CAT 2021 Slot 3 Question Paper

For a question created by externals(SME), the minimum number of reviews required is two and the maximum number of reviews is three to confirm the acceptance or the rejection.

For a question created by anyone among Amal, Bimal, Komal a question can be reviewed after one review or two reviews.

The information provided states :

Q02, Q06, Q09, Q11, and Q12 were rejected. Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13 were accepted.

For the questions :

Amal - Q02, Q03, Q04, Q06, Q08, Q10, Q11, and Q13.
Bimal - Q02, Q04, Q06, Q07, Q08, Q09, Q12, and Q13.

Komal - Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, and Q12.

Q01 is reviewed by only Komal and is accepted hence Bimal must have prepared the question as it was accepted in a single review. (Bimal)

Q02-is reviewed by Amal, Bimal, and Komal and was rejected this has a possibility of being prepared by an external person and faced rejection by one among Amal and Bimal and Komal. (SME).

Q03-Was accepted and reviewed by Amal, Komal and hence must have been prepared by Bimal(Bimal).

Q04- Accepted and reviewed by the three of them and hence must have been prepared by an external expert. (SME).

Q05-Only reviewed by Komal and is accepted and hence must be prepared by Bimal. (Bimal)

Q06-Rejected and reviewed by Amal and Bimal. Hence could have been prepared by Komal or External and rejected by Bimal and Amal. (Komal/SME).

Q07-Accepted and reviewed by Bimal and Komal. Hence must have been prepared by Amal. (Amal).

Q08- Reviewed by Amal, Bimal, and Komal and is accepted. Hence must have been prepared by SME. (SME)

Q09- Reviewed by Bimal and Komal and is rejected. It must have been prepared by Amal and was rejected by both of them. (Amal).

Q10- Was reviewed by Amal and was accepted. Must have been prepared by Komal. (Komal)

Q11- Reviewed by Amal and Komal and was rejected. Must have been prepared by Bimal and rejected by both Amal and Komal. (Bimal)

Q12- Reviewed by Bimal and Komal and was rejected. Must have been prepared by Amal and was rejected by both Bimal and Komal. (Amal).

Q13- Reviewed by Amal and Bimal and was accepted. The question could have been prepared by Komal or SME. (Komal/SME).

Q02, Q04, Q08 were prepared SME

For a question created by externals(SME), the minimum number of reviews required is two and the maximum number of reviews is three to confirm the acceptance or the rejection.

For a question created by anyone among Amal, Bimal, Komal a question can be reviewed after one review or two reviews.

The information provided states :

Q02, Q06, Q09, Q11, and Q12 were rejected. Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13 were accepted.

For the questions :

Amal - Q02, Q03, Q04, Q06, Q08, Q10, Q11, and Q13.
Bimal - Q02, Q04, Q06, Q07, Q08, Q09, Q12, and Q13.

Komal - Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, and Q12.

Q01 is reviewed by only Komal and is accepted hence Bimal must have prepared the question as it was accepted in a single review. (Bimal)

Q02-is reviewed by Amal, Bimal, and Komal and was rejected this has a possibility of being prepared by an external person and faced rejection by one among Amal and Bimal and Komal. (SME).

Q03-Was accepted and reviewed by Amal, Komal and hence must have been prepared by Bimal(Bimal).

Q04- Accepted and reviewed by the three of them and hence must have been prepared by an external expert. (SME).

Q05-Only reviewed by Komal and is accepted and hence must be prepared by Bimal. (Bimal)

Q06-Rejected and reviewed by Amal and Bimal. Hence could have been prepared by Komal or External and rejected by Bimal and Amal. (Komal/SME).

Q07-Accepted and reviewed by Bimal and Komal. Hence must have been prepared by Amal. (Amal).

Q08- Reviewed by Amal, Bimal, and Komal and is accepted. Hence must have been prepared by SME. (SME)

Q09- Reviewed by Bimal and Komal and is rejected. It must have been prepared by Amal and was rejected by both of them. (Amal).

Q10- Was reviewed by Amal and was accepted. Must have been prepared by Komal. (Komal)

Q11- Reviewed by Amal and Komal and was rejected. Must have been prepared by Bimal and rejected by both Amal and Komal. (Bimal)

Q12- Reviewed by Bimal and Komal and was rejected. Must have been prepared by Amal and was rejected by both Bimal and Komal. (Amal).

Q13- Reviewed by Amal and Bimal and was accepted. The question could have been prepared by Komal or SME. (Komal/SME).

Q06 irrespective of the question prepared by Komal/SME Bimal must reject the question.

Q07, Q09, Q12 are prepared by Amal and were reviewed by Bimal first and Komal next for this to happen Bimal must reject these questions.

Hence a total of four questions were rejected by Bimal.

For a question created by externals(SME), the minimum number of reviews required is two and the maximum number of reviews is three to confirm the acceptance or the rejection.

For a question created by anyone among Amal, Bimal, Komal a question can be reviewed after one review or two reviews.

