Algebra is one of the most important areas of the CAT Quantitative Aptitude section. However, there is no dependable fixed number of algebra questions in every paper. Recent post-exam analyses have classified approximately four to seven questions as algebra in different CAT 2025 slots, depending on the slot and whether mixed-topic questions were counted under algebra.
The best preparation strategy is therefore not to predict an exact question count, but to become comfortable with the recurring concepts: equations, polynomials, inequalities, modulus, functions, logarithms, exponents, progressions, and algebraic maxima and minima.
This practice set contains 32 original CAT-style algebra questions, followed by an answer key and concise solutions. These are independent practice questions and are not reproduced from an official CAT paper.
Reviewed by Kaushik Kapoor
Co-founder and Director of Erudite, a premium MBA entrance coaching institute headquartered in Kolkata. An alumnus of XLRI Jamshedpur and St. Xavier's College, Kolkata, he has over 25 years of experience in mentoring students, MBA entrance preparation, test-prep strategy, and academic training.
Divide the questions into four timed sets of eight.
Students who need to rebuild their fundamentals should first revise polynomials, linear equations, quadratic equations, and arithmetic progressions from a reliable school-level mathematics source such as the official NCERT textbooks.
If 2x + 3y = 17 and 3x + 2y = 18, find (x + y).
The real solutions of 1/(x−1) + 1/(x+1) = 3/4 are r1 and r2. Find r1 + r2.
If x + 1/x = 3, find x2 + 1/x2.
The two real roots of x2 − (k+2)x + 2k = 0 differ by 2. How many real values of k are possible?
The positive roots α and β of x2 − 10x + m = 0 satisfy α2 + β2 = 58. Find m.
Find the remainder when x5 − 3x3 + 2x − 7 is divided by (x − 2).
The polynomial p(x) = x3 − 4x2 − x + 4 has three real roots. Find the sum of its positive roots.
How many real roots does the equation x4 − 5x2 + 4 = 0 have?
If x + y = 7 and x2 + y2 = 29, find x3 + y3.
The equation x2 + (k−3)x + k = 0 has equal roots. Find the sum of all possible values of k.
Find the sum of all real solutions of |x−1| + |x−4| = 5.
Find the minimum value of |x−2| + |x+1| + |x−5|.
How many integers in the interval [−2, 7] satisfy (x−1)(x−3)(x−5) ≤ 0?
Find the least positive integer satisfying (x−2)/(x+1) > 1/2.
For x > 0, find the minimum value of x + 9/x.
Find the maximum value of −2x2 + 8x + 5.
If x > 0, y > 0, and xy = 16, find the minimum value of x + y + 4/x + 4/y.
How many integral values of x satisfy |x−2| + |x+2| ≤ 8?
Let f(x) = (3x−2)/(x+4). Find f−1(1).
Let f(x) = x + 2 and g(x) = x2. Find the real value of x for which f(g(x)) = g(f(x)).
A function satisfies f(x) + f(1−x) = x2 − x + 1 for every real x. Find f(1/2).
Let f(x) = x2 − 4x + 7. Find the minimum value of f(f(x)).
Solve: log2(x−1) + log2(x−3) = 3.
If logx16 = 4, find x.
Solve: log2x + log4x = 6.
The real solutions of 2x + 2−x = 5/2 are a and b. Find a2 + b2.
The arithmetic progression 7, 11, 15, … has a sum of 250 for its first n terms. Find n.
A geometric progression has positive terms. Its third term is 12 and its fifth term is 48. Find the sum of its first six terms.
Find the sum, for n = 1 to 10, of n(n+1).
A sequence is defined by a1 = 2 and an+1 = 3an + 1. Find a4.
Find the sum, for n = 1 to 9, of 1/(n(n+1)).
