XAT 2018 Question Paper - Quant

$$10 + 10^3 + 10^6 + 10^9 = 10 + 1000 + 1000000 + 1000000000$$
=$$1001001010$$
Therefore, option D is the right answer.

Let us assume the amount of work to be finished = LCM of {10, 12, 15, 18} = 180 units.

The amount of work which Abdul can complete in a day = $$\dfrac{180}{10}$$ = 18 units.

The amount of work which Bimal can complete in a day = $$\dfrac{180}{12}$$ = 15 units.

The amount of work which Charlie  can complete in a day = $$\dfrac{180}{15}$$ = 12 units.

The amount of work which Dilbar can complete in a day = $$\dfrac{180}{18}$$ = 10 units.

It is given that 50 percent of the total work gets completed after 3 days. Therefore, we can say that 90 units of work was completed in 3 days. 

Let us check options. 
Option A: Each of them worked for exactly 2 days.
In this case amount of work completed = 2*(10+15+12+18) = 110 units.

Option B: Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.
In this case amount of work completed = 1*(10+15)+2*(12)+3*(18) = 103 units.

Option C: Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.
In this case amount of work completed = 1*(10)+3*(15)+2*(18+12) = 115 units.

Option D: Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.
In this case amount of work completed = 1*(12)+3*(15)+2*(18+10) = 113 units.

Option E: Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.
In this case amount of work completed = 1*(18+12)+2*(15)+3*(10) = 90 units.

Therefore, we can say that option E is the correct answer. 

It is given that the length of the diagonals are in 3:4. Let '3x', and '4x' be the lengths of semi-diagonals as shown in the figure. We know that diagonals of a rhombus intersect each other perpendicularly.

In right angle triangle AOB,
$$AB^2=AO^2+BO^2$$

$$\Rightarrow$$ $$AB= \sqrt{AO^2+BO^2}$$

$$\Rightarrow$$ $$AB= \sqrt{(3x)^2+(4x)^2}$$

$$\Rightarrow$$ $$15= 5x$$

$$\Rightarrow$$ $$x= 3$$cm.

Therefore, we can say that the length of diagonals = 6x and 8x or 18 and 24 cm.

Hence, the area of the rhombus = $$\dfrac{1}{2}*18*24$$ = 216 cm$$^2$$. Therefore, option E is the correct answer. 

Let the cost price of the article be Rs.100.
The shopkeeper raises the price by x% and then decreases it by y%.
As a result, he reaches the cost price of the article. 
Also, it has been given that y is the difference between y% and x%.
$$y = x-y$$
$$2y = x$$
We know that $$(1+2y)(1-y)*100 = 100$$
$$(1+2y)(1-y) = 1$$
$$1-y+2y-2y^2 = 1$$
$$2y^2-y=0$$
$$2y = 1$$
$$y = 1/2$$ or $$0.5$$
$$x = 2y$$ 
=> $$x = 1$$ or $$x = 100$$%
Therefore, option D is the right answer. 

Liquids A and B are in the ratio 5:3. The volume of water is one-third the total mixture. 
Let us assume the volume of the total mixture to be 24x.

Volume of liquid A = 10x
Volume of liquid B = 6x
Volume of water = 8x

The mixture is passed through some medium that absorbs water completely and some quantity of liquid A. 
Water absorbed = 8x
Let the amount of liquid A absorbed be y.

8x+y = 200
=> y = 200-8x -----(1)
It has been given that (10x-y)/6x  = 7/9

Substituting (1), we get,
(10x-200+8x)/6x = 7/9
(18x-200)/6x = 7/9
162x-1800=42x
120x = 1800
=> x = 1800/120
Amount of water absorbed = 8*1800/120 = 120 ml.
Therefore, option E is the right answer. 

The radius of the coin is 3 cm. 
So, if the coin should not cross the edge, the centre of the coin should at least be 3 cm away from the edge of the tile. 

In the given diagram the center of the circle should lie in the blue region. As we can see, the area in which the centre of the coin can fall is a square of side 10-3-3 = 4 cm. 
Therefore, the area in which the centre of the coin can fall is 16 square cm. 
Area of the tile = 100 square cm.
Required probability = 16/100 = 0.16.
Therefore, option E is the right answer. 

