XAT 2025 Algebra Questions with Solutions

XAT 2025 Algebra Questions: The XAT 2025 exam will include roughly 28 questions in the Quantitative Ability and Data Interpretation section. This part of the exam can be very scoring since it covers topics that many students already know from their Class 11 and 12 Mathematics. Around 3-4 questions in this section will focus on Algebra, including topics like quadratic and linear equations. Since Algebra is one of the most important topics of the XAT 2025 syllabus, along with arithmetic, mensuration, and geometry, it's important to practice thoroughly with high-quality questions to perform well in the exam. Candidates can check here 15+ XAT 2025 algebra questions with solutions, based on previous year question papers.

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XAT 2025 Algebra Questions with Solutions

With around 3-4 questions expected from Algebra, thorough practice is very important. Below, you can find 15 XAT 2025 algebra questions with solutions, to help you practice and hone your skills:

Ques 1: What is the coefficient of x2 in the expansion of (5−x2/3)3?

1. -25
2. -25/3
3. 25
4. -5/3

Answer: Option A

Solution:

(5−x2/3)3= (5−x2/3) (5−x2/3)2

= (5−x2/3)(25+x4/9−10x2/3)

= 125+5x4/9−50x2/3−25x2/3−x6/27+10x4/9

= −x6/27+5x4/3−25x2+125

The coefficient of x2 in the expansion = -25

Ques 2: If y=2x+1, then what is the value of  (8x3–y3+6xy)?

1. 1
2. -1
3. 15
4. -15

Answer: Option B

Solution: 

y=2x+1

2x−y=−1…….(1)

Cubing on both sides, we get

8x3−y3−3.2x.y(2x−y)=−1

8x3−y3−6xy(−1)=−1 [From (1)]

8x3−y3+6xy=−1

Ques 3: If x–2/x=15, then what is the value of (x2+4/x2)?

1. 229
2. 227
3. 221
4. 223

Answer: Option A

Solution: 

x−2x=15

Squaring on both sides,

x2+4/x2−2.x.2x=225

x2+4/x2−4=225

x2+4/x2=229

Ques 4: If 2x+3y+1=0, then what is the value of (8x3+8+27y3–18xy)?

1. -7
2. 7
3. -9
4. 9

Answer: Option B

Solution:

2x+3y+1=0

2x+3y=−1……..(1)

Cubing on both sides,

8x3+27y3+3.2x.3y(2x+3y)=−1

8x3+27y3+18xy(−1)=−1

8x3+27y3−18xy+8=−1+8

8x3+27y3−18xy+8=7

Ques 5: If x+1/x=174,x>1, then what is the value of x-1/x

1. 9/4
2. 3/2
3. 8/3
4. 15/4

Answer: Option D

Solution:

x+1/x=174

(x+1/x)2=289/16

x2+1/x2+2=289/16

x2+1/x2=289/16−2

x2+1/x2=257/16

x2+1/x2−2=257/16−2

(x−1/x)2=257−32/16

(x−1/x)2=225/16

x−1/x=154

Ques 6: The roots of the polynomial P(x)=2x3−11x2+17x−6 are the radii of three concentric circles. The ratio of their area, when arranged from the largest to the smallest, is:

1. 6:2:1
2. 9:4:1
3. 16:6:3
4. 36:16:1`
5. None of the above

Solution: 

First, we have to find the roots of the cubical equation.

One root has to be figured out using the trial and error method.
Let's try 1.
P(x)=2x3−11x2+17x−6
P(1)=2∗13−11∗12+17−6=1
So, 1 is not a root.

Let's try 2,
P(1)=2∗23−11∗22+17 *2−6=1
Hence, two is one of the roots of this equation.
Now, using the synthetic division method cubical equation can be written as (x-2)(2x2−7x+3)=0
The roots of 2x2−7x+3 are 1/2,3
The area of circles with radii 3, 2, 1/2 are 9π, 4π, π/4
Their ratio of areas is 9:4:1/4
36:16:1 is the ratio.

Ques 7: Some members of a social service organization in Kolkata decide to prepare 2400 laddoos to give to children in various orphanages and slums in the city, during Durga puja. The plan is that each of them makes the same number of laddoos. However, on the ladoo-making day, ten members are absent, thus each remaining member makes 12  laddoos more than earlier decided.

How many members actually make the laddoos?

1. 100
2. 50
3. 90
4. 24
5. 40

Answer: Option E

Solution:

Initially considering the number of members = a

The number of ladoos each member is required to make as per the original plan = b.

Given : a*b = 2400.

Given that 10 members were absent and each member had to make an additional 12 ladoos :

(a-10)*(b+12) = 2400.

ab - 10b+12a-120 = 2400.

Since a*b = 2400.

Hence 12a-10b = 120.

Substituting b = 120/a.   

12a − 10⋅(2400/a) = 120

12a2 − 2400-120a=0

The roots are a = 50, a = -40.

Hence a = 50, b = 48.

The number of people who took part in making ladoos = a-10 = 40

Ques 8: The sum of the cubes of two numbers is 128, while the sum of the reciprocals of their cubes is 2.

What is the product of the squares of the numbers?

1. 64
2. 256
3. 16
4. 48
5. 32

Answer: Option C

Solution:

Considering the two numbers to a, b :

We were given that :

a3+b3 = 128

1/a3 + 1/b3 = 2

a3+b3/a3-b3= 2=120/k

k=64

Hence, a3.b3=64

a*b = 4 and a2 . b2 =16

Ques 9: Two different quadratic equations have a common root. Let the three unique roots of the two equations be A, B and C - all of them are positive integers. If (A + B + C) = 41 and the product of the roots of one of the equations is 35, which of the following options is definitely correct?

