XAT 2024 Question Paper

The number of item combinations Rohit and his friends could have ordered veg items were $$^{16}C_2$$
And the number of ways they could have ordered non-veg items would be $$^9C_3$$

The total number of item combinations that Rohit and his friends could choose from would then be $$^{16}C_2\times\ ^9C_3$$

Let's denote the number of veg gluten-free items as $$V\sim G$$ and the number of non-veg gluten-free items as $$NV\sim G$$

The total number of ways Bela and her friends could have chosen the items from the menu would then be 
$$NV\sim G\times\ ^{N\sim G}C_2$$

We are given that Rohit and his friend's number of combinations were twelve times this value, giving us the relation

$$^{16}C_2\times\ ^9C_3=12\times\ NV\sim G\times\ ^{N\sim G}C_2$$
$$^{16}C_2\times\ ^9C_3=12\times\ NV\sim G\times\ \frac{\left(N\sim G\right)\left(N\sim G\ -1\right)}{2}$$
$$2^4\times\ 3\times\ 5\times\ 7=\ NV\sim G\times\ \left(N\sim G\right)\left(N\sim G\ -1\right)$$

At this point, we need two consecutive numbers on the left-hand side to account for the $$\left(N\sim G\right)\left(N\sim G\ -1\right)$$ on the right-hand side. 

We know that $$NV\sim G$$ must be 9 or less; trying to put this value as values from 9 to 1 and get two consecutive digits from the remaining numbers, 

The value of $$NV\sim G$$ can be 7, giving the value of $$\left(N\sim G\right)\left(N\sim G\ -1\right)$$ as 16 x 15
OR
The value of $$NV\sim G$$ can be 8, giving the value of $$\left(N\sim G\right)\left(N\sim G\ -1\right)$$ as 15 x 14

In either of these cases, the number of items on the menu not having gluten would be 7+16 = 8+15 = 23
This means that 2 of the items on the menu will not have gluten in them. 

Therefore, Option B is the correct answer. 

In order for the line two be coincident, the ratio of the coefficients of the variables and the constants from both the equations must be the same. 

That is, $$\frac{3}{q}=\frac{21}{r}=\frac{p}{-7}$$

Let's consider each option individually.
Checking through the easier options first.

Option A: p and q must have opposite signs
In order for  $$\frac{3}{q}=\frac{p}{-7}$$ to be true, p and q must be of opposite signs as pq = -21
Hence, this option does not contradict the lines coinciding. 

Option D: r and q must have the same signs

Similar to option A, for $$\frac{3}{q}=\frac{21}{r}$$ to hold true, q and r must be of the same signs. 
Hence, this option too does not contradict the lines coinciding. 

Option E: p cannot be zero
In order for q and r to have real values, the fractions $$\frac{3}{q}=\frac{21}{r}=\frac{p}{-7}$$ must not be zero. 
Putting p=0 would give the value of p/(-7) as 0, which would then not give real values of q and r
Hence, p can never be zero. This statement, too, does not contradict the two lines being coincident. 

Now that A, D, and E are not our answers, we need to consider the more complex options. 

Option B: r is the smallest amongst p, q, r
We would want $$\frac{3}{q}=\frac{21}{r}$$ to hold true.
If we take q and r to be positive, we must take p to be negative, in that case p would be negative and the smallest. So, in order to consider the possibility of r being the smallest, we must take q and r as negative values.
Now, with q and r negative, a smaller number would have a bigger magnitude or a bigger absolute value.
We can try putting in values at this point to see if this would contradict the lines coinciding.
taking r as -21 and q as -3, we can take p as 7
These values would make the lines coincident, with r being the smallest value.
Hence, this options too does not necessarily contradcit the lines coinciding.

Option C: q is the largest amongst p, q, r
Follwoing the same logic as Optoin B, we cannot take q and r to be negative, as that would make p positive, making p the largest value.
Therefore, in order to consider the possibility of q being the largest, we must take positive values of q and r
Now, comparing $$\frac{3}{q}=\frac{21}{r}$$, if q is larger than r, then the numerator of the first term would be smaller than the numerator of the second term, and the denominator of the first term would be larger than the denominator of the second term
Making the fraction $$\frac{3}{q}$$ strictly smaller than $$\frac{21}{r}$$
If this option is true, the lines can never coincide, as the ratios of the coefficient can never be equal.

