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All questions of Quadratic Equations for SSC CGL Exam
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 16x + 55 = 0
II. y2 + y – 30 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 16x + 55 = 0
II. y2 + y – 30 = 0
I. x2 – 16x + 55 = 0
II. y2 + y – 30 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Krish Iyer answered |
Solving the Equations
To determine the relationship between x and y, we first need to solve both equations provided.
Equation I: x² - 16x + 55 = 0
1. Using the quadratic formula:
The roots are calculated using the formula x = [-b ± √(b² - 4ac)] / 2a, where a = 1, b = -16, and c = 55.
2. Calculating the discriminant:
Discriminant (D) = b² - 4ac = (-16)² - 4(1)(55) = 256 - 220 = 36.
3. Finding the roots:
x = [16 ± √36] / 2
x = [16 ± 6] / 2
- x₁ = (16 + 6) / 2 = 22 / 2 = 11
- x₂ = (16 - 6) / 2 = 10 / 2 = 5
So, the solutions for x are 11 and 5.
Equation II: y² + y - 30 = 0
1. Using the quadratic formula:
Here, a = 1, b = 1, and c = -30.
2. Calculating the discriminant:
D = 1² - 4(1)(-30) = 1 + 120 = 121.
3. Finding the roots:
y = [-1 ± √121] / 2
y = [-1 ± 11] / 2
- y₁ = (10) / 2 = 5
- y₂ = (-12) / 2 = -6
So, the solutions for y are 5 and -6.
Comparing x and y
- The values of x are 11 and 5.
- The values of y are 5 and -6.
Relationship Analysis
- If we compare the maximum value of x (which is 11) to the maximum value of y (which is 5), we find:
- 11 > 5 (x > y)
- Now comparing the minimum values:
- 5 (x) ≥ -6 (y)
Thus, it can be concluded that:
Final Conclusion
The relationship is:
- x ≥ y, as the maximum x (11) is greater than the maximum y (5) and the minimum x (5) is also greater than the minimum y (-6).
Therefore, the correct answer is option 'C': if x ≥ y.
To determine the relationship between x and y, we first need to solve both equations provided.
Equation I: x² - 16x + 55 = 0
1. Using the quadratic formula:
The roots are calculated using the formula x = [-b ± √(b² - 4ac)] / 2a, where a = 1, b = -16, and c = 55.
2. Calculating the discriminant:
Discriminant (D) = b² - 4ac = (-16)² - 4(1)(55) = 256 - 220 = 36.
3. Finding the roots:
x = [16 ± √36] / 2
x = [16 ± 6] / 2
- x₁ = (16 + 6) / 2 = 22 / 2 = 11
- x₂ = (16 - 6) / 2 = 10 / 2 = 5
So, the solutions for x are 11 and 5.
Equation II: y² + y - 30 = 0
1. Using the quadratic formula:
Here, a = 1, b = 1, and c = -30.
2. Calculating the discriminant:
D = 1² - 4(1)(-30) = 1 + 120 = 121.
3. Finding the roots:
y = [-1 ± √121] / 2
y = [-1 ± 11] / 2
- y₁ = (10) / 2 = 5
- y₂ = (-12) / 2 = -6
So, the solutions for y are 5 and -6.
Comparing x and y
- The values of x are 11 and 5.
- The values of y are 5 and -6.
Relationship Analysis
- If we compare the maximum value of x (which is 11) to the maximum value of y (which is 5), we find:
- 11 > 5 (x > y)
- Now comparing the minimum values:
- 5 (x) ≥ -6 (y)
Thus, it can be concluded that:
Final Conclusion
The relationship is:
- x ≥ y, as the maximum x (11) is greater than the maximum y (5) and the minimum x (5) is also greater than the minimum y (-6).
Therefore, the correct answer is option 'C': if x ≥ y.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 4x + 2y = 38
II. 3x – 5y = – 30- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 4x + 2y = 38
II. 3x – 5y = – 30
I. 4x + 2y = 38
II. 3x – 5y = – 30
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Talent Skill Learning answered |
I. 4x + 2y = 38
II. 3x – 5y = – 30
→ I × 5 + II × 2
4x × 5 + 3x × 2 = 38 × 5 – 30 × 2
26x = 130
x = 5
Answer: x < y
Hence, Option D is correct.