1. The common root is 29
2. The smallest among the roots is 1
3. One of the roots is 5
4. Product of the roots of the other equation is 5
5. All of the above are possible, but none are definitely correct.

Answer: Option C

Solution:

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. 

Ques 10: Consider the system of two linear equations as follows: 3x+21y+p=0; and qx+ry−7=0, where p, q, and r are real numbers. Which of the following statements DEFINITELY CONTRADICTS the fact that the lines represented by the two equations are coinciding?

1. P and q must have opposite signs
2. The smallest among p,q,r is r
3. The largest among p,q,r is q
4. r and q must have same signs
5. P cannot be 0

Answer: Option C

Solution:

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

That is, 3/q=21/r=p/-7

in order to consider the possibility of q being the largest, we must take positive values of q and r

Now, comparing 3/q=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 3/q strictly smaller than 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. 

Ques 11: The cost of running a movie theater is Rs. 10,000 per day, plus additional Rs. 5000 per show. The theater has 200 seats. A new movie released on Friday. There were three shows, where the ticket price was Rs. 250 each for the first two shows and Rs. 200 for the late-night show.

For all shows together, total occupancy was 80%. What was the maximum amount of profit possible?

1. Rs. 1,20,000
2. Rs. 87,000
3. Rs. 95,000
4. Rs. 91,000
5. Rs. 1,16,000

Answer: Option C

Solution:

The cost of running all three shows a day is 10,000+3*5,000=25,000.

There has to be 80% occupancy of total seats. In total there are 200*3 shows=600 seats.

Out of that 80% is 80/100 . 600 = 420

Since we need to maximize the profit. We shall take the vacant 120 seats in show 3 which cost less.

The income from first show 200 * 250= 50,000

The income from second show 200 * 250= 50,000

The income form third show 80*200= 16,000

The total income is 1,16,000-25,000=91,000

Ques 12: Aman has come to the market with Rs. 100. If he buys 5 kilograms of cabbage and 4 kilograms of potato, he will have Rs. 20 left; or else, if he buys 4 kilograms of cabbage and 5 kilograms of onion, he will have Rs. 7 left. The per kilogram prices of cabbage, onion and potato are positive integers (in rupees), and any type of these vegetables can only be purchased in positive integer kilogram, or none at all.

Aman decides to buy only onion, in whatever maximum quantity possible (in positive integer kilogram), with the money he has come to the market with. How much money will he be left with after the purchase?

1. Rs.12
2. Rs. 9
3. Rs. 7
4. Rs. 5
5. Rs. 1

Answer: Option E

Solution:

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

Ques 13: The Guava club has won 40% of their football matches in the Apple Cup that they have played so far. If they play another n match and win all of them, their winning percentage will improve to 50. Further, if they play 15 more matches and win all of them, their winning percentage will improve from 50 to 60. How many matches has the Guava club played in the Apple Cup so far? In the Apple Cup matches, there are only two possible outcomes, win or loss; draw is not possible.

1. 50
2. 40
3. 30
4. Cannot be determined, as the value of n is not given
5. 60

Answer: Option A

Solution:

Let 'm' be the number of matches Guava played till now. They won '0.4m' matches. 

After playing another 'n' match and winning all of them, their winning percentage will improve to 50.

i.e, 

0.4m+n/m+n = 0.5

m=5n

Playing 15 more matches and winning all of them, their winning percentage will improve from 50 to 60.

i.e 

0.4m+n+15/m+n+15 = 0.6

Solving, we get 6+0.4n = 0.2m

Substituting m=5n, we get 

6+0.4n = n

n=10 & m =50

Hence, Guava club played 50 matches so far. 

Ques 14: Let x and y be two positive integers and p be a prime number. If x (x - p) - y (y + p) = 7p, what will be the minimum value of x - y?

1. 1
2. 3
3. 5
4. 7
5. None of the above

Answer: Option E

Solution: 

The given equation is,

x (x - p) - y (y + p) = 7p

x2−px−y2-py=7p

x2−y2−px−y2-py=7p

(x+y)(x−y)−p(x+y)=7p

(x−y−p)(x+y)=7p

As '7' & 'p' both are prime numbers

(x−y−p)(x+y) can be expressed as 

(7× p) or (7p× 1)

Case (i) - 

(x+y) × (x−y−p)=7× p

 x+y+x−y−p=7+p

2x−p=7+p

x=7/2+p

But it's given that 'x' is a positive integer. This case is not possible.

Case (ii) -

(x+y) × (x−y−p)=7p× 1

x+y+x−y−p=7p+1                 

2x−p=7p+1   

x=1/2+p

But it's given that 'x' is a positive integer. This case is not possible.

The given equation is not possible with given conditions.

Ques 15: A shop sells bags in three sizes: small, medium and large. A large bag costs Rs.1000, a medium bag costs Rs.200, and a small bag costs Rs.50. Three buyers, Ashish, Banti and Chintu, independently buy some numbers of these types of bags. The respective amounts spent by Ashish, Banti and Chintu are equal. Put together, the shop sells 1 large bag, 15 small bags and some medium bags to these three buyers. What is the minimum number of medium bags that the shop sells to them?

1. 5
2. 9
3. 4
4. 10
5. 7

Answer: Option E

Solution:

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 medium bags = 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.

XAT 2025 Preparation Guide

By practicing the algebra questions provided in this article, candidates can sharpen their problem-solving skills and enhance their preparation. Best of luck with your XAT 2025 preparations! 

If you have further queries regarding the XAT exam, drop them in the CollegeDekho QnA zone . For assistance related to admission, fill out the Common Application Form (CAF) or call our toll-free number 1800-572-9877.

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