CAT 2023 Slot 2 Question Paper

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

Anjali completed a total of 4 rides, 3 of which were completed at 2. Therefore the answer is Option B: Ride-1, Ride 3, and Ride -2

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

Only Ride-1 was taken by all the visitors. Therefore the correct answer is Option A: Ride-1

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

Anjali took 4 rides, and Chitra took 2 rides. Therefore the correct answer is 6

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

As we can see from the table of Anjali she spent a total of 20+30+50+40= 140

Therefore the required answer is 140

It is given that $$a^m\cdot b^n\ =\ 144^{145}$$, where $$a>1\ $$ and $$b>1$$.

$$144$$ can be written as $$144\ =\ 2^4\times\ 3^2$$

Hence, $$a^m\cdot b^n\ =\ 144^{145}$$ can be written as $$a^m\cdot b^n\ =\ \left(2^4\times\ 3^2\right)^{^{145}}=2^{580}\times\ 3^{290}$$

We know that $$3^{290}$$ is a natural number, which implies it can be written as $$a^1$$, where $$a\ >\ 1$$

Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.

Hence, the largest value of (n-m) is (580-1) = 579

The correct option is D

It is given that $$\frac{x}{y}<\ \frac{\ x+3}{y-3}$$, which can be written as $$\frac{x}{y}-\frac{\ x+3}{y-3}<0$$

=> $$\ \frac{\ x\left(y-3\right)-y\left(x+3\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ xy-3x-xy-3y}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ -3\left(x+y\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$

From this inequality, we can say that, when $$y<0=>y(y-3)>0$$. Now to satisfy the given equation $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$, 

$$(x+y)$$ must be greater than zero Hence, $$x>0$$ and $$\left|x\right|\ >\left|y\right|$$

Therefore, the magnitude of $$x$$ is greater than the magnitude of $$y$$. 

Hence, $$x>y$$, and $$\left|x\right|\ >\ \left|y\right|$$ =>$$-x\ <\ y$$ (Since the magnitude of $$x$$ is greater than the magnitude of $$y$$.)

The correct option is B.

It is given that k divides m+2n and 3m+4n.

Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.

Similarly, we know that k divides 3m+4n.

We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.

By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.

Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n is also divisible by k.

Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.

Therefore, m, and 2n are also divisible by k.

The correct option is C

It is given that $$2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$$, which can be written as:

=>$$\left(2^{2x^2}\right)^2-2^{2x^2}\cdot2^{x+15}\cdot2^1+\left(2^{x+15}\right)^{^2}=0$$

=> $$\left(2^{2x^2}-2^{x+15}\right)^{^2}=0$$

=> $$2^{2x^2}-2^{x+15}=0$$ (Since $$\left(a-b\right)^2\ =\ 0\ =>\ a-b\ =0$$)

=> $$2x^2\ =\ x+15$$

=> $$2x^2-x-15=0$$ 

=> $$2x^2-6x+5x-15=0$$

=> $$2x\left(x-3\right)+5\left(x-3\right)=0$$

=> $$\left(2x+5\right)\left(x-3\right)\ =\ 0$$

Hence, the possible values of x are $$-\frac{5}{2}$$, and $$3$$, respectively.

Therefore, the sum of the possible values is $$\left(3-\frac{5}{2}\right)=\frac{1}{2}$$

The correct option is D

Since there are two distinct factors other than 1, and itself, which implies the total number of factors of N is 4.

It can be done in two ways. 

First case: $$N\ =\ p^3$$ (where p is a prime number)

Second case: $$N\ =\ p_1\times\ p_2$$ (Where $$\ p_1,\ p_2$$ are the prime numbers)

From case 1, we can see that the numbers which is a cube of prime and less than 50 are 8, and 27 (2 numbers).

From case 2, we will get the numbers in the form (2*3), (2*5), (2*7), (2*11), (2*13), (2*17), (2*19), (2*23), (3*5), (3*7), (3*11), (3*13), (5*7) {(13 numbers)}

Hence, the total number of numbers having two distinct factors is (13+2) = 15.

It is given that  $$\log_{\sqrt{3}}{(x)}+\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}}=\frac{16}{3}$$, which can be written as:

=> $$2\log_3x+\log_{0.008}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{8}{1000}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{1}{125}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{5^{-3}}\left(5\right)^2\ =\ \frac{16}{3}$$

=> $$2\log_3x-\frac{2}{3}=\ \frac{16}{3}$$

=> $$2\log_3x=\frac{16}{3}+\frac{2}{3}$$

=> $$2\log_3x=6$$

=> $$\log_3x^2=6\ =>\ x^2\ =\ 3^6$$

Hence, $$\log_3\left(3\cdot x^2\right)\ =\ \log_3\left(3\cdot3^6\right)\ =\log_33^7\ =\ 7$$