2 litres are required to paint the surface of a solid sphere. 
Surface area of the solid sphere = $$4\pi*r^2$$
Now, the solid sphere is cut into 4 identical parts. This is possible only when the sphere is cut into 4 quarter spheres. 
After making the first cut, 2 hemispheres will be formed. 2 circles of area $$\pi*r^2$$ will get exposed in addition to the surface area of the sphere (the base surface of the bottom hemisphere and the base of the top hemisphere).
Now, another perpendicular cut will be made along the diameter of the sphere. 2 additional surfaces will get exposed again (one on left hemisphere and the another on the right hemisphere).

Area exposed after making 4 identical pieces = $$4*\pi*r^2$$ + $$2*\pi*r^2$$ + $$2*\pi*r^2$$
                                                                   = $$8*\pi*r^2$$
2 litres of paint is required to paint an area of $$4\pi*r^2$$
=> 4 litres of paint will be required to paint an area of $$8*\pi*r^2$$.

Therefore, option D is the right answer.

The man walks with a speed of V1 for 30 minutes. 
=> Distance covered = 30*V1.
On a particular day, he walks for 10 minutes at V1, takes a rest of 5 minutes, and then covers the distance by walking at V2 for 30 minutes. 
=> 10*V1 + 30*V2 = 30*V1
20*V1 = 30*V2
V1 = 1.5V2---------(1)
=> Total distance = 30*1.5*V2 = 45V2.
Now, we know that the person took 10 + 5 + 30 = 45 minutes to cover the entire distance. 
=> Average speed = 45V2/45 = V2.
Ratio of V2 and the average speed = 1:1.
Therefore, option A is the right answer. 

One boat is stationed to the North of the light house and the other boat is stationed to the East of the light house. 

The boat stationed to the East subtends an angle of 45 degrees and the boat stationed to the North subtends an angle of 30 degrees. 
Now, distance between the boat stationed to the East and the light house,d1 = tan 45
$$300/d1 = 1$$
=> $$d1 = 300$$ feet

Distance between the boat stationed to the North and the light house,d2 = tan 30
$$300/d2 = 1/\sqrt{3}$$
=> $$d2=300\sqrt{3}$$

Shortest distance between the 2 boats = $$\sqrt{300^2+(300\sqrt{3})^2}$$
                                                           = $$300*\sqrt{4}$$
                                                           =  $$600$$ feet.
Therefore, option C is the right answer.

It has been given that A+B+C = 41.
Let the common root be B. 
All the roots are positive integers. 
The product of the roots of one of the equations is 35. 
35 can be obtained only in 2 ways - either as 5*7 or 35*1. 
A+B+C = 41.
If A and B are 5 and 7 in any order, then C = 41 - 5 - 7 = 29.
If A and B are 35 and 1 in any order, then C = 41 - 35 - 1 = 5. 
As we can see, in either case, 5 is one of the 3 roots. 
Therefore, option C is the right answer. 

Let us find out the difference between the times given to figure out the pattern. 
The times given are 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm.
The difference between 2 consecutive times given are  8 minutes, 8 minutes, 13 minutes, 21 minutes, 34 minutes, and 55 minutes.
We can observe that the difference between the times are in the Fibonacci series. 
8 + 13 = 21
21 + 13 = 34
34 + 21 = 55

The Fibonacci series is as follows:
1,1,2,3,5,8,13,21,34,55.
But the first difference in the times given is 8. 
Therefore, the missing time must be such that it divides the interval of 8 minutes into 3 minutes and 5 minutes. 
The missing time should be 1:58 pm and hence, option B is the right answer. 

Let the number of girls in the school be G. 
=> Number of boys = G+30.
Some girls joined the class and the number of boys and girls became 3:5. 
Let the number of girls who joined the class be 'X'.
It has been given that (G+30)/(G+X) = 3/5
5G + 150 = 3G + 3X
2G + 150 = 3X
=> X = (2G/3) + 50.
2G has to be divisible by 3. 
Therefore, the least value that G can take is 3. 
When G = 3, X = 2 + 50
X = 52. 
The least number of girls who could have joined is 52. 
Therefore, option E is the right answer.