II. 3x – 5y = – 30
→ I × 5 + II × 2
4x × 5 + 3x × 2 = 38 × 5 – 30 × 2
26x = 130
x = 5
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x2 – 21x + 36 = 0
II. 5y2 – 25y + 30 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x2 – 21x + 36 = 0
II. 5y2 – 25y + 30 = 0
I. 3x2 – 21x + 36 = 0
II. 5y2 – 25y + 30 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Target Study Academy answered |
I. 3x2 – 21x + 36 = 0
3x2 – 12x – 9x + 36 = 0
II. 5y2 – 25y + 30 = 0
5y2 – 10y – 15y + 30 = 0
Answer: x ≥ y
Hence, Option C is correct.
3x2 – 12x – 9x + 36 = 0
II. 5y2 – 25y + 30 = 0
5y2 – 10y – 15y + 30 = 0
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 7x + 12y = 13
II. 12x + 17y = 8- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 7x + 12y = 13
II. 12x + 17y = 8
I. 7x + 12y = 13
II. 12x + 17y = 8
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Ssc Cgl answered |
I. 7x + 12y = 13
II. 12x + 17y = 8
→ I × 12 – II × 7
12y × 12 – 17y × 7 = 13 × 12 – 8 × 7
25y = 100
y = 4
7x + 12 × 4 = 13
x = – 5
Answer: x < y
Hence, Option D is correct.
II. 12x + 17y = 8
→ I × 12 – II × 7
12y × 12 – 17y × 7 = 13 × 12 – 8 × 7
25y = 100
y = 4
7x + 12 × 4 = 13
x = – 5
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 31x + 168 = 0
II. y2 + 5y – 14 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 31x + 168 = 0
II. y2 + 5y – 14 = 0
I. x2 + 31x + 168 = 0
II. y2 + 5y – 14 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Tarun Nambiar answered |
Solving the Equations
To determine the relationship between x and y, we need to solve both equations separately.
Equation I: x² + 31x + 168 = 0
1. Finding Roots:
- Use the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a.
- Here, a = 1, b = 31, c = 168.
- Discriminant = b² - 4ac = 31² - 4(1)(168) = 961 - 672 = 289.
- Roots: x = [-31 ± √289] / 2 = [-31 ± 17] / 2.
2. Calculating the Roots:
- x₁ = (-31 + 17) / 2 = -14 / 2 = -7.
- x₂ = (-31 - 17) / 2 = -48 / 2 = -24.
Thus, the solutions for x are: x = -7 and x = -24.
Equation II: y² + 5y - 14 = 0
1. Finding Roots:
- Again use the quadratic formula: y = [-b ± √(b² - 4ac)] / 2a.
- Here, a = 1, b = 5, c = -14.
- Discriminant = b² - 4ac = 5² - 4(1)(-14) = 25 + 56 = 81.
- Roots: y = [-5 ± √81] / 2 = [-5 ± 9] / 2.
2. Calculating the Roots:
- y₁ = (-5 + 9) / 2 = 4 / 2 = 2.
- y₂ = (-5 - 9) / 2 = -14 / 2 = -7.
Thus, the solutions for y are: y = 2 and y = -7.
Comparing x and y
- Values of x: -7, -24.
- Values of y: 2, -7.
Analyzing Relationships
- Case 1: If x = -7 and y = 2, then x < y.="" -="">case="" 2:="" if="" x="-24" and="" y="-7," then="" x="">< y.="" in="" both="" cases,="" x="" is="" less="" than="" or="" equal="" to="" y.="">Final Conclusion
The correct answer is option 'B': if x ≤ y.
To determine the relationship between x and y, we need to solve both equations separately.
Equation I: x² + 31x + 168 = 0
1. Finding Roots:
- Use the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a.
- Here, a = 1, b = 31, c = 168.
- Discriminant = b² - 4ac = 31² - 4(1)(168) = 961 - 672 = 289.
- Roots: x = [-31 ± √289] / 2 = [-31 ± 17] / 2.
2. Calculating the Roots:
- x₁ = (-31 + 17) / 2 = -14 / 2 = -7.
- x₂ = (-31 - 17) / 2 = -48 / 2 = -24.
Thus, the solutions for x are: x = -7 and x = -24.
Equation II: y² + 5y - 14 = 0
1. Finding Roots:
- Again use the quadratic formula: y = [-b ± √(b² - 4ac)] / 2a.
- Here, a = 1, b = 5, c = -14.
- Discriminant = b² - 4ac = 5² - 4(1)(-14) = 25 + 56 = 81.
- Roots: y = [-5 ± √81] / 2 = [-5 ± 9] / 2.
2. Calculating the Roots:
- y₁ = (-5 + 9) / 2 = 4 / 2 = 2.
- y₂ = (-5 - 9) / 2 = -14 / 2 = -7.
Thus, the solutions for y are: y = 2 and y = -7.
