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

121. 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. 2x2+ 5x+3 = 0
II. y2+9y+14 = 0




SHOW ANSWER
Correct Ans:D
Explanation:
From I,
2x2+ 5x+3 = 0
(2x+3) (x+1) = 0
x = -3/2, -1
From II,
y2+9y+14 = 0
(y+7) (y+2) = 0
x>y
Workspace



122. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Statment: There are three positive numbers a, b and c. The average of a and b is less than the average of b and c by 1.
Quantity I: value of c.

Quantity II: value of a. 




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Given that, The average of a and b is less than the average of b and c by 1.
---> (a + b)/2 = [(b + c)/2] - 1
---> [(b + c)/2] - (a + b)/2 = 1
---> (b + c - a - b) / 2 = 1
---> (c - a) / 2 = 1
---> c - a = 2

---> Quantity I: c = a + 2 ---> which implies 'c' is 2 more than 'a'
and Quantity II: a = c - 2 ---> which implies 'a' is 2 less than 'c'

Hence, Quantity I > Quantity II
Workspace



123. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Quantity I: In how many ways the word 'SCOOTER' can be arranged such that 'S' and 'R' are always at two ends?

Quantity II: A tricolor flag is to be formed having three adjacent strips of three different colors chosen from six different colors. How many different colored flags can be formed with different design in which all the three strips are always in horizontal positions? 




SHOW ANSWER
Correct Ans:Quantity I = Quantity II or relation cannot be established
Explanation:
Quantity I:
In the word SCOOTER ---- the letter 'O' is repeated 2 times.
And S and R are considered as one element.
So, (S, C, 2O, T, E, R)
Required number of ways = (5! * 2!) / 2!
= 5!
= 5 * 4 * 3 * 2 * 1
= 120 ways

Quantity II:
First strip can be coloured in 6 ways;
second strip can be coloured in 5 ways; and
third strip can be coloured in 4 ways.

Hence, all the three strips can be coloured in 6 * 5 * 4 ways = 120 ways.

So, Quantity I = Quantity II
Workspace



124. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

A hemisphere of radius 4 cm is to be shipped in a shipping box of rectangular shape. The dimension of the box are consecutive odd numbers.
Quantity I: Minimum volume of box required to have the shipment possible.
Quantity II: 960 cm3 




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
The minimum length of any side of rectangular box should be greater than the diameter of the hemisphere.
Given Radius of hemisphere = 4 cm.
Diameter of hemisphere = 2 * Radius = 8 cm

Also given that, the dimension of the rectangular box are consecutive odd numbers.
So for odd number, minimum length= 9, next side =11, next side=13
Hence, Minimum Volume of box = length * width * height
= 9 * 11 * 13
= 1287 cm3

Given Quantity II = 960 cm3
Hence, Quantity I > Quantity II
Workspace



125. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

1 > a > 0 > b
Quantity I: Value of [(a + b)2 - a2 - b2] / [(a + b)2 - (a - b)2]

Quantity II: 1/ [2(ab3 + ab] 




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
[(a + b)2 - a2 - b2] / [(a + b)2 - (a - b)2]
---> [a2 + b2 + 2ab - a2 - b2] / [a2 + b2 + 2ab - a2 - b2 + 2ab]
---> 2ab / 4ab
---> 1 / 2

Quantity II:
1/ [2(ab3 + ab]
---> 1/ [2ab (b2 + 1)]
Given that 0 > b, so if we subsitute b = -1 and
From the statement 1 > a > 0, say a = 0.5, Then the above eqn becomes,
---> 1/ [2 * 0.5 * (-1) (-12 + 1)]
---> 1/ [-1 (1 + 1)]
---> 1/ [-1 (1 + 1)]
---> 1/ (-2)

So, Quantity I > Quantity II
Workspace



126. Compare the value of 2 quantities given in the question and give answer.

Quantity I: At simple Interest, a sum becomes 3 times in 20 year. Find the time, in which the sum will be double at the same rate of interest.
Quantity II: Simple interest for the sum of Rs. 1500 is Rs. 30 in 4 year and Rs. 60 in 8 year. Find the rate of simple interest. 




