CAT 2023 Slot 1 Question Paper

Given a and b are the distinct roots of the equation $$2x^{2} - 6x + k = 0$$

=> a + b = -(-6/2) = 3 (Sum of the roots)

=> ab = k/2 (Product of the roots)

Now, (a+b) and ab are the roots of the quadratic equation $$x^{2} + px + p = 0$$

=> a + b + ab = -p => 3 + k/2 = -p ---(1)

=> (a + b)(ab) = p => 3(k/2) = p ---(2)

$$3+\dfrac{k}{2}=-\dfrac{3k}{2}$$ => 2k = -3 => k = $$-\dfrac{3}{2}$$

p = $$\dfrac{3k}{2}=\dfrac{3}{2}\left(-\dfrac{3}{2}\right)=-\dfrac{9}{4}$$

=> 8(k-p) = $$8\left(-\frac{3}{2}+\frac{9}{4}\right)=-12+18=6$$

Given that in the final mixture, the ratio of coffee and cocoa is 16:9

Let us assume coffee is 16 units and cocoa is 9 units.

=> Initially, there are 25 units of coffee and 0 units of cocoa

Let's say x units of the mixture is removed and replaced with cocoa

=> Now, we have (25-x) coffee and x units of cocoa. => Mixture P

Now, if x units of the mixture is removed:

Amount of coffee present = (25-x) - $$\dfrac{\left(25-x\right)}{25}\times\ x$$

=> $$\left(25-x\right)\left(1-\dfrac{x}{25}\right)=16$$

=> $$\left(25-x\right)^2=16\times\ 25$$

=> 25 - x = 20 => x = 5.

In mixture P, cocoa = x = 5

In mixture Q, cocoa = 9 units.

=> Required ratio = 5:9

The given time is 8:48 AM.

Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.

Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.

=> The angle between them is 288 - 264 = 24 degrees.

This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle = (6 - 0.5)*m = $$\dfrac{11}{2}m=12$$ => m = $$\dfrac{24}{11}$$

Let us assume the initial selling prices of A and B is p.

Given, she made profit of 20% on A => 1.2 * c = p => c = 5p/6 => cost of A is $$\dfrac{5}{6}p$$

Given, she made a loss of 10% on B => 0.9 * c = p => c = 10p/9 => cost of B is $$\dfrac{10}{9}p$$

Now, she sold them at a price such that a 10% profit is made on B

=> Selling price = s = 11/10 * 10/9 p => $$\dfrac{11}{9}p$$

=> Profit % on A = $$\dfrac{\left(\dfrac{11}{9}-\dfrac{5}{6}\right)}{\left(\dfrac{5}{6}\right)}\times\ 100$$ = 46.66% = nearly 47%

Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.

If avg. speed of the car during the trip is 's' => the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.

=> From the options if s = 110 => The reading will be 26862 - 110*8 = 25982 (Not a palindrome)

=> If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.

=> s = 100 is the correct option.

Given that the average marks of 4 girls and 6 boys is 24.

Let us assume 'b' is the marks scored by a boy and 'g' is the marks scored by a girl.

=> 4g + 6b = 10*24 = 240 ---(1)

Given that, $$b\le g\le2b$$

We need to find the distinct possible values of 2g + 6b = 2g + 240 - 4g = 240 - 2g.

From (1)

when b = g => 10g = 240 => g = 24

when b = g/2 => 7g = 240 => g = 240/7

=> 240 - 2g varies from 240 - 2*24 to 240 - 2*240/7

=> 171.42 to 192 => Integer values of 172 to 192 => 21 values.

Given, the salaries of Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively, Let us assume their salaries are 5p, 6p and 7p.

They get salary hikes of 20%, 25% and 20%, respectively.

=> Their salaries are 6/5 * 5p, 5/4 * 6p and 6/5 * 7p => 6p, 7.5p, 8.4p

Now, Sita and Mita get salary hikes of 40% and 25%, respectively

=> Sita's salary = 1.4 * 6p = 8.4p and Mita's salary = 1.25 * 8.4p = 10.5p

Let Gita's salary be 'g' after hike

=> 3g = 8.4p + g + 10.5p => 2g = 18.9p => g = 9.45p

=> Hike % = $$\dfrac{\left(9.45-7.5\right)}{7.5}\times\ 100$$ = 26%

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => $$t^2=6\times\ 24$$ => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.

=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

Anil invested 22000 for 6 years at 4% interest compounded half-yearly

=> Amount = $$22000\left(1.02\right)^12$$

Let Sunil invest 'S' rupees for 5 years at 4% C.I. half-yearly and 10% S.I. for 1 additional year

=> Amount = $$S\left(1.02\right)^{10}\left(1.1\right)$$

Given that the both amounts are equal

=> $$22000\left(1.02\right)^{12}=S\left(1.02\right)^{10}\left(1.1\right)$$

=> $$S=\dfrac{22000\left(1.02\right)^2}{1.1}=20808$$

Let us assume the efficiencies of Amal, Sunil, and Kamal are a, s, and k, respectively.

Given that they are in H.P.

=> $$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{1}{k}$$ ---(1)

Also, given that Kamal takes twice as much time as Amal to do the same amount of job

=> a = 2k

Given that when Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job.

=> If W is the total work => 4a + 9s + 16k = W.

from (1)$$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{2}{a}$$ => $$a=\dfrac{3}{2}s$$ and $$k=\dfrac{3}{4}s$$

=> $$4\left(\dfrac{3s}{2}\right)+9s+16\left(\dfrac{3s}{4}\right)=W$$

=> $$6s+9s+12s=W$$

=> $$27s=W\ =>\ s\ =\ \dfrac{W}{27}$$

=> Sunil will take 27 days to finish the work when working alone.