CAT 2026 · Slot 2 · Expected paper

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CAT 2026 Slot 2 Quant: Algebra Questions with Solutions

9 questions: Q2, Q4, Q6, Q10, Q11, Q12, Q18, Q19, Q22. Solve each one here: tap the answer, press Check, then open the solution.

Question 2Functions

MCQ · +3 / −1 · medium

A function f maps the set of natural numbers to whole numbers such that f(xy) = f(x)f(y) + f(x) + f(y) for all natural numbers x, y, and f(p) = 1 for every prime number p. Then, the value of f(54000) is:

Answer and solution

Correct answer: B

Solution

Answer: B — 1023

1. The given rule becomes much easier if 1 is added to both sides.
2. f(xy) + 1 = f(x)f(y) + f(x) + f(y) + 1
3. So,
4. f(xy) + 1 = [f(x) + 1][f(y) + 1]
5. For every prime number, f(p) = 1. Therefore, each prime factor contributes 2 to the expression on the right.
6. Now, 54000 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5.
7. There are 10 prime factors when repeated factors are counted.
8. Hence,
9. f(54000) + 1 = 2¹⁰ = 1024
10. So, f(54000) = 1024 − 1 = 1023.
11. Answer: 1023.

Why this method works:
Introduce g(n)=f(n)+1 because the added 1 completes the product (f(x)+1)(f(y)+1). The rule is completely multiplicative, so repeated prime factors each contribute a factor of 2; only counting distinct primes would be incorrect.

Independent check and common trap:
The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.

G Strategy shortcut

Spot: Algebra – Functions.
Apply: Add 1: g=f+1 is multiplicative and g(p)=2. 54000=2⁴×3³×5³, so g=2¹⁰ and f=1023.
Check: The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.

G Strategy: Spot: Algebra – Functions. Apply: Add 1: g=f+1 is multiplicative and g(p)=2. 54000=2⁴×3³×5³, so g=2¹⁰ and f=1023. Check: The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.

Question 4Inequalities

MCQ · +3 / −1 · medium

If x and y satisfy the equations	x	+ x + y = 18 and x +	y	− y = 14, then the value of (x − y) is:

Answer and solution

Correct answer: C

Solution

Answer: C — 12

1. Write the original equations:	x	+ x + y = 18 and x +	y	− y = 14.

2. Establish the sign of x. If x ≤ 0, then	x	+ x = 0, so y = 18. The second equation then gives x = 14, a contradiction. Therefore x > 0.

3. Establish the sign of y. If y ≥ 0, the second equation gives x = 14. The first then gives y = −10, another contradiction. Therefore y < 0.

4. Substitute	x	= x and	y	= −y. The equations become 2x + y = 18 and x − 2y = 14.

5. From the first, y = 18 − 2x. Substitute into the second: x − 2(18 − 2x) = 14, so 5x = 50 and x = 10. Hence y = −2.

6. Compute x − y = 10 − (−2) = 12, option C.

7. Check both original equations: 10 + 10 − 2 = 18 and 10 + 2 − (−2) = 14. Both hold. Proving the signs before removing the modulus prevents lost or invalid branches.

G Strategy shortcut

Spot: Algebra – Modulus and Equations.
Apply: x≤0 would force y=18 and then x=14, impossible. y≥0 also forces x=14 and fails. Hence x>0, y<0: solve 2x+y=18 and x−2y=14 → x−y=12.
Check: Substitute x=10, y=−2 into both original equations: 10+10−2=18 and 10+2−(−2)=14. The requested subtraction is 10−(−2)=12; it is not the sum 8.

G Strategy: Spot: Algebra – Modulus and Equations. Apply: x≤0 would force y=18 and then x=14, impossible. y≥0 also forces x=14 and fails. Hence x>0, y<0: solve 2x+y=18 and x−2y=14 → x−y=12. Check: Substitute x=10, y=−2 into both original equations: 10+10−2=18 and 10+2−(−2)=14. The requested subtraction is 10−(−2)=12; it is not the sum 8.