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Formula for Simple Interest:
SI = PNR/t
SI = Simple Interest
P = Principle Amount
N = Number of times compounded in one t
R = Rate
t =Time period
Quantity-I
Let sum = P, then for 20 year
SI = 3P- 2P = 2P
2P= P* R *T/100
R=10%
For the sum to be double
SI = 2P-p= P
P = P*10*N/100
N = 10 years
Quantity-II
SI in 8 years= 60
SI in 4 years= 30
so,
1500* R * 8/100 - 1500*R*4/100 = 60 -30
1500*R*4/100 = 30
6000R/100 = 30
R = 30/60 = 0.5%
Hence, Quantity I > Quantity II
Workspace



127. Compare the value of 2 quantities given in the question and give answer.

Quantity I: Two numbers are respectively 20% and 50% more than a 3rd number. What is the percentage of 2nd with respect to 1st?
Quantity II: From 2008 to 2009, the sales of a book decreased by 80%. If the sales in 2010 was the same as in 2008, by what percent did it increase from 2009 to 2010?




SHOW ANSWER
Correct Ans:Quantity II > Quantity I
Explanation:
Quantity-I:
Let us assume that the 3rd number is 100, so that 1st and 2nd numbers are 20% and 50% more. so,
First number =120
Second number= 150
Third number = 100
Required percentage of 2nd with respect to 1st = (150/120)*100
= 125%
Quantity-II
Sale in 2008= 100
Sale in 2009= 20
Sale in 2010= 100
Required percentage increase= (80/20)*100
= 400%
Hence Quantity II> Quantity I
Workspace



128. Compare the value of 2 quantities given in the question and give answer.

Two trains of lengths 50m and 65m are moving in the same direction at 18 m/s and 17m/s respectively.
Quantity I: Time taken by the faster train to cross the slower train.
Quantity II: Time taken by the slower train to cross a platform of length 156 m.




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Quantity-I
Relative speed = 18-17= 1 m/s
Required time = (50+65)/1 = 115 seconds
Quantity-II
Time taken to cross a platform of length 156 m
= (65+156)/17
= 13 seconds
Hence Quantity I > Quantity II = 13 seconds
Workspace



129. Compare the value of 2 quantities given in the question and give answer.(Compare only numerical values)

Quantity A: The least prime number greater than 24
Quantity B: The greatest prime number less than 28 




SHOW ANSWER
Correct Ans:if quantity A > quantity B
Explanation:
For the integers greater than 24, note that 25, 26, 27, and 28 are not prime numbers, but 29 is a prime number, as are 31 and many other greater integers.
Thus, 29 is the least prime number greater than 24, and Quantity A is 29.
For the integers less than 28, note that 27, 26, 25, and 24 are not prime numbers, but 23 is a prime number, as are 19 and several other lesser integers.
Thus, 23 is the greatest prime number less than 28, and Quantity B is 23.
The correct answer is Choice A, Quantity A is greater.
Workspace



130. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I. 2x2 -7x +3 = 0
II. 2y2 - 7y + 6 = 0




SHOW ANSWER
Correct Ans:E
Explanation:
I. 2x2 -7x +3 = 0
2x2 -6x - x +3 = 0
(x-3) (2x-1) = 0
x = 3, 1/2
II. 2y2 - 7y + 6 = 0
2y2 - 4y - 3y + 6 = 0
(y-2) (2y-3) = 0
y = 2, 3/2
x and y are not related in any way. The relationship between x and y cannot be determined.
Workspace



131. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I. 2x2 + 11x + 14 = 0
II. 2y2 + 15y + 28 = 0




