CMAT 2021 Slot 1 Question Paper

We have :
$$-\frac{4}{5}\ ,\ -\frac{5}{12}\ ,\ -\frac{7}{18}\ ,\ -\frac{2}{3}$$
Now LCM is 180
so we get :
$$-\frac{144}{180}\ ,\ -\frac{75}{180}\ ,\ -\frac{70}{180}\ ,\ -\frac{120}{180}$$
Now arranging in ascending order
we get :
$$-\frac{144}{180}\ <\ -\frac{120}{180}<-\frac{75}{180}<-\frac{70}{180}$$
so order is A<D<B<C

Let the sample size of the above events be 6 units.

Since $$P(A)=\frac{2}{3}, P(B)=\frac{1}{2}$$ and $$P(A\cap B)=\frac{1}{3}$$,

Then event A will have 4 units($$\frac{2}{3}\times\ 6$$), event B will have 3 units and event $$A\cap B$$ will have 2 units.

This can be expressed in terms of the Venn diagram as follows:

Now, it can be observed that $$A\cap B^c\ =2$$ units.

Thus, $$P\left(A\cap B^c\right)\ =\frac{2}{6}=\frac{1}{3}$$.

Similarly, $$A\ U\ B^c\ =5$$ units($$A=4$$ and $$B^c-A=1$$ units)

So, $$P\left(A\ U\ B^c\right)\ =\frac{5}{6}$$

Since only option A has $$P\left(A\cap B^c\right)=\frac{1}{3}$$ and $$P\left(A\ U\ B^c\right)\ =\frac{5}{6}$$,

The correct option is A.

Production of unit B in April = 180
Production of unit B in June = 160
Now therefore percentage decrease = $$\frac{\left(180-160\right)}{180}\times\ 100\ \approx\ \ 11\%$$

The difference between C.I and S.I for a period of 2 years is given by : P(R/100)^2 
So we get :
P(R/100)^2 = 32
5000(R/100)^2 = 32
(R/100)^2 = 32/5000 = 16/2500
We get 
R/100 = 4/50 
R = 8%

We have :
30, 90, 182, 306, 462,P.
Now 90-30 =60  (1)
182-90 = 92 (2)
306-182 = 124   (3)
462 -306 = 156  
If you observe , the first difference of the terms form an arithmetic progression
So we can say the next term will be
462 +156+32 = 462 +188 = 650
So P is 650
So statement 1 is true .
Taking statement 2 :
we get 9^8
Now number of digits = [log9^8 +1]
we get [8log9+1]= 8
so number of digits in 9^8 is 8

On counting the total triangles comes out to be 13

Has an outcome can be related to skills acquired if person is socialized so I follows and
Has an outcome does not mean change in behaviour so II do not follow 

we have 8, 13, 10, 15, 12 
Here the pattern is 8+5 =13 , 13-3 =10 , 10+5 =15 ;15-3 =12 so next term will be 12+5 =17

30, 56, 90, _____ , 182, 240.
Here the first differences forms an A.P with a common difference of 8
so we get next term as 90+42 = 132.

As per C Swati will be the highest scorer 
so C alone is sufficient to determine the highest scorer .