Question 6Progressions – AP/GP

MCQ · +3 / −1 · medium

The sum of the infinite series (1/4)(1/4 − 1/6) + (1/4)^2[(1/4)^2 − (1/6)^2] + (1/4)^3[(1/4)^3 − (1/6)^3] + ⋯ is equal to:

Answer and solution

Correct answer: B

Solution

Answer: B — 8/345

1. Look at the pattern in each term. The first term is (1/4)² − (1/24), and the following terms follow the same two geometric patterns.
2. The complete series can be separated into:
3. (1/16) + (1/16)² + (1/16)³ + ...
minus
(1/24) + (1/24)² + (1/24)³ + ...
4. The sum of an infinite geometric series is first term ÷ (1 − common ratio).
5. First series:
6. (1/16) ÷ (1 − 1/16) = 1/15
7. Second series:
8. (1/24) ÷ (1 − 1/24) = 1/23
9. Required sum = 1/15 − 1/23
10. Taking the LCM:
11. = (23 − 15)/345
= 8/345
12. Answer: 8/345.

Why this method works:
For the nth term, distribute (1/4)ⁿ across the bracket: it becomes (1/16)ⁿ−(1/24)ⁿ. Both common ratios have absolute value below 1, so both infinite geometric sums converge and may be subtracted.

Independent check and common trap:
The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.

G Strategy shortcut

Spot: Algebra – Infinite Series.
Apply: Each term is 16⁻ⁿ−24⁻ⁿ. Sum=1/(16−1)−1/(24−1)=1/15−1/23=8/345.
Check: The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.

G Strategy: Spot: Algebra – Infinite Series. Apply: Each term is 16⁻ⁿ−24⁻ⁿ. Sum=1/(16−1)−1/(24−1)=1/15−1/23=8/345. Check: The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.

Question 10Surds & Indices

TITA (type the answer) · +3 / 0 · medium

If √(x + √5) − √(x − √5) = √2, then the value of x is:

Your answer

Answer and solution

Correct answer: 3

Solution

Answer: 3

1. Start by squaring both sides:
2. [√(x + √5) − √(x − √5)]² = 2
3. After expanding:
4. x + √5 + x − √5 − 2√(x² − 5) = 2
5. So,
6. 2x − 2√(x² − 5) = 2
x − √(x² − 5) = 1
7. Therefore,
8. √(x² − 5) = x − 1
9. Squaring again:
10. x² − 5 = (x − 1)²
x² − 5 = x² − 2x + 1
2x = 6
x = 3
11. Answer: 3 (TITA).
12. Domain and substitution: x must be at least √5. At x=3, both radicals are real and their difference is positive with square 2, so the original equation holds.

Why this method works:
The domain x≥√5 ensures both original square roots exist. Squaring the difference introduces a cross term −2√(x²−5); omitting this term is the main algebra trap. Isolate the remaining radical before squaring a second time.

Independent check and common trap:
At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.

G Strategy shortcut

Spot: Algebra – Surds and Equations.
Apply: Square once: x−√(x²−5)=1. Square the isolated radical: 2x=6 →x=3. Substitute to verify the positive square-root difference.
Check: At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.

G Strategy: Spot: Algebra – Surds and Equations. Apply: Square once: x−√(x²−5)=1. Square the isolated radical: 2x=6 →x=3. Substitute to verify the positive square-root difference. Check: At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.

Question 11Surds & Indices

TITA (type the answer) · +3 / 0 · easy

If m and n are natural numbers such that n > 1, and m^n = 2^15 × 3^10, then the value of m − n is.

Your answer

Answer and solution

Correct answer: 67

Solution

Answer: 67

1. We are given:
2. mⁿ = 2¹⁵ × 3¹⁰
3. Since m and n are natural numbers, the exponent n must divide both 15 and 10.
4. The only common factor of 15 and 10 that is greater than 1 is 5.
5. Therefore, n = 5.
6. Now take the fifth root:
7. m = 2³ × 3²
m = 8 × 9
m = 72
8. Hence,
9. m − n = 72 − 5 = 67.
10. Answer: 67 (TITA).

Why this method works:
In a perfect nth power, every prime exponent is divisible by n. Here n must divide gcd(15,10)=5. The condition n>1 eliminates 1, leaving n=5 uniquely; no other prime factors can appear in m.

Independent check and common trap:
With m=72=2³×3², raising to the fifth power gives precisely 2¹⁵×3¹⁰. The final quantity is m−n, so subtract 5 from 72 after finding the base, rather than stopping at either intermediate value.

