# 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

61. Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

M can do a work in 16 days. N is 60% more efficient than M.

M can do a work in 16 days. N is 60% more efficient than M.

**Quantity I:**Time taken by M and N together to do the work.**Quantity II:**Time taken by M and N to do the work together when M works at doubles his original efficiency and N works at half his original efficiency.SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

Quantity I:

M = 16 days; N = 16 * (100/160) =10 days

M + N together = 1/16 + 1/10

= (16 + 10)/(16*10)

= 26/160 =13/80

= 80/13 days

Quantity II:

M = 16 days; N = 10 days

M (double efficiency) = 8 day; N (half efficiency) = 20 days

M + N together = 1/8 + 1/20 = (20 + 8)/20*8

= 28/160 = 14/80

= 80/14 days

So, Quantity I > Quantity II.

M = 16 days; N = 16 * (100/160) =10 days

M + N together = 1/16 + 1/10

= (16 + 10)/(16*10)

= 26/160 =13/80

= 80/13 days

Quantity II:

M = 16 days; N = 10 days

M (double efficiency) = 8 day; N (half efficiency) = 20 days

M + N together = 1/8 + 1/20 = (20 + 8)/20*8

= 28/160 = 14/80

= 80/14 days

So, Quantity I > Quantity II.

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62. Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

A bag contains, 4 pink, 7 yellow and 5 black balls.

A bag contains, 4 pink, 7 yellow and 5 black balls.

**Quantity I:**If 3 balls are drawn randomly, then find the probability of getting at least one yellow ball?**Quantity II:**If 2 balls are drawn randomly, then find the probability of getting both the balls are either pink or black?SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

Total no of balls = 4 + 7 + 5 = 16 balls

Quantity I:

Balls drawn randomly = 16C

The probability of getting at least one yellow ball = 1- P(None is yellow ball)

P(None is yellow ball) = 9C

= [(9*8*7)/(1*2*3)] / [(16*15*14)/(1*2*3)] = 3/20

Required probability = 1 - 3/20 = 17/20

Quantity II:

Balls drawn randomly = 16C

Probability that both the balls are either pink or black = 4C

= (6 + 10)/120 = 16/120 = 2/15

So, Quantity I > Quantity II.

Quantity I:

Balls drawn randomly = 16C

_{3}= (16*15*14)/(1*2*3)The probability of getting at least one yellow ball = 1- P(None is yellow ball)

P(None is yellow ball) = 9C

_{3}/16C_{3}= [(9*8*7)/(1*2*3)] / [(16*15*14)/(1*2*3)] = 3/20

Required probability = 1 - 3/20 = 17/20

Quantity II:

Balls drawn randomly = 16C

_{2}= (16*15)/(1*2) = 120Probability that both the balls are either pink or black = 4C

_{2}or 5C_{2}/ 16C_{2}= (6 + 10)/120 = 16/120 = 2/15

So, Quantity I > Quantity II.

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63. Find the appropriate relation for quantity I and quantity II in the following question:

**Quantity I:**In a three digit number the digit in the unit's place is twice the digit in the ten's place and 1.5 times the digit in the hundred's place. If the sum of all three digits of the number is 13, what is the number?**Quantity II:**Sum of eight consecutive odd numbers is 656. Average of four consecutive even numbers is 87. What is the sum of the largest even number and the largest odd number?SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Quantity I:**

Let the three digit number be 100a + 10b + c

Let the digit in unit's place = c,

digit in ten's place = b

digit in hundred's place = a

Given, c = 2b ---> b = c/2

c = 1.5a ---> a = c/1.5

According to the question, sum of all three digits of the number = 13

---> (c/1.5) + (c/2) + c = 13

---> [(1/1.5) + (1/2) + 1] * c = 13

---> [(10/15) + (1/2) + 1] * c = 13

---> [(20 + 15 + 30)/30] * c = 13

---> [65/30] * c = 13

---> [5/30] * c = 1

---> c/6 = 1

---> c = 6 ---> which is third digit (unit digit) of the 3 digit number

And b = 3 ---> ten's digit

a = 4 ---> hundred's digit

Thus, the required 3 digit number is

**436**

**Quantity II:**

Let the eight consecutive odd numbers be x - 8, x - 6, x - 4, x - 2, x, x + 2, x + 4, x + 6

