CAT 2024 Slot 3 Question Paper

We are given that Sam completes a piece of work in 20 days. We are also given that Mohit is twice as fast, so he should take only 10 days. Mohit is thrice as fast as Ayna, so he would take 30 days. 

Let's take the total work to be 60 units; this would give the work done per day for Mohit, Sam, and Ayna to be 6, 3 and 2, respectively. 

On the first day Sam and Mohit work: doing 9 units
On the second day Sam and Ayan work: doing 5 units
On the third day, Mohit and Ayan work: doing 8 units

Essentially doing 22 units in a 3 days cycle. 
After two such cycles, there will be 60-44 = 16 units of work left

On day 7, Sam and Mohit would work 9 units, leaving 7 units 
On day 8, Sam and Ayan would work 5 units, leaving 2 units
And on day 9, Ayan and Mohit would complete the remaining work. 

So Sam worked for a total of 2+2+2= 6 days and on each day he did 3 units of work, completing 18 units of work.

The ratio of work done by Sam would be $$\frac{18}{60}=\frac{3}{10}$$

Therefore, Option B is the correct answer. 

This is the situation that is described in the question above, Angle AOB is 120 degrees and the chord AB is of length 10cm. 

Using Cosine rule we can find the length of AO and AB
$$\cos\left(AOB\right)=\frac{\left(r^2+r^2-300\right)}{2r^2}$$
$$-\frac{1}{2}=\frac{\left(2r^2-300\right)}{2r^2}$$
$$3r^2=300$$
$$r=10$$

Area of the Sector AOB is $$\frac{120}{360}\times\ \pi\ \times\ r^2$$
$$\frac{1}{3}\times\ \pi\ \times\ \frac{100}{1}$$ which is $$\frac{100\pi}{3}\ $$

Area of triangle AOB is $$\frac{1}{2}\times\ r^2\times\ \sin\left(120\right)$$

$$\frac{1}{2}\times\ \frac{100}{1}\times\ \frac{\sqrt{3}}{2}$$

Area of triangle AOB is $$\frac{75}{\sqrt{3}}$$

The smaller region will be, $$\frac{100\pi\ }{3}-\frac{75}{\sqrt{3}}$$
Taking 25 common we will get, 
$$25\left(\cfrac{4 \pi}{3} - \sqrt{3}\right)$$

Finding the terms in the sequence, we see that $$t_3=0$$, $$t_4=-\frac{1}{3}$$, $$t_5=0$$
We would notice that all the odd terms are 0, and we are also asked the sum of only even terms, so we do not need to consider those

$$t_6=-\frac{1}{5}$$
We see that the even terms are in an HP: $$-1,\ -\frac{1}{3},\ -\frac{1}{5},\ -\frac{1}{7},\ ...$$

The sum we are asked is the inverse of these terms, that is: -1, -3, -5, -7, up to 1012 terms

The sum of this AP would be $$\frac{\left[-\left(2\times\ 1\right)+\left(1012-1\right)\left(-2\right)\right]}{2}\times\ 1012$$
Which is equal to $$-1012\times\ 1012\ =\ -1024144$$

Therefore, Option A is the correct answer. 

The expression on the right-hand side will have two critical points: 2 and -2

For any value of x greater than equal to 2, the equation changes to x+2-(x-2) = 4

the value of min{x,2} would be 2, so we would want max{x,2} to be 4+2 = 6

Therefore, x=6 works. 

For any value of x less than equal to -2, the equation changes to -(x+2)+(x-2) = -4
the value of max{x,2} would be 2, so we would want min{x,2} to be 6 again; this cannot be the case.
In general, subtracting the smaller (min) of the two values from the bigger (max) can not lead to a negative number. The max it can lead to is a 0

When x lies between -2 and 2,  the equation becomes 2x
The maximum function will give back 2, and the minimum function will give back x, with the right-hand side giving 2x
Solving this we would get 2-x = 2x which is x=2/3

Therefore, there are two real values of x for which the given equation holds. 

Hence, 2 is the correct answer.

The graph of the function on the right hand side can be visualised as:

Hence, 2 is the correct answer. 

Let us assume the amount invested to be $$P$$, and the rate of interest to be $$r$$. 
Value of the investment after 3 years will be, $$P\left(1+r\right)^3$$
Value of the investment after 5 years will be, $$P\left(1+r\right)^5$$

$$\left(1+r\right)^2=\frac{36}{25}$$
$$\left(1+r\right)^2=1.44$$
$$r=0.2$$

We need to find the value of n for which $$4000\left(1+r\right)^n>20000$$
$$\left(1+r\right)^n>5$$
$$\left(1.2\right)^n>5$$
We see that, 
$$1.2^8=4.2999$$
$$1.2^9=5.15$$

Hence it takes 9 years to grow to over 20,000. 

