CAT 2020 Question Paper Slot 1

Let the speed of Car 2 be 'x' kmph and the time taken by the two cars to meet be 't' hours.

In 't' hours, Car 1 travels $$\left(60\ \times\ t\right)\ km$$ while Car 2 travels $$\left(x\ \times\ t\right)\ km$$

It is given that the time taken by Car 1 to travel $$\left(x\ \times\ t\right)\ km$$ is 45 minutes or (3/4) hours. $$\therefore\ \frac{\left(x\ \times\ t\right)}{60}\ =\ \frac{3}{4}\ $$ or $$t=\frac{180}{4x}$$....(i)

Similarly, the time taken by Car 2 to travel $$\left(60\ \times\ t\right)\ km$$ is 20 minutes or (1/3) hours. $$\therefore\ \frac{\left(60\times\ t\right)}{x}=\frac{1}{3}$$ or $$\therefore\ t=\frac{x}{180}$$....(ii)

Equating the values in (i) and (ii), and solving for x:

$$\therefore\ \frac{180}{4x}=\frac{x}{180}\ \ \longrightarrow\ \ \ x\ =90\ kmph$$

Hence, Option B is the correct answer.

Let the price of desktop and laptop be x,y respectively.

Given,

x+y=50000...(i)

12.x+0.9y=50000(1.02)=51000...(ii)

(ii)-0.9(i) gives

0.3x=6000=> x=20000.

Initially the amount of Dye and Water are 16,24 respectively.

To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye=> 16l of water is added.

Now, the Dye and Water arr 16,40 respectively.

After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.

To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye => 8l of Dye should be added.

Hence , 8 is correct answer. 

$$x=2^{12\left(7+4\sqrt{\ 3}\right)}$$.

$$x^{\frac{7}{2}}=2^{42\left(7+4\sqrt{\ 3}\right)}$$

$$x^{2\sqrt{\ 3}}=2^{24\sqrt{\ 3}\left(7+4\sqrt{\ 3}\right)}$$

$$\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$$ = $$2^{\left(7+4\sqrt{\ 3}\right)\left(42-24\sqrt{\ 3}\right)}=2^{\left(7+4\sqrt{\ 3}\right)\left(7-4\sqrt{\ 3}\right)6}$$ =$$2^6$$.

Hence C is correct answer.

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.'. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.

.'.`The weight, in kg. of the metal C is 42xy=84.

Let $$a=x+\frac{1}{x}$$
So, the given equation is $$a^2-3a+2=0$$
So, $$a$$ can be either 2 or 1.

If $$a=1$$, $$x+\frac{1}{x}=1$$ and it has no real roots. 
If $$a=2$$, $$x+\frac{1}{x}=2$$ and it has exactly one real root which is $$x=1$$

So, the total number of distinct real roots of the given equation is 1

$$\frac{\log\ 5}{2\log2}\ =\frac{\log\ y}{2\log2}\cdot\frac{\log\ 5}{2\log6}$$

$$\log\ 36\ =\ \log\ y;\ \therefore\ y\ =36$$

The difference in the time take to traverse the same distance $$'d'$$ at two different speeds is 35 minutes. Equating this: $$\frac{d}{8}-\frac{d}{15}\ =\ \frac{35}{60}$$

On solving, we obtain $$d = 10 kms$$. Let $$x kmph$$ be the speed at which Amal needs to travel to reach the office in 50 minutes; then

$$\frac{10}{x}=\frac{50}{60}\ or\ x\ =\ 12\ kmph$$.Hence, Option B is the correct answer.

Let the total distance be 'D' km and the speed of the train be 's' kmph. The time taken to cover D at speed d is 't' hours. Based on the information: on equating the distance, we get $$s\ \times\ t\ =\ \frac{s}{3}\times\ \left(t+\frac{1}{2}\right)$$ 
On solving we acquire the value of $$\ t\ =\frac{1}{4}$$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed 's' {distance traveled in terms of s = $$\ \frac{s}{12}$$}. Remaining distance in terms of 's' = $$\ \frac{s}{4}-\frac{s}{12}\ =\frac{s}{6}$$

The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $$15 - 9 = 6 mins$$ or {1/10 hours}. Thus, let 'x' kmph be the new value of speed. Based on the above, we get $$\frac{s}{\frac{6}{x}}\ =\frac{1}{10}\ or\ x\ =\frac{10s}{6}$$

Since the increase in speed needs to be calculated: $$\frac{\left(\frac{10s}{6}\ -s\right)}{s}\times\ 100\ =\frac{200}{3}\approx\ 67\%$$ increase.

Hence, Option C is the correct answer.

The four digit even numbers will be of form:

1100, 1122, 1144 ... 1188, 2200, 2222, 2244 ... 9900, 9922, 9944, 9966, 9988

Their sum 'S' will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+...)....+(9900+9900+22+9900+44+9900+66+9900+88)

=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)....+9900*5+(22+44+66+88)

=> S=5*1100(1+2+3+...9)+9(22+44+66+88)

=>S=5*1100*9*10/2 + 9*11*20

Total number of numbers are 9*5=45

.'. Mean will be S/45 = 5*1100+44=5544.

Option D