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Test: Quadratic Equations (April 23) - CAT MCQ


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10 Questions MCQ Test Daily Test for CAT Preparation - Test: Quadratic Equations (April 23)

Test: Quadratic Equations (April 23) for CAT 2024 is part of Daily Test for CAT Preparation preparation. The Test: Quadratic Equations (April 23) questions and answers have been prepared according to the CAT exam syllabus.The Test: Quadratic Equations (April 23) MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Quadratic Equations (April 23) below.
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Test: Quadratic Equations (April 23) - Question 1
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Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?

Detailed Solution for Test: Quadratic Equations (April 23) - Question 1

Explanation:

a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab 
= [(a + b)2 - 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)2 - 2(4)]/4 = 56/4 = 14.

Test: Quadratic Equations (April 23) - Question 2
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Detailed Solution for Test: Quadratic Equations (April 23) - Question 2

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Test: Quadratic Equations (April 23) - Question 3
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Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?

Detailed Solution for Test: Quadratic Equations (April 23) - Question 3

Explanation:

The quadratic equation whose roots are reciprocal of 2x2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3x2 + 5x + 2 = 0

Test: Quadratic Equations (April 23) - Question 4
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A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
Detailed Solution for Test: Quadratic Equations (April 23) - Question 4

Explanation:

Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x 
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.

Test: Quadratic Equations (April 23) - Question 5
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I. a2 - 7a + 12 = 0,
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?
Detailed Solution for Test: Quadratic Equations (April 23) - Question 5

Explanation:

I.(a - 3)(a - 4) = 0
=> a = 3, 4


II. (b - 2)(b - 1) = 0
=> b = 1, 2
=> a > b

Test: Quadratic Equations (April 23) - Question 6
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I. a2 - 9a + 20 = 0,
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?
Detailed Solution for Test: Quadratic Equations (April 23) - Question 6

Explanation:

I. (a - 5)(a - 4) = 0
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b

Test: Quadratic Equations (April 23) - Question 7
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 I. a2 + 11a + 30 = 0,
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?
Detailed Solution for Test: Quadratic Equations (April 23) - Question 7

Explanation:

I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b

Test: Quadratic Equations (April 23) - Question 8
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I. a2 + 8a + 16 = 0,
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?
Detailed Solution for Test: Quadratic Equations (April 23) - Question 8

Explanation:

I. (a + 4)2 = 0 => a = -4
II.(b - 3)(b - 1) = 0
=> b = 1, 3 => a < b

Test: Quadratic Equations (April 23) - Question 9
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I. a2 - 2a - 8 = 0,
II. b2 = 9 to solve both the equations to find the values of a and b?
Detailed Solution for Test: Quadratic Equations (April 23) - Question 9

Explanation:

I. (a - 4)(a + 2) = 0
=> a = 4, -2
II. b2 = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.

Test: Quadratic Equations (April 23) - Question 10
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I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?
Detailed Solution for Test: Quadratic Equations (April 23) - Question 10

I. x2 + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.

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