XAT 2025 Question Paper

Let's assume that all the dustbins are at equal distances and the distance between two dustbin's be l.

We are told that a minimum 800 steps to reach every second dustbin. 

800$$\le$$2l $$\Rightarrow$$ 400 $$\le$$ l

Similarly, we are also told that a maximum 1260 steps to reach every third dustbin.

So, 3l$$\le$$1260

400$$\le$$l$$\le$$420

The circumference will be 5l.

2000$$\le$$ 5l $$\le$$ 2100

2000 and 2100 are both in steps of Ramesh.  

We are told that each of his step is 0.77m.

So, 2000 steps = 2000*0.77= 1540m

2100 steps = 2100*0.77 = 1617

So, the circumference will lie between 1540 and 1617.

1540 $$\le 2\pi r \le$$ 1617.

$$\frac{1540}{2\pi}\le r\le\frac{1617}{2\pi}$$

$$245.22\le r\le\ 257.48$$

So, 250m is the correct option.

In the given equation $$\dfrac{x^2 \log_a(16)}{\log_a(32)} - \dfrac{\log_a(64)}{\log_a(32)} - x = 0 $$

We know the property of log that $$\dfrac{\log_ab}{\log_ac}=\log_cb$$

So, $$\frac{\log_a16}{\log_a32}=\log_{32}16=\log_{2^5}2^4=\frac{4}{5}$$

Similarly, $$\frac{\log_a64}{\log_a32}=\log_{32}64=\log_{2^5}2^6=\frac{6}{5}$$

Hence, the equation can be written as 

$$\frac{4}{5}x^2-\frac{6}{5}-x=0$$

or, $$4x^2-6-5x=0$$

Evaluating the discriminant we get

$$D=b^2-4ac=\left(-5\right)^2-4\left(4\right)\left(-6\right)$$

or, $$D=121>0$$

Hence, the equation has two distinct real roots.

But as the question mentions $$a>0$$, so it can take the value $$a=1$$, for which the log term will not be defined, hence there will be no defined solution.

Hence, the answer is Depends on the value of $$a$$

Considering statement 1: We can only get one of the sides lenght of the land. But that is not sufficient to get the area.

Considering statement 2: 

We are given the diagonal length.

With this information and the perimeter it is not possible to find the area of land.

Considering both 1 and 2 statements.

Now we have two right angle triangles. We know the hypotenues of both the triangles.

$$AB^2+AD^2=BD^2$$

$$110^2+AB^2= 255^2$$

AB$$\approx\ $$230

Let DC = x, BC = 700 - 110-230-x = 360-x.

$$x^2+(360-x)^2 = 255^2$$

Upon solving this we get x = 180$$\pm\frac{15}{\sqrt{2}}$$

360-x = 180$$\pm\frac{15}{\sqrt{2}}$$

Area of triangle ABD = $$\frac{1}{2}\cdot110\cdot230\ =\ 12650$$

Area of triangle BDC = $$\frac{1}{2}\left(180+\frac{15}{\sqrt{2}}\right)\left(180-\frac{15}{\sqrt{2}}\right)=16143.75$$

So, total area = 12650+16143.75 = 28793.75

Cost = 28793.75*400 = Rs. 11517500.

So, option C is the answer.

For the roots to be irrational, the value of D > 0 and D must not be a perfect square.

The value of D for the equation given is $$=\ b^2\ -\ 4\times\ a\ \times\ a\ =\ b^2\ -\ 4a^2$$

The pairs (a. b) that satisfy the above conditions are,

b =  1 no value of a exists

b = 2 no value of a exists

b = 3, a = 1

b = 4, a = 1

b = 5, a = 1

b = 6, a = 1, 2

b = 7, a = 1, 2, 3

b = 8, a = 1, 2, 3

b = 9, a = 1, 2, 3, 4

There are a total of 15 pairs of (a, b) that satisfy the above condition and out of them, the value of a + b is greater than 9 for 7 of them.

So, the probability that a + b is greater than 9 is given by $$\dfrac{7}{15}$$.

Hence, the correct answer is option D.

It is given that the average of the first five highest-rated employees is 38. So, the sum of the ratings of the top 5 highest-rated employees is 38 * 5 = 190.

It is given that the average of the first five lowest-rated employees is 25. So, the sum of the ratings of the top 5 lowest-rated employees is 25 * 5 = 125.

