# 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. 3x

II. 12y

I. 3x

^{2}- 10x + 8 = 0II. 12y

^{2}- 17y + 6 = 0SHOW ANSWER

Correct Ans:x > y

Explanation:

I. 3x

3x

3x(x - 2) - 4(x - 2) = 0

(3x - 4)(x - 2) = 0

x = 4/3, 2

II. 12y

12y

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

^{2}- 10x + 8 = 03x

^{2}- 6x - 4x + 8 = 03x(x - 2) - 4(x - 2) = 0

(3x - 4)(x - 2) = 0

x = 4/3, 2

II. 12y

^{2}- 17y + 6 = 012y

^{2}- 9y - 8y + 6 = 03y(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. 2x

II. 2y

I. 2x

^{2}+ 13x + 21 = 0II. 2y

^{2}+ 11y + 15 = 0SHOW ANSWER

Correct Ans:x ≤ y

Explanation:

I. 2x

2x

(x + 3)(2x + 7) = 0

x = -3, -7/2

II. 2y

2y

(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

^{2}+ 13x + 21 = 02x

^{2}+ 6x + 7x + 21 = 0(x + 3)(2x + 7) = 0

x = -3, -7/2

II. 2y

^{2}+ 11y + 15 = 02y

^{2}+ 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.

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?

**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?

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.

Workspace

84.

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.

**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.

SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

We know that if base of triangle is same as base of rectangle ,then

Where

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.

Workspace

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.

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 = (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.

Workspace

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. 50000SHOW 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

Workspace

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?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

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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.

Workspace

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

I. 28x

II. 28y

I. 28x

^{2}- 8x - 11 = 0II. 28y

^{2}+ 32y + 9 = 0SHOW ANSWER

Correct Ans:x ≥ y

Explanation:

I. 28x

28x

14x(2x + 1) - 11(2x + 1) = 0

(14x - 11) (2x + 1) = 0

x = 11/14; -1/2

II. 28y

28y

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.

^{2}- 8x - 11 = 028x

^{2}+ 14x - 22x - 11 = 014x(2x + 1) - 11(2x + 1) = 0

(14x - 11) (2x + 1) = 0

x = 11/14; -1/2

II. 28y

^{2}+ 32y + 9 = 028y

^{2}+ 18y + 14y + 9 = 02y(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. 100x

II. 10y

I. 100x

^{2}- 120x + 32 = 0II. 10y

^{2}- 17y + 6 = 0SHOW ANSWER

Correct Ans:x = y or no relation can be established

Explanation:

I. 100x

100x

20x(5x - 2) - 16(5x - 2) = 0

(20x -16) (5x - 2) = 0

x = 16/40; 2/5

x = 4/5; 2/5

II. 10y

10y

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.

^{2}- 120x + 32 = 0100x

^{2}- 40x - 80x + 32 = 020x(5x - 2) - 16(5x - 2) = 0

(20x -16) (5x - 2) = 0

x = 16/40; 2/5

x = 4/5; 2/5

II. 10y

^{2}- 17y + 6 = 010y

^{2}- 12y - 5y + 6 = 02y(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) 2x

II) 2y

I) 2x

^{2}- 11x + 12 = 0II) 2y

^{2}- 17y + 36 = 0SHOW ANSWER

Correct Ans:x ≤ y

Explanation:

I) 2x

2x

2x (x - 4) - 3 (x - 4) = 0

(2x - 3) (x - 4) = 0

x = 3/2, 4

II) 2y

2y

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

^{2}- 11x + 12 = 02x

^{2}- 8x - 3x + 12 = 02x (x - 4) - 3 (x - 4) = 0

(2x - 3) (x - 4) = 0

x = 3/2, 4

II) 2y

^{2}-17y +36 = 02y

^{2}- 9y - 8y + 36 = 0y (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) 2x

II) 5y

I) 2x

^{2}+ 3x - 20 = 0II) 5y

^{2}- 3y- 2 = 0SHOW ANSWER

Correct Ans:x = y or relation between x and y cannot be determined.

Explanation:

I) 2x

2x

2x(x + 4) - 5(x + 4) = 0

(2x - 5)(x + 4) = 0

x = 5/2, -4

II) 5y

5y

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.

^{2}+ 3x - 20 = 02x

^{2}+ 8x - 5x - 20 = 02x(x + 4) - 5(x + 4) = 0

(2x - 5)(x + 4) = 0

x = 5/2, -4

II) 5y

^{2}- 3y - 2 = 05y

^{2}- 5y + 2y - 2 = 05y(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.

Workspace

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.

Workspace

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. 2y

I. 21/√x + 11/√x = 7√x

II. 2y

^{2}– 11y + 12 = 0SHOW 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. 2y

2y

2y (y – 4) – 3 (y – 4) = 0

(2y – 3) (y – 4) = 0

y = 3/2, 4

(x, y

(x, y

So, x > y.

(21 + 11)/√x = 7√x

32/√x = 7√x

x = 32/7

II. 2y

^{2}– 11y + 12 = 02y

^{2}– 8y – 3y + 12 = 02y (y – 4) – 3 (y – 4) = 0

(2y – 3) (y – 4) = 0

y = 3/2, 4

(x, y

_{1}) = (32/7, 3/2) = x > y(x, y

_{2}) = (32/7, 4) = x > ySo, x > y.

Workspace

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

I. x

II. y

I. x

^{3}– 2744 = 0II. y

^{2}– 256 = 0SHOW ANSWER

Correct Ans:Relationship between x and y cannot be determined

Explanation:

I. x

x

x = 14

II. y

y

y = +16, -16

(x, y

(x, y

While comparing the values of x and y the relationship between x and y cannot be determined

^{3}– 2744 = 0x

^{3}= 2744x = 14

II. y

^{2}– 256 = 0y

^{2}= 256y = +16, -16

(x, y

_{1}) = (14, 16) = x < y_{1}(x, y

_{2}) = (14, -16) = x > y_{2}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) + (5/2x

II. 14y = (6/y) + 5

I. 4 = (3/x) + (5/2x

^{2})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) + (5/2x

---> 4 = (6x + 5)/2x

---> 4 * 2x

---> 8x

---> 8x

----> 2x (4x - 5) + (4x - 5) = 0

----> (4x - 5) + (2x + 1) = 0

---->

Equation II. 14y = (6/y) + 5

---> 14y - (6/y) = 5

---> (14y

---> 14y

---> 14y

---> 14y

---> 7y (2y + 1) -6 (2y + 1) = 0

---> (2y + 1) (7y - 6) = 0

--->

For x = 5/4, and y =-1/2OR 6/7;

For x = -1/2, and y = -1/2,

For x = -1/2 and y = 6/7,

Hence,

^{2})---> 4 = (6x + 5)/2x

^{2}---> 4 * 2x

^{2}= (6x + 5)---> 8x

^{2}- 6x - 5 = 0---> 8x

^{2}- 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

---> (14y

^{2}- 6) / y = 5---> 14y

^{2}- 6 -5y = 0---> 14y

^{2}-5y - 6 = 0---> 14y

^{2}+ 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 = âˆ›4913

II. 3y

I. x = âˆ›4913

II. 3y

^{2}= âˆ›216000SHOW ANSWER

Correct Ans:x > y

Explanation:

Given Equation I. x = âˆ›4913

--->

II. 3y

----> 3y

----> y

----> y

----> y = 2√5

----> y = 2 * 2.236

--->

Here,

--->

**x = 17**II. 3y

^{2}= âˆ›216000----> 3y

^{2}= 60----> y

^{2}= 60/3----> y

^{2}= 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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