SHOW ANSWER
Correct Ans:C
Explanation:
I. 2x2 + 11x + 14 = 0
2x2 + 4x + 7x + 14 = 0
(x+2) (2x+7) = 0
x = -2,-7/2
II. 2y2 + 15y + 28 = 0
2y2 + 8y + 7y + 28 = 0
(y+4) (2y+7) = 0
y = -4,-7/2
x ≥ y
Workspace



132. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined 

I. 9x2 - 45x + 56 = 0
II. 4y2 - 17y + 18 = 0 




SHOW ANSWER
Correct Ans:D
Explanation:
I. 9x2 - 45x + 56 = 0
9x2 - 45x + 56 = 0
3x(3x-8) - 7 (3x-8) = 0
(3x-8) (3x-7) = 0
x = 8/3, 7/3

II. 4y<sup>2</sup> — 17y + 18 = 0
4y<sup>2</sup> — 8y - 9y + 18 = 0
(y-2) (4y-9) = 0
y =2, 9/4
x>y

Workspace



133. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I. 4x2 + 16x + 15 = 0
II. 2y2 +3y +1 = 0




SHOW ANSWER
Correct Ans:A
Explanation:
I. 4x2 + 16x + 15 = 0
4x2 + 10x + 6x + 15 = 0
2x(2x+5) + 3(2x+5) = 0
(2x+5) + (2x+3) = 0
x = -5/2, -3/2
II. 2y2 + 3y + 1 = 0
2y2 + 2y + y + 1 = 0
(y+1) (2y+1) = 0
y=-1, -1/2
y>x
Workspace



134. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I.9x2 - 36x + 35 = 0
II. 2y2 - 15y- 17 = 0




SHOW ANSWER
Correct Ans:E
Explanation:
I. 9x2 - 36x + 35 = 0
9x2 - 21x - 15x + 35 = 0
3x (3x-7) - 5 (3x-7) = 0
(3x-7) (3x-5) = 0
x = 5/3 , 7/3
II. 2y2- 15y- 17 = 0
2y2 -17y + 2y - 17 = 0
(y+1) (2y-17) = 0
y = -1, 17/2
x and y are not related in any way. The relationship between x and y cannot be determined.
Workspace



135. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Quantity I: How many numbers greater than 1000 but less than 5000 can be made from digits 0, 1, 3, 5, 6?

Quantity II: There are 5 blue flags, 4 red flags and 3 green flags, in Debu’s wardrobe. He has to select 4 flags from this set. In how many ways can he select these four flags such that there is at least one blue flag and exactly one green flag in them (Do not consider that the flags are in pairs)? 




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
When not written about repetition, take that repetition to be allowed.
Numbers between 1000 to 5000 means 4 digit numbers. So 4 boxes.
Now 0, 5 and 6 cannot be placed in Box 1 (Thousandth place). Because, 0 will make it 3-digit number and 5 & 6 will make number greater than 5000.
So there are 2 choices (ie., numbers 1 and 3) for Box 1 (Thousandth place), and all 5 choices (ie., numbers 0, 1, 3, 5, 6) for Box 2, 3, and 4 (Hundredths, Tenths and units places).
So total numbers = 2C1 * 5C1 * 5C1 * 5C1 ---> (since 4 digit number)
= 2*5*5*5
= 250

Quantity II:
No. of Blue flags = 5;
No. of Red flags = 4;
No. of Green flags = 3

In the selected four flags ---> there is at least one blue flag and exactly one green flag
So, the possible selections are:
Way 1: 1 blue flag out of 5 * 1 green flag out of 3 * 2 Red flags out of 4
----> 5C1 * 3C1 * 4C2
----> 5 * 3 * 6
----> 90

Way 2: 2 blue flag out of 5 * 1 green flag out of 3 * 1 Red flags out of 4
----> 5C2 * 3C1 * 4C1
----> 10 * 3 * 4
----> 120

Way 3: 3 blue flag out of 5 * 1 green flag out of 3
----> 5C3 * 3C1
----> 10 * 3
----> 30

So, Total No. of ways = 90 + 120 + 30 = 240

here, Quantity I > Quantity II.
Workspace



136. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Ram invested Rs. P in scheme A and Rs. 2P in scheme B, for two years each. Scheme A offers simple interest p.a. Scheme B offers compound interest (compounded annually) at the rate of 10% per annum. Respective ratio between the interest earned from scheme A and that earned from scheme B was 8 : 21.
Quantity I: Rate of interest offered by scheme A.

