CAT MCQs on Maxima and Minima: CAT Questions for Practice with Solutions

The CAT QA section requires speed and accuracy, along with a thorough understanding of the Maxima and Minima. This article provides a set of MCQs on Maxima and Minima to help you understand the topic and improve your problem-solving skills with the help of detailed solutions by ensuring conceptual clarity, which will help you in the CAT 2025 exam preparation

Whether you're revising the basics or testing your knowledge, these MCQs will serve as a valuable practice resource.

The CAT 2025 exam is expected to follow a similar trend to the CAT 2024, with 22 questions from the QA section out of a total of 68 questions.

CAT MCQs on Maxima and Minima

1. Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

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2. Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

\(6\sqrt{2}\)

\(8\sqrt{2}\)

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3. In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

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4. The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3)x - (a + 5) = 0 is

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5. From a square piece of cardboard measuring \( 2a \) on each side, a box with no top is to be formed by cutting out from each corner a square with sides \( b \) and bending up the flaps. The value of \( b \) for which the box has the greatest volume is:

\( \frac{a}{5} \)

\( \frac{a}{4} \)

\( \frac{a}{6} \)

\( \frac{2a}{3} \)

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CAT Questions

1.
In a class of 100 students, 73 like coffee, 80 like tea, and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is

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2.
The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3)x - (a + 5) = 0 is

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3.
In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

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4.
Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

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5.
The minimum possible value of \(\frac{x^2−6x+10}{3−x}\),for x<3,is

\(\frac{1}{2}\)
\(-\frac{1}{2}\)
2
-2

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6.
From a square piece of cardboard measuring \( 2a \) on each side, a box with no top is to be formed by cutting out from each corner a square with sides \( b \) and bending up the flaps. The value of \( b \) for which the box has the greatest volume is:

\( \frac{a}{5} \)
\( \frac{a}{4} \)
\( \frac{a}{6} \)
\( \frac{2a}{3} \)

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CAT MCQs on Maxima and Minima: CAT Questions for Practice with Solutions

CAT MCQs on Maxima and Minima

CAT Questions

1.
In a class of 100 students, 73 like coffee, 80 like tea, and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is

2.
The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3)x - (a + 5) = 0 is

4.
Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

5.
The minimum possible value of \(\frac{x^2−6x+10}{3−x}\),for x<3,is

6.
From a square piece of cardboard measuring \( 2a \) on each side, a box with no top is to be formed by cutting out from each corner a square with sides \( b \) and bending up the flaps. The value of \( b \) for which the box has the greatest volume is:

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CAT MCQs on Maxima and Minima

CAT Questions

1.In a class of 100 students, 73 like coffee, 80 like tea, and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is

2.The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3)x - (a + 5) = 0 is

4.Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

5.The minimum possible value of \(\frac{x^2−6x+10}{3−x}\),for x<3,is

6.From a square piece of cardboard measuring \( 2a \) on each side, a box with no top is to be formed by cutting out from each corner a square with sides \( b \) and bending up the flaps. The value of \( b \) for which the box has the greatest volume is:

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1.
In a class of 100 students, 73 like coffee, 80 like tea, and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is

2.
The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3)x - (a + 5) = 0 is

4.
Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

5.
The minimum possible value of \(\frac{x^2−6x+10}{3−x}\),for x<3,is

6.
From a square piece of cardboard measuring \( 2a \) on each side, a box with no top is to be formed by cutting out from each corner a square with sides \( b \) and bending up the flaps. The value of \( b \) for which the box has the greatest volume is: