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

101. 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. 3x2 + 28x + 60 = 0
II. 3y2 + 37y + 114 = 0




SHOW ANSWER
Correct Ans:C
Explanation:
I. 3x2 + 28x + 60 = 0
x2 + 1/3(28x) + 180 = 0
x = -6, -10/3
II. 3y2 + 37y + 114 = 0
y2 + 1/3(37y) + 342 = 0
y = -19/3, -18/3
y = -19/3, -6
The relationship clearly shows that x≥y
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102. In the following question two equations numbered I and II are given. Solve both the equations and give answer if
(a) x=y; relationship between x and y cannot be established
(b) x>y
(c) x(d) x≥y
(e) x≤y

I. (x-18)2 = 0
II. y2 = 324




SHOW ANSWER
Correct Ans:D
Explanation:
I. (x-18)2 = 0
II. y2 = 324
(x-18) (x-18) = 0
x2 + 182 - 36x = 0
x2 -36x + 324 =0
x = 18
II. y2 = 324
y = ±18
x≥y
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103. In the following question two equations numbered I and II are given. Solve both the equations and give answer if
(a) x=y; relationship between x and y cannot be established
(b) x>y
(c) x(d) x≥y
(e) x≤y

I. 4y2 + 8y = 4y + 8
II. x2 + 9x = 2x - 12




SHOW ANSWER
Correct Ans:C
Explanation:
From I, 4y2 + 8y = 4y + 8
y2 + 2y = y + 2
y2 + 2y - y - 2 = 0
(y - 1)(y + 2) = 0
y = 1, -2
From II, x2 + 9x = 2x - 12
x2 + 7x + 12 = 0
(x + 4)(x + 3) = 0
x = -4, -3
Comparing the values of x and y, it can be determined that y>x.
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104. 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)
a) If Quantity I > Quantity II
b) If Quantity I ≥ Quantity II
c) If Quantity I < Quantity II
d) If Quantity I ≤ Quantity II
e) If Quantity I = Quantity II or the relation cannot be established

Quantity I: Find the amount on compound interest on a sum of Rs.55000 at the rate of 15% per annum after three years.
Quantity II: Find the simple interest on a sum of Rs.60000 at the rate 25% per annum after 5.5 years.




SHOW ANSWER
Correct Ans:A
Explanation:
Quantity I:
Find the amount on compound interest on a sum of Rs.55000 at the rate of 15% per annum after three years.
Required amount = 55000 * 115/100 * 115/100 * 115/100
Required amount= Rs.83648.125
Quantity II :
Find the simple interest on a sum of Rs.60000 at the rate 25% per annum after 5.5 years.
Required SI = 60000*25*5.5/ 100 = Rs.82500
Hence, Quantity I > Quantity II
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105. In the following question, two equations I and II are given. Solve both the equations and give answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established

I. x2 + 7x - 330 = 0
II. y = (194481)1/4




SHOW ANSWER
Correct Ans:C
Explanation:
I. x2 + 7x - 330 = 0
(x - 15) (x + 22) = 0
X = 15, -22
II. y = (194481)1/4
∜194481 = ∜21*21*21*21
y = ∜441*441
y = 21
x < y
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106. In the given question, two quantities are given, one as Quantity I and another as Quantity II. Determine the relationship between two quantities and choose the appropriate option: 

Quantity I: On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
Quantity II: A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. The selling price is: 




SHOW ANSWER
Correct Ans:Quantity II>Quantity I
Explanation:
Quantity I:
C.P. of 12 balls = S.P. of 17 balls = Rs.720.
CP of 1 ball = 720/12=Rs60.
Quantity II:
SP =85% of 1400
=85/100*1400
SP=Rs.1190
So, Quantity II is greater than Quantity I.
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107. In the given question, two quantities are given, one as ‘Quantity I’ and another as ‘Quantity II’. Determine the relationship between two quantities and choose the appropriate option:

Quantity I: The difference between SI and CI compounded annually on a certain sum of money for 2 years at 8% per annum is Rs.12.80. Find the principal.
Quantity II: A sum fetched a total simple interest of Rs.800 at the rate of 8 % per annum in 5 years. What is the sum? 




