Equations and Inequations Questions and Answers updated daily – Aptitude
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Equations and Inequations Questions
81. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. 3x2 - 10x + 8 = 0
II. 12y2 - 17y + 6 = 0
I. 3x2 - 10x + 8 = 0
II. 12y2 - 17y + 6 = 0
SHOW ANSWER
Correct Ans:x > y
Explanation:
I. 3x2 - 10x + 8 = 0
3x2 - 6x - 4x + 8 = 0
3x(x - 2) - 4(x - 2) = 0
(3x - 4)(x - 2) = 0
x = 4/3, 2
II. 12y2 - 17y + 6 = 0
12y2 - 9y - 8y + 6 = 0
3y(4y - 3) - 2(4y - 3) = 0
(3y - 2)(4y - 3) = 0
y = 2/3, 3/4
(x1, y1) = (4/3, 2/3) = x > y
(x1, y2) = (4/3, 3/4) = x > y
(x2, y1) = (2, 2/3) = x > y
(x2, y2) = (2, 3/4) = x > y
So, x > y
3x2 - 6x - 4x + 8 = 0
3x(x - 2) - 4(x - 2) = 0
(3x - 4)(x - 2) = 0
x = 4/3, 2
II. 12y2 - 17y + 6 = 0
12y2 - 9y - 8y + 6 = 0
3y(4y - 3) - 2(4y - 3) = 0
(3y - 2)(4y - 3) = 0
y = 2/3, 3/4
(x1, y1) = (4/3, 2/3) = x > y
(x1, y2) = (4/3, 3/4) = x > y
(x2, y1) = (2, 2/3) = x > y
(x2, y2) = (2, 3/4) = x > y
So, x > y
Workspace
82. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. 2x2 + 13x + 21 = 0
II. 2y2 + 11y + 15 = 0
I. 2x2 + 13x + 21 = 0
II. 2y2 + 11y + 15 = 0
SHOW ANSWER
Correct Ans:x ≤ y
Explanation:
I. 2x2 + 13x + 21 = 0
2x2 + 6x + 7x + 21 = 0
(x + 3)(2x + 7) = 0
x = -3, -7/2
II. 2y2 + 11y + 15 = 0
2y2 + 6y + 5y + 15 = 0
(y + 3)(2y + 5) = 0
y = -3, -5/2
(x1, y1) = (-3, -3) = (x = y)
(x1, y2) = (-3, -5/2) = x < y
(x2, y1) = (-7/2, -3) = x < y
(x2, y2) = (-7/2, -5/2) = x < y
So, x ≤ y
2x2 + 6x + 7x + 21 = 0
(x + 3)(2x + 7) = 0
x = -3, -7/2
II. 2y2 + 11y + 15 = 0
2y2 + 6y + 5y + 15 = 0
(y + 3)(2y + 5) = 0
y = -3, -5/2
(x1, y1) = (-3, -3) = (x = y)
(x1, y2) = (-3, -5/2) = x < y
(x2, y1) = (-7/2, -3) = x < y
(x2, y2) = (-7/2, -5/2) = x < y
So, x ≤ y
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83. In each of the following questions, read the given statement and compare the Quantity I and Quantity II on its basis. (only quantity is to be considered)
Quantity I: If 37a + 37b = 1036. What is the average of a and b?
Quantity II: The average of four consecutive numbers A, B, C and D is 27.5. What is the value of B?
Quantity I: If 37a + 37b = 1036. What is the average of a and b?
Quantity II: The average of four consecutive numbers A, B, C and D is 27.5. What is the value of B?
SHOW ANSWER
Correct Ans:Quantity I < Quantity II
Explanation:
Quantity I:
37a + 37b = 1036
a + b = 1036/37 = 28
Average Value = (a + b)/2
= 28/2 = 14
Quantity II:
Let the numbers A, B, C, D be x, x + 1, x + 2, x + 3.
or
[x + (x + 1) + (x + 2) + (x + 3)] / 4 = 27.5
4x + 6 = 27.5*4
4x = 104
x = 26
And value of B = x + 1 = 26 + 1 = 27
Hence Quantity II > Quantity I.