The overall average rating of all the employees is given as 30. So, the sum of the ratings of all the employees is given as 30 * 8 = 240.

The sum of the 3 highest rated employees' ratings can be obtained by subtracting the sum of the 5 lowest rated players' ratings from the overall rating, which is 240 - 125 = 115.

The sum of the 3 lowest rated employees' ratings can be obtained by subtracting the sum of the 5 highest rated players' ratings from the overall rating, which is 240 - 190 = 50.

So, the sum of the 4th and 5th highest-rated employees is 190 - 115 = 75.

Now let us look at the options to eliminate the wrong option,

Option A)

It is given that the second-highest rating is 38. It is not an incorrect option because there is a possible case of the second highest being 38 and satisfying all the above conditions. The first 5, in that case, can be 39, 38, 38, 38, 37, which satisfies the above conditions.

Option B)

Same as the above case, we can have the 4th and 5th ratings to be 38 and 37, and in that case, the median of the ratings of the employees becomes $$\dfrac{37\ +\ 38}{2}\ =\ 37.5$$. So, this is not an incorrect option.

Option C)

We know that the sum of the lowest three ratings is 50, and in that case, there are possibilities of the lowest rating being 1 and the sum of the three is 50. For example, 37, 12, 1 is one of the cases. Hence, this is not an incorrect option.

Option D)

The highest rating cannot be 40 because if it is 40, then the sum of the 2nd and 3rd becomes 75. In that case, the 3rd rated person's rating has to be less than or equal to 37, and we know that the 4th person's rating must be greater than or equal to 38. So, if the third person rating is 37 or less, then there is no possibility as the 4th person rating must be less than 3rd person rating. So, option D is incorrect.

Option E)

As explained in option C, the set of the last three rated employees can be 37, 12 and 1, and in this case, we can see that the third lowest player's rating is 37. So, E is not an incorrect option.

Hence, the correct answer is option D.

We know that if the greatest common divisor of two numbers is 1, they are co-prime to each other.

Given that the greatest common divisor of the selected integer $$x$$ and any other integer between 2 and 40, both inclusive, is 1.

So, the selected number has to be a prime number, and that too greater than $$\dfrac{40}{2}$$ or $$20$$ because prime numbers less than 20, like 17, will have the greatest common divisor = 17 with their multiple, like 34 which is also in the range of selected integers.

Hence, the only possible values of $$x$$ are 23, 29, 31 and 37 i.e. a total of 4 values.

According to the set, there are 8 tests and each test can have a maximum of 4 marks so any student can get a maximum of 32 marks. 

Now, the number of marks received by any student in a test can be 0/1/2/3/4.

We are given information about two students, Ravi and Sumana. 

In the first test, they both scored the same marks. Let the marks scored by both is $$x$$.

Further, from the second to eighth tests, Ravi scored the exact same non-zero marks. Let the marks scored by Ravi from second to eighth test is $$y$$ where y can be 1/2/3/4. Also, Sumana scored the same marks as Ravi from the fifth test onwards.

With this, we get the following table:

Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests.

Let the marks of Sumana in the second test be $$z$$.

We are given that, $$x\ +\ 2y=x+z$$

Therefore, we can say, z = 2y.

We don't have any other information about their marks in specific tests so let the marks of Sumana in test 3 and test 4 be $$a$$ and $$b$$ respectively. 

With this, we have:

At last, we are given that Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.

i.e. x, (x + 2y) and (x + a + b + 6y) are in a geometric progression. 

With this, we can infer that x can't be 0 as if x is 0, any number multiplied with it will be 0 and (x+2y) and (x + a + b + 6y) can't be in a geometric progression.

Hence, x can be 1 / 2 / 3 / 4.

Let's try to make all possible geometric progression with this information:

1. (1, 3, 9) when x = 1, y = 1 and a + b = 2

2. (2, 4, 8) when x = 2, y = 1 and a + b = 0

3. (2, 6, 8) when x = 2, y = 2 and a + b = 4

4. (4, 8, 16) when x = 4, y = 2 and a + b = 0

In the question, we need to answer which of the mentioned statements can be true:

A: Ravi scored 4 marks in the third test

y can't be 4 in any of the possible case. Hence, the statement can't be true. 

B: Sumana scored 2 marks in the first test

Sumana scored x marks in the first test and x can be 2. Hence, this can be true. 