How many real solutions does the equation |x2 − 5x + 6| = x − 2 have?
| Question | Answer | Question | Answer | Question | Answer | Question | Answer |
|---|---|---|---|---|---|---|---|
| 1 | 7 | 9 | 133 | 17 | 10 | 25 | 16 |
| 2 | 8/3 | 10 | 10 | 18 | 9 | 26 | 2 |
| 3 | 7 | 11 | 5 | 19 | 3 | 27 | 10 |
| 4 | 2 | 12 | 6 | 20 | −1/2 | 28 | 189 |
| 5 | 21 | 13 | 7 | 21 | 3/8 | 29 | 440 |
| 6 | 5 | 14 | 6 | 22 | 4 | 30 | 67 |
| 7 | 5 | 15 | 6 | 23 | 5 | 31 | 9/10 |
| 8 | 4 | 16 | 13 | 24 | 2 | 32 | 2 |
Add the two equations: (2x+3y) + (3x+2y) = 17 + 18. Therefore, 5x + 5y = 35 and x + y = 7.
Answer: 7
Combine the fractions: [(x+1)+(x−1)] / (x2−1) = 3/4. Thus, 2x / (x2−1) = 3/4. Cross-multiplying, 8x = 3x2 − 3, so 3x2 − 8x − 3 = 0. By the sum-of-roots relation, r1 + r2 = 8/3. Neither root is 1 or −1, so both satisfy the original domain.
Answer: 8/3
Square the given relation: (x + 1/x)2 = 9. Therefore, x2 + 2 + 1/x2 = 9, which gives x2 + 1/x2 = 7.
Answer: 7
For a quadratic ax2 + bx + c = 0, the difference between its roots has magnitude √D / |a|, where D is the discriminant. Here, D = (k+2)2 − 8k = k2 − 4k + 4 = (k−2)2. The roots differ by 2, so √(k−2)2 = 2. Hence, |k−2| = 2, giving k = 0 or k = 4. There are two possible values.
Answer: 2
Using the sum and product of roots, α + β = 10 and αβ = m. Now, α2 + β2 = (α+β)2 − 2αβ. Therefore, 58 = 100 − 2m. Thus, 2m = 42 and m = 21.
Answer: 21
By the Remainder Theorem, the required remainder is p(2): 25 − 3(23) + 2(2) − 7 = 32 − 24 + 4 − 7 = 5.
Answer: 5
Factor by grouping: x3 − 4x2 − x + 4 = x2(x−4) − 1(x−4) = (x2−1)(x−4) = (x−1)(x+1)(x−4). The positive roots are 1 and 4. Their sum is 1 + 4 = 5.
Answer: 5
Let y = x2. Then y2 − 5y + 4 = 0, which factors as (y−1)(y−4) = 0. Thus, x2 = 1 or x2 = 4. The real roots are −2, −1, 1, 2.
Answer: 4
First find xy: x2 + y2 = (x+y)2 − 2xy. Therefore, 29 = 49 − 2xy, so xy = 10. Now, x3 + y3 = (x+y)3 − 3xy(x+y). Hence, x3 + y3 = 73 − 3(10)(7) = 343 − 210 = 133.
Answer: 133
Equal roots require the discriminant to be zero: (k−3)2 − 4k = 0. Expanding, k2 − 10k + 9 = 0. Thus, (k−1)(k−9) = 0. The possible values are 1 and 9, whose sum is 1 + 9 = 10.
Answer: 10
The expression represents the sum of the distances of x from 1 and 4. For 1 ≤ x ≤ 4, the sum is always 3, so there is no solution in that interval. For x ≤ 1: |x−1| + |x−4| = (1−x) + (4−x) = 5−2x. Thus, 5−2x = 5, giving x = 0. For x ≥ 4: |x−1| + |x−4| = (x−1) + (x−4) = 2x−5. Thus, 2x−5 = 5, giving x = 5. The sum of the solutions is 0 + 5 = 5.
Answer: 5
For a sum of distances from three points on a number line, the minimum occurs at the median point. The three points are −1, 2, and 5. Their median is 2. At x = 2: |2−2| + |2+1| + |2−5| = 0 + 3 + 3 = 6.
Answer: 6
The critical points are 1, 3, and 5. A sign check gives (x−1)(x−3)(x−5) ≤ 0 for (−∞, 1] ∪ [3, 5]. Within [−2, 7], the valid integers are −2, −1, 0, 1, 3, 4, 5. There are seven such integers.