37a + 37b = 1036
a + b = 1036/37 = 28
Average Value = (a + b)/2
= 28/2 = 14
Quantity II:
Let the numbers A, B, C, D be x, x + 1, x + 2, x + 3.
or
[x + (x + 1) + (x + 2) + (x + 3)] / 4 = 27.5
4x + 6 = 27.5*4
4x = 104
x = 26
And value of B = x + 1 = 26 + 1 = 27
Hence Quantity II > Quantity I.
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84. In each of the following questions, read the given statement and compare the Quantity I and Quantity II on its basis. (only quantity is to be considered)
Quantity I : In the given figure, ABCD is a rectangle with dimensions 12 cm and 5 cm. What is the sum of the area of the triangle ADE and the triangle BCE?
Quantity II : 25 sq. cm.
Quantity I : In the given figure, ABCD is a rectangle with dimensions 12 cm and 5 cm. What is the sum of the area of the triangle ADE and the triangle BCE?
Quantity II : 25 sq. cm.
SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
We know that if base of triangle is same as base of rectangle ,then
Area of Triangle = Area of Rectangle / 2
Where Area of Rectangle = Length * Breadth
Now consider Triangle DEC
Area of DEC = (12*5) / 2
= 30 sq. cm.
Area of Triangle ADE and Area of Triangle BCE = Area of the Rectangle - Area of the Triangle DEC.
= (12*5) - 30
= 60 - 30
= 30 sq. cm
Quantity II : 25 sq. cm
Therefore Quantity I > Quantity II.
Area of Triangle = Area of Rectangle / 2
Where Area of Rectangle = Length * Breadth
Now consider Triangle DEC
Area of DEC = (12*5) / 2
= 30 sq. cm.
Area of Triangle ADE and Area of Triangle BCE = Area of the Rectangle - Area of the Triangle DEC.
= (12*5) - 30
= 60 - 30
= 30 sq. cm
Quantity II : 25 sq. cm
Therefore Quantity I > Quantity II.
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85. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
Train A crosses another train B in 15 seconds and they are running in the opposite directions. The length of train A and train B is 320m and 280m respectively and the ratio of the speed of train A and train B is 3 : 5 respectively.
Quantity I: Speed of train B.
Quantity II: 30 m/s.
Train A crosses another train B in 15 seconds and they are running in the opposite directions. The length of train A and train B is 320m and 280m respectively and the ratio of the speed of train A and train B is 3 : 5 respectively.
Quantity I: Speed of train B.
Quantity II: 30 m/s.
SHOW ANSWER
Correct Ans:Quantity I < Quantity II
Explanation:
Given:
Length of train A = 320 m
Length of train B = 380 m
Let the speed of train A be 'x' m/s and speed of the train B be 'y'
m/s.
WKT, Time = Distance/Speed
Time = (Length of train A + Length of train B)/ (Speed of train A + Speed of train B)
15 = (320 + 380)/(x + y)
x + y = 600/15
x + y = 40
So, Speed of the train A and B together = 40 m/s
Also given, ratio of the speed of train A and train B = 3 : 5
Speed of train B = (5/8)*40
= 25 m/s.
Hence, Quantity I < Quantity II.
Length of train A = 320 m
Length of train B = 380 m
Let the speed of train A be 'x' m/s and speed of the train B be 'y'
m/s.
WKT, Time = Distance/Speed
Time = (Length of train A + Length of train B)/ (Speed of train A + Speed of train B)
15 = (320 + 380)/(x + y)
x + y = 600/15
x + y = 40
So, Speed of the train A and B together = 40 m/s
Also given, ratio of the speed of train A and train B = 3 : 5
Speed of train B = (5/8)*40
= 25 m/s.
Hence, Quantity I < Quantity II.
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86. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
Quantity I: A and B started a business by investing in the ratio of 5: 8. If 20% of the total profit goes to Charity and the remaining profit will be shared by both of them and A's share is Rs. 20000, then find the total profit?
Quantity II: Rs. 50000
Quantity I: A and B started a business by investing in the ratio of 5: 8. If 20% of the total profit goes to Charity and the remaining profit will be shared by both of them and A's share is Rs. 20000, then find the total profit?