SHOW ANSWER
Correct Ans:Quantity I = Quantity II or Relation cannot be established
Explanation:
Quantity I:
SI-CI = Pr2/100
P*8r2/100 = 12.8
64P/(100*100) =12.8
P=Rs.2000
Quantity II:
SI=Pnr/100
800=P*8*5/100
P=800*100/40
P =Rs.2000
Here, Quantity I is equal to Quantity II.
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108. In the given question, two equations numbered I and II are given. You have to solve both the equations and Give answer:

I. 6x² + 11x + 4 = 0
II. 4y² - 7y – 2 = 0 




SHOW ANSWER
Correct Ans:If x < y
Explanation:
Equation I:
6x² + 11x + 4 = 0
6x² + 8x + 3x + 4 = 0
(3x+4) (2x+1) = 0
x= -4/3, -1/2
Equation II:
4y² - 7y – 2 = 0
4y² - 8y + y – 2 = 0
(y-2) (4y+1) = 0
y = 2, -1/4
x < y
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109. In the given question, two equations numbered I and II are given. You have to solve both the equations and Give answer: 
 
I. 2x² - 7x+ 3 = 0
II. 2y² - 7y + 6 = 0 




SHOW ANSWER
Correct Ans:If relationship between x and y cannot be established.
Explanation:
Equation I:
2x² - 7x+ 3 = 0
2x² - 6x - x + 3 = 0
(x-3) (2x-1) = 0
x = 3, 1/2
Equation II:
2y² - 7y + 6 = 0
2y² - 4y -3y + 6 = 0
(y-2)(2y-3) = 0
y = 2, 3/2
Hence no relation can be established
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110. In the question, two equations numbered I and II are given. Solve both the equations and give answer.

Quantity I: There are three numbers in the ratio 5:6:10. The sum of the largest and the smallest numbers is 126 more than the other number. Find the largest number? 
Quantity II: 12% of first number is equal to 25% second number. The difference of these two numbers is 78. Then find the largest number?

a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or Relation cannot be established  




SHOW ANSWER
Correct Ans:B
Explanation:
Quantity I:
Let the three numbers will be 5x, 6x and 10x respectively
10x+5x−6x=126
9x=126
X=14
So the largest number =10x=10×14=140
Quantity II:
let x and y are two numbers
It is given that:
12% of x=25%of y
x/y = 25/12
It is given that their difference is 78
So first no.=150
2nd number=72
Hence, Quantity I < Quantity II
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111. In the question, two equations numbered I and II are given. Solve both the equations and give answer:

Quantity I: A square plot area is equal to a rectangular plot of length 75m and 64m width. Find the length of the diagonal of the square?
Quantity II: x2 – (40√6 + 70)x + 2800√6 = 0

a) Quantity I > Quantity II
b) Quantity 1 < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or Relation cannot be established 




SHOW ANSWER
Correct Ans:C
Explanation:
Quantity I:
Formula:

Area of the rectangle = L * B
Area of square= a2
Given:
Length of the rectangular plot = 75m
width of the rectangular plot = 64 m
Area of the rectangular plot = 75 × 64 = 4800
It is given that it is equal to the area of square
a2 = 4800
Side = 40 √3 m
Now, To find the the diagonal of the square
Diagonal of the Square = √2a2
D2 = a√2
a=40√3
Diagonal of the Square = 40√3 * √2
Diagonal of the Square = 40√6
Hence, the diagonal of the square is 40√6 m
Quantity II:
x2 – (40√6 + 70)x + 2800√6 = 0
⇒x2 - 40√6x – 70x + 2800√6 = 0
⇒x(x - 40√6) – 70(x - 40√6) = 0
⇒(x - 40√6)(x – 70) = 0
⇒x = 70 and 40√6
Now comparing
40√6 ≥ 70
Therefore, Quantity I ≥ Quantity II
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112. Directions: In the following question two quantities are given. Compare the numeric value of both the quantities and answer accordingly.

In a two digit number, digit at unit place exceeds the digit in its tens place by 2 and the product of the required number with the sum of its digit is equal to 144.
Quantity I: Value of two digit number.

Quantity II: 26 




SHOW ANSWER
Correct Ans:Quantity I < Quantity II
Explanation:
Quantity I:
Let the units digit be y
And tens digit be x
Then, Required Number = 10x +y

According to the question,
y = x + 2
and product of required number and sum of its digits = 144
---> (10x + y) * (x + y) = 144
---> Substitute y = x + 2 in the above eqn
---> (10x + x + 2) * (x + x + 2) = 144
---> (11x + 2) * (2x + 2) = 144
---> (11x + 2) * 2(x + 1) = 144
---> (11x + 2) * (x + 1) = 144 / 2
---> (11x + 2) * (x + 1) = 72
---> (11x *x + 11x + 2x + 2) = 72
---> 11x2 + 13x + 2 - 72 = 0
---> 11x2 + 13x - 70 = 0
---> 11x2 + 13x - 70 = 0
On solving, we get
---> 11x2 -22x + 35x - 70 = 0
---> 11x (x - 2) + 35 (x - 2) = 0
---> (x - 2) (11x + 35) = 0
So, (x - 2) = 0
and x = 2 which is the tens digit
units digit y = x + 2 = 2 + 2 = 4
Hence, the required 2 digit number is 24

Given Quantity II: = 26
Thus, Quantity I < Quantity II
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113. Directions: In the following question two quantities are given. Compare the numeric value of both the quantities and answer accordingly.