Comparing x and y
- Values of x: -7, -24.
- Values of y: 2, -7.
Analyzing Relationships
- Case 1: If x = -7 and y = 2, then x < y.="" -="">case="" 2:="" if="" x="-24" and="" y="-7," then="" x="">< y.="" in="" both="" cases,="" x="" is="" less="" than="" or="" equal="" to="" y.="">Final Conclusion
The correct answer is option 'B': if x ≤ y.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 4x – 3y = 16
II. 7x – 8y = – 16- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x = y or relationship between x and y can't be established
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 4x – 3y = 16
II. 7x – 8y = – 16
I. 4x – 3y = 16
II. 7x – 8y = – 16
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x = y or relationship between x and y can't be established
|
Iq Funda answered |
I. 4x – 3y = 16
II. 7x – 8y = – 16
4x – 3y = 8y – 7x
11x = 11y
x = y
Hence, Option D is correct.
II. 7x – 8y = – 16
4x – 3y = 8y – 7x
11x = 11y
x = y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 11x + 28 = 0
II. 2y2 + 30y + 112 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 11x + 28 = 0
II. 2y2 + 30y + 112 = 0
I. x2 + 11x + 28 = 0
II. 2y2 + 30y + 112 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Aspire Academy answered |
I. x2 + 11x + 28 = 0
x2 + 4x + 7x + 28 = 0
x = – 4 and – 7
x2 + 4x + 7x + 28 = 0
x = – 4 and – 7
II. 2y2 + 30y + 112= 0
2y2 + 14y + 16y + 112 = 0
Answer: x ≥ y
Hence, Option C is correct.
2y2 + 14y + 16y + 112 = 0
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x – √196 = 0
II. y2 – 196 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x – √196 = 0
II. y2 – 196 = 0
I. x – √196 = 0
II. y2 – 196 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Aim It Academy answered |
I. x – √196 = 0
x = 14
x = 14
II. y2 – 196 = 0
y = ±14
Answer: x ≥ y
Hence, Option C is correct.
y = ±14
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 144x2 = 16 + 20
II. 20y = √169 – √9- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 144x2 = 16 + 20
II. 20y = √169 – √9
I. 144x2 = 16 + 20
II. 20y = √169 – √9
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Talent Skill Learning answered |
I. 144x2 = 16 + 20
II. 20y = √169 – √9
20y = 13 – 3
20y = 10
Answer: x ≤ y
Hence, Option B is correct.
20y = 13 – 3
20y = 10
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 8x + 2y = 90
II. 3x + 7y = 90- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x = y or relationship between x and y can't be established
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 8x + 2y = 90
II. 3x + 7y = 90
I. 8x + 2y = 90
II. 3x + 7y = 90
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x = y or relationship between x and y can't be established
|
|
Aspire Academy answered |
8x + 2y = 3x + 7y
8x – 3x = 7y – 2y
x = y
Answer: x = y
Hence, Option D is correct.
8x – 3x = 7y – 2y
x = y
Answer: x = y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 13x + 22 = 0
II. y2 – 28y + 187 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 13x + 22 = 0
II. y2 – 28y + 187 = 0
I. x2 – 13x + 22 = 0
II. y2 – 28y + 187 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Learnpro Institute answered |
I. x2 – 13x + 22 = 0
x2 – 11x – 2x + 22 = 0
x = 11 and 2
x2 – 11x – 2x + 22 = 0
x = 11 and 2
II. y2 – 28y + 187 = 0
y2 – 11y – 17y + 187 = 0
y = 11 and 17
Answer: x ≤ y
Hence, Option B is correct.
y2 – 11y – 17y + 187 = 0
y = 11 and 17
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 19x – 42 = 0
II. y2 + 15y + 26 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 19x – 42 = 0
II. y2 + 15y + 26 = 0
I. x2 – 19x – 42 = 0
II. y2 + 15y + 26 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Learnpro Institute answered |
I. x2 – 19x – 42 = 0
x2 – 21x + 2x – 42= 0
x = 21 and – 2
x2 – 21x + 2x – 42= 0
x = 21 and – 2
II. y2 + 15y + 26 = 0
y2 + 2y + 13y + 26 = 0
y = – 2 and – 13
Answer: x ≥ y
Hence, Option C is correct.
y2 + 2y + 13y + 26 = 0
y = – 2 and – 13
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 38x + 72 = 0
II. y2 – 3y + 2 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 38x + 72 = 0
II. y2 – 3y + 2 = 0
I. x2 – 38x + 72 = 0
II. y2 – 3y + 2 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. x2 – 38x + 72 = 0
x2 – 36x – 2x + 72 = 0
x = 36 and 2
II. y2 – 3y + 2 = 0
y2 – y – 2y + 2 = 0
y = 1 and 2
Answer: x ≥ y
Hence, Option C is correct.