We draw BD perpendiculat to AC.
In right angled triangle BDC,  BD / BC = sin 30°
or, BD = (V2 * T)/2 ......(i)
In right angled triangle BDA, BD / BA = sin 45°
Or, BD = (V1 * T)/$$\sqrt{2}$$ ......(ii)
From (i) and (ii), we get
V2 / V1 = $$\sqrt{2}$$
Total distance to be travelled from C to A = CD + DA  = $$\sqrt{3}$$BD + BD
= BD($$1 + \sqrt{3}$$)
Replacing BD = (V2 * T)/2 in the avove equation,
CA = $$\dfrac{\text{(V2 * T)}}{2} (1 + \sqrt{3}$$)
Time taken at speed V2 = $$0.5(\sqrt{3}+1)$$T
Hence, option C is the correct answer.

The number of candidates registered for 3 of the 4 courses is 45, 55, and 70.
There are 2 cases.
70 can be the total number of students registered in an elective or the mandatory course. 
Case (i):
70 students have registered in an elective. 
Each student will choose 2 elective.
=> Total number of electives chosen = 2* total number of students.
45+55+70 = 2*number of students.
Number of students = 170/2 = 85.
In this case, at least 45 students would have selected an elective. 

Case (ii):
70 students have registered in the mandatory class.
=> 45+55+x = 2*70
=> x =40.
As we can see, the number of students registered in an elective can be 40 or 45. Therefore, option E is the right answer. 

All numbers from 1 to 9 repeat their last digits over a cycle of 4. 
63 can be written as 4k+3.
85 can be written as 4k+1. 
Some number's third power should yield the first digit as some number's first power.
$$2^3$$ will yield $$8$$ as the last digit (2 and 8 is a possible solution).X+Y = 10
$$3^3$$ will yield $$7$$ as the last digit (3 and 7 is a possible solution).X + Y = 10
$$4^3$$ will yield $$4$$ as the last digit and hence, can be eliminated. 
$$5$$ and $$6$$ yield $$5$$ and $$6$$ respectively as the last digit for any power and hence, can be eliminated.
$$7^3$$ will yield $$3$$ as the last digit (7 and 3 is a possible solution). X+Y=10.
$$8^3$$ will yield $$2$$ as the last digit. 8+2 =10.
As we can see, X+Y = 10 in all the cases. Therefore, option B is the right answer.

2 ≤ |x - 1| × |y + 3| ≤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x - 1| will be 2 and the maximum value of |x - 1| will be 5 as per the question.

When, |x - 1| = 2, |y + 3| can be either 1 or 2
So, for x =  -1, y can be - 4 or - 2 or - 5 or -1.
Thus, we get 4 pairs of (x, y) 

When |x - 1| = 3, |y + 3| can be 1 only
So, for x = - 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x - 1| = 4, |y + 3| can be 1 only
So, for x = - 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x - 1| = 5, |y + 3| can be 1 only
So, for x = - 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option E is the correct answer.

The sum of 4 consecutive numbers should be a power of 2. 
The powers of 2 are 1, 2, 4, 8, 16, 32, 64, 128. The maximum possible value that 4 consecutive days can take is 28 + 29 + 30 + 31 = 118.
We can eliminate 1, 2, 4, and 8 since sum of 4 consecutive integers is always greater than 9. 
Let the first day be x. 
x+x+1+x+2+x+3 = 16
=> 4x = 10
x = 2.5
We can eliminate 16.
Let us check for 32.
x+x+1+x+2+x+3 = 32
4x = 26.
x is not an integer. 
Let us check the case in which x is the last day of the month. 
Even in the month of February, the least value that x can take is 28.
28+1+2+3 = 34 > 32.
We can eliminate 32 as well. 
Let us check for 64.
4x+6 = 64
4x = 58
The sum of no 4 consecutive days in the same month can be expressed as 4k+6. 
Let us check the cases in which 2 months are involved.
29+30+1+2 = 62 < 64.
30 + 31 + 1 + 2 = 64. This is a possible combination. 
There are 7 months with 31 days in a year. We have to eliminate December since 1 will spill over to the 1st of January of the next year. 
Therefore, in a year, there will be 6 such instances. Therefore, option B is the right answer. 

