XAT 2020 Question Paper - Quant

Let four friends Anita, Bina, Chaitra and Divya be represented as A,B,C,D respectively. 

From the given information, Messages sent by A are 150 out of which 90 are Jokes

Messages sent by B are 180 out of which 72 are Jokes

Messages sent by C are 150 out of which 120 are Jokes

Messages sent by D are 120 out of which 60 are Jokes

.'. Percentage of jokes that were neither sent by D or B is 210*100/342=61.4

Given that :

(A O B) = (A + B) * B.

For (5 O 2) = ( 5+2 )*2 = 14.

((14) O 5 )= ( 14 + 5 )*5 = 19*5 = 95.

2500=$$\left(5+10t\right)^2+\left(10+20t\right)^2$$ => $$\left(1+2t\right)^2=20$$

$$AB^2=25+100\left(1+2t\right)^2$$ => AB= 45km.

Triangle ACC' is similar to triangle ABB'

Considering AC = a, BC = b, CC' = h, AA' is given as 60, BB' is given to be 40.

AC/AB = CC'/BB' = h/40.

$$\left(\frac{a}{a+b}\right)\ =\ \frac{h}{40}$$  (1)

Similarly triangle BCC' is similar to BAA'.

BC/AB = CC'/AA' = h/60.

$$\left(\frac{b}{a+b}\right)\ =\ \frac{h}{60}$$  (2)

Adding (1) and (2).

$$\frac{h}{40}+\frac{h}{60\ }=\ 1$$

1/h = $$\left(\frac{1}{40}+\frac{1}{60}\right)$$

h = 24

Using crossed ladder theorem 

$$\frac{1}{CC'}=\frac{1}{AA'}+\frac{1}{BB'}$$=1/60 +1/40 =5/120 =24.  

On one side, to place 1st rock, he had to travel 10m. For 2nd rock he had to travel 20m..similarly, till last rock he had to travel 400.

Total sum would be 10+20+30+...+400 = $$\frac{40}{2}\left(410\right)=8200$$.

Similarly, on the other side it will be 8200.

Total 8200+8200=16400. 

The volume of cuboid will be 50*25*1=1250

The volume of the portion below cuboid will be Area of triangle*width =$$\frac{1}{2}\cdot3\cdot50\cdot25=1875$$

.'. Total volume = 1875+1250= 3125.

Let the amount spent by Ashish is 'a', Banti is 'b' and Chintu is 'c'. Given, a = b = c.

One of them has bought a large bag. So, he must have spent at least 1000 rupees. It means, everyone has spent at least a thousand rupees. Or, a +b + c ≥ 3000.

Revenue from small bags = 50*15 = 750.

Revenue from large bag = 1000*1 = 1000. Total revenue excluding from medium bag = 750 + 1000 = 1750.

If 4 medium bags are sold, total revenue = 200*4 + 1750 = 2550, which is less than 3000. Hence, not the right answer.

If 5 medium bags are sold, total revenue = 200*5 + 1750 = 2750, which is less than 3000. Hence, not the right answer.

If 7 medium bags are sold, total revenue = 200*7 + 1750 = 3150, which is more than 3000. Hence, 7 is the correct answer.

In triangle OAP, since AP=6, OA will be 12.

Area of AQBMA=Area of Triangle ABQ- (Area of minor arc AMB-Area of OAB)

Length of PQ=MQ+PM = 12+(12-$$6\sqrt{\ 3}$$)=24-$$6\sqrt{\ 3}$$

Area of Triangle ABQ=$$\frac{1}{2}\cdot12.\left(24-6\sqrt{\ 3}\right)=6\left(24-6\sqrt{\ 3}\right)$$.=81.64

Area of minor arc AMB-Area of OAB=$$\frac{60}{360}\cdot\pi\ \cdot144-\frac{1}{2}\cdot12\cdot6\sqrt{\ 3}$$ = $$24\pi\ -36\sqrt{\ 3}$$.=13.07

.'.Area of AQBMA=68.57$$\approx\ 69$$

By noticing the table we can come to conclusion that either of P in 2018 or Q in 2017 must have highest yoy growth.

YOY of P in 2018=50/200 *100=25%

YOY of Q in 2017 = 40/140 *100 = 28.57%

Hence D is correct answer.

If P's revenue in 2018 is 225 and S's revenue in 2016 is 160 then their revenues will be in A.P. 

Let us first compute the individual changes in value in both years before zeroing in on the companies that fall into the 25% increase and 50% increase groups. 

For 25% increase:

For 50% increase: 

We can move forth with the assumption that P saw a 50% increase while the rest of the companies saw a 25% increase. In this case, the total increase in revenue, for all four companies put together, will be Rs. 125 lakhs. Same is the case when we consider company R to have seen a 50% increase, while the rest had a 25% increase. {Hence, Option D is true} We observe that Options B, C and E hold true as well. The increase observed for each of the companies is distinct, based on the given condition. Thus, Option A cannot be true.

