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.

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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

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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

(Y-3) (Y-5) = 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

(Y-3) (Y-5) = 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

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