The ratio of the curved surface area of the upper cone to the lower frustum is 1:2.
=> the ratio of the curved surface area of the upper cone to the total cone = 1:3.
Curved surface area (CSA) of a cone = $$\pi*r*l$$
For the given cone, the slant height, $$l=12$$cm

CSA of the cone = $$48*\pi$$
CSA of the smaller cone = $$16*\pi$$
Both the slant height and the radius would have been reduced by the same ratio. Let that ratio be $$x$$.
$$x^2*48*\pi$$=$$16\pi$$
=>$$x^2=\frac{1}{3}$$
$$x=\frac{1}{\sqrt{3}}$$
Slant height of the smaller cone = $$\frac{12}{\sqrt{3}}$$
Slant height of the frustum = $$12-\frac{12}{\sqrt{3}}$$
= $$12*\frac{\sqrt{3}-1}{\sqrt{3}}$$
=$$12*\frac{(3-\sqrt{3})}{3}$$
=$$12-4\sqrt{3}$$
Therefore, option D is the right answer.

Let us draw the diagram according to the given info,

We can see that AD = AO*cos60° = 2R*$$\dfrac{1}{2}$$ = R

In triangle, ACD

$$\Rightarrow$$ $$sinACD=\dfrac{AC}{AC}$$

$$\Rightarrow$$ $$sinACD=\dfrac{R}{\sqrt{2}*R}$$ = $$\dfrac{1}{\sqrt{2}}$$

$$\Rightarrow$$ $$\angle$$ ACD = 45°

By symmetry we can say that $$\angle$$ BCD = 45°

Therefore we can say that $$\angle$$ ACB = 90°

Hence, the area colored by green color = $$\dfrac{270°}{360°}*\pi*(\sqrt{2}R)^2$$ = $$\dfrac{3}{2}*\pi*R^2$$ ... (1)

Area of triangle ACB = $$\dfrac{1}{2}*R*2R$$ = $$R^2$$ ... (2) 

Area shown in blue color = $$\dfrac{60°}{360°}*\pi*(2R)^2-\dfrac{\sqrt{3}}{4}*(2R)^2$$ = $$\dfrac{2}{3}*\pi*R^2-\sqrt{3}*R^2$$ ... (3)

By adding (1) + (2) + (3) 

Therefore, the area of the common region between two circles = $$\dfrac{3}{2}*\pi*R^2$$ + $$R^2$$ + $$\dfrac{2}{3}*\pi*R^2-\sqrt{3}*R^2$$

$$\Rightarrow$$ $$(\dfrac{13\pi}{6}+1-\sqrt{3})R^2$$

Hence, option C is the correct answer. 

Sum of the ages of the 6 friends is the square of a prime number.

Statement I states that all the members were aged between 50 and 85 years. There are 6 friends in total. Therefore, the sum of the ages can vary from 6*50 to 6*85.
Sum of the ages of the 6 persons should lie between 300 and 510. 
361 is the only square of a prime number (19) in the given range. Therefore, statement I alone is sufficient to answer the question. 

Statement II states that the standard deviation of their ages is 4.6. SD gives us how much the values deviate from the mean. We cannot obtain the mean using the SD. Therefore, statement II alone is insufficient to answer the question without knowing the values in the data set. 

Therefore, option A is the right answer. 

Harry and Sunny have cards numbered from 1 to 10.
Sum of the number of cards picked by them = 1+2+3+...+10 = 55.

Sum of the numbers on the cards picked by Harry is 5 more than that of Sunny.
Let x be the sum of the numbers picked by Sunny. 
x+x+5 = 55
x = 25
25 can be written as 10+9+1+2+3 or 9+7+5+1+3 etc. (There are many such combinations). We cannot determine the person who has card number 2. 
Statement I alone is insufficient. 

Statement II states that one of the 2 persons has exactly 4 even-numbered cards.
The person with 4 even-numbered cards might or might not contain card number 2. Therefore, statement II alone is insufficient. 

Combining both the statements, we know that the sum of the numbers on the cards with Harry is 5 more than the sum of the number on the cards with Sunny. Therefore, the sum of the number of cards with Sunny should be 25 and Harry should be 30.