The information provided states :

Q02, Q06, Q09, Q11, and Q12 were rejected. Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13 were accepted.

For the questions :

Amal - Q02, Q03, Q04, Q06, Q08, Q10, Q11, and Q13.
Bimal - Q02, Q04, Q06, Q07, Q08, Q09, Q12, and Q13.

Komal - Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, and Q12.

Q01 is reviewed by only Komal and is accepted hence Bimal must have prepared the question as it was accepted in a single review. (Bimal)

Q02-is reviewed by Amal, Bimal, and Komal and was rejected this has a possibility of being prepared by an external person and faced rejection by one among Amal and Bimal and Komal. (SME).

Q03-Was accepted and reviewed by Amal, Komal and hence must have been prepared by Bimal(Bimal).

Q04- Accepted and reviewed by the three of them and hence must have been prepared by an external expert. (SME).

Q05-Only reviewed by Komal and is accepted and hence must be prepared by Bimal. (Bimal)

Q06-Rejected and reviewed by Amal and Bimal. Hence could have been prepared by Komal or External and rejected by Bimal and Amal. (Komal/SME).

Q07-Accepted and reviewed by Bimal and Komal. Hence must have been prepared by Amal. (Amal).

Q08- Reviewed by Amal, Bimal, and Komal and is accepted. Hence must have been prepared by SME. (SME)

Q09- Reviewed by Bimal and Komal and is rejected. It must have been prepared by Amal and was rejected by both of them. (Amal).

Q10- Was reviewed by Amal and was accepted. Must have been prepared by Komal. (Komal)

Q11- Reviewed by Amal and Komal and was rejected. Must have been prepared by Bimal and rejected by both Amal and Komal. (Bimal)

Q12- Reviewed by Bimal and Komal and was rejected. Must have been prepared by Amal and was rejected by both Bimal and Komal. (Amal).

Q13- Reviewed by Amal and Bimal and was accepted. The question could have been prepared by Komal or SME. (Komal/SME).

Amal reviewed the questions :

Q02, Q03, Q04, Q06, Q08, Q10, Q11, and Q13.

For Q2 - Amal can either approve or reject the question.

For Q3- Amal must approve the question.

For Q4 -Amal can either approve or reject the question.

For Q06 - Amal must reject the question.

For Q08 - Amal can either approve or reject the question.

For Q10 - Amal must approve the question.

For Q11 - Amal must reject the question.

For Q13-  Amal can either approve or reject the question.

The approval ratio has a minimum of 2 questions only Q3, Q10, and a maximum of 6 questions Q02, Q3, Q04, Q08, Q10, Q13.

Hence 2/8 = 0.25 to 6/8 = 0.75

For a question created by externals(SME), the minimum number of reviews required is two and the maximum number of reviews is three to confirm the acceptance or the rejection.

For a question created by anyone among Amal, Bimal, Komal a question can be reviewed after one review or two reviews.

The information provided states :

Q02, Q06, Q09, Q11, and Q12 were rejected. Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13 were accepted.

For the questions :

Amal - Q02, Q03, Q04, Q06, Q08, Q10, Q11, and Q13.
Bimal - Q02, Q04, Q06, Q07, Q08, Q09, Q12, and Q13.

Komal - Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, and Q12.

Q01 is reviewed by only Komal and is accepted hence Bimal must have prepared the question as it was accepted in a single review. (Bimal)

Q02-is reviewed by Amal, Bimal, and Komal and was rejected this has a possibility of being prepared by an external person and faced rejection by one among Amal and Bimal and Komal. (SME).

Q03-Was accepted and reviewed by Amal, Komal and hence must have been prepared by Bimal(Bimal).

Q04- Accepted and reviewed by the three of them and hence must have been prepared by an external expert. (SME).

Q05-Only reviewed by Komal and is accepted and hence must be prepared by Bimal. (Bimal)

Q06-Rejected and reviewed by Amal and Bimal. Hence could have been prepared by Komal or External and rejected by Bimal and Amal. (Komal/SME).

Q07-Accepted and reviewed by Bimal and Komal. Hence must have been prepared by Amal. (Amal).

Q08- Reviewed by Amal, Bimal, and Komal and is accepted. Hence must have been prepared by SME. (SME)

Q09- Reviewed by Bimal and Komal and is rejected. It must have been prepared by Amal and was rejected by both of them. (Amal).

Q10- Was reviewed by Amal and was accepted. Must have been prepared by Komal. (Komal)

Q11- Reviewed by Amal and Komal and was rejected. Must have been prepared by Bimal and rejected by both Amal and Komal. (Bimal)

Q12- Reviewed by Bimal and Komal and was rejected. Must have been prepared by Amal and was rejected by both Bimal and Komal. (Amal).

Q13- Reviewed by Amal and Bimal and was accepted. The question could have been prepared by Komal or SME. (Komal/SME).

Amal could have possibly created the questions :

Q07, Q09, Q12 of which all three of them were reviewed by two people each, and hence they must have been rejected by the first reviewer.

Bimal could have created the questions :

Q01, Q03, Q05, Q 11.