Taking $$10^x=a$$
we get $$a+\frac{4}{a}=\frac{81}{2}$$
This would give the quadratic equation: $$2a^2-81a+8=0$$

We want to find the sum of possible values of x, let the value of x be x1 and x2

these would correspond to log a1, and log a2

The sum of log a1 + log a2 would be log (a1 x a2)

From the quadratic equation we got above, we can see that the product of the possible values of a would-be 8/2 = 4

Threfore, the sum of values of x would be log (4) which would be $$2\ \log_{10}2$$

Therefore, Option A is the correct answer. 

Let us assume that the distance is D, speed of the train is S and time taken by the train is t. 

t is nothing but $$\frac{D}{S}$$

Statement 1: Had the speed been 6 km per hour more, it would have needed 4 hours less
$$\frac{D}{S+6}=t-4$$

$$\frac{D}{S+6}=\frac{D}{S}-4$$

$$4=\frac{D}{S}-\frac{D}{S+6}$$

$$\frac{S+6-S}{S\left(S+6\right)}=\frac{4}{D}$$

$$\frac{6}{S\left(S+6\right)}=\frac{4}{D}$$

$$D=\frac{2S\left(S+6\right)}{3}$$

Statement 2:  Had the speed been 6 km per hour less, it would have needed 6 hours more
$$\frac{D}{S-6}=t+6$$

$$D\left[\frac{1}{S-6}-\frac{1}{S}\right]=6$$

$$\frac{S-S+6}{S\left(S-6\right)}=6$$

$$D=S\left(S-6\right)$$

Equating the two equations for distance, 
$$S\left(S-6\right)=\frac{2S\left(S+6\right)}{3}$$

$$3S-18=2S+12$$

$$S=30$$

Hence the speed is 30 kmph

Taking a log for each of the expressions, we get the following:

$$\log_34=a,\ \log_45=b,\ \log_56=c,\ \log_67=d,\ \log_78=e,\ \log_89=f$$

The expression $$abcef$$ would then be: $$\log_34\times\ \log_45\times\ \log_56\times\ \log_67\times\ \log_78\times\ \log_89$$

Next, we can use this property of log: $$\frac{\log_ba}{\log_bc}=\log_ca$$
Using this, we get:

$$\frac{\log\ 4}{\log\ 3}\times\ \frac{\log\ 5}{\log\ 4}\times\ \frac{\log\ 6}{\log\ 5}\times\ \frac{\log\ 7}{\log\ 6}\times\ \frac{\log\ 8}{\log\ 7}\times\ \frac{\log\ 9}{\log\ 8}$$

All the terms will cancel out except: $$\frac{\log\ 9}{\log\ 3}=\log_39=2$$

Therefore, 2 is the correct answer. 

Let us say the cost price of an item is X
It is said that it is marked to make a profit of 20%. 
That means it is marked at 1.2X

Ravi gets a 10% discount on the marked price, 
$$0.9\left(1.2X\right)=1.08X$$

Saves 15 rupees, so 1.2X-1.08X
0.12X=15
X=125

Profit made by Gopi is 0.08(125)=10 rupees. 

Let us assume the amount of Milk in the container to be X and the amount of water in the container to be Y. 
We are told that X+Y=300. 

Now, an operation is given where "an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again" 
Volume of the water initially is Y. If twice that amount is taken out, the percentage of the solution that is taken out will be, $$\frac{2Y}{X+Y}$$
That means the quantity of milk that will remain in the solution will be, $$X\left(1-\frac{2Y}{X+Y}\right)$$
This value is given to be 72%, 72% of 300 will be 216

$$X\left(1-\frac{2Y}{X+Y}\right)=216$$
$$X\left(\frac{X+Y-2Y}{X+Y}\right)=216$$
Writing $$X=300-Y$$
$$\left(300-Y\right)\left(300-2Y\right)=64800$$

Expanding this we have, 
$$2Y^2-900Y+25200=0$$
Factorising this equation we have, 
$$2\left(Y-30\right)\left(Y-420\right)=0$$

Y is either 30 or 420. 

Given that the capacity of the container itself is 300, Y has to be 30. 

Hence the amount of water initially is 30 Litres.