Mission IBPS Clerk Mains 2017: Quadratic Equation Questions Day 5
Quadratic Equation Questions for IBPS Clerk Mains 2017. Welcome to the www.letsstudytogether.co online English section. If you are preparing for IBPS PO/Clerk 2017 and other competitive exams, you will come across a section of the Quant Section. Here we are providing you a Quant Quiz on Quadratic Equation for IBPS Clerk Mains Exam 2017″ based on the latest pattern for your daily practice.
LST Team created a Very special 30day IBPS Clerk Mains preparation Plan to crack IBPS Clerk Mains 2017, to help you study smarter and move closer to success. Here we provide you a Mission IBPS Clerk Mains 2017: Quadratic Equation Questions Day 5.
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Quadratic Equation Questions for IBPS Clerk Mains Exam 2017: Set – 7
Mission IBPS Clerk Mains 2017: Quadratic Equation Quiz Day 5
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Mission IBPS Clerk Mains 2017: Quadratic Equation Quiz Day 5
Quantative Aptitude Section
No Of Question: 10
Test Time: 10 Mins
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Question 1 of 10
1. Question
Category: QuantIn the following questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x^{2} +14x 207 = 0
II. y^{2}33y +266 = 0Correct
From equation I:
x^{2} +14x 207 = (x + 23)(x 9)= 0
=> x = 23, 9
From equation II:
y^{2}33y +266 = (y 19)(y 14) = 0
=> y = 19, 14X = 23 X = 9 Y =19 x < y x < y Y = 14 x < y x < y So, x < y
Incorrect
From equation I:
x^{2} +14x 207 = (x + 23)(x 9)= 0
=> x = 23, 9
From equation II:
y^{2}33y +266 = (y 19)(y 14) = 0
=> y = 19, 14X = 23 X = 9 Y =19 x < y x < y Y = 14 x < y x < y So, x < y

Question 2 of 10
2. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x^{2}44x +483 = 0
II. 25y^{2} +15y 54 = 0Correct
From equation I:
x^{2}44x +483 = (x 21)(x 23)= 0
=> x = 21, 23
From equation II:
25y^{2} +15y 54 = (5y 6)(5y + 9) = 0
=> y = 6/5, 9/5X = 21 X = 23 Y = 6/5 x > y x > y Y = 9/5 x > y x > y So, x > y
Incorrect
From equation I:
x^{2}44x +483 = (x 21)(x 23)= 0
=> x = 21, 23
From equation II:
25y^{2} +15y 54 = (5y 6)(5y + 9) = 0
=> y = 6/5, 9/5X = 21 X = 23 Y = 6/5 x > y x > y Y = 9/5 x > y x > y So, x > y

Question 3 of 10
3. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x^{2}31x +184 = 0
II. y^{2}22y +96 = 0Correct
From equation I:
x^{2}31x +184 = (x 8)(x 23)= 0
=> x = 8, 23
From equation II:
y^{2}22y +96 = (y 6)(y 16) = 0
=> y = 6, 16X = 8 X = 23 Y =6 x > y x > y Y = 16 x < y x > y So, relationship cannot be established between x and y
Incorrect
From equation I:
x^{2}31x +184 = (x 8)(x 23)= 0
=> x = 8, 23
From equation II:
y^{2}22y +96 = (y 6)(y 16) = 0
=> y = 6, 16X = 8 X = 23 Y =6 x > y x > y Y = 16 x < y x > y So, relationship cannot be established between x and y

Question 4 of 10
4. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. 15x^{2} +22x +8 = 0
II. y^{2} +y 462 = 0Correct
From equation I:
15x^{2} +22x +8 = (5x + 4)(3x + 2)= 0
=> x = 4/5, 2/3
From equation II:
y^{2} +y 462 = (y + 22)(y 21) = 0
=> y = 22, 21X = 4/5 X = 2/3 Y =22 x > y x > y Y = 21 x < y x < y So, relationship cannot be established between x and y
Incorrect
From equation I:
15x^{2} +22x +8 = (5x + 4)(3x + 2)= 0
=> x = 4/5, 2/3
From equation II:
y^{2} +y 462 = (y + 22)(y 21) = 0
=> y = 22, 21X = 4/5 X = 2/3 Y =22 x > y x > y Y = 21 x < y x < y So, relationship cannot be established between x and y

