Hello and welcome to exampundit. Today we are sharing an Quantitative Aptitude Quiz for IBPS PO 2018 Prelims Exam.
The following set consists 10 questions on Quadratic Equation and Data Sufficiency with 10 minutes in hand.
Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
The following set of quiz is created as per the standards of upcoming IBPS PO Prelims exam 2018.
Quiz Name: Quantitative Aptitude Quiz for IBPS PO Prelims 2018
Time: 10 Minutes
Difficulty Level: Moderate
All the best.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Not categorized 0%
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
Each of the questions below consists of a question and two statements numbered I and II given below. You have to decide whether data provided are sufficient to answer the question.
Question: What is the annual income of Rahul Saxena from April 2014 to March 2015?
(A) The annual income of Rahul is 80% more than that of his wife Vijaya.
(B) In April 2014 the income of Rahul was Rs. 25000 and his income increases 5% per month.Correct
From I, Let the annual income of Rahul’s wife = x
Annual income of Rahul= x + 80% of x
The value of x is not clear.
I is not sufficient.
From II, Initial income= 25000
Income in May= 25000 (1+ 5/100)
Hence II is sufficient to answer.
Incorrect
From I, Let the annual income of Rahul’s wife = x
Annual income of Rahul= x + 80% of x
The value of x is not clear.
I is not sufficient.
From II, Initial income= 25000
Income in May= 25000 (1+ 5/100)
Hence II is sufficient to answer.

Question 2 of 10
2. Question
Each of the questions below consists of a question and two statements numbered I and II given below. You have to decide whether data provided are sufficient to answer the question.
Question: What is a three digit number?
(A) The three digit number is exact multiple of 17.
(B) The first and third digits are 7 and 1 respectively.Correct
From I and II, by trial and error the number will be 731. So both statements are required.
Incorrect
From I and II, by trial and error the number will be 731. So both statements are required.

Question 3 of 10
3. Question
Each of the questions below consists of a question and two statements numbered I and II given below. You have to decide whether data provided are sufficient to answer the question.
Question: What is Shalini’s age?
(A) Shalini is half as old as Rohit.
(B) Shalini’s age is 2/5 of her mother’s age. The mother’s age is 55 years.Correct
From I, Age of Shalini= ½ * age of Rohit.
Hence I is not sufficient.
From II, Age of Shalini = 2/5 * Mother’s age
= 2/5 * 55
= 22 years
So, II is sufficient to answer.
Incorrect
From I, Age of Shalini= ½ * age of Rohit.
Hence I is not sufficient.
From II, Age of Shalini = 2/5 * Mother’s age
= 2/5 * 55
= 22 years
So, II is sufficient to answer.

Question 4 of 10
4. Question
Each of the questions below consists of a question and two statements numbered I and II given below. You have to decide whether data provided are sufficient to answer the question.
Question: In how many years can a simple interest of Rs. 7500 be obtained on an amount of Rs. 37500?
(A) The rate of interest is 6% per annum.
(B) The difference between simple interest and compound interest is Rs. 505.50.Correct
From I,
T= 7500 * 100/ 37500 * 6
= 10/3
= 3 1/3 years
Hence I is sufficient to answer the question.
Incorrect
From I,
T= 7500 * 100/ 37500 * 6
= 10/3
= 3 1/3 years
Hence I is sufficient to answer the question.

Question 5 of 10
5. Question
Each of the questions below consists of a question and two statements numbered I and II given below. You have to decide whether data provided are sufficient to answer the question.
Question: What is the area of a circle?
(A) The circumference of the circle is 154 cm.
(B) The diameter of the circle is equal to the side of the square having area 441 cm².
Correct
From I, 2 * 22/7 * r= 154
r= 49/2
Area= 22/7 * r²= 22/7 * 49/2 * 49/2
= 1886.5 cm²
From II, Diameter= Side of square
= √441
= 21
Radius = 21/2 cm
Area= 22/7 * r² = 22/7 * 21/2 * 21/2
= 346.5
Either the statements are required.
Incorrect
From I, 2 * 22/7 * r= 154
r= 49/2
Area= 22/7 * r²= 22/7 * 49/2 * 49/2
= 1886.5 cm²
From II, Diameter= Side of square
= √441
= 21
Radius = 21/2 cm
Area= 22/7 * r² = 22/7 * 21/2 * 21/2
= 346.5
Either the statements are required.