Quantity II: Rs. 50000
SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
The investment ratio of A and B = 5 : 8
A's share = 20,000
5x = 20,000
x = 4000
B's share = 8x = 8*4000 = 32,000
Let profit be X,
The remaining 80% of profit will be shared by both A and B.
80% of X = 52,000
X = (52,000*100)/80 = Rs. 65,000
Quantity I > Quantity II
The investment ratio of A and B = 5 : 8
A's share = 20,000
5x = 20,000
x = 4000
B's share = 8x = 8*4000 = 32,000
Let profit be X,
The remaining 80% of profit will be shared by both A and B.
80% of X = 52,000
X = (52,000*100)/80 = Rs. 65,000
Quantity I > Quantity II
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87. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
Quantity I: If the ratio of ages of Vicky and Teena, 5 years ago is 5: 8 and difference of their age is 15 years, then find the sum of the present age of Vicky and Teena?
Quantity II: If the average age of A, B, C and D is 27 years, then find the sum of the ages of all of them?
Quantity I: If the ratio of ages of Vicky and Teena, 5 years ago is 5: 8 and difference of their age is 15 years, then find the sum of the present age of Vicky and Teena?
Quantity II: If the average age of A, B, C and D is 27 years, then find the sum of the ages of all of them?
SHOW ANSWER
Correct Ans:Quantity I < Quantity II
Explanation:
Quantity I:
The ratio of ages of Vicky and Teena, 5 years ago = 5 : 8 (5x, 8x)
8x - 5x = 15
3x = 15
x = 5
Required sum = 13x + 5 + 5 = 13*5 + 10
= 65 + 10 = 75 years
Quantity II:
Required sum = 27 * 4 = 108 years
Quantity I < Quantity II
The ratio of ages of Vicky and Teena, 5 years ago = 5 : 8 (5x, 8x)
8x - 5x = 15
3x = 15
x = 5
Required sum = 13x + 5 + 5 = 13*5 + 10
= 65 + 10 = 75 years
Quantity II:
Required sum = 27 * 4 = 108 years
Quantity I < Quantity II
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88. Out of 1 5 applicants for a job, there are 7 women and 8 men. It is desired to select 2 persons for the job.
Quantity l: Probability of selecting no woman
Quantity II: Probability of selecting at least one woman
Quantity l: Probability of selecting no woman
Quantity II: Probability of selecting at least one woman
SHOW ANSWER
Correct Ans:Quantity I < Quantity II
Explanation:
Given:
Total applicants = 15; Number of women = 7; Number of men = 8
Quantity I:
Probability of selecting no woman = Probability of selecting both men
Therefore, probability of selecting both men = 8C2/15C2
= (8 x 7)/(15 x 14)
= 4/15
Quantity II:
Probability of selecting at least one woman = 1 - Probability of both men selected
= 1 - (4/15)
= 11/15
Hence, Quantity II < Quantity I.
Total applicants = 15; Number of women = 7; Number of men = 8
Quantity I:
Probability of selecting no woman = Probability of selecting both men
Therefore, probability of selecting both men = 8C2/15C2
= (8 x 7)/(15 x 14)
= 4/15
Quantity II:
Probability of selecting at least one woman = 1 - Probability of both men selected
= 1 - (4/15)
= 11/15
Hence, Quantity II < Quantity I.
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89. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. 28x2 - 8x - 11 = 0
II. 28y2 + 32y + 9 = 0
I. 28x2 - 8x - 11 = 0
II. 28y2 + 32y + 9 = 0
SHOW ANSWER
Correct Ans:x ≥ y
Explanation:
I. 28x2 - 8x - 11 = 0
28x2 + 14x - 22x - 11 = 0
14x(2x + 1) - 11(2x + 1) = 0
(14x - 11) (2x + 1) = 0
x = 11/14; -1/2
II. 28y2 + 32y + 9 = 0
28y2 + 18y + 14y + 9 = 0
2y(14y + 9) + 1(14y + 9) = 0
(2y + 1) (14y + 9) = 0
y = -1/2; -9/14
(x1, y1) = (11/14, -1/2) = x > y
(x1, y2) = (11/14, -9/14) = x > y
(x2, y1) = (-1/2, -1/2) = x = y
(x2, y2) = (-1/2, -9/14) = x > y
Hence, x ≥ y.