Sum of the 5 even numbers = 2+4+6+8+10 = 30.
Sum of the 5 odd numbers = 1+3+5+7+9 = 25.
By replacing one of the odd numbers with even numbers, we have to make the sums 30 and 25. We cannot replace an odd number with an even number and still get the sum as 25. Therefore, the set with 4 odd numbers must add up to 30 and the set with 4 even numbers must add up to 25. We can interchange 1 with 6 or 3 with 8 or 5 with 10. The card with number 2 will always remain with the person with 4 even cards. The card with number 25 will always remain with Sunny. Therefore, we can determine the answer using both the statements together and hence, option D is the right answer.

Team A has won both its matches and scored 5 goals while conceding 1. Thus, the possible scorelines for team A are (4-1) & (1-0), (3-1) & (2-0), (2-1) & (3-0).

Similarly, the possible outcomes for others teams are as follows:

B: (5-1) & (0-0) and (4-0) & (1-1)

C: (2-0) & (0-0) and (1-0) & (1-1)

D: (0-0) & (1-1)

E: (0-2) & (1-2) and (0-1) & (1-3)

F: (0-1) & (0-6), (0-2) & (0-5) and so on.

By carefully observing the above scorelines, the following table can be obtained.

Thus, the matches Team A - Team B and Team E - Team F are yet to take place.

Hence, option E is the answer.

Team A has won both its matches and scored 5 goals while conceding 1. Thus, the possible scorelines for team A are (4-1) & (1-0), (3-1) & (2-0), (2-1) & (3-0).

Similarly, the possible outcomes for others teams are as follows:

B: (5-1) & (0-0) and (4-0) & (1-1)

C: (2-0) & (0-0) and (1-0) & (1-1)

D: (0-0) & (1-1)

E: (0-2) & (1-2) and (0-1) & (1-3)

F: (0-1) & (0-6), (0-2) & (0-5) and so on.

By carefully observing the above scorelines, the following table can be obtained.

The above table shows that Team A - Team E match has the score-line (2 - 1).

Hence, option B is the answer.

Team A has won both its matches and scored 5 goals while conceding 1. Thus, the possible scorelines for team A are (4-1) & (1-0), (3-1) & (2-0), (2-1) & (3-0).

Similarly, the possible outcomes for others teams are as follows:

B: (5-1) & (0-0) and (4-0) & (1-1)

C: (2-0) & (0-0) and (1-0) & (1-1)

D: (0-0) & (1-1)

E: (0-2) & (1-2) and (0-1) & (1-3)

F: (0-1) & (0-6), (0-2) & (0-5) and so on.

By carefully observing the above scorelines, the following table can be obtained.

From the above table, we can see that options B, C and D are possible and option A is not.

Hence, option A is the answer.

Let the required percentage be P.

Option A: P_A = (6.5 - 4) * 100/ 4 = 62.5%

Option B: P_B = (8 - 5) * 100/ 5 = 60%

Option C: P_C = (7 - 4) * 100/ 4 = 75%

Option D: P_D = (9 - 5.5) * 100/ 5.5 = 63.63%

Thus, option C shows the maximum year-on-year growth and is the answer.

Prof. Arithmetic: There are two such instances, during 2013-14 and 2015-16

Prof. Algebra: There are three such instances, during 2011-12, 2014-15 and 2015-16

Prof. Geometry: There are two such instances, during 2010-11 and 2013-14

Prof. Calculus: There are two such instances, during 2010-11 and 2014-15

Thus, there are 2+3+2+2 = 9 such instances.

Hence, option C is the answer.

Research efficiency = (total number of publications)/(total time spent in research)

Prof. Arithmetic: RE = (8) / [90*(0.7 + 0.5 + 0.6)] = 8 / (18 * 9)

Prof. Algebra: RE = (8) / [90*(0.3 + 0.4 + 0.25)] = 8 / (9.5 * 9)

Prof. Geometry: RE = (8) / [90*(0.75 + 0.25 + 2)] = 8 / (12 * 9)

Prof. Calculus: RE = (8) / [90*(0.55 + 0.55 + 0.4)] = 8 / (15 * 9)

Thus, we can observe that Prof. Arithmetic has the lowest efficiency.

Hence, option A is the answer.