SBI PO 2019 Notification will be expected soon. It is one of the most expected recruitment among the banking aspirants. Every year the exam pattern for SBI PO has been changing. Depends upon the changing of exam pattern the questions are quite harder compare to the previous year. So the questions are in high level than the candidate’s assumption.
As per the latest trend, our IBPS Guide is providing the updated New Exam Pattern Quantitative Aptitude questions for SBI PO 2019 Day 17. Our Skilled experts were mounting the questions based on the aspirant’s needs. So candidates shall start your preparation and practice on daily basis with our SBI PO pattern quantitative aptitude questions 2019 day 19. Start your effective preparation from the right beginning to get success in upcoming SBI PO 2019.
“Be not afraid of growing slowly; be afraid only of standing still”
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Directions (Q. 1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
1) I) 2x2 + 19x + 45 = 0
II) y2 – 17y + 72 = 0
2) I) 3x2 – 19x + 28 = 0
II) (256)1/4 y + (216)1/3 = 0
3) I) 3/√x + 8/√x = √x
II) y2 = 95/2 / √y
4) I) 2x2 +29x +50 =0
II) 12y2 – 7y +1 =0
5) I) 5x2 + 23x +12 = 0
II) 15y2 +14y +3 = 0
Directions (Q. 6 – 10): Study the following information carefully and answer the given questions:
The following table shows the speed of boat in still water and stream and the distance covered by the boat and time taken to reach the place. Some values are missing
6) Find the difference between the distance covered by boat A to that of boat B?
a) 55 Km
b) 62 Km
c) 48 Km
d) 69 Km
e) None of these
7) Find the ratio between the time taken by boat C to that of boat E to reach the place (Both Downstream and Upstream)?
a) 7: 11
b) 6: 11
c) 5: 11
d) 8: 11
e) None of these
8) Find the average speed of all the given boats in still water, if the time taken by boat D to reach the place is 9 hrs?
a) 24 Km/hr
b) 20 Km/hr
c) 18 Km/hr
d) 16 Km/hr
e) None of these
9) The sum of the speed of stream in all the given boats is approximately what percentage of the sum of the speed of all the given boats in still water if the time taken by boat D to reach the place is 9 hrs?
a) 52 %
b) 58 %
c) 44 %
d) 65 %
e) None of these
10) Find the distance covered by boat E?
a) 92.25 Km
b) 86.5 Km
c) 106.25 Km
d) 100.75 Km
e) None of these
Answers :
Direction (1-5) :
1) Answer: c)
I) 2x2 + 19x + 45 = 0
2x2 + 10x + 9x + 45 = 0
2x(x + 5) + 9 (x + 5) = 0
(2x + 9) (x + 5) = 0
X = -9/2, -5 = -4.5, -5
II) y2 – 17y + 72 = 0
(y – 8) (y – 9) = 0
Y = 8, 9
X < y
2) Answer: a)
I) 3x2 – 19x + 28 = 0
3x2 – 12x – 7x + 28 = 0
3x (x – 4) – 7 (x – 4) = 0
(x – 4) (3x – 7) = 0
x = 4, 7/3
II) (256)1/4 y = – (216)1/3
(44)1/4 y = – (63)1/3
4y = -6
Y = – (6/4) = -1.5
X > y
3) Answer: a)
I) 3/√x + 8/√x = √x
11/√x = √x
11= x
II) y2 = 95/2/ √y
Y2 × y1/2 = 95/2
Y2 + (1/2) = 95/2
Y5/2 = 95/2
Y = 9
X > y
4) Answer: c)
I) 2x2 +29x + 50 = 0
2x2 + 4x + 25x + 50 = 0