28x2 + 14x - 22x - 11 = 0
14x(2x + 1) - 11(2x + 1) = 0
(14x - 11) (2x + 1) = 0
x = 11/14; -1/2
II. 28y2 + 32y + 9 = 0
28y2 + 18y + 14y + 9 = 0
2y(14y + 9) + 1(14y + 9) = 0
(2y + 1) (14y + 9) = 0
y = -1/2; -9/14
(x1, y1) = (11/14, -1/2) = x > y
(x1, y2) = (11/14, -9/14) = x > y
(x2, y1) = (-1/2, -1/2) = x = y
(x2, y2) = (-1/2, -9/14) = x > y
Hence, x ≥ y.
Workspace
90. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. 100x2 - 120x + 32 = 0
II. 10y2 - 17y + 6 = 0
I. 100x2 - 120x + 32 = 0
II. 10y2 - 17y + 6 = 0
SHOW ANSWER
Correct Ans:x = y or no relation can be established
Explanation:
I. 100x2 - 120x + 32 = 0
100x2 - 40x - 80x + 32 = 0
20x(5x - 2) - 16(5x - 2) = 0
(20x -16) (5x - 2) = 0
x = 16/40; 2/5
x = 4/5; 2/5
II. 10y2 - 17y + 6 = 0
10y2 - 12y - 5y + 6 = 0
2y(5y - 6) - 1(5y - 6) = 0
(2y - 1) (5y - 6) = 0
y = 1/2, 6/5
(x1, y1) = (4/5, 1/2) = x > y
(x1, y2) = (4/5, 6/5) = y > x
(x2, y1) = (2/5, 1/2) = x < y
(x2, y2) = (2/5, 6/5) = x < y
Hence, x = y or no relation can be established.
100x2 - 40x - 80x + 32 = 0
20x(5x - 2) - 16(5x - 2) = 0
(20x -16) (5x - 2) = 0
x = 16/40; 2/5
x = 4/5; 2/5
II. 10y2 - 17y + 6 = 0
10y2 - 12y - 5y + 6 = 0
2y(5y - 6) - 1(5y - 6) = 0
(2y - 1) (5y - 6) = 0
y = 1/2, 6/5
(x1, y1) = (4/5, 1/2) = x > y
(x1, y2) = (4/5, 6/5) = y > x
(x2, y1) = (2/5, 1/2) = x < y
(x2, y2) = (2/5, 6/5) = x < y
Hence, x = y or no relation can be established.
Workspace
91. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I) 2x2 - 11x + 12 = 0
II) 2y2 - 17y + 36 = 0
I) 2x2 - 11x + 12 = 0
II) 2y2 - 17y + 36 = 0
SHOW ANSWER
Correct Ans:x ≤ y
Explanation:
I) 2x2 - 11x + 12 = 0
2x2 - 8x - 3x + 12 = 0
2x (x - 4) - 3 (x - 4) = 0
(2x - 3) (x - 4) = 0
x = 3/2, 4
II) 2y2 -17y +36 = 0
2y2 - 9y - 8y + 36 = 0
y (2y - 9) - 4 (2y - 9) = 0
(y - 4) (2y - 9) = 0
y = 4, 9/2
(x1, y1) = (3/2, 4) = x < y
(x1, y2) = (3/2, 9/2) = x < y
(x2, y1) = (4, 4) = (x = y)
(x2, y2) = (4, 9/2) = x < y
So, x ≤ y
2x2 - 8x - 3x + 12 = 0
2x (x - 4) - 3 (x - 4) = 0
(2x - 3) (x - 4) = 0
x = 3/2, 4
II) 2y2 -17y +36 = 0
2y2 - 9y - 8y + 36 = 0
y (2y - 9) - 4 (2y - 9) = 0
(y - 4) (2y - 9) = 0
y = 4, 9/2
(x1, y1) = (3/2, 4) = x < y
(x1, y2) = (3/2, 9/2) = x < y
(x2, y1) = (4, 4) = (x = y)
(x2, y2) = (4, 9/2) = x < y
So, x ≤ y
Workspace
92. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I) 2x2 + 3x - 20 = 0
II) 5y2 - 3y- 2 = 0
I) 2x2 + 3x - 20 = 0
II) 5y2 - 3y- 2 = 0
SHOW ANSWER
Correct Ans:x = y or relation between x and y cannot be determined.