Quantity II: Rate of interest offered by scheme C (simple interest per annum), when Rs. 1600 is invested for 3 years earns an interest of Rs. 384. 




SHOW ANSWER
Correct Ans:Quantity I = Quantity II or Relation cannot be established
Explanation:
Let the rate of interest offered by scheme A ie., simple interest = R% per annum.
Given Time, (n) = 2 years.

Then, Simple interest, S.I = (p x n x r)/100
= (P * 2 * R)/100
= 2PR/100

Now, Compound Interest, C.I = Amount - Principal
Amount = P x [1+ (r/100)]n
C.I = P{[1+ (r/100)]n - 1}
So, Compound Interest received by Scheme B, C.I = 2P{[1+ (10/100)]2 - 1}
= 2P{[1+ (1/10)]2 - 1}
= 2P{[(10+ 1)/10]2 - 1}
= 2P{[11/10]2 - 1}
= 2P{121/100 - 1}
= 2P{121 - 100}/100
= 2P{21/100}
= 42P/100

Given that, S.I from scheme A : C.I from scheme B = 8 : 21
----> (2PR/100) : (42P/100) = 8 : 21
----> R : 21 = 8 : 21
----> R = 8
Hence, Quantity I: Rate of interest offered by scheme A, R = 8% per annum.

Now, Quantity II:
Simple interest --> Principal, P = Rs. 1600
n = 3 years
S.I = Rs. 384
Then, Simple interest, S.I = (p x n x r)/100
---> 384 = (1600 * 3 * r)/100
---> 384 = 48r
---> r = 8
Hence, Rate of interest offered by scheme C = 8% per annum.

Here Quantity I = Quantity II.
Workspace



137. In the following question, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly
I. 3x2 "“ 25x + 52 = 0
II. 3y2 "“ 17y + 20= 0




SHOW ANSWER
Correct Ans:If x≥y
Explanation:
I. 3x2 – 25x + 52 = 0
3x2 - 12x - 13x + 52 = 0
3x(x - 4) - 13(x - 4) = 0
(x - 4)(3x - 13) = 0
x = 4, 13/3 = 4, 4.3

II. 3y2 – 17y + 20= 0
3y2 - 12y - 5y + 20= 0
3y(y - 4) - 5(y - 4) = 0
(y - 4)(3y - 5) = 0
y = 4, 5/3 = 4, 1.7
Hence, x≥y
Workspace



138. If x = 332, y = 333, z = 335 then the value of x3 + y3 + z3 "“ 3xyz = ?




SHOW ANSWER
Correct Ans:7000
Explanation:
(x3 + y3 + z3 – 3xyz) = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Also we know that,
(x3 + y3 + z3 – 3xyz) = (x + y + z) * 1/2[(x – y)2 + (y – z)2 + (z – x)2]
On putting the values, we get
= 1000 × 1/2[(332 – 333)2 + (333 – 335)2 + (335 – 332)2
= 1000 × 1/2[1 + 4 + 9]
= 7000
Hence, option C is correct.
Workspace



139. Quantity 1: Perimeter of a Square having its side is equal to hypotenuse of a right-angled triangle having its sides 8mtrs and 6mtrs respectively.
Quantity 2: Perimeter of a Rhombus having its diagonals d1 and d2 are 12mtr and 16mtr respectively.