A group consist of 4 couples in which each of the 4 persons have one wife.
Quantity I: Number of ways in which they could be arranged in a straight line such that the men and women occupy alternate positions.

Quantity II: Eight times the number of ways in which they be seated around circular table such that men and women occupy alternate position. 




SHOW ANSWER
Correct Ans:Quantity I = Quantity II or no relation
Explanation:
Quantity I:
Let first we arrange all 4 men in 4! ways then,
we arrange 4 women in 4P4 ways at 4 places either left of the man or right of the man.
Required Number of ways = 4! * 4P4 + 4! * 4P4
= 4! * 4! + 4! * 4!
= 24 * 24 + 24 * 24
= 576 + 576
= 1152

Quantity II:
First arrange the 4 men around circular table in alternate chairs in (4 - 1)! = 3! ways. Now four alternate chairs are vacant.
Each of the 4 women can now be seated in the vacant chairs in 4! ways.
Total number of ways = 3! * 4! = 144 ways.
Now, 144 * 8 = 1152
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114. Directions: In the following question two quantities are given. Compare the numeric value of both the quantities and answer accordingly.

Radius of conical tent and cylindrical pipe is equal. Ratio of radius & height of cylindrical pipe is 1:4. Volume of conical tent and cylindrical pipe is 1232 cm3 and 4312cm3 respectively.
Quantity I: Slant height of conical tent.

Quantity II: Height of cylindrical pipe.




SHOW ANSWER
Correct Ans:Quantity I < Quantity II
Explanation:
Given that, Ratio of radius & height of cylindrical pipe is 1 : 4.
----> Let radius and height of cylindrical pipe be 'r' cm and '4r' cm respectively.
Given that, Volume of cylindrical pipe = 4312cm3
---> π r2 h = 4312
---> (22/7) * r2 * 4r = 4312
---> (22/7) * r3 * 4 = 4312
---> r3 = (4312 * 7)/ (22 * 4)
---> r3 = (4312 * 7)/ (22 * 4)
---> r3 = 343
---> r = ∛343
---> r = 7

Thus, Radius of conical tent = Radius of cylindrical pipe = 7 cm
Let Height of conical tent be 'h' cm
Given, Volume of conical tent = 1232cm3
---> (1/3) π r2 h = 1232
---> (1/3) * (22/7) * 72 * h = 1232
---> [(22 * 7) / 3] * h = 1232
---> h = (1232 * 3) / (22 * 7)
---> h = 24 cm = Height of conical tent

Quantity I:
Slant height of conical tent = √ (r2 + h2)
= √ (72 + 242)
= √ (49 + 576)
= √ (625)
= 25 cm

Quantity II:
Height of cylindrical pipe = 4r = 4 * 7 = 28 cm

Hence, Quantity I < Quantity II
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115. Directions: In the following question two quantities are given. Compare the numeric value of both the quantities and answer accordingly.

Quantity I: Kishan and Bhavya appear in an interview for a vacancy. The probability of Kishan’s selection is 1/7 and that of Bhavya’s selection is 1/5. What is the probability that one of the them will be selected?

Quantity II: Chiru goes to a shop to buy some bananas but some-how he managed to save Rs. 3 per 4 bananas and thus purchased 8 dozen bananas instead of 5 dozen banana. Then, find the amount he has initially with him? 




SHOW ANSWER
Correct Ans:Quantity II > Quantity I
Explanation:
Quantity I:
Case I: probability of Kishan selected * probability of Bhavya not selected
= (1/7) * (1 - 1/5)
= (1/7) * (4/5)
= 4/35

Case 2: probability of Bhavya selected * probability of Kishan not selected
= (1/5) * (1 - 1/7)
= (1/5) * (6/7)
= 6/35

Required Probability = probability of Kishan selected or probability of Bhavya selected
= (4/35) + (6/35)
= 10/35
= 2/7

Quantity II:
Let initially Chiru has Rs. x with him
He managed to save Rs. 3 per 4 bananas
---> He saves Rs. 9 per dozen
So, according to the question,
Amount spent by him to buy 8 dozen bananas + saved Rs. 9 = Amount spent by him to buy 5 dozen bananas
---> (x/8) + 9 = (x/5)
---> (x/5) - (x/8) = 9
---> (8x - 5x)/40 = 9
---> (3x)/40 = 9
---> x = 3 * 40
---> x = Rs. 120

So, Quantity I < Quantity II.
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116. Directions: In the following question two quantities are given. Compare the numeric value of both the quantities and answer accordingly.