Therefore, Option C would be the correct answer. 

abbb can be written as (1000*a+100*b+10*b+1*b) = 1000a+111b

Now, we need to check to the necessary and sufficient condition for which (1000a+111b) is divisible by a.

We know that 1000a is always divisible by 'a', hence, we need to check for which condition, 111b is always divisible by a.

111b can be written as (3*37*b) => (3b*37) must be divisible by a.

a can't be a factor of 37, which implies 'a' is a factor of 3b => 3b is divisible by 'a'

The correct option is E

It is given that ABC is a right-angled triangle, where $$\angle\ B\ =\ 90^{\circ\ }$$

Now, we can use Pythagoras theorem to calculate the value of AC, which is equal to $$\sqrt{\ 18^2+24^2}=\sqrt{\ 324+576}=\sqrt{\ 900}\ =\ 30$$ cm.

Here, we have drawn a common tangent MN, which is perpendicular to BC => Triangle MNC is a right-angled triangle.

It is also be concluded that traingle MNC is similar to traingle ABC

Hence, the ratio of the sides of triangle MNC is the same as the ratio of the sides of triangle ABC.

Thus, AB: BC: AC = MN: NC: MC = 18:24:30 = 3: 4: 5

Let the length of side MN be 3x => the length of NC = 4x, and the length of MC = 5x.

In triangle MNC, the circle inside it is the in circle of triangle MNC.

Thus, the in radius (r) of the circle in triangle MNC = (3x+4x-5x)/2 = x.

We know that BC = BN+NC, where BN is the diameter of the first circle, which is equal to 2r = 2x (since, r = x), and NC = 4x

=> 24 = 2x+4x 

=> 6x = 24 => x = 4 cm.

Hence, the radius of the circle (x = r) is 4 cm.

The correct option is B

Let the total number of jewels be X. 
The jewels in the first box would be X/3
The jewels in the second box would be kX/5
And the number of jewels in the third box is 66

The jewels of the second and third boxes should add up to 2X/3 jewels, giving us the relation. 
$$\frac{kX}{5}+66=\frac{2X}{3}$$
$$66=\frac{X\left(10-3k\right)}{15}$$
$$X\left(10-3k\right)=2\times\ 3^2\times\ 5\times\ 11$$

We are given that k is a positive integer. 
Taking k= 1, we get (10 - 3k) to be 7, which is not present on the right-hand side. 
Taking k = 2, we get (10 - 3k) to be 4, which is again not present on the right-hand side. 
Taking k = 3, we get (10 - 3k) to be 1, which is possible on the right-hand side and would give the value fo X to be 990. 

Taking any further value of k would give a negative value of (10 - 3k), which would not be possible. 

Therefore, the king must have had 990 jewels. 

Hence, Option A is the correct answer.  

Let the price of one kg of potato be P, one kg of cabbage be C, and one kg of onion be O. 
we are given the equations, 
$$5C+4P=80$$    (i)
$$4C+5O=93$$   (ii)

In equation (ii), we get 3 as the unit digit on the right-hand side. This is only possible with 8 + 5 on the left-hand side. 
So, the unit place value of C must be 2, which means that C can have values 2, 12, 22, 32, and so on. The value of O must be an odd number. 

Using this information in equation (i), we can see that the value of 5C will always have 0 as its unit digit. Meaning that the value of 4P must also have its unit digit as zero. 
This would tell that P is a multiple of 5. 
4P can then only be 20, 40 or 60 for C to also have a positive integer value. 

If P is 5, C would be 12, giving the value of O to be 9
If P is 10, C would be 8, and this is consistent with our criteria that the unit digit of C must be 2
If P is 15, C would be 4, which is again not consistent with the same criteria. 

Therefore, the price of one kg of potato would be Rs. 5, one of of cabbage would be Rs. 12 and one kg of onion would be Rs. 9

Aman can buy a maximum of 11 kg of onions for Rs. 99, leaving him with Rs. 1

Therefore, Option E is the correct answer. 

