CAT Maxima and Minima Bible Top 10 Must Solve Maxima and Minima Questions covering 90% of Syllabus – Page 2
All 40 Maxima and Minima questions from CAT papers. Tap Show answer, or open the detailed solution (free login). Page 2 of 9.
If log₆₄(x²) + log₈(√y) + 3 log₅₁₂( √(y)z) = 4, where x, y and z are positive real numbers, then the minimum possible value of (x + y + z) is
Answer: B. 48
Solution — first steps
Step 1: Convert all logs to the same base (base 8)
64 = 8²; 512 = 8³
Convert log₆₄(x²) to base 8: log₆₄(x²) = (1/2) log₈(x²) = log₈x
Convert 3 log₅₁₂(√(y)z) to base 8: 3 log₅₁₂(√(y)z) = 3 (1/3) log₈(√(y)z) = log₈(√(y)z)
Substitute the converted terms back into the original equation: log₈x + log₈(√y) + log₈(√(y)z) = 4 Combine the logarithmic terms using the property logₐA + logₐB = logₐ(AB): log₈(x ⋅ √y ⋅ √y ⋅ z) = 4 => log₈(xyz) = 4
Convert the logarithmic equation to an exponential equation: xyz = 8⁴ => xyz = 4096
Apply the Arithmetic Mean - Geometric Mean (AM-GM) inequality.
For positive real numbers x, y, and z, the minimum value of their sum occurs when x = y = z.
(x + y + z) / 3 ≥ ³√(xyz)
Substitute the value of xyz: (x + y + z) / 3 ≥ ³√(4096)
Calculate the cube root: Since 16³ = 4096, we have: (x + y + z) / 3 ≥ 16
Solve for the minimum value of (x + y + z): x + y + z ≥ 3 × 16 x + y + z ≥ 48
64 = 8²; 512 = 8³
Convert log₆₄(x²) to base 8: log₆₄(x²) = (1/2) log₈(x²) = log₈x
Convert 3 log₅₁₂(√(y)z) to base 8: 3 log₅₁₂(√(y)z) = 3 (1/3) log₈(√(y)z) = log₈(√(y)z)
Substitute the converted terms back into the original equation: log₈x + log₈(√y) + log₈(√(y)z) = 4 Combine the logarithmic terms using the property logₐA + logₐB = logₐ(AB): log₈(x ⋅ √y ⋅ √y ⋅ z) = 4 => log₈(xyz) = 4
Convert the logarithmic equation to an exponential equation: xyz = 8⁴ => xyz = 4096
Apply the Arithmetic Mean - Geometric Mean (AM-GM) inequality.
For positive real numbers x, y, and z, the minimum value of their sum occurs when x = y = z.
(x + y + z) / 3 ≥ ³√(xyz)
Substitute the value of xyz: (x + y + z) / 3 ≥ ³√(4096)
Calculate the cube root: Since 16³ = 4096, we have: (x + y + z) / 3 ≥ 16
Solve for the minimum value of (x + y + z): x + y + z ≥ 3 × 16 x + y + z ≥ 48
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In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is equal to half the number of students who chose both mathematics and physics. What is the maximum possible number of students who chose physics but not mathematics?