Given their sum = 656

---> x - 8 + x - 6 + x - 4 + x - 2 + x + x + 2 + x + 4 + x + 6 = 656

---> 8x - 20 + 12 = 656

---> 8x - 8 = 656

---> 8 (x - 1) = 656

---> x - 1 = 82

---> x = 83

Thus the largest odd number = x + 6 = 83 + 6 = 89

Given, Average of four consecutive even numbers is 87

Let the four consecutive even numbers be y - 2, y, y + 2, y + 4

---> sum of these numbers = 87 * 4

---> y - 2 + y + y + 2 + y + 4 = 87 * 4

---> 4y + 4 = 87 * 4

---> y + 1 = 87

---> y = 86

Thus the largest even number = y + 4 = 86 + 4 = 90

Hence, the sum of the largest odd number and the largest even number = 89 + 90 =

**179**

Here,

**Quantity I > Quantity II**.

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64. Find the appropriate relation for quantity I and quantity II in the following question:

**Quantity I:**In how many different ways can the letters of the word "COMPLAINT" be arranged in such a way that the vowels occupy only the odd positions?**Quantity II:**Ankit borrowed a certain sum of money at simple interest for 3 years at 8% per annum and he pays Rs. 4200 as interest. Find the corresponding compound interest?SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Quantity I:**

There are 9 different letters in the given word "COMPLAINT", out of which there are 3 vowels and 6 consonants.

Let us mark these positions as under:

[1] [2] [3] [4] [5] [6] [7] [8] [9]

Now, 3 vowels can be placed at any of the three places out of 5 marked 1, 3, 5, 7 and 9.

Number of ways of arranging the vowels = 5P3 = 5 * 4 * 3 = 60 ways.

Also, the 6 consonants at the remaining positions may be arranged in 6P6 ways = 6! ways = 720 ways.

Therefore, required number of ways = 60 x 720 =

**43200 ways.**

**Quantity II:**

S.I = PNR/100

---> 4200 = (P*3*8)/100

---> P = (4200 * 100) /24

---> P = 17500

Corresponding compound interest, C.I:

P* (r/100) = 17500 * (8/100) = 1400

---> [17500 + 1400]*(8/100) = 18900 * (8/100) = 1512

---> [18900 + 1512]*(8/100) = 20412*(8/100) = 1632.96

C.I = 1400 + 1512 + 1632.96 =

**4544.96**

Hence,

**Quantity I > Quantity II**

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65. Read the following information carefully & establish a relation between quantity I & quantity II:

Quantity I : 3x

Quantity II : 3y

Quantity I : 3x

^{2}- 13x + 14 = 0Quantity II : 3y

^{2}- 14Y+ 15 = 0SHOW ANSWER

Correct Ans:Quantity I , Quantity II No relationship can't be established.

Explanation:

Let find the value of Quantity I , Quantity II

Quantity I : 3x

3x

(x - 2),(3x -7)

x = ( 7/3), 2;

Quantity II : 3y

3y

(3y - 5),(y - 3)

y = (5/3),3;

No relation

Quantity I : 3x

^{2}- 13x + 14 = 03x

^{2}- 13x + 14 = 0(x - 2),(3x -7)

x = ( 7/3), 2;

Quantity II : 3y

^{2}- 14Y+ 15 = 03y

^{2}- 14Y+ 15 = 0(3y - 5),(y - 3)

y = (5/3),3;

No relation

**Hence, Quantity I , Quantity II No relation**
Workspace

66. Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.