Answer: 7
Move all terms to one side: (x−2)/(x+1) − 1/2 > 0. Combining the fractions, [2x−4−x−1] / [2(x+1)] > 0, so (x−5)/(x+1) > 0. This expression is positive when x < −1 or x > 5. The least positive integer satisfying it is 6.
Answer: 6
By AM–GM, for x > 0, x + 9/x ≥ 2√(x · 9/x). Therefore, x + 9/x ≥ 2√9 = 6. Equality occurs when x = 9/x, or x = 3.
Answer: 6
Complete the square: −2x2 + 8x + 5 = −2(x2−4x) + 5 = −2[(x−2)2−4] + 5 = −2(x−2)2 + 13. Since (x−2)2 ≥ 0, the maximum value is 13.
Answer: 13
Since xy = 16, 4/x + 4/y = 4[(x+y)/(xy)] = (x+y)/4. The expression becomes x + y + (x+y)/4 = (5/4)(x+y). By AM–GM, x + y ≥ 2√(xy) = 8. Hence, (5/4)(x+y) ≥ (5/4)×8 = 10. Equality occurs at x = y = 4.
Answer: 10
For −2 ≤ x ≤ 2, |x−2| + |x+2| = 4, so every integer in this interval works. For x ≥ 2, the expression equals 2x, and the inequality becomes 2x ≤ 8, so x ≤ 4. For x ≤ −2, the expression equals −2x, giving x ≥ −4. Therefore, −4 ≤ x ≤ 4. There are nine integers in this interval.
Answer: 9
To find f−1(1), solve f(x) = 1: (3x−2)/(x+4) = 1. Thus, 3x−2 = x+4, so 2x = 6 and x = 3.
Answer: 3
We have f(g(x)) = f(x2) = x2+2. Also, g(f(x)) = g(x+2) = (x+2)2. Equating them, x2+2 = x2+4x+4. Therefore, 4x = −2 and x = −1/2.
Answer: −1/2
Substitute x = 1/2. Then 1−x = 1/2, so 2f(1/2) = 1/4 − 1/2 + 1 = 3/4. Therefore, f(1/2) = 3/8.
Answer: 3/8
Rewrite f(x): f(x) = x2−4x+7 = (x−2)2+3. Thus, f(x) ≥ 3. Let t = f(x). Then t ≥ 3, and f(f(x)) = f(t) = (t−2)2+3. For t ≥ 3, this is minimized at t = 3. Therefore, f(f(x))min = (3−2)2+3 = 4. The value t = 3 is attainable when x = 2.
Answer: 4
The domain requires x > 3. Combine the logarithms: log2[(x−1)(x−3)] = 3. Therefore, (x−1)(x−3) = 8. Expanding, x2−4x+3 = 8, so x2−4x−5 = 0. Factoring, (x−5)(x+1) = 0. The possible algebraic values are 5 and −1, but only x = 5 satisfies the domain.
Answer: 5
The equation logx16 = 4 means x4 = 16. A logarithmic base must be positive and cannot equal 1. Therefore, x = −2 is invalid and x = 2.
Answer: 2
Let y = log2x. Since log4x = log2x / log24 = y/2, the equation becomes y + y/2 = 6. Thus, 3y/2 = 6 and y = 4. Therefore, x = 24 = 16.
Answer: 16
Let t = 2x. Then t > 0 and 2−x = 1/t. The equation becomes t + 1/t = 5/2. Multiplying by 2t: 2t2−5t+2 = 0. Factoring, (2t−1)(t−2) = 0. Thus, t = 1/2 or t = 2. Therefore, x = −1 or x = 1. Hence, a2+b2 = (−1)2+12 = 2.
Answer: 2
Here, a = 7, d = 4. The sum of the first n terms is Sn = (n/2)[2a+(n−1)d]. Thus, 250 = (n/2)[14+4(n−1)]. Therefore, 250 = (n/2)(4n+10) = n(2n+5). So, 2n2+5n−250 = 0. Factoring or using the quadratic formula gives the positive solution n = 10.
