Test: Quadratic Equations- 2 - CAT MCQ

# Test: Quadratic Equations- 2 - CAT MCQ

Test Description

## 10 Questions MCQ Test - Test: Quadratic Equations- 2

Test: Quadratic Equations- 2 for CAT 2024 is part of CAT preparation. The Test: Quadratic Equations- 2 questions and answers have been prepared according to the CAT exam syllabus.The Test: Quadratic Equations- 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Quadratic Equations- 2 below.
Solutions of Test: Quadratic Equations- 2 questions in English are available as part of our course for CAT & Test: Quadratic Equations- 2 solutions in Hindi for CAT course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt Test: Quadratic Equations- 2 | 10 questions in 10 minutes | Mock test for CAT preparation | Free important questions MCQ to study for CAT Exam | Download free PDF with solutions
Test: Quadratic Equations- 2 - Question 1

### 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- 2 - 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- 2 - Question 2
Detailed Solution for Test: Quadratic Equations- 2 - Question 2

 1 Crore+ students have signed up on EduRev. Have you?
Test: Quadratic Equations- 2 - Question 3

### Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?

Detailed Solution for Test: Quadratic Equations- 2 - 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- 2 - Question 4

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- 2 - 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- 2 - Question 5

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- 2 - 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- 2 - Question 6

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- 2 - 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- 2 - Question 7

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- 2 - 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- 2 - Question 8

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- 2 - Question 8

Explanation:

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

Test: Quadratic Equations- 2 - Question 9

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- 2 - 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- 2 - Question 10

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- 2 - 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.

In this test you can find the Exam questions for Test: Quadratic Equations- 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Quadratic Equations- 2, EduRev gives you an ample number of Online tests for practice