Question 5 of 10
5. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x^{2}5x 300 = 0
II. y^{2} +46y +528 = 0Correct
From equation I:
x^{2}5x 300 = (x + 15)(x 20)= 0
=> x = 15, 20
From equation II:
y^{2} +46y +528 = (y + 24)(y + 22) = 0
=> y = 24, 22X = 15 X = 20 Y =24 x > y x > y Y = 22 x > y x > y So, x > y
Incorrect
From equation I:
x^{2}5x 300 = (x + 15)(x 20)= 0
=> x = 15, 20
From equation II:
y^{2} +46y +528 = (y + 24)(y + 22) = 0
=> y = 24, 22X = 15 X = 20 Y =24 x > y x > y Y = 22 x > y x > y So, x > y

Question 6 of 10
6. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
187x^{2} + 6x – 1 = 0
153y^{2}– 26y + 1 = 0Correct
First equation is –
187x^{2} + 6x – 1 = 0
or, 187x^{2}– 11x + 17x – 1= 0
or, 11x (17x – 1) + 1 (17x – 1) = 0
or, (11x + 1) (17x – 1) = 0
x = – 1/11, 1/17
Second equation is –
153y^{2}– 26y + 1 = 0
or, 153y^{2}– 17y – 9y + 1 = 0
or, 17y (9y – 1) – 1 (9y – 1) = 0
or, (17y – 1) (9y – 1) = 0
y = 1/17, 1/9
y is always more than x or equal to x.Incorrect
First equation is –
187x^{2} + 6x – 1 = 0
or, 187x^{2}– 11x + 17x – 1= 0
or, 11x (17x – 1) + 1 (17x – 1) = 0
or, (11x + 1) (17x – 1) = 0
x = – 1/11, 1/17
Second equation is –
153y^{2}– 26y + 1 = 0
or, 153y^{2}– 17y – 9y + 1 = 0
or, 17y (9y – 1) – 1 (9y – 1) = 0
or, (17y – 1) (9y – 1) = 0
y = 1/17, 1/9
y is always more than x or equal to x. 
Question 7 of 10
7. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. 10x^{2} +5x 5 = 0
II. y^{2}16y +15 = 0Correct
From equation I:
10x^{2} +5x 5 = (5x + 5)(2x 1)= 0
=> x = 5/5, 1/2
From equation II:
y^{2}16y +15 = (y 1)(y 15) = 0
=> y = 1, 15
So, x < yIncorrect
From equation I:
10x^{2} +5x 5 = (5x + 5)(2x 1)= 0
=> x = 5/5, 1/2
From equation II:
y^{2}16y +15 = (y 1)(y 15) = 0
=> y = 1, 15
So, x < y 
Question 8 of 10
8. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. x = √484
II. 7y – 4x = 45Correct
From I, x=22
From II, 7y – 88 = 45
=> 7y = 133
=> y = 19
Hence x > yIncorrect
From I, x=22
From II, 7y – 88 = 45
=> 7y = 133
=> y = 19
Hence x > y 
Question 9 of 10
9. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I. 3.5x^{2}– 28x + 42 = 0
II. 1.5y^{2} + 3y – 12 = 0Correct
From I, 3.5(x2)(x6) = 0
=> x = 2, 6
From II, 1.5(y2)(y+4) = 0
=> y = 2, 4
Hence x ≥ yIncorrect
From I, 3.5(x2)(x6) = 0
=> x = 2, 6
From II, 1.5(y2)(y+4) = 0
=> y = 2, 4
Hence x ≥ y 
Question 10 of 10
10. Question
Category: QuantIn the following questions, two equations numbered I and II are given. You have to solve both the equations and choose the correct option.
I y^{6} = 729
II x^{5} = 243Correct
From I, y = ±3
From II, x = 3
Hence x ≥ yIncorrect
From I, y = ±3
From II, x = 3
Hence x ≥ y
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