# 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.
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.

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.
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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.
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?

Correct Ans:Quantity I > Quantity II
Explanation:
Total no of balls = 4 + 7 + 5 = 16 balls
Quantity I:
Balls drawn randomly = 16C3 = (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) = 9C3/16C3
= [(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 = 16C2 = (16*15)/(1*2) = 120
Probability that both the balls are either pink or black = 4C2 or 5C2 / 16C2
= (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?

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?

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 : 3x2 - 13x + 14 = 0

Quantity II : 3y2 - 14Y+ 15 = 0

Correct Ans:Quantity I , Quantity II No relationship can't be established.
Explanation:
Let find the value of Quantity I , Quantity II
Quantity I : 3x2 - 13x + 14 = 0
3x2 - 13x + 14 = 0
(x - 2),(3x -7)
x = ( 7/3), 2;

Quantity II : 3y2 - 14Y+ 15 = 0
3y2 - 14Y+ 15 = 0
(3y - 5),(y - 3)
y = (5/3),3;

No relation
Hence, Quantity I , Quantity II No relation
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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.

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
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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 + 858

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)2 * 91/7 + (8)3 + 61 =(x)3 + 858
841*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 cm2. Value of percent by which area increased.

Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
Let length of rectangle = L
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
---> πr2 = (1/2) * L * B
---> π(B/2)2 = (1/2) * L * B
----> π(B2/4) = (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

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:
31 = 3
32 = 9
33 = 27 --> Here last digit is '7'
34 = 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:
35 = 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:
71 = 7
72 = 49 --> Here last digit is '9'
73 = 343 --> Here last digit is '3'
74 = 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:
75 = 16807 --> Here last digit is '7'
76 = 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.

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

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] / (x2 + 5x - 3x -15) = 1/3
---> [2x + 2] / (x2 + 2x -15) = 1/3
---> [2x + 2] * 3 = (x2 + 2x -15)
---> 6x + 6 = (x2 + 2x -15)
---> x2 + 2x -15 - 6x - 6 = 0
---> x2 - 4x - 21 = 0
---> (x -7) (x + 3) = 0
---> x = 7, -3

Given expression: II. (y + 2)(27 - y) = 210
---> 27y - y2 + 54 - 2y - 210 = 0
---> 25y - y2 -156 = 0
---> y2 - 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(1/2)
II. q10 - 365 = 0

Correct Ans:p = q or the relationship cannot be established
Explanation:
Given equation I. (15/√p) - (9/√p) = p(1/2)
---> (15-9) / √p = √p
---> 6 = √p * √p
---> 6 = p

Given equation II. q10 - 365 = 0
---> q10 - (62)5 = 0
---> q10 - (6)10 = 0
---> q10 = (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)1/2 = 0

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.
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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.

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

Correct Ans:Quantity I < Quantity II
Explanation:
Quantity I: Given Sum of four consecutive even numbers is 76.
--> 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

Correct Ans:x = y or the relation cannot be established
Explanation:
I. 5x2 + 59x + 44 = 0
5x2 + 4x + 55x + 44 = 0
x = -4/5, -11

II. 2y2 + 13y + 15 = 0
2y2 + 3y + 10y +15 = 0
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.
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75. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 6x2 + 31x - 77 = 0
II. 9y2 - 52y + 64 = 0

Correct Ans:x = y or the relation cannot be established
Explanation:
I. 6x2 + 31x â€“ 77 = 0
â‡’ 6x2 + 42x â€“ 11x â€“ 77 = 0
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.
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76. In the following question, two equations are given. Solve the equations and answer accordingly.

I. x2 - 12x + 32 = 0
II. y2 - 20y + 96 = 0

Correct Ans:x ≤ y
Explanation:
Given, expression I. x2 - 12x + 32 = 0
---> x2 - 4x - 8x + 32 = 0
---> x (x - 4) - 8 (x - 4) = 0
---> (x - 4) (x - 8) = 0
---> x = 4, 8

II. y2 - 20y + 96 = 0
---> y2 - 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. 4x2 - 3x - 52 = 0
II. 3y2 - 4y - 55 = 0

Correct Ans:x = y or the relation cannot be established
Explanation:
I. 4x2 - 3x - 52 = 0
4x2 - 16x + 13x - 52 = 0
4x(x - 4) + 13(x - 4) = 0
(4x + 13) (x - 4) = 0
x = -13/4; 4
x = -3.25, 4

II. 3y2 - 4y - 55 = 0
3y2 - 15y + 11y - 55 = 0
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.
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78. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 3x2 = âˆ›(216000) + 48
II. 11y - y2 = 24

Correct Ans:x < y
Explanation:
I. 3x2 = âˆ›(216000) + 48
3x2 = 60 + 48
3x2 = 108
x2 = 36
x = 6

II. 11y - y2 = 24
y2 - 11y + 24 = 0
y2 - 8y - 3y + 24 = 0
y(y - 8) - 3(y - 8) = 0
(y - 3) (y - 8) = 0
y = 3, 8

Hence, x < y.
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79. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 15√x/√x - 9/√x = x1/2
II. y10 - 495 = 0

Correct Ans:x = y or the relationship cannot be established
Explanation:
(15√x/x) - (9/√x) = x1/2
[(15 - 9)√x]/x = √x
6/√x = √x
6 = x
y10 - 495 = 0
(y2)5 - (72)5 = 0
(y2)5 = (72)5
y2 = 72
y = ±7

Hence, relationship cannot be established.
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80. In the following question, two equations are given. Solve the equations and answer accordingly.

I. 721x2 – 657x2 = 256
II. √256y3 – 14y3 = 16

Correct Ans:x ≤ y
Explanation:
721x2 – 657x2 = 256
64x2 = 256
x2 = 256/64
x2 = 4
x = ±2

√256y3 – 14y3 = 16
16y3 – 14y3 = 16
2y3 = 16
y3 = 8
y = 2
Hence, x ≤ y.
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