CAT 2022 Question Paper Slot 1

Priya can reach S from T via R or V.

In the east-west direction, the first train from P arrives at T at time = 6 am + $$\left(4\times2\right)+\left(3\times1\right)=\ 11$$ minutes = 6:11 am

Since T is a junction so this train will halt for 2 minutes at T and leave at 6:13.

Since every 10 minutes, a train starts from P in the east-west direction so the latest by which Priya will be able to board such a train is at 10:33 am.

In the north-south direction, the first train from B arrives at T at time = 6 am + $$\left(3\times3\right)+\left(2\times1\right)=\ 11$$ minutes = 6:11 am

Since T is a junction so this train will halt for 2 minutes at T and leave at 6:13.

Since every 15 minutes a train starts from P in the east-west direction so the latest by which Priya will be able to board such a train is at 10:28 am.

Now since she will be able to board a north-south train earlier than the east-west train so Priya will board a train for R from T at 10:28 am.

There are 3 stations between T and R

Travelling time between T and R = $$\left(4\times3\right)+\left(\left(3\times1\right)\right)=\ 15$$ minutes

Therefore, Priya will reach R latest by 10:43 am

In the east-west direction, the first train from M arrives at R at time = 6 am + $$\left(4\times2\right)+\left(3\times1\right)=\ 11$$ minutes = 6:11 am

Since R is a junction so this train will halt for 2 minutes at R and leave at 6:13.

Since every 10 minutes, a train starts from M in the east-west direction, so the latest by which Priya will be able to board such a train is at 10:43 am.

There are 9 stations between R and S

Travelling time between R and S = $$\left(10\times2\right)+\left(9\times1\right)=29\ $$ minutes

Time by which she reaches S = 10:43 +29 minutes = 11:12 am

If Priya boards a east-west train then Priya will board a train for V from T at 10:33 am.

There are 9 stations between T and V

Travelling time between T and V = $$\left(10\times2\right)+\left(9\times1\right)=29\ $$ minutes

Therefore, Priya will reach V latest by 10:33 am + 29 minutes = 11:02 am

In the north-south direction, the first train from D arrives at V at time = 6 am + $$\left(3\times3\right)+\left(2\times1\right)=\ 11$$ minutes = 6:11 am

Since V is a junction so this train will halt for 2 minutes at V and leave at 6:13.

Since every 15 minutes, a train starts from D in the north-south direction, so the latest by which Priya will be able to board such a train from V is at 11:13 am.

There are 3 stations between V and S

Travelling time between R and S =$$\left(4\times3\right)+\left(\left(3\times1\right)\right)=\ 15$$ minutes

Time by which she reaches S = 11:13 +15 minutes = 11:28 am

Travelling time between S and R = $$\left(10\times2\right)+\left(9\times1\right)=29\ $$ minutes

There is a stoppage of 2 minutes at R 

Travelling time between R and B = $$\left(7\times3\right)+\left(1\times2\right)+\left(5\times1\right)=28$$ minutes

In the north-south direction, the first train from A arrives at R at time = 6 am +$$\left(3\times3\right)+\left(2\times1\right)$$ = 6:11 am.

Since R is a junction so this train will halt for 2 minutes at R and leave at 6:13.

Every 15 minutes, a train starts from A in the north-south direction.

The last train that leaves A will be at 12:00 am and it will leave R at 12:13 am, so Haripriya must reach R till 12:13 am.

Travelling time between S and R = $$\left(10\times2\right)+\left(9\times1\right)=29\ $$ minutes

So Haripriya must board the train at S by 11:44 pm

In the east-west direction, the first train from N arrives at S at time = 6 am +$$\left(6\times2\right)+\left(5\times1\right)$$ = 6:17 am.

Since S is a junction so this train will halt for 2 minutes at S and leave at 6:19.

Every 10 minutes, a train starts from N in the east-west direction.

Therefore, Haripriya should board the train which leaves S at 11:39.

Travel time between A and B = $$\left(10\times3\right)+\left(7\times1\right)+\left(2\times2\right)=41$$ minutes

After completing a journey, a train must rest for 15 minutes at least before starting again.

So if a train starts from 6 am from A to B, then the latest by which that train will start from B to A will be at 7 am, as in the north-south direction, a train starts from A and B every 15 minutes.