Answer: B. 35
Solution — first steps
Given Information: Physics students = 75, Mathematics students = 111 , Chemistry students = 40 Assume: x - number of students who chose both Physics and Mathematics , y - students who chose both Mathematics and Chemistry , z - student who chose both Physics and Chemistry , t - number of students who chose all three subjects And z = y , y = x/2 (Given) Using Set formula (the total number of students who chose at least one subject) |P ∪ M ∪ C| = |P| + |M| + |C| − |P ∩ M| − |M ∩ C| − |P ∩ C| + |P ∩ M ∩ C| Substitute the values: 150 = 75 + 111 + 40 - x - y - z + t Now, substitute z = y and y = x/2: 150 = 75 + 111 + 40 - x - (x/2) - (x/2) + t 150 = 226 - x - x + t 150 = 226 - 2x + t 2x - 76 = t (Equation 1) We want to maximize 75 - x , (students who chose only Physics but not mathematics). 30 , 75 - x = 30 , x = 45 , t = 14 Since 75 - x is maximum at option 2 thus option 2 is the answer
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Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose values lies between 150 and 500.If p is less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is: _____
Answer: 397
Solution — first steps
3 × 320 = 96 Pmax + P min = 301 + 96 = 397 Answer 397
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For a 4 digit number greater than 1000 , sum of the digits in thousands , hundreds and tens places is 15.And the sum of digits in hundreds , tens and units place is 16.Also the digits in tens place is 6 more than the digit in the unit place. The difference between the largest possible and smallest possible value of the number is :
Answer: D. 811
Solution — first steps
Given conditions for a 4-digit number: a + b + c = 15 —-- I b + c + d = 16 —-- II c = d + 6 —---- III Solving the equations: b = 10 - 2d a = d - 1 Now using values for d: If d = 3: a = 2, b = 4, c = 9, number = 2493 If d = 2: a = 1, b = 6, c = 8, number = 1682 Largest number = 2493, Smallest number = 1682 Difference = 2493 - 1682 = 811 Answer 811
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In a school with 1500 students , each students chooses one of the streams out of science , arts and commerce by paying a fee of rs 1100 , 1000 and 800 respectively. The total fees paid by all the students is RS. 15,50000.If the number of science students is not more than the number of arts students , then the maximum possible number of science students in school is : [TITA] _______
Answer: 700
Solution — first steps
Correct Answer: 700 Given S - science students , A - arts students , C - commerce students S + A + C = 1500 —- I 1100S + 1000A + 800C = 1550000 becomes 11S + 10A + 8C = 15500 —---- II Where S ≤ A , since we have to maximise S this S = A Multiplying I by 8 : 8S + 8A + 8C = 12000 —--- III Subtracting III from I 3S + 2A = 3500 , Since S = A 5S = 3500 , S = 700 Answer : 700
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Last day solving checklist
- Name the unknown. Include its unit and state whether it is a count, amount, rate or parameter.
- Form the relationships. Translate each given condition into an equation or restriction.
- Choose the method. Use the formula, substitution, elimination, a sign chart or a bound appropriate to this concept.
- Check restrictions. Check integer counts, signs, nonzero denominators, domains and all original conditions.
- Verify the answer. Substitute into the original question; for an extreme, show a feasible example reaches it.
All 16 repeating CAT PYQs and how much each adds
| CAT PYQ | Repeating CAT PYQ | Questions | Representative | Cumulative questions | Cumulative share |
|---|---|---|---|---|---|
| 1 | Integer allocation from linear totals and bounds | 9 | CAT 2022 S1 Q8 | 9 | 16.67% |
| 2 | Divisibility and factor constraints with feasible candidates | 9 | CAT 2022 S1 Q4 | 18 | 33.33% |
| 3 | Ordered or balanced integer values with fixed totals | 6 | CAT 2021 S3 Q7 | 24 | 44.44% |
| 4 | Smallest integer crossing an exponential threshold | 6 | CAT 2021 S3 Q20 | 30 | 55.56% |
| 5 | AM GM minimum or maximum with equality | 4 | CAT 2021 S3 Q19 | 34 | 62.96% |
| 6 | Set overlap extrema with attainability | 3 | CAT 2022 S1 Q17 | 37 | 68.52% |
| 7 | Digit restrictions and positional extrema | 3 | CAT 2021 S2 Q12 | 40 | 74.07% |
| 8 | Compare finitely many algebraic parameter cases | 3 | CAT 2023 S2 Q21 | 43 | 79.63% |
| 9 | Quadratic vertex by completing the square | 2 | CAT 2021 S3 Q22 | 45 | 83.33% |
| 10 | Extreme integer parameter from a discriminant | 2 | CAT 2023 S1 Q7 | 47 | 87.04% |
| 11 | Endpoint extrema of piecewise or linear expressions | 2 | CAT 2022 S1 Q22 | 49 | 90.74% |
| 12 | Largest common term of APs | 1 | CAT 2023 S2 Q22 | 50 | 92.59% |
| 13 | Bound a logarithmic condition and construct equality | 1 | CAT 2024 S2 Q9 | 51 | 94.44% |
| 14 | Maximise an integer representation of a surd | 1 | CAT 2024 S1 Q2 | 52 | 96.30% |
| 15 | Minimum labour cost with integer paid days | 1 | CAT 2025 S1 Q18 | 53 | 98.15% |
| 16 | Maximum admissible percentage decline | 1 | CAT 2023 S3 Q10 | 54 | 100% |
Based on CAT Actual Quant Papers.