Explanation:
I) 2x2 + 3x - 20 = 0
2x2 + 8x - 5x - 20 = 0
2x(x + 4) - 5(x + 4) = 0
(2x - 5)(x + 4) = 0
x = 5/2, -4
II) 5y2 - 3y - 2 = 0
5y2 - 5y + 2y - 2 = 0
5y(y - 1) + 2(y -1) = 0
(5y + 2) (y - 1) = 0
y = -2/5, 1
(x1, y1) = (5/2, -2/5) = x > y
(x1, y2) = (5/2, 1) = x > y
(x2, y1) = (-4, -2/5) = x < y
(x2, y2) = (-4, 1) = x < y
So, relation between x and y cannot be determined.
2x2 + 8x - 5x - 20 = 0
2x(x + 4) - 5(x + 4) = 0
(2x - 5)(x + 4) = 0
x = 5/2, -4
II) 5y2 - 3y - 2 = 0
5y2 - 5y + 2y - 2 = 0
5y(y - 1) + 2(y -1) = 0
(5y + 2) (y - 1) = 0
y = -2/5, 1
(x1, y1) = (5/2, -2/5) = x > y
(x1, y2) = (5/2, 1) = x > y
(x2, y1) = (-4, -2/5) = x < y
(x2, y2) = (-4, 1) = x < y
So, relation between x and y cannot be determined.
Workspace
93. Find the appropriate relation for quantity1 and quantity2 in the following question.
Quantity I: 13(2/5) - [4(1/2) + 3(1/6) + 2(1/3)]
Quantity II: √11.56
Quantity I: 13(2/5) - [4(1/2) + 3(1/6) + 2(1/3)]
Quantity II: √11.56
SHOW ANSWER
Correct Ans:Quantity I = Quantity II or Relation cannot be established
Explanation:
Quantity I: 13(2/5) - [4(1/2) + 3(1/6) + 2(1/3)]
= 13(2/5) - [4(1/2) + 3(1/6) + 2(1/3)]
= 67/5 - [(9/2) + (19/6) + (7/3)]
= 67/5 - [(27 + 19 + 14)/6]
= 67/5 - 60/6
= 67/5 - 10
= (67 - 50)/10
= 17/10 = 3.4
Quantity II: √11.56
√11.56 = 3.4
Hence, Quantity I = Quantity II.
= 13(2/5) - [4(1/2) + 3(1/6) + 2(1/3)]
= 67/5 - [(9/2) + (19/6) + (7/3)]
= 67/5 - [(27 + 19 + 14)/6]
= 67/5 - 60/6
= 67/5 - 10
= (67 - 50)/10
= 17/10 = 3.4
Quantity II: √11.56
√11.56 = 3.4
Hence, Quantity I = Quantity II.
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94. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
The ratio of A's income to B's income is 4 : 5 and the difference between their income is Rs. 10000.
Quantity I: A saves 30% of his income then what is his expenditure?
Quantity II: B spends 45% of his income then what is his saving?
The ratio of A's income to B's income is 4 : 5 and the difference between their income is Rs. 10000.
Quantity I: A saves 30% of his income then what is his expenditure?
Quantity II: B spends 45% of his income then what is his saving?
SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Given:
Ratio of A’s income to B’s income = 4 : 5
Let's A's income be 4X and B's income be 5X.
Difference of their income = Rs. 10,000
5X - 4X = 10000
X = 10,000
Hence, A's income = 4X = 4(10000) = Rs. 40,000
B's income = 5X = 5(10000) = Rs. 50,000
Quantity I:
A's expenditure = (100 - 30)% of 40,000
= (70/100)*40000
= Rs. 28,000
Quantity II:
B’s saving = (100 - 45)% of 50,000
= (55/100)*50000
= Rs. 27,500
Hence, Quantity I > Quantity II.
Ratio of A’s income to B’s income = 4 : 5
Let's A's income be 4X and B's income be 5X.
