XAT 2020 Question Paper

Each circle on the end of the diagonal will touch sides of the rectangular field

Using Pythagoras' theorem, the distance between the vertex of the rectangle and center of the first circle drawn on the diagonal = $$1.25\sqrt{\ 2}$$

Distance between the vertex of the rectangle and circumference of the first circle drawn on the diagonal = $$1.25\sqrt{\ 2}$$ - 1.25 = 0.51 meters

Space that cannot be used to draw circle otherwise they will go outside rectangle on every diagonal = 0.51 * 2 = 1.02 meters

Space that can be used to draw circles = length of diagonal - unused space = 50 - 1.02 = 48.98 meters

On every diagonal, maximum number of such circles = usable length/diameter of each circle = 48.98/2.5 = 19

Or, on every diagonal, one circle will be at the center (intersection of diagonals) and 9 circles will be on each half of the diagonal

.’. The circle in center will be common for both diagonals, and 9 circles can be drawn on each half of the diagonal. So total circles = 9*4 + 1 = 37

For hare and tortoise speed is given. Tortoise will take 12 hours to complete 1 round. And during this, hare will make 12 rounds of OP. In the first round, both has started from point O. After some time, hare will cross tortoise and distance between them will be 1 km. After some more time, when hare is returning back from P to O, before and after crossing tortoise, hare will be two more times 1km apart from tortoise. So, in first round, there are three such occurrences.

In the second round, when the hare has started from point O, while going and returning back, there will be four occurrences when before and after crossing the tortoise, the hare will be exactly 1 km apart. But the first occurrence of round 2 is already counted in round 1. So, in second round as well, there will be total 3 occurrences.

In the third, fourth and fifth rounds, there will be 4 such occurrences.

In the sixth round, because the tortoise will be at point P, there will be only 2 cases.

Now, till round 6 there are 20 such occurrences. And from round 7 to 12, it will be exactly the same but in reverse order of 2, 4, 4, 4, 3, 3. Hence, total such occurrences = 20 * 2 = 40.

To maximize Z, B has to be minimized and A,C,D are to be maximised.

The value of B can be 0 or 1.

Case 1:

B=0 => A=1=> C=8 and D=12.

Z= 15+8+6=29.

Case 2:

B=1 => A=0=> C=12 and D=15.

Z=-3+12+7.5=16.5.

.'. D=12 is correct answer. 

If we join YP and PZ, XYPZ will become a cyclic quadrilateral. YZ and XP will be diagonals of this quadrilateral.

Ptolemy's theorem states that the product of the length of the diagonals is equal to the sum of products of measures of the pairs of opposite sides.

As per this theorem for the quadrilateral :

XY*PZ +XZ*PY = YZ*XP.

For the equilateral triangle.

Now, in equilateral triangle XYZ, YZ = XZ = XY.

Hence equation 1 becomes, XP = YP + PZ.

Let the part of the amount divided between Y and Z be 5k => Y gets 2k and Z gets 3k.

The overall profit is Rs 200000.

Hence the remaining profit is Rs 200000 - 5k. =

Left over profit of 2-5k is divided in the ratio 2.5:3.5:4

The final profit distribution among X, Y and Z.

=> Finally, X gets $$\frac{2.5}{10}\left(200000-5k\right)\ $$, Y gets $$2k+\frac{3.5}{10}\left(200000-5k\right)\ $$ and Z gets $$3k+\frac{4}{10}\left(200000-5k\right)\ $$.

Given the ratio of profit distribution of X and Z is 1 : 4

Given, $$3k+\frac{4}{10}\left(200000-5k\right)\ $$ = 4($$\frac{2.5}{10}\left(200000-5k\right)\ $$) => 3k=$$\frac{6}{10}\left(200000-5k\right)\ $$ =>10k=400000-10k => 20k=400000 => k=20000.

.'. Share of Y = $$2k+\frac{3.5}{10}\left(200000-5k\right)\ $$  = 75000. 

The two circles are symmetric about the diagonal.

NC1=$$\frac{2\sqrt{\ 2}}{3}$$ = $$\sqrt{\ \left(\frac{2}{3}\right)^{^2}+\left(\frac{2}{3}\right)^2}=\ \frac{\left(2\sqrt{\ 2}\right)}{3}$$

The lengths FC1, EC1 are the radius of the circle which is 2km/3.

The length C1P is the radius of the circle.

Because of symmetry C1O = C2O and C1N = C2B.

2*(C1N+ C1O) = $$2\sqrt{\ 2}$$ the length of the diagonal of the square.