So the total no. of trains required = $$\left(\frac{60}{15}\right)\times2=8$$

Travel time between A and B = $$\left(10\times3\right)+\left(7\times1\right)+\left(2\times2\right)=41$$ minutes

After completing a journey, a train must rest for 15 minutes at least before starting again.

So if a train starts from 6 am from A to B, then the latest by which that train will start from B to A will be at 7 am, as in the north-south direction, a train starts from A and B every 15 minutes.

So the total no. of trains required for the north-south lines =$$\left(\frac{60}{15}\right)\times2\times2=16$$

Travel time between M and N = $$\left(20\times2\right)+\left(17\times1\right)+\left(2\times2\right)=61$$ minutes

After completing a journey, a train must rest for 15 minutes at least before starting again.

So if a train starts from 6 am from M to N, then the latest by which that train will start from N to M will be at 7:20 am, as in the east-west direction, a train starts from M and N every 15 minutes.

So the total no. of trains required for the east-west lines= $$\left(\frac{80}{10}\right)\times2\times2=32$$

Total no. of trains required to service the city = 16+32 = 48

In the question, it is given that the ratio of number of literate males to literate females is 2 : 3.

Given, the number of literate males = 3600

The number of literate females = $$\frac{3600}{2}\times3$$ = 5400

Males : Females = 5 : 4

Let the number of males be 5y and the number of females be 4y

Illiterate males = 5y - 3600

Illiterate females = 4y - 5400

It is given,

$$\ \frac{\ 5y-3600}{4y-5400}=\frac{4}{3}$$

15y - 10800 = 16y - 21600

y = 10800

Number of females = 4y = 4*10800 = 43200

Let the original number of students be 'n' whose average weight is 'x'

Let the number of students added be 'm' and the average weight will be x + 3

We need to find the value of n : m

It is given, average weight of students in a class increased by 0.6 after new students are added.

Therefore,

$$\ \frac{\ nx+m\left(x+3\right)}{n+m}=x+0.6$$

$$\ \ nx+mx+3m=mx+nx+0.6n+0.6m$$

$$2.4m=0.6n$$

$$4m=n$$

$$\frac{n}{m}=\frac{4}{1}$$

The answer is option C.

It is given,

$$S_n=2n^2+n$$

$$S_{n-1}=2\left(n-1\right)^2+\left(n-1\right)$$

$$S_{n-1}=2n^2-3n+1$$

$$T_n=S_n-S_{n-1}=2n^2+n-2n^2+3n-1=4n-1$$

$$T_n=4n-1$$

The terms are 3, 7, 11, 15, 19, 23, 27,......

27 is the first term in the series divisible by 9.

27 is the 7th term.

Therefore, the least possible value of n is 7.

The answer is option C.

x>=a, so |x-a| = x-a

x<100, so |x-100| = 100-x

f(x) = (x-a) + (100-x) + |x-a-50| =100

or, |x-a-50| = a

From the graph we can can see that when x=a then

|x-a-50|=a

or, a= 50

Similarly when x=a+100

|x-a-50|=a

or, a= 50

So value of a is 50 when f(x) is 100.

M - First meeting point

Let the speeds of trains A and B be 'a' and 'b', respectively.

$$\frac{x}{a}=\ \frac{\ D-x}{b}$$

It is given,

$$\frac{D}{a}=10$$ and $$\frac{x}{b}=9$$

$$\frac{x}{\frac{D}{10}}=\ \frac{\ D-x}{\frac{x}{9}}$$

$$\frac{10x}{D}=\ \frac{\ 9D-9x}{x}$$

$$10x^2=\ \ 9D^2-9Dx$$

$$10x^2+9Dx-9D^2=\ 0$$

Solving, we get $$x=\frac{3D}{5}$$

$$\frac{x}{b}=9$$

$$\frac{3D}{b\times5}=9$$

$$\frac{D}{b}=15$$ 

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B

CD = $$\sqrt{\ y^2+25}$$

$$11+y+\sqrt{y^2+25}=36$$

$$\sqrt{y^2+25}=25-y$$

$$y^2+25=25^2+y^2-50y$$

2y = 24

y = 12

Area of trapezium = $$3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$$