Given, z/4 > x/6 > y/5 ...(i)

and (x+y+z)/15= 2*y/5 => x+z=5y

The value of y can only be 1.

=> x+z=5.

If z = 1, then z/4 is less than x/6.

If z = 2, then z/4 is equal to x/6.

If z = 4 or 5 then y/5 is greater than x/6.

The only possible value to satisfy (i) condition is z=3. and x=2.

Option A: $$\left(\frac{\pi}{2}\right) \left[\left(\frac{1}{\pi}\right) + 1\right] - \frac{\pi}{2}$$ =1/2

Option B: $$\sin^2 1^\circ + \sin^2 2^\circ + .... + \sin^2 {89}^\circ$$ =44+1/2 ($$\sin^2\left(89\right)=\cos^2\left(1\right)\ \&\ \sin^2\left(1\right)+\cos^2\left(1\right)=1$$

Option C: $$\sqrt{2}\left(3\sqrt{2} - \frac{4}{\sqrt{2}}\right) + \sqrt{3}$$ = $$6-4+\sqrt{3}$$=$$2+\sqrt{3}$$ which is non-terminating and non repeating.

Option D: $$\frac{\left(\sqrt[3]{729}\right)}{3} + \frac{22}{7}$$ =3+22/7

Option E: $$\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^3 + ...$$ (infinite terms)= $$\frac{1}{1-\frac{\pi\ }{4}}=\frac{4}{4\ -\pi\ }\ $$ => $$(4 - \pi)[1 + \left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^3 + ...$$ (infinite terms)] =4

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.

Using Fermat's theorem :

If p is a prime number and a, p are co primes $$\left(a^{p-1}\right)\ mod\ p=1$$

Remainder when $$19^{20}$$ is divided by 7 = $$19^2$$ mod 7 =4. ( Here $$19^{20\ }=\ \left(\left(19\right)^6\right)^3\cdot\left(19\right)^2$$

Since the remainder for $$19^6$$ is 1 the remainder for $$\ 19^{20}$$ is  equivalent to the $$\frac{19^2}{7}$$ = 4.

Remainder when $$20^{19}$$ is divided by 7 = $$20^1$$ mod 7 =6.( Here $$\frac{20^6}{7}$$ the remainder is 1 and since 

$$20^{19}=\ \left(20^6\right)^3\cdot\left(20\right)^1\ =\ \frac{\left(1\cdot20\right)}{7}$$. The remainder is 6.

Remainder when $$19^{20} - 20^{19}$$ is divided by 7=4-6=-2=> 5. 

Let's five small drums have a capacity of 1 unit capacity each and one bigger drum has that of 2 units capacity. We need to consider two cases here. One with minimum volume and the other with maximum volume.

Case 1: Minimum value is possible if the bigger drum is half filled. So, total volume of water = 1 + 2 * (1/2) + 2 * (2/3) + 1 = 26/6 ~ 4.3

Case 2: Maximum value is possible if the bigger drum is completely full. So, total volume of water = 2 + 3 * (1/2) + 2 * (2/3) = 29/6 = 4.833

In any case volume of water is more than 4 units and less than 5 units. Hence, exactly 5 smaller drums are adequate to store the water.

FRom ii, China is F.

FRom iv and v, India must be A, Brazil must be G and Russia must be B.

India's GDP due to agriculture = 17% of 2 trillion =0.34 trillion.

FRom i, USA GDP due to agriculture> 0.34 trillion and <1 trillion i.e. it's percentage must be >2% and less than 5.8%

So, USA can be C/E/H.

From point 3, UK must be D and France must be H=> U.S.A must be C/E.

Now, Germany and Japan can be C or I.

6 nations can be uniquely identified. 

FRom ii, China is F.

FRom iv and v, India must be A, Brazil must be G and Russia must be B.

India's GDP due to agriculture = 17% of 2 trillion =0.34 trillion.

FRom i, USA GDP due to agriculture> 0.34 trillion and <1 trillion i.e. it's percentage must be >2% and less than 5.8%

So, USA can be C/E/H.

From point 3, UK must be D and France must be H=> U.S.A must be C/E.

Now, Germany and Japan can be C or I or E.

.'. Everything except option D can be ruled out.

FRom ii, China is F.

FRom iv and v, India must be A, Brazil must be G and Russia must be B.

India's GDP due to agriculture = 17% of 2 trillion =0.34 trillion.

FRom i, USA GDP due to agriculture> 0.34 trillion and <1 trillion i.e. it's percentage must be >2% and less than 5.8%

So, USA can be C/E/H.

From point 3, UK must be D and France must be H=> U.S.A  must be C/E.

Now, Germany and Japan can be C or I.

Option A: If Germany is C, then its GDP will be 1 but given its gdp is 1.2 So, C is not Germany. If Germany is I, then its GDP will be 1.2trillion. Hence, Germany is I => Japan is C/E and USA is C/E. If USA is C then its GDP due to agriculture is not 3rd highest. So, USA is E and Japan is C. 

Option A is correct.