BASED on New Pattern – Practice Quantitative Aptitude Questions Quadratic Equation for IBPS Exam 2017

BASED on New Pattern – Practice Quantitative Aptitude Questions (Quadratic Equation) for IBPS Exam 2017:

Dear Readers, Important Practice Aptitude Questions for IBPS Exams 2017 was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this material.

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Direction (1-10): In each of the following question, read the given statement and compare the two given quantities on its basis. Give answer
If Quantity I> Quantity II
If Quantity I< Quantity II
If Quantity I ≤ Quantity II
If Quantity I ≥ Quantity II
If Quantity I = Quantity II or No relationship can be established.

1). Quantity I.Twelve students compete in a race. In how many ways first three prizes are given?
Quantity II. There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 5 choices each and the next three have 4 choices each?

a)
b)
c)
d)
e)

Quantity I:
The first prize is given in 12 ways and the second prize is given in 11 ways and the third prize is given in 10 ways.
∴ required number of ways = 12× 11× 10= 1320
Quantity II:
The first 3 questions can be answered in 5 ways each and next 3 questions can be answered in 4 ways each.
∴ The required number of sequences = 4× 4 × 4 × 5 × 5 × 5 = 8000
Hence, Quantity I< Quantity II

2). The diameter is 14 cm.
Quantity I.Total surface area of the sphere
Quantity II.Curved surface area of the hemi sphere

a)
b)
c)
d)
e)
Quantity I:
Total surface area of the sphere = 4 π r^2
r = 7cm
4 × 22/7 × 7 × 7 = 616 sq cm
Quantity II:
Curved surface area of the hemi sphere = 2 π r^2
2 × 22/7 × 7 × 7 = 308 sq cm
Hence, Quantity I> Quantity II

3).Quantity I: The average age of 11 boys is 30. If the average age increases by 0.5 when kamal, one of the boys leaves the group. Find the age of kamal.
Quantity II:The average of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by 3 months. Find the age of teacher.

a)
b)
c)
d)
e)
Quantity I:
The age of 11 boys = 11 × 30 = 330
The age of boys after excluding Kamal = 10 × 30.5 = 305
∴ the age of kamal = 330 – 305 = 25 years
Quantity II:
Total age of 39 students = 39 × 15 = 585
Average age of 40 persons (39 students + 1 teacher) = 40 × 61/4 = 610
∴ the age of teacher = 610 – 585 = 25 years
Hence, Quantity I = Quantity II

4).

Quantity I:
Quantity II:Circumference of the circle

a)
b)
c)
d)
e)
Quantity I:
Area of the shaded region = Area of the square – Area of the circle
= (42 × 42) – (π ×21 ×21) [since, r =21]
= 1764 – (22/7 × 21 × 21) = 1764 – 1386
= 378 sq cm
Quantity II:
Circumference of the circle = 2 × 22/7 × 21 = 132 cm
Hence, Quantity I> Quantity II

5).Quantity I: 30x^2 + 28x + 6 = 0
Quantity II: x^2 + 8x + 15 = 0

a)
b)
c)
d)
e)
Quantity I: 30x² + 28x + 6 = 0
15x² + 14x + 3 = 0
5x (3x+1) + 3 (3x+1) =0
(5x+3) (3x+1) =0
x = -3/5, -1/3
Quantity II: x² + 8x + 15 = 0
(x+5) (x+3) = 0
x = -5, -3
Hence, Quantity I> Quantity II

6). A basket contains 3 Blue, 2 Yellow and 5 Red balls.
Quantity I: If four balls are picked at random, what is the probability that two are Yellow and two are Blue.
Quantity II: If three balls are picked at random, what is the probability that at least one is Red.

a)
b)
c)
d)
e)
Quantity I:
(3C₂× 2C₂)/ 10C₄ = 3 × 1 / 210 = 1 /70
Quantity II:
5C₃ /10C₃ + (5C₂× 5C₁)/10C₃+(5C₁× 5C₂)/10C₃
= 1/120 × [10+50+50] = 110/120 = 11/12
Hence, Quantity I< Quantity II

7). A shopkeeper allows a discount of 4% on his article. If the cost price of the article is Rs.100 and he wants to get a profit of 20%.
Quantity I:Selling price of the article
Quantity II: Marked price of the article

a)
b)
c)
d)
e)
Quantity I: Selling price of the article for 20% profit
= 100 × (100 + 20)/ 100 = 120
Quantity II: Marked price of the article for 4% discount
= 120 × 100 / (100-4) =120 × 100 / 96 = 125
Hence, Quantity I< Quantity II

8). Quantity I: [{ (- ½ )²}^(-1)]^(-2)
Quantity I: 1/{[₅√(16^(-⅗)]^(-5/3)}⁵

a)
b)
c)
d)
e)

9).

Quantity I:Area of the shaded portion
Quantity II: 170 sq cm

a)
b)
c)
d)
e)
Quantity I:
Area of the shaded portion = Area of a square - 4 × Area of a quadrant
= a² – 4 × ¼ ×π× r²
= (28×28) – (4 × ¼ × 22/7 × 14 × 14)
= 784 – 616 = 168 sq cm
Quantity II:170 sq cm
Hence, Quantity I< Quantity II

10). Quantity I: 2/x – 5/3x = 1/9
Quantity II:9/2y + 1/y =22

a)
b)
c)
d)
e)
Quantity I: 2/x – 5/3x = 1/9
6-5/3x = 1/9
1/3x = 1/9
x = 3
Quantity II: 9/2y + y =22
11/2 y = 22
Y = 1/4
Hence, Quantity I> Quantity II