**Quantity I:**Average of the present age of Sunny and Surya is 36 years. Ratio of the present ages of Sunny and Surya is 7: 5 respectively. Find the age of Sunny 4 years ago.**Quantity II:**Average age of a class of 18 students is 18 years. Average age of the class is increased by 1 year if the age of the class teacher is also included. Find the age of the class teacher.SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

Quantity I:

Sum of the present ages of Sunny and Surya = 36*2 = 72 years

So, the present age of Sunny = 72 * 7/12 = 42 years

Age of Sunny 4 years ago = 42 - 4 = 38 years

Quantity II:

Sum of the ages of students in the class = 18*18 = 324 years

Sum of the ages of students along with the class teacher = 19*19 = 361 years

So, the age of the class teacher = 361 - 324 = 37 years

So, Quantity I > Quantity II

Sum of the present ages of Sunny and Surya = 36*2 = 72 years

So, the present age of Sunny = 72 * 7/12 = 42 years

Age of Sunny 4 years ago = 42 - 4 = 38 years

Quantity II:

Sum of the ages of students in the class = 18*18 = 324 years

Sum of the ages of students along with the class teacher = 19*19 = 361 years

So, the age of the class teacher = 361 - 324 = 37 years

So, Quantity I > Quantity II

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67. Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.

**Quantity I:**A and B can do a piece of work in 25 days and 30 days respectively. They start the work together but after some days B leaves the job. A alone does the remaining work in 3 days. Find after how many days does B leave the job?**Quantity II:**Value of ?: (29)^{2}* 91 Ã· 7 + (8)^{3}+ 61 =(?)^{3}+ 858SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Quantity I:

LCM (25, 30) = 150

Let total work = 150 unit

A efficiency = 150/25 = 6 unit

B efficiency = 150/30 = 5 unit

A work alone in last three days = 3*6 = 18 unit

Remaining work = 150 - 18 = 132

Days taken by A&B for this work = 132/(6+5) = 132/11 = 12 days

So, B left the work after 12 days

Quantity II:

(29)

841*13 + 512 + 61 - 858 = (x)

x = ∛10648 = 22

So, Quantity II Ëƒ Quantity I

LCM (25, 30) = 150

Let total work = 150 unit

A efficiency = 150/25 = 6 unit

B efficiency = 150/30 = 5 unit

A work alone in last three days = 3*6 = 18 unit

Remaining work = 150 - 18 = 132

Days taken by A&B for this work = 132/(6+5) = 132/11 = 12 days

So, B left the work after 12 days

Quantity II:

(29)

^{2}* 91/7 + (8)^{3}+ 61 =(x)^{3}+ 858841*13 + 512 + 61 - 858 = (x)

^{3}x = ∛10648 = 22

So, Quantity II Ëƒ Quantity I

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68. Find the appropriate relation for quantity I and quantity II in the following question:

**Quantity I:**Area of circle, given in figure, is half of the area of rectangle. Value of percent by which length of rectangle is more than breadth.**Quantity II:**A pair of opposite sides of a square when increase by 10 cm, then area of the above figure increased by 400 cm^{2}. Value of percent by which area increased.SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

__Quantity I:__Let length of rectangle = L

Breadth of rectangle = B

Then, In the given figure, Diameter of the circle = breadth of the rectangle

---> Radius of circle = half of the breadth of the rectangle = B/2

Given, Area of circle = (1/2) * Area of rectangle

---> πr

^{2}= (1/2) * L * B

---> π(B/2)

^{2}= (1/2) * L * B

----> π(B

^{2}/4) = (1/2) * L * B

----> π(B/4) = (1/2) * L

----> (22/7) * (B/4) * 2 = L

---> L = (11/7)B

**Percentage by which length of rectangle is more than its breadth**= {[(11B/7) - B]/B} * 100

= {(11B - 7B)/7B} * 100

= {4B/7B} * 100

= (400/7)%

=

**57 (1/7)%**

__Quantity II:__When two opposite sides of a square is increased by 10 cm, then Square gets changed into the rectangle.

By increasing 10 cm in two opposite sides, Area increased = 400

---> Side = 400/10 = 40 cm

Area of square = 40 * 40= 1600 square cm

**Percentage by which area increase**= (400/1600) * 100

= (100/4)%

=

**25%**

Here,

**Quantity I > Quantity II**

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69. Find the appropriate relation for quantity I and quantity II in the following question:

**Quantity I:**Unit digit of the number [(1333)^{27}]^{55}**Quantity II:**Unit digit of the number [(127)^{562581}]SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

**Quantity I:**

[(1333)

^{27}]

^{55}

---> (1333)

^{1485}

Here, last digit in power value is '5'

Number whose last digit is '3' has a power cycle of 4 different numbers, as follows:

3

^{1}= 3

3

^{2}= 9

3

^{3}= 27 --> Here last digit is '7'

3

^{4}= 81 --> Here last digit is '1'

So, the cyclicity of 3 has 4 different numbers 3, 9, 7, 1.

and after that 3, 9, 7, 1 starts repeating, as follows:

3

^{5}= 243 --> Here last digit is '3'

Now, Step 1: We know that the cyclicity of 3 is 4.

Step 2: Divide the power 1485 by 4.

By doing that, we get a remainder = 1.

Step 3: 1st power in the power cycle of 3 is 3.

Hence, the

**answer ie., Unit digit of (1333)**

^{1485}is 3.**Quantity II:**

[(127)

^{562581}]

Here, last digit in power value is '6'

Number whose last digit is '7' has a power cycle of 4 different numbers, as follows:

7

^{1}= 7

7

^{2}= 49 --> Here last digit is '9'

7

^{3}= 343 --> Here last digit is '3'

7

^{4}= 2401 --> Here last digit is '1'

So, the cyclicity of 7 has 4 different numbers 7, 9, 3, 1.

and after that 7, 9, 3, 1 starts repeating, as follows:

7

^{5}= 16807 --> Here last digit is '7'

7

^{6}= 117649â€¬ --> Here last digit is '9'

Now, Step 1: We know that the cyclicity of 7 is 4.

Step 2: Divide the power 562581 by 4.

By doing that, we get a remainder = 1.

Step 3: 1st power in the power cycle of 7 is 7.

Hence, the

**answer is 7.**

Here

**Quantity I < Quantity II**

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70. In the following question two equations numbered I and II are given. You have to solve both the equations and give the answer.

I. [1/(x - 3)] + [1/(x + 5)] = 1/3

II. (y + 2)(27 - y) = 210

I. [1/(x - 3)] + [1/(x + 5)] = 1/3

II. (y + 2)(27 - y) = 210

SHOW ANSWER

Correct Ans:x < y

Explanation:

Given expression: I. [1/(x - 3)] + [1/(x + 5)] = 1/3

---> [(x + 5) + (x - 3)] / [(x - 3) * (x + 5)] = 1/3

---> [2x + 2] / (x

---> [2x + 2] / (x

---> [2x + 2] * 3 = (x

---> 6x + 6 = (x

---> x

---> x

---> (x -7) (x + 3) = 0

--->

Given expression: II. (y + 2)(27 - y) = 210

---> 27y - y

---> 25y - y

---> y

---> (y - 13) (y - 12) = 0

--->

Here,

---> [(x + 5) + (x - 3)] / [(x - 3) * (x + 5)] = 1/3

---> [2x + 2] / (x

^{2}+ 5x - 3x -15) = 1/3---> [2x + 2] / (x

^{2}+ 2x -15) = 1/3---> [2x + 2] * 3 = (x

^{2}+ 2x -15)---> 6x + 6 = (x

^{2}+ 2x -15)---> x

^{2}+ 2x -15 - 6x - 6 = 0---> x

^{2}- 4x - 21 = 0---> (x -7) (x + 3) = 0

--->

**x = 7, -3**Given expression: II. (y + 2)(27 - y) = 210

---> 27y - y

^{2}+ 54 - 2y - 210 = 0---> 25y - y

^{2}-156 = 0---> y

^{2}- 25y + 156 = 0---> (y - 13) (y - 12) = 0

--->

**y = 13, 12**Here,

**x < y**
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71. In the following question two equations numbered I and II are given. You have to solve both the equations and give the answer.

