CRACK IBPS Clerk Prelims – Quadratic Equation Day 25
Dear Bankersdaily Zealot ,
IBPS Clerk preliminary examination will be held in the month of December this year and you know the preparations are to be in sky level to reach the perfect destination of achieving the feat. Daily Quizzes enrich your preparations to the next level and you could also learn the mistakes at a daily pace.
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IBPS Clerk Study Planner
Quadratic Equation day 25
Time : 10 Minutes
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Question 1 of 10
1. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x^{2} – 29x + 56 = 0
II. 3y^{2} – 5y – 8 = 0
Correct
(b)
3x^{2 }– 29x + 56 = 0
or 3x^{2 }– 21x – 8x + 56 = 0
or 3x(x – 7) – 8(x – 7) = 0
or (3x – 8) (x – 7) = 0
x = , 7
II. 3y^{2 }– 5y – 8 = 0
or 3y^{2 }+ 3y – 8y – 8 = 0
or 3y(y + 1) – 8(y + 1) = 0
or (3y – 8) (y + 1) = 0
or (3y – 8) (y + 1) = 0
y = 1,
x≥ y
Incorrect
(b)
3x^{2 }– 29x + 56 = 0
or 3x^{2 }– 21x – 8x + 56 = 0
or 3x(x – 7) – 8(x – 7) = 0
or (3x – 8) (x – 7) = 0
x = , 7
II. 3y^{2 }– 5y – 8 = 0
or 3y^{2 }+ 3y – 8y – 8 = 0
or 3y(y + 1) – 8(y + 1) = 0
or (3y – 8) (y + 1) = 0
or (3y – 8) (y + 1) = 0
y = 1,
x≥ y

Question 2 of 10
2. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 5x^{2} + 26x – 24 = 0
II. 5y^{2} – 34y + 24 = 0
Correct
(d)
5x^{2 }+ 26x – 24 = 0
or 5x^{2 }+ 30x – 4x – 24 = 0
or 5x(x + 6) – 4(x + 6) = 0
or (5x – 4) (x + 6) = 0
x = 4/5, 6
II. 5y^{2 }– 30y – 4y + 24 = 0
or 5y(y – 6) – 4(y – 6) = 0
or (5y – 4) (y – 6) = 0 4
y =4/5 6
x≤y
Incorrect
(d)
5x^{2 }+ 26x – 24 = 0
or 5x^{2 }+ 30x – 4x – 24 = 0
or 5x(x + 6) – 4(x + 6) = 0
or (5x – 4) (x + 6) = 0
x = 4/5, 6
II. 5y^{2 }– 30y – 4y + 24 = 0
or 5y(y – 6) – 4(y – 6) = 0
or (5y – 4) (y – 6) = 0 4
y =4/5 6
x≤y

Question 3 of 10
3. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x^{2}– 7x = 0
II. 2y^{2} + 5y + 3 = 0
Correct
(a)
x^{2 }– 7x = 0
or x (x – 7) = 0
x = 0, 7
II. 2y^{2 }+ 5y + 3 = 0
or 2y^{2 }+ 2y + 3y + 3 = 0
or 2y(y + 1) + 3(y + 1) = 0
or (2y + 3) (y + 1) = 0
y = 1, 3/2
x > y
Incorrect
(a)
x^{2 }– 7x = 0
or x (x – 7) = 0
x = 0, 7
II. 2y^{2 }+ 5y + 3 = 0
or 2y^{2 }+ 2y + 3y + 3 = 0
or 2y(y + 1) + 3(y + 1) = 0
or (2y + 3) (y + 1) = 0
y = 1, 3/2
x > y

Question 4 of 10
4. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 7x – 4y = 40
II. 8x + 8y = 8
Correct
(a)
7x – 4y = 40 …(i)
and 8x + 8y = 8
or x+y= 1 …(ii)
Solving (i) and (ii), we have
x = 4, y = 3
x > y
Incorrect
(a)
7x – 4y = 40 …(i)
and 8x + 8y = 8
or x+y= 1 …(ii)
Solving (i) and (ii), we have
x = 4, y = 3
x > y