2x(x + 2) + 25(x + 2) = 0
(2x + 25) (x + 2) = 0
X = -25/2, -2
II) 12y2 – 7y +1 = 0
12y2 – 4y-3y +1 = 0
4y (3y – 1) -1(3y – 1) = 0
(4y-1) (3y-1) =0
Y= ¼, 1/3
X < y
5) Answer: d)
I) 5x2 + 23x + 12 = 0
5x2 + 20x + 3x + 12 = 0
5x(x + 4) + 3 (x + 4) = 0
(5x + 3) (x + 4) = 0
X = -3/5, -4
II) 15y2 +14y + 3 = 0
15y2 + 5y + 9y + 3 = 0
5y (3y + 1) + 3 (3y + 1) = 0
(5y + 3) (3y + 1) = 0
Y= -3/5, -1/3
X ≤ y
Direction (6-10) :
6) Answer: a)
The speed of downstream = Speed of boat in Still water + Speed of stream
= > 20 + 12 = 32 Km/hr
The speed of upstream = Speed of boat in Still water – Speed of stream
= > 20 – 12 = 8 Km/hr
Time = Distance/Speed
10 = (x/32) + (x/8)
10 = 40x/(32*8)
X = 64 Km
The distance covered by boat B = 119 Km
Required difference = 119 – 64 = 55 km
(Or)
Distance = Time*[(Speed of still water2 – Speed of Stream2)/(2*Speed of still water)]
The distance covered by boat A
= > 10*[(202 – 122)/(2*20)] = 64 Km
The distance covered by boat B = 119 Km
Required difference = 119 – 64 = 55 km
7) Answer: d)
Speed of boat C in still water = 16 Km/hr
Speed of stream (Boat C) = 5 Km/hr
Speed of downstream = (16 + 5) = 21 Km/hr
Speed of upstream = 16 – 5 = 11 Km/hr
The time taken by boat C to reach the place
= > Time = (57.75/21) + (57.75/11)
= > 2.75 + 5.25 = 8 hr
The time taken by boat E to reach the place = 11 hr
Required ratio = 8: 11
8) Answer: b)
Let the Speed of boat D in still water be x,
Time = Distance/Speed
9 = 65/(x + 8) + 65/(x – 8)
9 = [65x – 520 + 65x + 520]/[(x + 8) (x – 8)] [130x / (x2 – 64)] = 9
130x = 9x2 – 576
9x2 – 130x – 576 = 0
9x2 – 162x + 32x – 576 = 0
9x(x – 18) + 32(x – 18) = 0
(9x + 32) (x – 18) = 0
X = -32/9, 18 (negative value will be eliminated)
The average speed of all the given boats in still water
= > (20 + 24 + 16 + 18 + 22)/5 = 100/5
= > 20 Km/hr
9) Answer: c)
Let the speed of stream in boat B be x,
= > 12 = [119/(24 + x) + 119/(24 – x)]
= > 12 = (2856 – 119x + 2856 + 119x) / (24 + x) (24 – x)
= > 12 = [5712 / (576 – x2)
= > 12*(576 – x2) = 5712
= > 6912 – 12x2 = 5712
= > 1200 = 12x2
= > x2 = 100
= > x = 10, -10 (negative value will be eliminated)
So, Speed of stream in boat B = 10 Km/hr
The sum of the speed of stream in all the given boats
= > 12 + 10 + 5 + 8 + 9 = 44 Km/hr
Let the Speed of boat D in still water be y,
Time = Distance/Speed
9 = 65/(y + 8) + 65/(y – 8)
9 = [65y – 520 + 65y + 520]/[(y + 8) (y – 8)] [130y / (y2 – 64)] = 9
130y = 9y2 – 576
9y2 – 130y – 576 = 0
9y2 – 162y + 32y – 576 = 0
9y (y – 18) + 32(y – 18) = 0
(9y + 32) (y – 18) = 0
Y = -32/9, 18 (negative value will be eliminated)
So, Speed of boat D in still water = 18 Km/hr
The sum of the speed of all the given boats in still water
= > 20 + 24 + 16 + 18 + 22 = 100 Km/hr
Required % = (44/100)*100 = 44 %
10) Answer: d)
Distance = Time*[(Speed of still water2 – Speed of Stream2)/(2*Speed of still water)]
The distance covered by boat A
= > 11*[(222 – 92)/(2*22)]
= > 11*[(484 – 81)/44] = 100.75 Km