Difference of their income = Rs. 10,000
5X - 4X = 10000
X = 10,000
Hence, A's income = 4X = 4(10000) = Rs. 40,000
B's income = 5X = 5(10000) = Rs. 50,000
Quantity I:
A's expenditure = (100 - 30)% of 40,000
= (70/100)*40000
= Rs. 28,000
Quantity II:
B’s saving = (100 - 45)% of 50,000
= (55/100)*50000
= Rs. 27,500
Hence, Quantity I > Quantity II.
Workspace
95. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.
Quantity I: Cost Price of Item X
Quantity II: Selling Price of item Y after two successive discount of 10% and 15% is provided on the Marked Price of Rs 10,000.
The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.
Quantity I: Cost Price of Item X
Quantity II: Selling Price of item Y after two successive discount of 10% and 15% is provided on the Marked Price of Rs 10,000.
SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Given: MP = Rs. 10,000; Discount = 4%; Gain = 20%
Quantity I:
WKT, SP = [(100 - Discount)/100]*MP
SP = [(100 - 4)/100]*10000
SP = (96/100)*10000
SP = RS. 9600
Wkt, CP = SP[100/(100 + Profit)]
CP = 9600*[100/(100 + 20)]
CP = 9600*(100/120)
CP = Rs. 8000
Quantity II:
WKT, SP = [(100 - Discount)/100]*MP
For successive discount,
SP = 10000*[(100 - 10)/100]*[(100 - 15)/100]
SP = 10000*(90/100)*(85/100)
SP = Rs. 7650
Hence, Quantity I > Quantity II.
Quantity I:
WKT, SP = [(100 - Discount)/100]*MP
SP = [(100 - 4)/100]*10000
SP = (96/100)*10000
SP = RS. 9600
Wkt, CP = SP[100/(100 + Profit)]
CP = 9600*[100/(100 + 20)]
CP = 9600*(100/120)
CP = Rs. 8000
Quantity II:
WKT, SP = [(100 - Discount)/100]*MP
For successive discount,
SP = 10000*[(100 - 10)/100]*[(100 - 15)/100]
SP = 10000*(90/100)*(85/100)
SP = Rs. 7650
Hence, Quantity I > Quantity II.
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96. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. 21/√x + 11/√x = 7√x
II. 2y2 – 11y + 12 = 0
I. 21/√x + 11/√x = 7√x
II. 2y2 – 11y + 12 = 0
SHOW ANSWER
Correct Ans:x > y
Explanation:
I. 21/√x + 11/√x = 7√x
(21 + 11)/√x = 7√x
32/√x = 7√x
x = 32/7
II. 2y2 – 11y + 12 = 0
2y2 – 8y – 3y + 12 = 0
2y (y – 4) – 3 (y – 4) = 0
(2y – 3) (y – 4) = 0
y = 3/2, 4
(x, y1) = (32/7, 3/2) = x > y
(x, y2) = (32/7, 4) = x > y
So, x > y.
(21 + 11)/√x = 7√x
32/√x = 7√x
x = 32/7
II. 2y2 – 11y + 12 = 0
2y2 – 8y – 3y + 12 = 0
2y (y – 4) – 3 (y – 4) = 0
(2y – 3) (y – 4) = 0
y = 3/2, 4
(x, y1) = (32/7, 3/2) = x > y
(x, y2) = (32/7, 4) = x > y
So, x > y.