x2 – 36x – 2x + 72 = 0
x = 36 and 2
II. y2 – 3y + 2 = 0
y2 – y – 2y + 2 = 0
y = 1 and 2
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 17x + 7y = 323
II. 13x – 14y = – 82- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 17x + 7y = 323
II. 13x – 14y = – 82
I. 17x + 7y = 323
II. 13x – 14y = – 82
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Spectrum Coaching Institute answered |
I. 17x + 7y = 323
II. 13x – 14y = – 82
→ 2 × I + II
34x + 13x = 646 – 82
47x = 564
x = 12
Put the value of x = 12 in I
y = 17
Answer: x < y
Hence, Option D is correct.
II. 13x – 14y = – 82
→ 2 × I + II
34x + 13x = 646 – 82
47x = 564
x = 12
Put the value of x = 12 in I
y = 17
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 31x + 210 = 0
II. y2 + 3y + 2 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 31x + 210 = 0
II. y2 + 3y + 2 = 0
I. x2 + 31x + 210 = 0
II. y2 + 3y + 2 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. x2 + 31x + 210 = 0
x2 + 21x + 10x + 210 = 0
x = – 21 and – 10
x2 + 21x + 10x + 210 = 0
x = – 21 and – 10
II. y2 + 3y + 2 = 0
y2 + 2y + y + 2 = 0
y = –2 and –1
Answer: x < y
Hence, Option D is correct.
y2 + 2y + y + 2 = 0
y = –2 and –1
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 = 81
II. y2 – 32y + 240 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 = 81
II. y2 – 32y + 240 = 0
I. x2 = 81
II. y2 – 32y + 240 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aspire Academy answered |
I. x2 = 81
x = 9 and – 9
II. y2 – 32y + 240 = 0
y2 – 12y – 20y + 240 = 0
y = 12 and 20
Answer: x < y
Hence, Option D is correct.
x = 9 and – 9
II. y2 – 32y + 240 = 0
y2 – 12y – 20y + 240 = 0
y = 12 and 20
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 48x – 153 = 0
II. y2 = 32y – 156- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 48x – 153 = 0
II. y2 = 32y – 156
I. x2 + 48x – 153 = 0
II. y2 = 32y – 156
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Iq Funda answered |
I. x2 + 48x – 153 = 0
x2 + 51x – 3x – 153 = 0
x = – 51 and 3
x2 + 51x – 3x – 153 = 0
x = – 51 and 3
II. y2 = 32y – 156
y2 – 32y + 156 = 0
y2 – 26y – 6x + 156 = 0
y = 26 and 6
Answer: x < y
Hence, Option D is correct.
y2 – 32y + 156 = 0
y2 – 26y – 6x + 156 = 0
y = 26 and 6
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 32x + 247 = 0
II. y2 – 24y + 143 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 32x + 247 = 0
II. y2 – 24y + 143 = 0
I. x2 – 32x + 247 = 0
II. y2 – 24y + 143 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. x2 – 32x + 247 = 0
x2 – 19x – 13x + 247 = 0
x = 19 and 13
II. y2 – 24y + 143 = 0
y2 – 13y – 11y + 143 = 0
y = 13 and 11
Answer: x ≥ y
Hence, Option C is correct.
x2 – 19x – 13x + 247 = 0
x = 19 and 13
II. y2 – 24y + 143 = 0
y2 – 13y – 11y + 143 = 0
y = 13 and 11
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 18x + 65 = 0
II. 2y2 – 11y + 14 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'A'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 18x + 65 = 0
II. 2y2 – 11y + 14 = 0
I. x2 – 18x + 65 = 0
II. 2y2 – 11y + 14 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Talent Skill Learning answered |
I. x2 – 18x + 65 = 0
x2 – 5x – 13x + 65 = 0
x = 5 and 13
II. 2y2 – 11y + 14 = 0
2y2 – 7y – 4y + 14 =0
Answer: x > y
Hence, Option A is correct.
x2 – 5x – 13x + 65 = 0
x = 5 and 13
II. 2y2 – 11y + 14 = 0
2y2 – 7y – 4y + 14 =0
Answer: x > y
Hence, Option A is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x = ∛2197
II. 2y2 – 56y + 390 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x = ∛2197
II. 2y2 – 56y + 390 = 0
I. x = ∛2197
II. 2y2 – 56y + 390 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
Wizius Careers answered |
I. x = ∛2197
x = 13
II. 2y2 – 56y + 390 = 0
2y2 – 30y – 26y + 390 = 0
Answer: x ≤ y
Hence, Option B is correct.