I. (15/√p) - (9/√p) = p

II. q

I. (15/√p) - (9/√p) = p

^{(1/2)}II. q

^{10}- 36^{5}= 0SHOW ANSWER

Correct Ans:p = q or the relationship cannot be established

Explanation:

Given equation I. (15/√p) - (9/√p) = p

---> (15-9) / √p = √p

---> 6 = √p * √p

--->

Given equation II. q

---> q

---> q

---> q

Since power values are equal,

--->

Here,

^{(1/2)}---> (15-9) / √p = √p

---> 6 = √p * √p

--->

**6 = p**Given equation II. q

^{10}- 36^{5}= 0---> q

^{10}- (6^{2})^{5}= 0---> q

^{10}- (6)^{10}= 0---> q

^{10}= (6)^{10}Since power values are equal,

--->

**q = 6**Here,

**p = q = 6**
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72. In the following question, two equations are given. Solve the equations and answer accordingly.

I. √500x + √402 = 0

II. √360y + (200)

I. √500x + √402 = 0

II. √360y + (200)

^{1/2}= 0SHOW ANSWER

Correct Ans:x < y

Explanation:

I. 10√5x + 6√67 = 0

x = - 6√67/10√5

= - 3√67/5√5

= - (9√67) / (15√5)

II. 6√10y + 10√2 = 0

y = - 10√2/6√10

= - 5√2/3√10

= - 5√2/3√2*5

= - 5/3√5 = - 25 / (15√5)

So, x < y.

x = - 6√67/10√5

= - 3√67/5√5

= - (9√67) / (15√5)

II. 6√10y + 10√2 = 0

y = - 10√2/6√10

= - 5√2/3√10

= - 5√2/3√2*5

= - 5/3√5 = - 25 / (15√5)

So, x < y.

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73. In the following question, two Quantities numbered 1 and 2 are given. You have to solve both the Quantities and find out the relationship between them. Then give answer accordingly.

Sum of four consecutive even numbers is 76.

66(2/3)% of y of (9/2) = 996

**Quantity 1:**Average of numbers.Sum of four consecutive even numbers is 76.

**Quantity 2:**The value of y.66(2/3)% of y of (9/2) = 996

SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Quantity I: Given Sum of four consecutive even numbers is 76.

--> Averge of these numbers = 76/4 =

Quantity II: 66(2/3)% of y of (9/2) = 996

---> (200/300) * y * (9/2) = 996

---> y * 3 = 996

---> y = 996 / 3

--->

Here,

--> Averge of these numbers = 76/4 =

**19**Quantity II: 66(2/3)% of y of (9/2) = 996

---> (200/300) * y * (9/2) = 996

---> y * 3 = 996

---> y = 996 / 3

--->

**y = 332**Here,

**Quantity I < Quantity II.**
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74. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 5xÂ² + 59x + 44 = 0

II. 2yÂ² + 13y + 15 = 0

I. 5xÂ² + 59x + 44 = 0

II. 2yÂ² + 13y + 15 = 0

SHOW ANSWER

Correct Ans:x = y or the relation cannot be established

Explanation:

I. 5x

5x

x = -4/5, -11

II. 2y

2y

y = -3/2, -5

(x1, y1) = (-4/5, -3/2) = x > y

(x1, y2) = (-4/5, -5) = x > y

(x2, y1) = (-11, -3/2) = x < y

(x2, y2) = (-11, -5) = x < y

So, relation cannot be established.