G Strategy shortcut

Spot: Number System – Exponents.
Apply: n divides both 15 and 10, with n>1: n=5. m=2³×3²=72; m−n=67.
Check: With m=72=2³×3², raising to the fifth power gives precisely 2¹⁵×3¹⁰. The final quantity is m−n, so subtract 5 from 72 after finding the base, rather than stopping at either intermediate value.

G Strategy: Spot: Number System – Exponents. Apply: n divides both 15 and 10, with n>1: n=5. m=2³×3²=72; m−n=67. Check: With m=72=2³×3², raising to the fifth power gives precisely 2¹⁵×3¹⁰. The final quantity is m−n, so subtract 5 from 72 after finding the base, rather than stopping at either intermediate value.

Question 12Equations

MCQ · +3 / −1 · medium

The roots α, β of the equation 2x² + λx − 1 = 0 satisfy 1/α² + 1/β² = 20. The value of (α³ + β³)² is:

Answer and solution

Correct answer: B

Solution

Answer: B — 121

1. For the quadratic equation, the sum of the roots is:
2. α + β = −λ/2
3. The product of the roots is:
4. αβ = −1/2
5. We are given:
6. 1/α² + 1/β² = 20
7. Combining the fractions:
8. [(α + β)² − 2αβ]/(αβ)² = 20
9. Since αβ = −1/2:
10. [(α + β)² + 1] ÷ (1/4) = 20
11. So,
12. (α + β)² + 1 = 5
(α + β)² = 4
13. Now use:
14. α³ + β³ = (α + β)³ − 3αβ(α + β)
15. = (α + β)[(α + β)² − 3αβ]
16. = (α + β)[4 + 3/2]
17. Squaring:
18. (α³ + β³)² = 4 × (11/2)²
= 4 × 121/4
= 121
19. Answer: 121.

Why this method works:
Use symmetric expressions instead of solving for the individual roots. Since αβ=−1/2, neither root is zero, so the reciprocals in the question are defined. Writing s=α+β retains exactly the information needed.

Independent check and common trap:
The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.

G Strategy shortcut

Spot: Algebra – Quadratic Equations.
Apply: Use s=α+β, p=−1/2. (s²−2p)/p²=20 gives s²=4. Required s²(s²−3p)²=4(11/2)²=121.
Check: The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.

G Strategy: Spot: Algebra – Quadratic Equations. Apply: Use s=α+β, p=−1/2. (s²−2p)/p²=20 gives s²=4. Required s²(s²−3p)²=4(11/2)²=121. Check: The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.

Question 18Inequalities

MCQ · +3 / −1 · medium

All the values of x satisfying the inequality 1/(x + 4) ≤ 1/(3x − 2) are:

Answer and solution

Correct answer: A

Solution

Answer: A — (−∞, −4) ∪ (2/3, 3]

1. First, note the values that are not allowed:
x ≠ −4 and x ≠ 2/3, because they make a denominator zero.
2. Bring both fractions to the left:
1/(x + 4) − 1/(3x − 2) ≤ 0
3. Take the common denominator:
[(3x − 2) − (x + 4)] / [(x + 4)(3x − 2)] ≤ 0
2(x − 3) / [(x + 4)(3x − 2)] ≤ 0
4. The critical values are −4, 2/3 and 3. They divide the number line into four intervals.
5. Now check the sign in each interval:
1. For x < −4, take x = −5. The expression is negative, so this interval is included.
2. For −4 < x < 2/3, take x = 0. The expression is positive, so this interval is excluded.
3. For 2/3 < x < 3, take x = 1. The expression is negative, so this interval is included.
4. For x > 3, take x = 4. The expression is positive, so this interval is excluded.
6. At x = 3, the numerator becomes zero while the denominators remain non-zero. Since the inequality includes equality (≤), x = 3 is included.
x = −4 and x = 2/3 are excluded because the original expression is undefined there.
7. Therefore, the solution is:
(−∞, −4) ∪ (2/3, 3]
8. Answer: Option A.

Why this method works:
Do not cross-multiply an inequality by (x+4)(3x−2) before checking its sign: the inequality reverses where that product is negative. The sign chart handles all intervals and denominator exclusions at once.

Independent check and common trap:
The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.