Question 5 of 10
5. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 15x^{2} – 41x + 14 = 0
II. 2y^{2} – 13y + 20 = 0
Correct
(c)
15x^{2 }– 4x + 14 = 0
or 15x^{2 }– 6x – 35x + 14 = 0
or 3x(5x – 2) – 7(5x – 2) = 0
or (3x – 7)(5x – 2) = 0
x = 7/3, 2/5
II. 2y^{2 }– 13y + 20 = 0
or 2y^{2 }– 8y – 5y + 20 = 0
or 2y(y – 4) – 5(y – 4) = 0
or (2y – 5) (y – 4) = 0
y = 4, 5/2
x < y
Incorrect
(c)
15x^{2 }– 4x + 14 = 0
or 15x^{2 }– 6x – 35x + 14 = 0
or 3x(5x – 2) – 7(5x – 2) = 0
or (3x – 7)(5x – 2) = 0
x = 7/3, 2/5
II. 2y^{2 }– 13y + 20 = 0
or 2y^{2 }– 8y – 5y + 20 = 0
or 2y(y – 4) – 5(y – 4) = 0
or (2y – 5) (y – 4) = 0
y = 4, 5/2
x < y

Question 6 of 10
6. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x^{2}8√3x+45=0
II. y^{2}√2y24=0
Correct
(e)
I.x^{2 }8√3x + 45= 0
or x^{2} – 5√3x + 3√3 (x – 5√3) = 0
or, (x + 3√3) (x – 5√3) = 0
X = 3√3, 5√3
II. y^{2} – √2y – 24 = 0
Or y^{2} – 4√2y + 3 √2y – 24 = 0
Or (y4√2y) (y + 2√2)
y = 3 √2, 4√2
Hence relation cannot be established between x and y.
Incorrect
(e)
I.x^{2 }8√3x + 45= 0
or x^{2} – 5√3x + 3√3 (x – 5√3) = 0
or, (x + 3√3) (x – 5√3) = 0
X = 3√3, 5√3
II. y^{2} – √2y – 24 = 0
Or y^{2} – 4√2y + 3 √2y – 24 = 0
Or (y4√2y) (y + 2√2)
y = 3 √2, 4√2
Hence relation cannot be established between x and y.

Question 7 of 10
7. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x7√2x+24=0
II. y5√2y+12=0
Correct
x – 7 √2x + 24 = 0
Or x – 4√2x – 3 √2x + 24 = 0
Or √x (√x – 4√2) – 3√2 (√x – 4√2) = 0
Or (√x – 3√2) (√x – 4√2) = 0
Now, if √x 3√2 = 0
then √x = 3√2
x = 9 × 2 = 18
If √x – 4√2 = 0
then √x = 4√2
x = 16 × 2 = 32
II. y – 5√2y + 12 = 0
y 3√2y – 2√2y + 12 = 0
Or √y (√y – 3√2) 2√2y + 12 = 0
Or (√y – 2√2) – (√y 3√2) = 0
If (√y 2√2) = 0
Then √y = 2√2
y = 4 2 = 18
If √y 3√2 = 0
Then, √y √2
y = 9 × 2 = 18
x ≥ y
Incorrect
x – 7 √2x + 24 = 0
Or x – 4√2x – 3 √2x + 24 = 0
Or √x (√x – 4√2) – 3√2 (√x – 4√2) = 0
Or (√x – 3√2) (√x – 4√2) = 0
Now, if √x 3√2 = 0
then √x = 3√2
x = 9 × 2 = 18
If √x – 4√2 = 0
then √x = 4√2
x = 16 × 2 = 32
II. y – 5√2y + 12 = 0
y 3√2y – 2√2y + 12 = 0
Or √y (√y – 3√2) 2√2y + 12 = 0
Or (√y – 2√2) – (√y 3√2) = 0
If (√y 2√2) = 0
Then √y = 2√2
y = 4 2 = 18
If √y 3√2 = 0
Then, √y √2
y = 9 × 2 = 18
x ≥ y