C1N+C1O = $$\sqrt{\ 2}$$

C1O = $$\sqrt{\ 2}-\frac{2\sqrt{\ 2}}{3}=\ \frac{\sqrt{\ 2}}{3}$$

OP = $$\frac{\left(2-\sqrt{\ 2}\right)}{3}$$

The diagonal perpendicularly bisects the line GH. Hence C1OH is 90 degrees.$$C1H^2\ =\ C1O^2+OH^2$$

OH = $$\frac{\sqrt{\ 2}}{3}$$. Similarly OG = $$\frac{\sqrt{\ 2}}{3}$$

HC1, C1G both are equal to $$\frac{2}{3}$$ each. H1G is $$\frac{2\sqrt{\ 2}}{3}$$. HC1G is a raight angled traingle with angle HC1G is 90 degrees.

Area of the region required is 2*(Area of segment OGH) 

.'. Area of required region=$$2\left(\frac{90}{360}\cdot\frac{4\pi\ }{9}-\frac{1}{2}\cdot\frac{2\sqrt{\ 2}}{3}\cdot\frac{\sqrt{\ 2}}{3}\right)$$=$$\frac{2(\pi - 2)}{9}$$

Given, a+b+c=50 and a+b+c+x+y+z=190 => x+y+z=140.

Also, let E1=k => E2=2k and E3=3k

E1+E2+E3= 6k=190+50=240 => k=40.

Option A,  the number of students choosing E1 is 40.

Option B, number of students choosing either E1 or E2 or both, but not E3 = total - E3 = 190-120 = 70.

Option C, number of students choosing both E1 and E2 => this can not be obtained.

Option D, number of students choosing E3 = 3x = 120.

Option E, number of students choosing exactly one elective = Out of 190, 50 are choosing two electives, hence 190-50 = 140 are choosing exactly one elective.

Given, a+b+c=50 and a+b+c+x+y+z=190 => x+y+z=140.

Also, let E1=k => E2=2k and E3=3k

E1+E2+E3= 6k=190+50=240 => k=40.

Option A: If the number of students choosing only E2, the number of students choosing both E2 and E3, are given then the number of students who choose E2 and E1, E1 and E3 can be found. From this only E1, only E3 can be calculated.

Option B: knowing the number of students choosing both E1 and E2, the number of students choosing both E2 and E3, and a number of students choosing both E3 and E1 is insufficient. This information is not enough to calculate the number of students who choose only E1, only E2, and only E3.

Option C: If x and c are known, we can't find y and z.

Option E is not sufficient. 

The number of ways of selecting one card from Ashok's bag and other from Shilpa bag = $$40_{C_1}\times\ 5_{C_1}$$ = 200
Now, if the card taken from Shilpa's bag shows 1, then 1 will divide all the numbers on Ashok's card. Hence, the number of ways = 40
If the card taken from Shilpa's bag shows 2, then the remainder will be either 0 or 1. Hence, the number of ways = 40
If the card taken from Shilpa's bag shows 3, then the remainder will be 0, 1 or 2. Hence, the number of ways = 40
If the card taken from Shilpa's bag shows 4, then the remainder will be 0, 1, 2 or 3. So the numbers having 3 as remainder will be rejected. So the number of form 4n+3 will be rejected. Total number of such numbers = $$\frac{\left(39-3\right)}{4}+1$$ = 10
If the card taken from Shilpa's bag shows 5, then the remainder will be 0, 1, 2, 3 or 4. So the numbers having 3 or 4 as remainder will be rejected. So the number of form 5n+3, 5n+4 will be rejected. Total number of such terms = $$\frac{\left(39-3\right)}{4}+1$$ = 10
The numbers left = 40-10 = 30
The total numbers having 5n+3 form = $$\frac{\left(39-4\right)}{5}+1\ =\ 8$$
The total numbers having 5n+4 form = $$\frac{\left(38-3\right)}{5}+1\ =\ 8$$
The numbers left = 40-8-8=24

Hence, the probability = $$\frac{\left(40+40+40+30+24\right)}{200}=\frac{174}{200}=0.87$$

Let the work done per day by X, Y and Z is respectively 'x', 'y' and 'z' units.

According to the first statement, out of 6 days, X works for 5 days, and Y works for 5 days. Total work done = 5x + 5y = 1...(i)

According to the second statement, out of 8 days, Z works for 7 days and X works for 6 days and they complete two jobs. 7z + 6x = 2...(ii)

According to the third statement, out of 12 days, Y works for 10 days and Z works for 10 days and they complete three jobs. 10y + 10z = 3...(iii)

Solving, we get x = 1/10, y = 1/10, z = 2/10.

=> every day, X does 10% of the job, Y does 10% of the job and Z does 20% of the job.

Together, every day they can do 40% of the job. Hence to complete 100% of the job, they will take 100/40 = 2.5 days.