CMAT 2021 Slot-1

We have :
$$f\left(x\right)\ =\ x^4-px^2+q$$
Now $$x^2-3x+2\ =\ \left(x-2\right)\left(x-1\right)\ $$ is a factor of f(x) 
so we can say f(2) = f(1) =0 
substituting we get
1-p+q =0
p-q =1  
and 16-4p+q =0
4p-q =16
solving we get
p=5 and q =4

Now 
When A covers 100 meters 
B covers 90 meters 
so we can say Va :Vb = 100:90 = 10:9   (1) 
Similarly Vb:Vc = 100:95 = 20:19             (2)
Now multiplying (1) and (2)
we get Va : Vc = 200:171
Now therefore when A covers 100 meters
C covers 85.5 meters
And so A beats C by 14.5 meters

It is given that Mohan bought a trouser at 10% discount and sold it to Sohan at a loss of 10%
Now Sohan paid Rs 729
So S.P for Mohan = 729 
This is sold at 10% loss 
so cost price of Mohan = 729/0.9 =810
Now Mohan bought for 810 at 10% discount
so original price (1-0.1) =810
We get undiscounted  price = 900

Graph A and D represent the function of x, because a vertical line is drawn in (a) meets the graph at only one point i.e., for one x, in the domain there exist only one f(x) in the codomain.

We know
A+B+C+D+E+F =100
So we get F = 15%
Now let income be I
so we get I(0.15) = 45000
We get
I = 30,000

The distance covered by both Neha and Monica is same .
Now Neha takes 40 minutes to cover the distance which Monica covers in 30 minutes
So we can say their speed ratio = 3:4
Let speed of Neha be 3v and speed of Monica be 4v
Now in 5 minutes Neha would cover 15v of distance
Now Neha is ahead by 15v when Monica starts
so relative distance = 15v
And relative speed = 4v-3v =v
Therefore time to overtake = 15v/v = 15minutes

Now we have 5 men and 3 women
committee of 4 to be made such that it has at least 1 woman
Case 1 :
1W and 3M
We get $$^5C_3\times\ ^3C_{1\ }=\ 30$$
Case 2 : 2W 2M
$$^5C_2\times\ ^3C_{2\ }=\ 30$$
Case 3 : 3W 1 M
$$^5C_1\times\ ^3C_{3\ }=\ 5$$
So total cases = 65.
so 1 is true
Now
ARRANGE
If vowels to be at even position :
we get cases as $$\frac{3!}{2!}\times\ \frac{4!}{2!}$$ = 36
So statement 2 is true

Let the ratio be K
so we can say A=BK
and C = Dk
Now $$\frac{A^2+B^2}{C^2+D^2}=\ \frac{A^2\left(1+k^2\right)}{C^2\left(1+k^2\right)}=\ \frac{A^2}{C^2}$$
Now checking the options
A,B,C gets eliminated
And AB/CD = $$\ \frac{A^2}{C^2}$$
so D is the correct

We have dimensions : (24*18)
LCM (24,18) =72

HCF ( 24, 18) = 6

Now Area of floor: 432

In order to minimise the number of tiles considering a tile with a dimension equivalent to the HCF,  we have the dimension of the tile to 6*6 = 36 units.

A total of $$\frac{432}{36}$$  = 12 tiles are required.

Minimum number of tiles that pave the entire floor :(LCM)^2/Area = 12

We have :
$$6x^2-5x-4\ =0\ $$
=$$6x^2-8x+3x-4\ =0\ $$
(2x+1)(3x-4)=0
so solution set is {-1/2 ,4/3}

$$9x^2+6x+1\ =0$$
D = 36-36 =0
So we can say roots are real and equal

$$4x^2-3x+k$$=0
let roots be p and 3p
we get 4p =3/4
p =3/16
Now k/4 = 3p^2
we get k = 12p^2
we get k = 27/64
So A and B are true and C is false