Answer: 10
Let the first term be a and the common ratio be r. The third and fifth terms give ar2 = 12 and ar4 = 48. Dividing, r2 = 4. Because all terms are positive, r = 2. Then a = 12/22 = 3. The sum of the first six terms is S6 = a(r6−1)/(r−1). Therefore, S6 = 3(26−1)/(2−1) = 3(63) = 189.
Answer: 189
Expand the summand: n(n+1) = n2+n. Therefore, the sum for n = 1 to 10 of n(n+1) equals the sum of n2 plus the sum of n. Using the standard sums, the sum of n2 from 1 to 10 is 385, and the sum of n from 1 to 10 is 55. Hence, 385 + 55 = 440.
Answer: 440
Calculate the terms successively: a1 = 2, a2 = 3(2)+1 = 7, a3 = 3(7)+1 = 22, and a4 = 3(22)+1 = 67.
Answer: 67
Use partial fractions: 1/[n(n+1)] = 1/n − 1/(n+1). Therefore, the sum telescopes: (1−1/2) + (1/2−1/3) + … + (1/9−1/10). All intermediate terms cancel, leaving 1−1/10 = 9/10.
Answer: 9/10
The right-hand side must be non-negative, so x ≥ 2. Factor the quadratic: x2−5x+6 = (x−2)(x−3). For 2 ≤ x ≤ 3, the quadratic is non-positive. Thus, |(x−2)(x−3)| = (x−2)(3−x). The equation becomes (x−2)(3−x) = x−2. This gives x = 2. For x ≥ 3, the quadratic is non-negative: (x−2)(x−3) = x−2. Therefore, (x−2)(x−4) = 0. Within x ≥ 3, this gives x = 4. The two real solutions are 2 and 4.
Answer: 2 real solutions
Domain restrictions are especially important in logarithmic, rational, radical, and inverse-function questions. An algebraically obtained value is not automatically a valid solution.
When a question gives information about the roots of a quadratic, avoid solving for the roots unless necessary. For ax2+bx+c = 0, with roots α and β, α+β = −b/a and αβ = c/a.
For expressions containing several absolute-value terms or polynomial factors:
This is safer than expanding cases without a clear interval structure.
Expressions involving x4 and x2 often become quadratic after substituting y = x2. Similarly:
The standard two-variable form a+b ≥ 2√(ab) requires a, b ≥ 0. Equality occurs when a = b.
A correct solution that required five or six minutes still indicates a strategy gap. During review, search for:
CAT does not have a fixed topic-wise allocation that candidates can safely rely on. Recent 2025 post-exam analyses classified approximately four to seven questions as algebra in different slots, although the number changes according to the slot and the treatment of mixed-topic questions. Prepare algebra as a major QA area, but do not build your strategy around one predicted question count.
No. The 32 questions in this set are original CAT-style practice questions. They are designed to test concepts commonly associated with CAT algebra but are not represented as official IIM questions.
A practical order is: linear equations; algebraic identities and factorization; quadratic equations; polynomials; inequalities; modulus; functions; logarithms and exponents; arithmetic and geometric progressions; and maxima and minima. Students with weak fundamentals should not skip introductory exercises. Basic fluency is necessary before moving to timed mixed-concept problems.
Eight to ten carefully reviewed questions are usually more useful than a large set completed without analysis. Record why each incorrect answer occurred and reattempt the question later without looking at the solution.
Memorize frequently used identities and standard results, but also understand their conditions and derivations. CAT questions often disguise familiar relationships, so recognition and application matter more than maintaining an unusually long formula list.
Improve speed by reducing unnecessary work:
CAT algebra preparation should combine conceptual revision, timed practice, and detailed error analysis. Use this set to identify the topics in which your approach is still slow or unreliable, and then practise those areas through focused sectional tests and mixed Quantitative Aptitude sets.
Do not judge progress only by the number of questions completed. Track whether you are selecting the right method, respecting domain restrictions, and reaching the answer with fewer avoidable steps.