^{2}+ 59x + 44 = 05x

^{2}+ 4x + 55x + 44 = 0x = -4/5, -11

II. 2y

^{2}+ 13y + 15 = 02y

^{2}+ 3y + 10y +15 = 0y = -3/2, -5

(x1, y1) = (-4/5, -3/2) = x > y

(x1, y2) = (-4/5, -5) = x > y

(x2, y1) = (-11, -3/2) = x < y

(x2, y2) = (-11, -5) = x < y

So, relation cannot be established.

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75. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 6x

II. 9y

I. 6x

^{2}+ 31x - 77 = 0II. 9y

^{2}- 52y + 64 = 0SHOW ANSWER

Correct Ans:x = y or the relation cannot be established

Explanation:

I. 6x

â‡’ 6x

â‡’ 6x (x + 7) â€“ 11 (x + 7) = 0

â‡’ (6x â€“ 11) (x + 7) = 0

x = 11/6, -7

II. 9y2 â€“ 52y + 64 = 0

â‡’ 9y2 â€“ 36y â€“ 16y + 64 = 0

â‡’ 9y(y â€“ 4) â€“ 16(y â€“ 4) = 0

â‡’ (9y â€“ 16) (y â€“ 4) = 0

â‡’ y = 16/9, 4

(x1, y1) = (11/6, 16/9) = x > y

(x1, y2) = (11/6, 4) = x < y

(x2, y1) = (-7, 16/9) = x < y

(x2, y2) = (-7, 4) = x < y

Hence, relationship between x and y cannot be determined.

^{2}+ 31x â€“ 77 = 0â‡’ 6x

^{2}+ 42x â€“ 11x â€“ 77 = 0â‡’ 6x (x + 7) â€“ 11 (x + 7) = 0

â‡’ (6x â€“ 11) (x + 7) = 0

x = 11/6, -7

II. 9y2 â€“ 52y + 64 = 0

â‡’ 9y2 â€“ 36y â€“ 16y + 64 = 0

â‡’ 9y(y â€“ 4) â€“ 16(y â€“ 4) = 0

â‡’ (9y â€“ 16) (y â€“ 4) = 0

â‡’ y = 16/9, 4

(x1, y1) = (11/6, 16/9) = x > y

(x1, y2) = (11/6, 4) = x < y

(x2, y1) = (-7, 16/9) = x < y

(x2, y2) = (-7, 4) = x < y

Hence, relationship between x and y cannot be determined.

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76. In the following question, two equations are given. Solve the equations and answer accordingly.

I. x

II. y

I. x

^{2}- 12x + 32 = 0II. y

^{2}- 20y + 96 = 0SHOW ANSWER

Correct Ans:x ≤ y

Explanation:

Given, expression I. x

---> x

---> x (x - 4) - 8 (x - 4) = 0

---> (x - 4) (x - 8) = 0

--->

II. y

---> y

---> y (y - 8) - 12 (y - 8) = 0

---> (y - 8) (y - 12) = 0

--->

Now, (x1, y1) = (4, 8) ---> Here x < y

(x1, y2) = (4, 12) ---> Here x < y

(x2, y1) = (8, 8) ---> Here x = y

(x2, y2) = (8, 12) ---> Here x < y

Hence, the correct relation between x and y is

^{2}- 12x + 32 = 0---> x

^{2}- 4x - 8x + 32 = 0---> x (x - 4) - 8 (x - 4) = 0

---> (x - 4) (x - 8) = 0

--->

**x = 4, 8**II. y

^{2}- 20y + 96 = 0---> y

^{2}- 8y - 12y + 96 = 0---> y (y - 8) - 12 (y - 8) = 0

---> (y - 8) (y - 12) = 0

--->

**y = 8, 12**Now, (x1, y1) = (4, 8) ---> Here x < y

(x1, y2) = (4, 12) ---> Here x < y

(x2, y1) = (8, 8) ---> Here x = y

(x2, y2) = (8, 12) ---> Here x < y

Hence, the correct relation between x and y is

**x ≤ y**
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77. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 4x