Workspace
97. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. x3 – 2744 = 0
II. y2 – 256 = 0
I. x3 – 2744 = 0
II. y2 – 256 = 0
SHOW ANSWER
Correct Ans:Relationship between x and y cannot be determined
Explanation:
I. x3 – 2744 = 0
x3 = 2744
x = 14
II. y2 – 256 = 0
y2 = 256
y = +16, -16
(x, y1) = (14, 16) = x < y1
(x, y2) = (14, -16) = x > y2
While comparing the values of x and y the relationship between x and y cannot be determined
x3 = 2744
x = 14
II. y2 – 256 = 0
y2 = 256
y = +16, -16
(x, y1) = (14, 16) = x < y1
(x, y2) = (14, -16) = x > y2
While comparing the values of x and y the relationship between x and y cannot be determined
Workspace
98. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. 4 = (3/x) + (5/2x2)
II. 14y = (6/y) + 5
I. 4 = (3/x) + (5/2x2)
II. 14y = (6/y) + 5
SHOW ANSWER
Correct Ans:x=y or Relationship between x and y cannot be determined
Explanation:
Given Equation I. 4 = (3/x) + (5/2x2)
---> 4 = (6x + 5)/2x2
---> 4 * 2x2 = (6x + 5)
---> 8x2 - 6x - 5 = 0
---> 8x2 - 10x + 4x - 5 = 0
----> 2x (4x - 5) + (4x - 5) = 0
----> (4x - 5) + (2x + 1) = 0
----> x = 5/4, -1/2
Equation II. 14y = (6/y) + 5
---> 14y - (6/y) = 5
---> (14y2 - 6) / y = 5
---> 14y2 - 6 -5y = 0
---> 14y2 -5y - 6 = 0
---> 14y2 + 7y -12y - 6 = 0
---> 7y (2y + 1) -6 (2y + 1) = 0
---> (2y + 1) (7y - 6) = 0
---> y = -1/2, 6/7
For x = 5/4, and y =-1/2OR 6/7; x > y
For x = -1/2, and y = -1/2, x = y
For x = -1/2 and y = 6/7, x < y
Hence, Relationship between x and y cannot be determined.
---> 4 = (6x + 5)/2x2
---> 4 * 2x2 = (6x + 5)
---> 8x2 - 6x - 5 = 0
---> 8x2 - 10x + 4x - 5 = 0
----> 2x (4x - 5) + (4x - 5) = 0
----> (4x - 5) + (2x + 1) = 0
----> x = 5/4, -1/2
Equation II. 14y = (6/y) + 5
---> 14y - (6/y) = 5
---> (14y2 - 6) / y = 5
---> 14y2 - 6 -5y = 0
---> 14y2 -5y - 6 = 0
---> 14y2 + 7y -12y - 6 = 0
---> 7y (2y + 1) -6 (2y + 1) = 0
---> (2y + 1) (7y - 6) = 0
---> y = -1/2, 6/7
For x = 5/4, and y =-1/2OR 6/7; x > y
For x = -1/2, and y = -1/2, x = y
For x = -1/2 and y = 6/7, x < y
Hence, Relationship between x and y cannot be determined.
Workspace
99. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.
I. x = ∛4913
II. 3y2 = ∛216000
I. x = ∛4913
II. 3y2 = ∛216000
SHOW ANSWER
Correct Ans:x > y
Explanation:
Given Equation I. x = ∛4913
---> x = 17
II. 3y2 = ∛216000
----> 3y2 = 60
----> y2 = 60/3
----> y2 = 20
----> y = 2√5
----> y = 2 * 2.236
---> y = 4.47
Here, x > y.
---> x = 17
II. 3y2 = ∛216000
----> 3y2 = 60
----> y2 = 60/3
----> y2 = 20
----> y = 2√5
----> y = 2 * 2.236
---> y = 4.47
Here, x > y.
Workspace
100. In the following question two equations numbered I and II are given. Solve both the equations and give answer if
A) x > y
B) x < y
C) x≥y
D) x≤y
E) x = y or relationship cannot be determined
I. 63x - 194√x + 143 = 0
II. 99y - 255√y + 150 = 0
A) x > y
B) x < y
C) x≥y
D) x≤y
E) x = y or relationship cannot be determined
I. 63x - 194√x + 143 = 0
II. 99y - 255√y + 150 = 0
SHOW ANSWER
Correct Ans:E
Explanation:
I. 63x - 194√x + 143 = 0
63x - 117√x -77√x + 143 = 0
x = 169/49 , 121/81
II. 99y - 255√y + 150 = 0
99y - 90√y - 165√y + 150 = 0
(11√y -10) (9√y - 15) = 0
y = 100/121, 225/81
Therefore relation cannot be established between x and y.
63x - 117√x -77√x + 143 = 0
x = 169/49 , 121/81
II. 99y - 255√y + 150 = 0
99y - 90√y - 165√y + 150 = 0
(11√y -10) (9√y - 15) = 0
y = 100/121, 225/81
Therefore relation cannot be established between x and y.
Workspace
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