# Quant Test for IBPS RRB 2018 Prelim Exam Set – 12

Institute of Banking Personnel Selection (IBPS) had released the official notification for the Common Recruitment Process for RRBs (CRP RRBs VII) for the recruitment of Group “A”-Officers (Scale-I, II & III) and Group “B”-Office Assistant (Multipurpose)

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The examination will be held in two phases i.e. Preliminary Examination and Main Examination. The RRB Scale I Preliminary Exam is scheduled on 11th, 12th & 18th of August 2018. And RRB Assistant Preliminary Exam is scheduled on 19th, 25th August & 1st September 2018Details of the exam are as under: Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam.

Directions(1-5): Find the relation between x and y and choose a correct option.

1. I. 6x² + 19x + 15 = 0
II. 24y² + 11y + 1 = 0

x>y
y>=x
y>x
x>=y
x=y or relation cannot be established.
Option C
I. 6x² + 19x + 15 = 0
=>6x² + 9x + 10x + 15 = 0
=> (2x + 3) (3x + 5) = 0
=> x = –3/2, –5/3
II. 24y² + 11y + 1 = 0
=>24y² + 8y + 3y + 1= 0
=> (3y + 1) (8y + 1) = 0
=> y = –1/3, –1/8
y > x

2. I. 2x² + 11x + 15 = 0
II. 4y² + 22y + 24 = 0

x=y or relation cannot be established.
y>x
x>=y
x>y
y>=x
Option A
I. 2x² + 11x + 15 = 0
=> 2x² + 6x + 5x + 15 = 0
=> (x + 3) (2x + 5) = 0
=> x = – 3, –5/2
II. 4y² + 22y + 24 = 0
=> 2y² + 11y + 12 = 0
=> 2y² + 8y + 3y + 12 = 0
=> (y + 4) (2y + 3) = 0
=> y = –4, –3/2
No relation

3. I. 2x² + 9x + 9 = 0
II. 2y² + 17y + 36 = 0

y>=x
y>x
x>y
x=y or relation cannot be established.
x>=y
Option C
I. 2x² + 9x + 9 = 0
=> 2x² + 6x + 3x + 9 = 0
=> (x + 3) (2x + 3) = 0
=> x = –3, –3/2
II. 2y² + 17y + 36 = 0
=> 2y² + 8y + 9y + 36 = 0
=> (y + 4) (2y + 9) = 0
y = – 4, –9/2
x > y

4. I. 5x² + 29x + 20 = 0
II. 25y² + 25y + 6 = 0

x=y or relation cannot be established.
x>=y
y>=x
y>x
x>y
Option D
I. 5x² + 29 + 20 = 0
=> 5x² + 25x + 4x + 20 = 0
=> (x + 5) (5x + 4) = 0
=> x = –5, –4/5
II. 25y² + 25y + 6 = 0
=> 25y² + 15y + 10y + 6 = 0
=> (5y + 3) (5y + 2) = 0
=> y = – 3/5, –2/5
y > x

5. I. 3x² – 16x + 21 = 0
II. 3y² – 28y + 65 = 0

x>=y
x>y
x=y or relation cannot be established.
y>=x
y>x
Option E
I. 3x² – 16x + 21 = 0
=> 3x² – 9x – 7x + 21 = 0
=> (x – 3) (3x – 7) = 0
=>x = 3, 7/3
II. 3y² – 28y + 65 = 0
=>3y² – 15y – 13y + 65 = 0
=> (y – 5) (3y – 13) = 0
=> y = 5, 13/3
y > x

6. A boat covers 48 km in upstream and 72 km in downstream in 12 hours, while it covers 72 km in upstream and 48 km in downstream in 13 hours. Find the speed of the stream.
5 kmph.
2 kmph.
4 kmph.
6 kmph.
1 kmph.
Option B
Let downstream speed be u and Upstream speed be v.
Then (48/u+72/v) = 12 ——– (1) and (48/u+72/v) = 13 ———— (2)
On solving the eqn (1) and (2), we get u = 12 kmph and v = 8 kmph
Thus, speed of the current
= (u-v)/2
= 4/2 = 2 kmph.

7. Presennt age of A is 37.5% more than the present age of B. Present Ages of C and D are in the ratioof 4:3 resp. A is two years younger than C. If the present average age of A. B and D is 20 years , the find the value of B.
21
16
13
20
15
Option B
Le the present age of B be x years .
Sum of the present ages = x+1.375x + 1.375x + 2 +3/4 (1.375x+2) = 4*20
=> x = 16

8. Somesh’s age is 1/6th of his father’s age. Somesh’s father’s age will be twice Shankar’s age after 10 years. If Shankar’s eight birthdays was celebrated two years before, then what is Somesh ‘s present age.
7
13
8
11
5
Option E
Let Somesh ‘s present age = x
Then, her father’s age = 6x
Given that Somesh ‘s father’s age will be twice Shankar’s age after 10 years.
Therefore, Shankar’s age after 10 years
=> 1/2(6x+10) = 3x+5
Also given that Shankar’s eight birthdays was celebrated Two years before. Therefore, Shankar’s age after 10 years = 8+12 = 20
Therefore,3x+5 = 20
=> x=15/3 = 5

9. The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
Rs.9000
Rs.7000
Rs.2000
Rs.3000
Rs.4000
Option C
let cp 100% profit and loss = x
100+x = 1920
100-x = 1280
200% = 3200
cp = 1600
Required S.P =125% of 1600 = (125/100*1600) = Rs.2000

10. A fruit basket contains 10 Guavas and 20 Bananas out of which 3 Guavas and 5 Bananas are defective. If two fruits selected at random, what is the probability that either both are Bananas or both are non-defective?
316/435
311/430
303/445
355/437
317/400
Option A
P(A) = 20c2 / 30c2
P(B) = 22c2 / 30c2
P(A∩B) = 15c2 / 30c2
P(A∪B) = P(A) + P(B) – P(A∩B)
=> (20c2/30c2)+(22c2/30c2)-(15c2/30c2)
= 316/435