x = 13
II. 2y2 – 56y + 390 = 0
2y2 – 30y – 26y + 390 = 0
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 13x + 40 = 0
II. y2 + 8y + 15 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 13x + 40 = 0
II. y2 + 8y + 15 = 0
I. x2 + 13x + 40 = 0
II. y2 + 8y + 15 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Learnpro Institute answered |
I. x2 + 13x + 40 = 0
x2 + 8x + 5x + 40 = 0
x = –8 and –5
II. y2 + 8y + 15 = 0
y2 + 3y + 5y + 15 = 0
y = – 3 and –5
Answer: x ≤ y
Hence, Option B is correct.
x2 + 8x + 5x + 40 = 0
x = –8 and –5
II. y2 + 8y + 15 = 0
y2 + 3y + 5y + 15 = 0
y = – 3 and –5
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + x – 20 = 0
II. 4y2 – 34y + 72 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + x – 20 = 0
II. 4y2 – 34y + 72 = 0
I. x2 + x – 20 = 0
II. 4y2 – 34y + 72 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. x2 + x – 20 = 0
x2 + 5x – 4x – 20 = 0
x = 4 and – 5
II. 4y2 – 34y + 72 = 0
4y2 – 18y – 16y + 72 = 0
Answer: x ≤ y
Hence, Option B is correct.
x2 + 5x – 4x – 20 = 0
x = 4 and – 5
II. 4y2 – 34y + 72 = 0
4y2 – 18y – 16y + 72 = 0
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 + 26x + 84 = 0
II. y2 + 11y + 30 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 + 26x + 84 = 0
II. y2 + 11y + 30 = 0
I. 2x2 + 26x + 84 = 0
II. y2 + 11y + 30 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Talent Skill Learning answered |
I. 2x2 + 26x + 84 = 0
2x2 + 14x + 12x + 84 = 0
II. y2 + 11y + 30 = 0
y2 + 5y + 6y + 30 = 0
y = – 5 and – 6
Answer: x ≤ y
Hence, Option B is correct.
2x2 + 14x + 12x + 84 = 0
II. y2 + 11y + 30 = 0
y2 + 5y + 6y + 30 = 0
y = – 5 and – 6
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 13x + 36 = 0
II. y2 – 7y + 12 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 13x + 36 = 0
II. y2 – 7y + 12 = 0
I. x2 – 13x + 36 = 0
II. y2 – 7y + 12 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. x2 – 13x + 36 = 0
x2 – 9x – 4x + 36 = 0
x = 9 and 4
x2 – 9x – 4x + 36 = 0
x = 9 and 4
II. y2 – 7y + 12= 0
y2 – 4y – 3y + 12 = 0
y = 4 and 3
Answer: x ≥ y
Hence, Option C is correct.
y2 – 4y – 3y + 12 = 0
y = 4 and 3
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x2 – 17x – 28 = 0
II. 2y2 – 41y + 210 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x2 – 17x – 28 = 0
II. 2y2 – 41y + 210 = 0
I. 3x2 – 17x – 28 = 0
II. 2y2 – 41y + 210 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. 3x2 – 17x – 28 = 0
3x2 – 21x + 4x – 28= 0
II. 2y2– 41y + 210 = 0
2y2 – 20y – 21y + 210 = 0
II. 2y2– 41y + 210 = 0
Answer: x < y
Hence, Option D is correct.
3x2 – 21x + 4x – 28= 0
II. 2y2– 41y + 210 = 0
2y2 – 20y – 21y + 210 = 0
II. 2y2– 41y + 210 = 0
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 27x + 140 = 0
II. y2 – 23y + 126 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 27x + 140 = 0
II. y2 – 23y + 126 = 0
I. x2 + 27x + 140 = 0
II. y2 – 23y + 126 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Talent Skill Learning answered |
I. x2 + 27x + 140 = 0
x2 + 20x + 7x + 140 = 0
x = – 20 and – 7
x2 + 20x + 7x + 140 = 0
x = – 20 and – 7
II. y2 – 23y + 126 = 0
y2 – 14y – 9y + 126 = 0
y = 14 and 9
x < y
y2 – 14y – 9y + 126 = 0
y = 14 and 9
x < y
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 – 4x + 2 = 0
II. 2y2 – 15y + 27 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 – 4x + 2 = 0
II. 2y2 – 15y + 27 = 0
I. 2x2 – 4x + 2 = 0
II. 2y2 – 15y + 27 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Wizius Careers answered |
I. 2x2 – 4x + 2 = 0
2x2 – 2x – 2x + 2 = 0
II. 2y2 – 15y + 27 = 0
2y2 – 6y – 9y + 27 = 0
Answer: x < y
Hence, Option D is correct.