C: Ravi scored 0 marks in the fifth test

y can't be 0 as we discussed above. Hence, this statement can't be true. 

D: Sumana scored 3 marks in the second test

Sumana scored 2y in the second test and if 2y = 3, y will be a non integer number which is not possible. 

E: Sumana scored 4 marks in the eighth test

y can't be 4 in any of the possible case. Hence, this can't be true. 

According to the set, there are 8 tests and each test can have a maximum of 4 marks so any student can get a maximum of 32 marks. 

Now, the number of marks received by any student in a test can be 0/1/2/3/4.

We are given information about two students, Ravi and Sumana. 

In the first test, they both scored the same marks. Let the marks scored by both is $$x$$.

Further, from the second to eighth tests, Ravi scored the exact same non-zero marks. Let the marks scored by Ravi from second to eighth test is $$y$$ where y can be 1/2/3/4. Also, Sumana scored the same marks as Ravi from the fifth test onwards.

With this, we get the following table:

Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests.

Let the marks of Sumana in the second test be $$z$$.

We are given that, $$x\ +\ 2y=x+z$$

Therefore, we can say, z = 2y.

We don't have any other information about their marks in specific tests so let the marks of Sumana in test 3 and test 4 be $$a$$ and $$b$$ respectively. 

With this, we have:

At last, we are given that Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.

i.e. x, (x + 2y) and (x + a + b + 6y) are in a geometric progression. 

With this, we can infer that x can't be 0 as if x is 0, any number multiplied with it will be 0 and (x+2y) and (x + a + b + 6y) can't be in a geometric progression.

Hence, x can be 1 / 2 / 3 / 4.

Let's try to make all possible geometric progression with this information:

1. (1, 3, 9) when x = 1, y = 1 and a + b = 2

2. (2, 4, 8) when x = 2, y = 1 and a + b = 0

3. (2, 6, 8) when x = 2, y = 2 and a + b = 4

4. (4, 8, 16) when x = 4, y = 2 and a + b = 0

According to the question, if Ravi scored 4 marks in the first test i.e. if x = 4, the possible case is case 4,

Hence, sumana scored $$a$$ marks in the third test and we know that in case 4, a + b = 0. Hence, both $$a$$ and $$b$$ are 0.

According to the set, there are 8 tests and each test can have a maximum of 4 marks so any student can get a maximum of 32 marks. 

Now, the number of marks received by any student in a test can be 0/1/2/3/4.

We are given information about two students, Ravi and Sumana. 

In the first test, they both scored the same marks. Let the marks scored by both is $$x$$.

Further, from the second to eighth tests, Ravi scored the exact same non-zero marks. Let the marks scored by Ravi from second to eighth test is $$y$$ where y can be 1/2/3/4. Also, Sumana scored the same marks as Ravi from the fifth test onwards.

With this, we get the following table:

Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests.

Let the marks of Sumana in the second test be $$z$$.

We are given that, $$x\ +\ 2y=x+z$$

Therefore, we can say, z = 2y.

We don't have any other information about their marks in specific tests so let the marks of Sumana in test 3 and test 4 be $$a$$ and $$b$$ respectively. 

With this, we have:

At last, we are given that Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.

i.e. x, (x + 2y) and (x + a + b + 6y) are in a geometric progression. 

With this, we can infer that x can't be 0 as if x is 0, any number multiplied with it will be 0 and (x+2y) and (x + a + b + 6y) can't be in a geometric progression.

Hence, x can be 1 / 2 / 3 / 4.

Let's try to make all possible geometric progression with this information:

1. (1, 3, 9) when x = 1, y = 1 and a + b = 2

2. (2, 4, 8) when x = 2, y = 1 and a + b = 0

3. (2, 6, 8) when x = 2, y = 2 and a + b = 4

4. (4, 8, 16) when x = 4, y = 2 and a + b = 0

If Ravi scored 1 mark in the second test i.e. y = 1, it can be either case 1 and case 2. 

The total marks scored by Sumana across the tests according to case 1: 

x + a + b + 6y = 1 + 2 + 6 = 9

The total marks scored by Sumana across the tests according to case 2:

x + a + b + 6y = 2 + 0 + 6 = 8

Hence, the maximum possible marks that can be scored by Sumana is 9.

Difference between the gap in 2015 and 2005 is least in I. That is 0.

Hence I is the answer.