SHOW ANSWER
Correct Ans:Quantity 1 = Quantity 2 or Relationship can’t be established.
Explanation:
Quantity 1:
Hypotenuse of a triangle = √(a²+b²)
= √(8²+6²)
= √(64+36)
= √100
= 10 mtrs
And Perimeter of the square = 4a = 4 × 10 = 40mtrs.

Quantity 2:
Diagonals of rhombus intersect at 90°angle.
The diagonals and the sides of the rhombus form right triangle.
One side of the right triangle is 16/2= 8mtrs
Other side of the right triangle is 12/2= 6mtrs
Then the third side of the triangle = √(a²+b²)
= √(8²+6²)
= 10mtrs
And Perimeter of the rhombus = 4a = 4 × 10 = 40mtrs.
So Quantity 1 is equal to Quantity 2.
Workspace



140. Study the following question carefully and choose the right answer.
If x = √3 + √4 + √5 than x4 "“ 8x3 + 8x2 + 32x = ?




SHOW ANSWER
Correct Ans:44
Explanation:
x = √3 + √4 + √5
x – 2 = √3 + √5
Squaring on both sides we get,
x2 – 4x + 4 = 8 + 2√15
x2 – 4x – 4 = 2√15
Squaring on both sides we get,
(a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ac
x4 + 16x2 + 16 – 8x3 + 32x – 8x2 = 60
x4 – 8x3 + 8x2 + 32x = 44
Hence, option C is correct.
Workspace



Are you seeking for good platform for practicing Equations and Inequations questions in online. This is the right place. The time you spent in Fresherslive will be the most beneficial one for you.

Online Test on Equations and Inequations @ Fresherslive

This page provides important questions on Equations and Inequations along with correct answers and clear explanation, which will be very useful for various Interviews, Competitive examinations and Entrance tests. Here, Most of the Equations and Inequations questions are framed with Latest concepts, so that you may get updated through these Equations and Inequations Online tests. Equations and Inequations Online Test questions are granted from basic level to complex level.

Why To Practice Equations and Inequations Test questions Online @ Fresherslive?

Equations and Inequations questions are delivered with accurate answer. For solving each and every question, very lucid explanations are provided with diagrams wherever necessary.
Practice in advance of similar questions on Equations and Inequations may improve your performance in the real Exams and Interview.
Time Management for answering the Equations and Inequations questions quickly is foremost important for success in Competitive Exams and Placement Interviews.
Through Fresherslive Equations and Inequations questions and answers, you can acquire all the essential idea to solve any difficult questions on Equations and Inequations in short time and also in short cut method.
Winners are those who can use the simplest method for solving a question. So that they have enough time for solving all the questions in examination, correctly without any tense. Fresherslive provides most simplest methods to answer any tough questions. Practise through Fresherslive test series to ensure success in all competitive exams, entrance exams and placement tests.

Why Fresherslive For Equations and Inequations Online Test Preparation?

Most of the job seekers finding it hard to clear Equations and Inequations test or get stuck on any particular question, our Equations and Inequations test sections will help you to success in Exams as well as Interviews. To acquire clear understanding of Equations and Inequations, exercise these advanced Equations and Inequations questions with answers.
You're Welcome to use the Fresherslive Online Test at any time you want. Start your beginning, of anything you want by using our sample Equations and Inequations Online Test and create yourself a successful one. Fresherslive provides you a new opportunity to improve yourself. Take it and make use of it to the fullest. GOODLUCK for Your Bright Future.


Online Test for Data Interpretation
Online Test for C Language

FreshersLive - No.1 Job site in India. Here you can find latest 2023 government as well as private job recruitment notifications for different posts vacancies in India. Get top company jobs for both fresher and experienced. Job Seekers can get useful interview tips, resume services & interview Question and answer. Practice online test free which is helpful for interview preparation. Register with us to get latest employment news/rojgar samachar notifications. Also get latest free govt and other sarkari naukri job alerts daily through E-mail...
DMCA.com Protection Status