Q01, Q05 are reviewed only by Komal and hence did not have any rejection.

Q03, Q11 faced at least one rejection.

Hence a total of five questions.

Let the number of large shirts be l and the number of small shirts be s.

Let the price of a small shirt be x and that of a large shirt be x + 50.

Now, s + l = 64

l (x+50) = 5000

sx = 1800

Adding them, we get,

lx + sx + 50l = 6800

64x +50l = 6800

Substituting l = (6800 - 64x) / 50, in the original equation, we get

$$\frac{\left(6800-64x\right)}{50}\left(x+50\right)=5000$$

(6800 - 64x)(x + 50) = 250000

$$6800x+340000-64x^2-3200x = 250000$$

$$64x^2-3600x-90000=0$$

Solving, we get, x= $$\frac{225\pm\ 375}{8}=\frac{600}{8}or-\frac{150}{8}$$

SO, x = 75

x + 50 = 125

Answer = 75 + 125 = 200.

Alternate approach: By options.

Hint: Each option gives the sum of the costs of one large and one small shirt. We know that large = small + 50

Hence, small + small + 50 = option.

SMALL = (Option - 50)/2

LARGE = Small + 50 = (Option + 50)/2

Option A and Option D gives us decimal values for SMALL and LARGE, hence we will consider them later. 

Lets start with Option B.

Large = 150 + 50 / 2 = 100

Small = 150 - 50 / 2 = 50

Now, total shirts = 5000/100 + 1800/50 = 50 + 36 = 86 (X - This is wrong)

Option C - 

Large = 200 + 50 / 2 = 125

Small = 200 - 50 / 2 = 75

Total shirts = 5000/125 + 1800/75 = 40 + 24 = 64 ( This is the right answer)

Let Rahul work at a units/hr and Gautam at b units/hour
Now as per the condition :
8a+6b =7.5a+7.5b
so we get 0.5a=1.5b
or a=3b
Therefore total work = 8a +6b = 8a +2a =10a
Now Rahul alone takes 10a/10 = 10 hours.

Initially number of matches = 40 
Now matches won = 12 
Now let remaining matches be x 
Now number of matches won = 0.6x
Now as per the condition :
$$\frac{\left(12+0.6x\right)}{40+x}=\frac{1}{2}$$
24 +1.2x=40+x
0.2x=16
x=80
Now when they won 90% of remaining = 80(0.9) =72
So total won = 84

Let us break this up into 2 inequations [ Let us assume x as m and y as n ]

| 1 + mn | < | m + n |

| m + n | < 5

Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.

Let us try out with the first quadrant and extend the results to the other quadrants.

We will also consider the +X and +Y axes along with the quadrant.

So, the first inequality becomes,

1 + mn < m + n

1 + mn - m - n < 0

1 - m + mn - n < 0

(1-m) + n(m-1) < 0

(1-m)(1-n) < 0

(m - 1)(n - 1) < 0

Let us try to plot the graph.

If we consider only mn < 0, then we get 

But, we have (m - 1)(n - 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.

So, we have,

But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.

So, if we look for only integer values, we get

(0,2), (0,3),.......

(0,-2), (0, -3),......

(2,0), (3,0), ......

(-2,0), (-3,0), .......

Now, let us consider the other inequation as well, in which |x + y| < 5

Since one of the values is always zero, the modulus of the other value is less than or equal to 4.

Hence, we get 

(0,2), (0,3), (0,4)

(0,-2), (0, -3), (0, -4)

(2,0), (3,0), (4,0)

(-2,0), (-3,0), (-4,0)

Hence, a total of 12 values.

We have :$$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$
We get $$\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$$
we get $$\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$$
we get $$\left(\log32\ +\log\ 15\right)=4\log a$$
=$$\log480=\log a^4$$
=$$a^4\ =480$$
so we can say a is between 4 and 5 .

Let us solve this question by assuming values(multiples of 100) and not variables(x).

Since we know that the female population was twice the male population in 1990, let us assume their respective values as 200 and 100.

Note that while assuming numbers, some of the population values might come out as a fraction(which is not possible, since the population needs to be a natural number). However, this would not affect our answer, since the calculations are in ratios and percentages and not real values of the population in any given year.

Now, we know that the female population became 1.25 times itself in 1990 from what it was in 1980.

Hence, the female population in 1980 = 200/1.25 = 160

Also, the female population became 1.2 times itself in 1980 from what it was in 1970.

Hence, the female population in 1970 = 160/1.2 = 1600/12 = 400/3

Let the male population in 1970 be x. Hence, the male population in 1980 is 1.4x.

Now, the total population in 1980 = 1.25 times the total population in 1970.

Hence, 1.25 (x + 400/3) = 1.4x + 160

Hence, x = 400/9.

Population change = 300 - 400/9 - 400/3 = 300 - 1600/9 = 1100/9

percentage change = $$\frac{\frac{1100}{9}}{\frac{1600}{9}}\times\ 100\ =\ \frac{1100}{16}\%=68.75\%$$