G Strategy shortcut

Spot: Algebra – Inequalities.
Apply: Combine to 2(x−3)/[(x+4)(3x−2)]≤0. Critical points −4,2/3,3; test−5,0,1,4 gives−,+,−,+. Include 3, exclude poles.
Check: The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.

G Strategy: Spot: Algebra – Inequalities. Apply: Combine to 2(x−3)/[(x+4)(3x−2)]≤0. Critical points −4,2/3,3; test−5,0,1,4 gives−,+,−,+. Include 3, exclude poles. Check: The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.

Question 19Logarithms

MCQ · +3 / −1 · medium

If a, b, c are positive real numbers such that a > 8 ≥ b ≥ c ≥ 2, and log₂(a + b) + log₂c = 6, then the greatest possible integer value of a is:

Answer and solution

Correct answer: B

Solution

Answer: B — 30

1. Combine the two logarithms:
2. log₂(a + b) + log₂c = log₂[c(a + b)]
3. Therefore,
4. log₂[c(a + b)] = 6
5. So,
6. c(a + b) = 2⁶ = 64
7. Rearranging:
8. a = 64/c − b
9. We need the greatest value of a. Since c and b are both at least 2, the smallest possible values are c = 2 and b = 2.
10. Putting these values in the expression:
11. a = 64/2 − 2
a = 32 − 2
a = 30
12. This also satisfies the condition a > 8.
13. Answer: 30.

Why this method works:
The upper bound follows without calculus: c≥2 gives 64/c≤32, and b≥2 subtracts at least 2. Therefore a≤30. A bound alone is insufficient unless a legal example reaches it.

Independent check and common trap:
The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.

G Strategy shortcut

Spot: Algebra – Logarithms.
Apply: c(a+b)=64. Since b, c≥2, a=64/c−b≤32−2=30. Equality at b=c=2 is legal.
Check: The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.

G Strategy: Spot: Algebra – Logarithms. Apply: c(a+b)=64. Since b, c≥2, a=64/c−b≤32−2=30. Equality at b=c=2 is legal. Check: The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.

Question 22Progressions – AP/GP

MCQ · +3 / −1 · medium

Consider the sequence t₁ = 1, t₂ = −1, and tₙ = [(n − 3)/(n − 1)]tₙ₋₂ for n ≥ 3. Then, the value of 1/t₂ + 1/t₄ + 1/t₆ + ⋯ + 1/t₂₀₂₄ + 1/t₂₀₂₆ is:

Answer and solution

Correct answer: B

Solution

Answer: B — −1,026,169

1. Focus on the even-numbered terms because the sum contains t₂, t₄, t₆ and so on.
2. For even n = 2k, the recurrence becomes:
3. t₂ₖ = [(2k − 3)/(2k − 1)] × t₂ₖ₋₂
4. We already know t₂ = −1.
5. Now calculate the next few terms:
6. t₄ = (1/3) × (−1) = −1/3
t₆ = (3/5) × (−1/3) = −1/5
t₈ = (5/7) × (−1/5) = −1/7
7. The pattern is clear:
8. t₂ = −1, t₄ = −1/3, t₆ = −1/5, t₈ = −1/7, and so on.
9. Taking reciprocals, the terms become:
10. −1, −3, −5, −7, ... , −2025
11. There are 1013 terms in this series.
12. The sum of the first n odd numbers is n². Therefore, the sum of these positive odd numbers is:
13. 1013² = 1,026,169
14. Since every term in our series is negative, the required sum is:
15. −1,026,169
16. Answer: −1,026,169.

Why this method works:
The recurrence connects terms two positions apart, so odd and even subsequences are independent. Only the even subsequence enters the sum. Its successive factors cancel: (1/3)(3/5)(5/7)…=1/(2k−1), with the initial negative sign retained.

Independent check and common trap:
The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.

G Strategy shortcut

Spot: Algebra – Sequences and Recurrence.
Apply: t₂k=−1/(2k−1), so reciprocals are−1,−3,…,−2025. Count 1013; sum=−1013²=−1026169.
Check: The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.

G Strategy: Spot: Algebra – Sequences and Recurrence. Apply: t₂k=−1/(2k−1), so reciprocals are−1,−3,…,−2025. Count 1013; sum=−1013²=−1026169. Check: The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.

An expected paper by Cetking, not the actual CAT paper. CAT is conducted by the IIMs.