# Equations and Inequations Questions and Answers updated daily – Aptitude

Equations and Inequations Questions: Solved 316 Equations and Inequations Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## 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

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
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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

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
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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?

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.
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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.

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.
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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.

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.
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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

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
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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?

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
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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

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.
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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

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.
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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

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.
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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

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
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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

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.
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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

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.
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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?

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.
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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.

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.
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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

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.
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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

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
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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

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.
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99. Two equations (I) and (II) are given. You have to solve both the equations and give the answer.

I. x = âˆ›4913
II. 3y2 = âˆ›216000

Correct Ans:x > y
Explanation:
Given Equation I. x = âˆ›4913
---> x = 17

II. 3y2 = âˆ›216000
----> 3y2 = 60
----> y2 = 60/3
----> y2 = 20
----> y = 2√5
----> y = 2 * 2.236
---> y = 4.47

Here, x > y.
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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

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.
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