Quantity I: Maximum aggregate marks that a student can get. In an examination it is required to get 65% of the aggregate marks to pass. A student gets 684 marks and is declared failed by 8% marks.

Quantity II: No. of different ways in which the letters of the word 'VIRTUAL' will be arranged such that all the vowels come together.




SHOW ANSWER
Correct Ans:quantity I > quantity II
Explanation:
Quantity 1:
Let the maximum marks be x
So, Minimum marks required to pass = 65% of x
But the student got 684 marks , which is 8% less than 65% of x
---> (65 - 8)% of x = 684
---> 57% of x = 684
---> (57/100) * x = 684
---> x = (684/57) * 100
----> x = 12 * 100
---> x = 1200 marks
Therefore, Maximum aggregate marks = 1200

Quantity II:
The word VIRTUAL consists of 7 distinct letters in which
vowels are A, I, U
Hence, VRTL(AIU)
Required number of arrangements = 5! * 3!
= 5 × 4 × 3 × 2 × 1 × 3 × 2 × 1
= 720

Hence, Quantity I > Quantity II.
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117. Directions: In the following question two quantities are given. Compare the numeric value of both the quantities and answer accordingly.

Cost price of one bat is 5x and that of one ball is x/2.
Quantity I: Profit earned on bat if he sold it at the price of 6.2x
Quantity II: Discount % on ball if he mark up the ball by 80% of cost price and earned a profit of (3/20)x.




SHOW ANSWER
Correct Ans:Quantity II > Quantity I
Explanation:
Quantity I:
Cost price of one bat = 5x,
Selling Price of one bat = 6.2x
Profit = {[Selling Price - Cost price] / Cost price} * 100
= {[6.2x - 5x] / 5x} * 100
= {1.2x / 5x} * 100
= 120/5
= 24%

Quantity II:
Cost price of one ball = x/2
Marked Price of one ball = Cost price + 80% of Cost price
= (x/2) + (80/100) * (x/2)
= (x/2) + (4/5) * (x/2)
= (x/2) + (2/5) * (x)
= (x/2) + (2x/5)
= (5x + 4x)/10
= 9x/10

Selling price of one ball = Cost price + profit earned
= (x/2) + (3/20)x
= (10x + 3x) / 20
= 13x/20

Discount = Marked Price - Selling Price
= (9x/10) - (13x/20)
= (18x- 13x) / 20
= 5x/20
= x/4

Required Percentage = (Discount amount / Marked Price) * 100
= {(x/4) / (9x/10)} * 100
= {(x/4) * (10/9x)} * 100
= {(1/2) * (5/9)} * 100
= {5/18} * 100
= 27.78%

Hence, Quantity II > Quantity I
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118. In the question, two equations numbered I and II are given. Solve both the equations and give answer
Quantity I: By selling an article for Rs. 720, a man loses 100/0. At what price should he sell it to gain 40 %?
Quantity II: The difference between the CP and SP of an article is Rs. 240. If the profit is 20%, find the selling price?
a) Quantity I > Quantity II
b) Quantity 1 < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or Relation cannot be established




SHOW ANSWER
Correct Ans:B
Explanation:
Quantity I:
SP = 720 and loss= 10%
CP = 100/90*720 = Rs 800
SP of article at 5% gain = 140/100*800 = Rs.1120
Quantity II:
Difference between SP and CP=240
Gain%= SP- CP/CP*100
CP=1200
SP=1200+240 = Rs.1440
Hence Quantity II > Quantity I
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119. In the question, two equations numbered I and II are given. Solve both the equations and give answer.
(a) x≥y
(b) x≤y
(c) x<y
(d) x>y
(e) Relationship between x and y cannot be established

I. 6x2-7x+2=0
II. 20y2-31y+12=0  




SHOW ANSWER
Correct Ans:C
Explanation:
From I,
6x2-7x+2=0
6x2-4x-3x+2=0
(3x-2)(2x-1)=0
x=2/3, 1/2
From II,
20y2-31y+12=0
20y2-16y-15y+12=0
(5y-4)(4y-3)=0
y=4/5,3/4
x < y
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120. In the question, two equations numbered I and II are given. Solve both the equations and give answer
(a) x≥y
(b) x≤y
(c) x<y
(d) x>y
(e) Relationship between x and y cannot be established

I.88x2-19x+1=0
II. 132y2-23y+1=0 




SHOW ANSWER
Correct Ans:A
Explanation:
From I,
88x2 19x+1=0
(8x-1) (11x-1) =0
x=1/8,1/11
From II,
132y2-23y+1=0
(11y-1)(12y-1)=0
y=1/11,1/12
x≥y
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