2x2 – 2x – 2x + 2 = 0
II. 2y2 – 15y + 27 = 0
2y2 – 6y – 9y + 27 = 0
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 44x + 475 = 0
II. y2 – 34y + 285 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 44x + 475 = 0
II. y2 – 34y + 285 = 0
I. x2 – 44x + 475 = 0
II. y2 – 34y + 285 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. x2 – 44x + 475 = 0
x2 – 25x – 19x + 475 = 0
x = 25 and 19
II. y2 – 34y + 285 = 0
y2 – 19y – 15y + 285 = 0
y = 19 and 15
Answer: x ≥ y
Hence, Option C is correct.
x2 – 25x – 19x + 475 = 0
x = 25 and 19
II. y2 – 34y + 285 = 0
y2 – 19y – 15y + 285 = 0
y = 19 and 15
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x + 3y = √1764
II. 7x – 5y = √1444- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'A'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 3x + 3y = √1764
II. 7x – 5y = √1444
I. 3x + 3y = √1764
II. 7x – 5y = √1444
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Talent Skill Learning answered |
I. 3x + 3y = √1764
3x + 3y = 42
x + y = 14
3x + 3y = 42
x + y = 14
II. 7x – 5y = √1444
7x – 5y = 38
5 × I + II
12x = 70 + 38
x = 9
y = 14 – 9 = 5
x > y
Hence, Option A is correct.
7x – 5y = 38
5 × I + II
12x = 70 + 38
x = 9
y = 14 – 9 = 5
x > y
Hence, Option A is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 31x + 234 = 0
II. y2 + 22y + 117 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 31x + 234 = 0
II. y2 + 22y + 117 = 0
I. x2 + 31x + 234 = 0
II. y2 + 22y + 117 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Learnpro Institute answered |
I. x2 + 31x + 234 = 0
x2 + 18x + 13x + 234 = 0
x = –13 and –18
x2 + 18x + 13x + 234 = 0
x = –13 and –18
II. y2 + 22y + 117 = 0
y2 + 13y + 9y + 117 = 0
y = –13 and –9
Answer: x ≤ y
Hence, Option B is correct.
y2 + 13y + 9y + 117 = 0
y = –13 and –9
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. (x – 15)2 = 0
II. y2 = 196- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'A'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. (x – 15)2 = 0
II. y2 = 196
I. (x – 15)2 = 0
II. y2 = 196
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Spectrum Coaching Institute answered |
I. (x – 15)2 = 0
x = 15
x = 15
II. y2 = 196
y = ±14
Answer: x > y
Hence, Option A is correct.
y = ±14
Answer: x > y
Hence, Option A is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 56 = 15x
II. 2y2 – 23y + 60 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x = y or relationship between x and y can't be established
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 56 = 15x
II. 2y2 – 23y + 60 = 0
I. x2 + 56 = 15x
II. 2y2 – 23y + 60 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x = y or relationship between x and y can't be established
|
|
Iq Funda answered |
I. x2 + 56 = 15x
x2 – 15x + 56 = 0
x2 – 7x – 8x + 56 = 0
x = 7 and 8
x2 – 15x + 56 = 0
x2 – 7x – 8x + 56 = 0
x = 7 and 8
II. 2y2 – 23y + 60 = 0
2y2 – 15y – 8y + 60 = 0
Answer: The relation cannot be established.
Hence, Option D is correct.
2y2 – 15y – 8y + 60 = 0
Answer: The relation cannot be established.
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 31x + 228 = 0
II. y2 – 19y + 84 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 31x + 228 = 0
II. y2 – 19y + 84 = 0
I. x2 – 31x + 228 = 0
II. y2 – 19y + 84 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Wizius Careers answered |
I. x2 – 31x + 228 = 0
x2 – 12x – 19x + 228 = 0
x = 12 and 19
x2 – 12x – 19x + 228 = 0
x = 12 and 19
II. y2 – 19y + 84 = 0
y2 – 12y – 7y + 84 = 0
y = 12 and 7
Answer: x ≥ y
Hence, Option C is correct.
y2 – 12y – 7y + 84 = 0
y = 12 and 7
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 = 64
II. 2y2 + 25y + 72 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x = y or relationship between x and y can't be established
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 = 64
II. 2y2 + 25y + 72 = 0
I. x2 = 64
II. 2y2 + 25y + 72 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x = y or relationship between x and y can't be established
|
|
Learnpro Institute answered |
I. x2 = 64
x = ±8
II. 2y2 + 25y + 72 = 0
2y2 + 9y + 16y + 72 = 0
Answer: The relation cannot be established.
