) = f(4)*f(2) =$$[f(2)]^3$$

f(24) = 54

$$[f(2)]^3$$ * $$[f(3)]$$ = $$3^3*2$$

On comparing LHS and RHS, we get 

f(2) = 3 and f(3) = 2

Now we have to find the value of f(18)

f(18) = $$[f(2)]$$ * $$[f(3)]^2$$

= 3*4=12

Let the total work be LCM of 20, 40 = 40 units 

Efficiency of Anil and Sunil is 2 units and 1 unit per day respectively.

Anil works alone for 3 days, so Anil must have completed 6 units.

Bimal completes 10% of the work while working along with Anil and Sunil.

Bimal must have completed 4 units.

The remaining 30 units of work is done by Anil and Sunil

Number of days taken by them 30/3=10

The total work is completed in 3+10=13 days

Let the scores of Rama, Anjali and Mohan be r, a, m.

It is given that Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali

r=$$\ \frac{\ m+a}{12}$$ ———-(1)

The scores of Rama, Anjali and Mohan after review = r+6, a+6, m+6

a+6:m+6:r+6 = 11:10:3

Let a+6 = 11x => a= 11x-6

      m+6=10x => m=10x-6

      r+ 6 =3x => r = 3x-6 
   Substituting these values in equation (1), we get   

3x-6=$$\ \frac{\ 21x-12}{12}$$

12(3x-6) = 21x-12

x=4

Anjali's score exceeds Rama's score by (a-r)=8x=32