Question 6 of 10
6. Question
In each of these questions, two equations are given. You have to solve both equations and give answer:
 X² – √√ ( 4096 ) = 56
 (y) ^4/3 * (y) ^ 5/3 – 523 = 206
Correct
 x² – √4096 = 56
X² – 8 = 56
X = √64
X= +8, 8
 (y) ^4/3 * (y) ^ 5/3 – 523 = 206
Y³ = 206 + 523
Y= 3√729
Y= 9
So, X < y
Incorrect
 x² – √4096 = 56
X² – 8 = 56
X = √64
X= +8, 8
 (y) ^4/3 * (y) ^ 5/3 – 523 = 206
Y³ = 206 + 523
Y= 3√729
Y= 9
So, X < y

Question 7 of 10
7. Question
In each of these questions, two equations are given. You have to solve both equations and give answer:
 3X + 4Y = 8
 4X + 2Y = 4
Correct
3x + 4y = 8……… * 4
4x + 2y = 4………… * 3
12x + 16y = 32
12x + 6y = 12
By solving equation,
X= 0 y= 2
So, X < y
Incorrect
3x + 4y = 8……… * 4
4x + 2y = 4………… * 3
12x + 16y = 32
12x + 6y = 12
By solving equation,
X= 0 y= 2
So, X < y

Question 8 of 10
8. Question
In each of these questions, two equations are given. You have to solve both equations and give answer:
 X² + 16 = 8X
 Y² + 15 = 8Y
Correct
 X² + 16 = 8X
X² – 8X + 16= 0
X² – 4X – 4X +16 = 0
(X – 4)² = 0
X= 4
 Y² + 15 = 8Y
Y² – 8Y + 15= 0
Y² – 5Y – 3Y + 15= 0
(Y3) (Y5) = 0
Y= 3, 5
SO, No RELATION BETWEEN X AND Y.
Incorrect
 X² + 16 = 8X
X² – 8X + 16= 0
X² – 4X – 4X +16 = 0
(X – 4)² = 0
X= 4
 Y² + 15 = 8Y
Y² – 8Y + 15= 0
Y² – 5Y – 3Y + 15= 0
(Y3) (Y5) = 0
Y= 3, 5
SO, No RELATION BETWEEN X AND Y.

Question 9 of 10
9. Question
In each of these questions, two equations are given. You have to solve both equations and give answer:
 √ 64 + √ X + 20 = √ 256
 Y² – 460 = 381
Correct
 √ 64 + √ X + 20 = √ 256
8 + √ X + 20 = 16
√ X +20 = 8
Squaring both side,
X + 20 = 64
X= 44
 Y² – 460 = 381
Y² = 841
Y = +29, 29
X > Y
Incorrect
 √ 64 + √ X + 20 = √ 256
8 + √ X + 20 = 16
√ X +20 = 8
Squaring both side,
X + 20 = 64
X= 44
 Y² – 460 = 381
Y² = 841
Y = +29, 29
X > Y

Question 10 of 10
10. Question
In each of these questions, two equations are given. You have to solve both equations and give answer:
 X² – (15)^5/2 / √X = 0
 28/ √Y – √Y = 7/√Y
Correct
 X² – (15)^5/2 / √X = 0
X^5/2 = (15)^5/2
X = 15
 28/ √Y – √Y = 7/√Y
28 – Y = 7
Y= 21
SO, X< Y
Incorrect
 X² – (15)^5/2 / √X = 0
X^5/2 = (15)^5/2
X = 15
 28/ √Y – √Y = 7/√Y
28 – Y = 7
Y= 21
SO, X< Y
Let us know in the comment section if you have liked our quiz. Your comments are extremely motivating for us. Also, do share the quizzes with your Friends and coaspirants.
All the best.
Regards
Team Exampundit
Average rating 0 / 5. Vote count: 0
No votes so far! Be the first to rate this post.
We are sorry that this post was not useful for you!
Let us improve this post!
Tell us how we can improve this post?