II. 3y

I. 4x

^{2}- 3x - 52 = 0II. 3y

^{2}- 4y - 55 = 0SHOW ANSWER

Correct Ans:x = y or the relation cannot be established

Explanation:

I. 4x

4x

4x(x - 4) + 13(x - 4) = 0

(4x + 13) (x - 4) = 0

x = -13/4; 4

x = -3.25, 4

II. 3y

3y

3y(y - 5) + 11(y - 5) = 0

(3y + 11) (y - 5) = 0

y = -11/3; 5

y = -3.6; 5

(x1, y1) = (-3.2, -3.6) = x < y

(x1, y2) = (-3.2, 5) = x < y

(x2, y1) = (4, -3.6) = x > y

(x2, y2) = (4, 5) = x < y

Hence, relationship between x and y can't be determined.

^{2}- 3x - 52 = 04x

^{2}- 16x + 13x - 52 = 04x(x - 4) + 13(x - 4) = 0

(4x + 13) (x - 4) = 0

x = -13/4; 4

x = -3.25, 4

II. 3y

^{2}- 4y - 55 = 03y

^{2}- 15y + 11y - 55 = 03y(y - 5) + 11(y - 5) = 0

(3y + 11) (y - 5) = 0

y = -11/3; 5

y = -3.6; 5

(x1, y1) = (-3.2, -3.6) = x < y

(x1, y2) = (-3.2, 5) = x < y

(x2, y1) = (4, -3.6) = x > y

(x2, y2) = (4, 5) = x < y

Hence, relationship between x and y can't be determined.

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78. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 3x

II. 11y - y

I. 3x

^{2}= âˆ›(216000) + 48II. 11y - y

^{2}= 24SHOW ANSWER

Correct Ans:x < y

Explanation:

I. 3x

3x

3x

x

x = 6

II. 11y - y

y

y

y(y - 8) - 3(y - 8) = 0

(y - 3) (y - 8) = 0

y = 3, 8

Hence, x < y.

^{2}= âˆ›(216000) + 483x

^{2}= 60 + 483x

^{2}= 108x

^{2}= 36x = 6

II. 11y - y

^{2}= 24y

^{2}- 11y + 24 = 0y

^{2}- 8y - 3y + 24 = 0y(y - 8) - 3(y - 8) = 0

(y - 3) (y - 8) = 0

y = 3, 8

Hence, x < y.

Workspace

79. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 15√x/√x - 9/√x = x

II. y

I. 15√x/√x - 9/√x = x

^{1/2}II. y

^{10}- 49^{5}= 0SHOW ANSWER

Correct Ans:x = y or the relationship cannot be established

Explanation:

(15√x/x) - (9/√x) = x

[(15 - 9)√x]/x = √x

6/√x = √x

6 = x

y

(y

(y

y

y = ±7

Hence, relationship cannot be established.

^{1/2}[(15 - 9)√x]/x = √x

6/√x = √x

6 = x

y

^{10}- 49^{5}= 0(y

^{2})^{5}- (7^{2})^{5}= 0(y

^{2})^{5}= (7^{2})^{5}y

^{2}= 7^{2}y = ±7

Hence, relationship cannot be established.

Workspace

80. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 721x

II. √256y

I. 721x

^{2}– 657x^{2}= 256II. √256y

^{3}– 14y^{3}= 16SHOW ANSWER

Correct Ans:x ≤ y

Explanation:

721x

64x

x

x

x = ±2

√256y

16y

2y

y

y = 2

Hence, x ≤ y.

^{2}– 657x^{2}= 25664x

^{2}= 256x

^{2}= 256/64x

^{2}= 4x = ±2

√256y

^{3}– 14y^{3}= 1616y

^{3}– 14y^{3}= 162y

^{3}= 16y

^{3}= 8y = 2

Hence, x ≤ y.

Workspace

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