Hence, Option D is correct.
x = ±8
II. 2y2 + 25y + 72 = 0
2y2 + 9y + 16y + 72 = 0
Answer: The relation cannot be established.
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x3 – 9x2 + 18x = 0
II. y2 + 22y + 85 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'A'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x3 – 9x2 + 18x = 0
II. y2 + 22y + 85 = 0
I. x3 – 9x2 + 18x = 0
II. y2 + 22y + 85 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Spectrum Coaching Institute answered |
I. x3 – 9x2 + 18x = 0
x(x2 – 9x + 18) = 0
x2 – 9x + 18 = 0
x2 – 6x – 3x + 18 = 0
x = 0, 6 and 3
II. y2 + 22y + 85 = 0
y2 + 17y + 5y + 85 = 0
y = – 17 and – 5
x > y
Hence, Option A is correct.
x(x2 – 9x + 18) = 0
x2 – 9x + 18 = 0
x2 – 6x – 3x + 18 = 0
x = 0, 6 and 3
II. y2 + 22y + 85 = 0
y2 + 17y + 5y + 85 = 0
y = – 17 and – 5
x > y
Hence, Option A is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 2x – 15 = 0
II. y2 – 7y + 12 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 + 2x – 15 = 0
II. y2 – 7y + 12 = 0
I. x2 + 2x – 15 = 0
II. y2 – 7y + 12 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Wizius Careers answered |
I. x2 + 2x – 15 = 0
x2 + 5x – 3x – 15 = 0
x= 3 and – 5
II. y2 – 7y + 12 = 0
y2 – 4x – 3y + 12 = 0
y = 4 and 3
Answer: x ≤ y
Hence, Option B is correct.
x2 + 5x – 3x – 15 = 0
x= 3 and – 5
II. y2 – 7y + 12 = 0
y2 – 4x – 3y + 12 = 0
y = 4 and 3
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 – 21x + 49 = 0
II. 2y2 – 29y + 102 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x = y or relationship between x and y can't be established
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 – 21x + 49 = 0
II. 2y2 – 29y + 102 = 0
I. 2x2 – 21x + 49 = 0
II. 2y2 – 29y + 102 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x = y or relationship between x and y can't be established
|
|
Spectrum Coaching Institute answered |
I. 2x2 – 21x + 49 = 0
2x2 – 7x – 14x + 49 = 0
II. 2y2 – 29y + 102 = 0
2y2 – 12y – 17y + 102 = 0
Answer: The relationship cannot be established.
Hence, Option D is correct.
2x2 – 7x – 14x + 49 = 0
II. 2y2 – 29y + 102 = 0
2y2 – 12y – 17y + 102 = 0
Answer: The relationship cannot be established.
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 9x + 14 = 0
II. y2 – 14y + 49 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 9x + 14 = 0
II. y2 – 14y + 49 = 0
I. x2 – 9x + 14 = 0
II. y2 – 14y + 49 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aspire Academy answered |
I. x2 – 9x + 14 = 0
x2 – 7x – 2x + 14 = 0
x = 7 and 2
II. y2 – 14y + 49 = 0
y2 – 7y – 7y + 49 = 0
y = 7 and 7
Answer: x ≤ y
Hence, Option B is correct.
x2 – 7x – 2x + 14 = 0
x = 7 and 2
II. y2 – 14y + 49 = 0
y2 – 7y – 7y + 49 = 0
y = 7 and 7
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 – 17x + 36 = 0
II. 2y2 – 21y + 54 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 2x2 – 17x + 36 = 0
II. 2y2 – 21y + 54 = 0
I. 2x2 – 17x + 36 = 0
II. 2y2 – 21y + 54 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aspire Academy answered |
I. 2x2 – 17x + 36 = 0
2x2 – 8x – 9x + 36 = 0
2x2 – 8x – 9x + 36 = 0
II. 2y2 – 21y + 54 = 0
2y2 – 9y – 12y + 44 = 0
Answer: x ≤ y
Hence, Option B is correct.
