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Our IBPS Guide is providing High-Level New Pattern Quantitative Aptitude Questions for SBI PO 2019 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these high-level questions daily to familiarize with the exact exam pattern. We wish that your rigorous preparation leads you to a successful target of becoming SBI PO.
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Direction (1-5): Study the following information carefully and answer the given questions:
The given bar graph shows the speeds of stream of three rivers on 6 different days.
1) The speed of still water on day 1 is 18 kmph and the speed of stream of river A travelled certain distance upstream and returned to starting point on day1 in 9 hours. How much total upstream and downstream distance on river A covered on day1?
a) 115 km
b) 112 km
c) 110 km
d) 118 km
e) None of these
2) The Chennai to Delhi is a downstream journey on a river C in day2. Boat S and T run a shuttle service between the two cities that are 600km apart. Boat S which starts from Chennai has the still water speed of 50kmph, while boat T which starts from Delhi at the same time has the still water speed 30kmph. When and where will they meet for the second time?
a) Second time meet after 20 hours and 400km apart from Delhi.
b) Second time meet after 10 hours and 200km apart from Chennai.
c) Second time meet after 15 hours and 400km apart from Delhi.
d) Second time meet after 18 hours and 200km apart from Chennai.
e) None of these
3) The boat takes 76 hours for travelling downstream from Chennai to Delhi and coming back to Bangalore midway between Chennai and Delhi. If the speed of river B in day6 and the speed of boat in still water is 75% more than that of the speed of river B in day5, what is the distance between Chennai and Delhi?
a) 1440km
b) 1260km
c) 1690km
d) 630km
e) None of these
4) The speed of boat S on day7 is 15kmph and the speed of river A in day7 is average of the speed of river A in day4 and day8. The ratio of the upstream time to downstream time to cover the same distance on day7 is 1:5. If the speed of boat T in still water on day4 and day8 is 12kmph and 18kmph respectively, then how much time will boat T take to cover 90km downstream on day8?
a) 3 hours
b) 2 hours
c) 4 hours
d) 1 hour
e) None of these
5) The boat S travels 135 km upstream in river B in day3 and return to its starting point and its entire journey time is 6 hours. The downstream speed of boat T in river B in day3 is three-fourth of the speed of boat S in still water. How much time will boat T take to cover same upstream and downstream distances together in same river in day3?
a) 5(7/16) hours
b) 6(7/16) hours
c) 7(7/16) hours
d) 9(7/16) hours
e) None of these
6) Ram, Geeta and Meena can do a work alone in 12, 24 and 16 days respectively. They started working together after ____ days, Meena leaves and the remaining work completed by Ram and Geeta. Total amount is Rs._____ and Geeta and Meena gets equal wage.
Which of the following satisfies the two blanks given in the questions?
a) 4, Rs.19200
b) 4, Rs.16000
c) 4, Rs.12000
d) 4, Rs.48000
e) All the above
7) Two bikes A and B start from Delhi and Mumbai towards each other, they take 16 hours and ______ hours to reach Mumbai and Delhi respectively after they meet each other. If bike B speed is 60kmph and then the speed of bike A is______kmph.
a) 36, 90
b) 64, 100
c) 25, 80
d) a or c
e) None of these
8) The height of the cylinder is equal to the perimeter of the square whose diagonal is ______m and the radius of the cylinder is equal to the side of the square and the volume of the cylinder is_____.
a) 14√2, 4782
b) 21√2, 7284
c) 7√2, 4312
d) Either b or c
e) None of these
9) A and B started a business with the investment of A is twice of that invested by B. At the end of _____ months from the start of the business, B doubled his investment. If at the end of the year, the total profit earned is Rs.2800 and then the share of B is Rs_____.
a) 6, 1200
b) 4, 1800
c) 8, 1100
d) Either a or c
e) None of these
10) A school bus covers a distance from a city to school at the speed of _____kmph and reaches the school 4 minutes late. The next day the bus covers the same distance at the speed of 10kmph and reaches the school 6 minutes early. The distance between the City and School is _____km.
a) 6, 2.5
b) 4, 1.5
c) 5, 2
d) 3, 1.4
e) None of these
Answers :
Direction (1-5) :
1) Answer: b)
Speed of still water=18 kmph
Upstream speed=18-10 = 8 kmph
Downstream speed=18+10=28 kmph
Total distance=2d
d/28 + d/8 = 9
36d / (28*8) = 9
d=56 km
Total distance =2*56= 112 km
2) Answer: a)
Speed of downstream boat S=50+10=60kmph
Speed of upstream boat T=30-10=20kmph
1st meeting time=600/80=7.5 hours
1st 7.5 hours boat S reached=60*7.5=450km
Boat T (upstream) = 7.5*20 = 150km
After 10 hours boat T reached=20*10=200 km
After 10 hours boat S reached Delhi=10*60=600km
Then boat S started upstream speed=50-10=40kmph
After more 10 hours boat S reached=40*10=400km
Boat T reached=20*20=400km
So, both the boats meets 2nd time, after 20 hours and 400km apart from Delhi.