Question 8 of 10
8. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 12x^{2} – 17x + 6 = 0
II. 20y^{2} – 31y + 12 = 0
Correct
(d)
12x^{2 }– 17x + 6 = 0
or 12x^{2 }– 9x – 8x + 6 = 0
or 3x(4x – 3) – 2(4x – 3) = 0
or (3x – 2) (4x – 3) = 0
If 3x – 2 = 0
then 3x = 2
x= 2/3
If 4x – 3 = 0
then x = 3/4
II. 20y^{2} – 31y + 12 = 0
or 20y^{2} – 16y – 15y + 12 = 0
or 4y (5y – 4) – 3 (5y – 4) = 0
or (4y – 3) (5y – 4) = 0
y=3/4,4/5
Hence x ≤ y
Incorrect
(d)
12x^{2 }– 17x + 6 = 0
or 12x^{2 }– 9x – 8x + 6 = 0
or 3x(4x – 3) – 2(4x – 3) = 0
or (3x – 2) (4x – 3) = 0
If 3x – 2 = 0
then 3x = 2
x= 2/3
If 4x – 3 = 0
then x = 3/4
II. 20y^{2} – 31y + 12 = 0
or 20y^{2} – 16y – 15y + 12 = 0
or 4y (5y – 4) – 3 (5y – 4) = 0
or (4y – 3) (5y – 4) = 0
y=3/4,4/5
Hence x ≤ y

Question 9 of 10
9. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x^{2} – 8x + 4 = 0
II. 4y^{2} – 15y + 9 = 0
Correct
(e)
3x^{2 }– 8x + 4 = 0
or 3x^{2 }– 6x – 2x + 4 = 0
or (3x – 2) (x – 2) = 0
x=2,2/3
II. 4y^{2} – 15y + 9 = 0
or 4y^{2} – 12y – 3y + 9 = 0
or 4y(y – 3) – 3(y – 3) = 0
or (4y – 3) (y – 3) = 0
y = 3/4, 3
Relation cannot be established between x and y.
Incorrect
(e)
3x^{2 }– 8x + 4 = 0
or 3x^{2 }– 6x – 2x + 4 = 0
or (3x – 2) (x – 2) = 0
x=2,2/3
II. 4y^{2} – 15y + 9 = 0
or 4y^{2} – 12y – 3y + 9 = 0
or 4y(y – 3) – 3(y – 3) = 0
or (4y – 3) (y – 3) = 0
y = 3/4, 3
Relation cannot be established between x and y.

Question 10 of 10
10. Question
1 pointsCategory: AptitudeD.15) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x^{2} 16x + 63 = 0
II. y^{2} – 2y – 35 = 0
Correct
(a)
I. x^{2 }– 16x + 63 = 0
or x^{2 } – 9x – 7x + 63 = 0
or x(x – 9) – 7(x – 9) = 0
or (x – 7) (x – 9) = 0
x = 7, 9
II. y^{2} – 2y – 35 = 0
or y^{2} – 17y + 5y – 35 = 0
or y(y – 7) + 5(y – 7) = 0
or (y + 5) (y – 7) = 0
y = 5, 7
Hence, x ≥ y
Incorrect
(a)
I. x^{2 }– 16x + 63 = 0
or x^{2 } – 9x – 7x + 63 = 0
or x(x – 9) – 7(x – 9) = 0
or (x – 7) (x – 9) = 0
x = 7, 9
II. y^{2} – 2y – 35 = 0
or y^{2} – 17y + 5y – 35 = 0
or y(y – 7) + 5(y – 7) = 0
or (y + 5) (y – 7) = 0
y = 5, 7
Hence, x ≥ y
IBPS CLERK Prelims Study Planner 2017
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