XAT 2018 Question Paper

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.