2y2 – 9y – 12y + 44 = 0
Answer: x ≤ y
Hence, Option B is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 15 = 14x
II. 17y = – 14 – 3y2- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 15 = 14x
II. 17y = – 14 – 3y2
I. x2 – 15 = 14x
II. 17y = – 14 – 3y2
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Aim It Academy answered |
I. x2 – 15 = 14x
x2 – 14x – 15 = 0
x2 – 15x + x – 15 = 0
x = 15 and – 1
x2 – 14x – 15 = 0
x2 – 15x + x – 15 = 0
x = 15 and – 1
II. 17y = – 14 – 3y2
3y2 + 17y + 14 = 0
3y2 + 3y + 14y + 14 = 0
Answer: x ≥ y
Hence, Option C is correct.
3y2 + 17y + 14 = 0
3y2 + 3y + 14y + 14 = 0
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 4x + 6y = 82
II. 21x + 4y = 183- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. 4x + 6y = 82
II. 21x + 4y = 183
I. 4x + 6y = 82
II. 21x + 4y = 183
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Spectrum Coaching Institute answered |
I. 4x + 6y = 82
II. 21x + 4y = 183
→ 1.50 × II – I
31.50x – 4x = 274.50 – 82
27.50x = 192.50
x = 7
Put the value of x = 7 in I
4 × 7 + 6y = 82
28 + 6y = 82
6y = 54
y = 9
Answer: x < y
Hence, Option D is correct.
II. 21x + 4y = 183
→ 1.50 × II – I
31.50x – 4x = 274.50 – 82
27.50x = 192.50
x = 7
Put the value of x = 7 in I
4 × 7 + 6y = 82
28 + 6y = 82
6y = 54
y = 9
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 4x – 96 = 0
II. 4y2 – 109y + 741 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'D'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 4x – 96 = 0
II. 4y2 – 109y + 741 = 0
I. x2 – 4x – 96 = 0
II. 4y2 – 109y + 741 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Iq Funda answered |
I. x2 – 4x – 96 = 0
x2 – 12x + 8x – 96 = 0
x = 12 and – 8
x2 – 12x + 8x – 96 = 0
x = 12 and – 8
II. 4y2 – 109y + 741 = 0
4y2 – 52y – 57y + 741 = 0
Answer: x < y
Hence, Option D is correct.
4y2 – 52y – 57y + 741 = 0
Answer: x < y
Hence, Option D is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 14x + 45 = 0
II. y2 + 2y – 35 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 14x + 45 = 0
II. y2 + 2y – 35 = 0
I. x2 – 14x + 45 = 0
II. y2 + 2y – 35 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Learnpro Institute answered |
I. x2 – 14x + 45 = 0
x2 – 9x – 5x + 45 =0
x = 9 and 5
II. y2 + 2y – 35 = 0
y2 + 7y – 5y – 35 = 0
y = – 7 and 5
Answer: x ≥ y
Hence, Option C is correct.
x2 – 9x – 5x + 45 =0
x = 9 and 5
II. y2 + 2y – 35 = 0
y2 + 7y – 5y – 35 = 0
y = – 7 and 5
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 7x + 12 = 0
II. y2 + 5y – 24 = 0- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'C'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 7x + 12 = 0
II. y2 + 5y – 24 = 0
I. x2 – 7x + 12 = 0
II. y2 + 5y – 24 = 0
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Talent Skill Learning answered |
I. x2 – 7x + 12 = 0
x2 – 4x – 3x + 12 = 0
x = 4 and 3
x2 – 4x – 3x + 12 = 0
x = 4 and 3
II. y2 + 5y – 24= 0
y2 + 8y – 3y – 12 = 0
y = – 8 and 3
Answer: x ≥ y
Hence, Option C is correct.
y2 + 8y – 3y – 12 = 0
y = – 8 and 3
Answer: x ≥ y
Hence, Option C is correct.
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 1200 = 169
II. y + 122 = 159- a)if x > y
- b)if x ≤ y
- c)if x ≥ y
- d)if x < y
Correct answer is option 'B'. Can you explain this answer?
Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
I. x2 – 1200 = 169
II. y + 122 = 159
I. x2 – 1200 = 169
II. y + 122 = 159
a)
if x > y
b)
if x ≤ y
c)
if x ≥ y
d)
if x < y
|
|
Spectrum Coaching Institute answered |
I. x2 – 1200 = 169
x2 = 1200 + 169
x2 = 1369 = √(1369) = ± 37
x2 = 1200 + 169
x2 = 1369 = √(1369) = ± 37
II. y + 122 = 159
y = 159 – 122 = 37
Answer: x ≤ y
Hence, Option B is correct.
y = 159 – 122 = 37
Answer: x ≤ y
Hence, Option B is correct.
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Quantitative Aptitude for SSC CGL
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