Let the price of one kg of potato be P, one kg of cabbage be C, and one kg of onion be O.
we are given the equations,
$$5C+4P=80$$ (i)
$$4C+5O=93$$ (ii)

In equation (ii), we get 3 as the unit digit on the right-hand side. This is only possible with 8 + 5 on the left-hand side.
So, the unit place value of C must be 2, which means that C can have values 2, 12, 22, 32, and so on. The value of O must be an odd number.

Using this information in equation (i), we can see that the value of 5C will always have 0 as its unit digit. Meaning that the value of 4P must also have its unit digit as zero.
This would tell that P is a multiple of 5.
4P can then only be 20, 40 or 60 for C to also have a positive integer value.

If P is 5, C would be 12, giving the value of O to be 9
If P is 10, C would be 8, and this is consistent with our criteria that the unit digit of C must be 2
If P is 15, C would be 4, which is again not consistent with the same criteria.

Therefore, the price of one kg of potato would be Rs. 5, one of cabbage would be Rs. 12, and one kg of onion would be Rs. 9

The question says that Aman buys only potatoes and onions. The money left after purchasing these must be less than Rs. 5 so that he cannot buy an additional kg of potatoes. 
Now, we can buy a maximum of 10 kg of onions, which would cost us Rs. 90, forcing Aman to buy 2 kg of potatoes to meet the above conditions. 

So, the number of kg of opinion bought can vary from 1 to 10, giving us a total of 10 possible scenarios. We need to check where the number of potatoes bought in kg outweighs the onions bought. 

We see that when Aman buys 7 kg of onions, he also has to buy 7 kg of potatoes. So only when he buys 10, 9 and 8 kg of onions is the weight of onions bought more than the weight of the potatoes bought. 

So, the probability that more onions are bought than potatoes would be $$\frac{3}{10}$$

Therefore, Option A is the correct answer.    

Here, we need to represent all the given data in a three sets venn-diagram that is Data Analytics (A), Database Handling (H), and Coding (C).

If we write all the data that is given in the table, we will get the following diagram:

Now we need to infer the characteristics of x, which can be concluded from the given information.

We know that $$2-x\ge\ 0$$, hence the possible values of x is 0, 1, 2

It is also known that $$x-2\ge\ 0$$, which implies the possible values of x is 2, 3, 4, ....

Now, from both equation, the only common value of x is 2.

Hence, x can take only 2 as the value.

The correct option is B

Here, we need to represent all the given data in a three sets venn-diagram that is Data Analytics (A), Database Handling (H), and Coding (C).

If we write all the data that is given in the table, we will get the following diagram:

Now we need to infer the characteristics of x, which can be concluded from the given information.

We know that $$2-x\ge\ 0$$, hence the possible values of x is 0, 1, 2

It is also known that $$x-2\ge\ 0$$, which implies the possible values of x is 2, 3, 4, ....

Now, from both equation, the only common value of x is 2.

Hence, x can take only 2 as the value.

Now the total number of people who know atleast one of the three is (x+4)+(2-x)+x+x+(6-x)+(3-x)+(x-2)=x+13=2+13=15.

 We know the total is 41 members. So, the number of people who don't know atleast 1 is 41-15=26.

The first clue is the minimum value provided:

That is not one of the eleven values that we already have; this must mean that one of the missing values, A or B, must be 2

Since we are given that A is less than B, it must be A, which is equal to 2. 

Now, writing down the twelve known numbers, 
2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

we are also provided that the median is of the data set was 12, that would be possible if the value 12 was shifter from 6th position to 7th position.
This must mean that the other missing value would also be in the first 6 values of the data set, lying in the range $$2\le B\le12$$

Now, narrowing down our focus to the lower hinge, the value is given to be 6.5
The first six values would be 2, 5, 6, 7, 8 and one unknown value

The lower hinge would be 6.5 only if the missing value appears after 7
Narrowing down the possible values of B to be 7, 8, 9, 10, 11, 12

For the first question, of all the given options, only option C lies in this range and hence would be the answer. 

Therefore, option C is the correct answer.