3) Answer: a)
Speed of river B in day6= 8kmph
Speed of boat in still water = (175/100)*16=28kmph
Downstream speed=8+28=36kmph
Upstream speed=28-8=20kmph
d/36+(d/2)/20=76
d/36 + d/40=76
10d+9d=27360
d=1440km
4) Answer: a)
River A speed on day 7=S
Speed of boat S on day7=15kmph
Upstream speed on day7=15-S
Downstream speed on day7=15+S
Time is inversely proportional to speed,
15+S / 15-S = 5/1
75-5S = 15+S
6S = 60
S =10kmph
10=8 + river A in day8 / 2
20=8 + river A in day8
Speed of river A in day8=12 kmph
Downstream speed on boat T in day 8=18 + 12=30kmph
Required Time =90/30=3 hours
5) Answer: e)
Speed of still water in boat S=x
Downstream speed of boat S in river B in day3=x+12
Upstream speed of boat S in river B in day3=x-12
135/(x+12) + 135/(x-12) = 6
2x/ (x2-144) = 1/22.5
x2-144=45x
x2-45x-144=0
(x + 3)(x – 48)=0
X=-3, 48(negative value will be neglected)
x=48 kmph
Speed of boat T in still water = (3/4)*48=36 kmph
Required time=135/(36+12) + 135/(36-12)
= > 135/48+135/24
= > 405/48 = 8(21/48) = 8(7/16) hours
6) Answer: e)
LCM of 12, 24 and 16 = 48
Total work = 48 units
Ram per day work = 48/12 = 4 units
Geeta per day work = 48/24 = 2 units
Meena per day work = 48/16 = 3 units
Ram, Geeta and Meena one day work = 4 + 2 + 3 = 9 units
Option (a):
(Ram, Geeta and Meena)’s 4 days work = 9 * 4 = 36 units
Remaining = 48 – 36 = 12 units
Remaining completed by Ram and Geeta = 12/(4+2) = 12/6 = 2 days
Share ratio = (4*6): (2*6): (3*4)
= 24: 12: 12
= 2: 1: 1
Geeta’s share = 19200 * (1/4) = Rs.4800
Meena’s share = 19200 * (1/4) = Rs.4800
This satisfies the given condition.
Option (b):
(Ram, Geeta and Meena)’s 4 days work = 9 * 4 = 36 units
Remaining = 48 – 36 = 12 units
Remaining completed by Ram and Geeta = 12/(4+2) = 12/6 = 2 days
Share ratio = (4*6): (2*6): (3*4)
= 24: 12: 12
= 2: 1: 1
Geeta’s share = 16000 * (1/4) = Rs.4000
Meena’s share = 16000 * (1/4) = Rs.4000
This satisfies the given condition.
Option (c):
(Ram, Geeta and Meena)’s 4 days work = 9 * 4 = 36 units
Remaining = 48 – 36 = 12 units
Remaining completed by Ram and Geeta = 12/(4+2) = 12/6 = 2 days
Share ratio = (4*6): (2*6): (3*4)
= 24: 12: 12
= 2: 1: 1
Geeta’s share = 12000 * (1/4) = Rs.3000
Meena’s share = 12000 * (1/4) = Rs.3000
This satisfies the given condition.
Option (d):
(Ram, Geeta and Meena)’s 4 days work = 9 * 4 = 36 units
Remaining = 48 – 36 = 12 units
Remaining completed by Ram and Geeta = 12/(4+2) = 12/6 = 2 days
Share ratio = (4*6): (2*6): (3*4)
= 24: 12: 12
= 2: 1: 1
Geeta’s share = 48000 * (1/4) = Rs.12000
Meena’s share = 48000 * (1/4) = Rs.12000
This satisfies the given condition.
7) Answer: a)
From Option (a)
x/60 =√(36/16)
x/60 =6/4
x=90 kmph
This satisfies the given condition.
From Option (b)
x/60=√(64/16)
x=120 kmph
This not satisfies the given condition.
From Option (c)
x/60=√(25/16)
x/60 = 5/4
x=75 kmph
This not satisfies the given condition.
8) Answer: c)
From Option (a)
Diagonal=a√2
14√2= a√2
a=14
Perimeter of square=4*14=56
Volume of the cylinder= (22/7)*14*14*56=34496
This not satisfies the given condition.
From Option (b)
Diagonal=a√2
21 √2= a√2
a=21
Perimeter of square =4*21=84
Volume of the cylinder= (22/7)*21*21*84=116424
This not satisfies the given condition.
From Option (c)
Diagonal=a√2
a√2=7√2
a=7
Perimeter of the square=4*7=28
Volume of the cylinder= (22/7)*7*7*28=4312
This satisfies the given condition.
9) Answer: a)
From Option (a)
B’s investment=x
A’s investment=2x
Profit share of A and B= (2x*12) : ((x*6)+(2x*6)) =24x : 18x = 4:3
Total profit=2800
B’s share= (3/7)*2800=1200
This satisfies the given condition.
From Option (B)
B’s investment=x
A’s investment=2x
Profit share of A and B = (2x*12) : ((x*4)+(2x*8)) = 24x : 20x = 6:5
B’s share= (5/11)*2800=1272.72
This not satisfies the given condition.
From Option (C)
B’s investment=x
A’s investment=2x
Profit share of A and B = (2x*12) : ((x*8)+(2x*4)) = 24x : 16x = 3:2
B’s share=(2/5)*2800=1120
This not satisfies the given condition.
10) Answer: a)
From Option (a)
x/6 – x/10 = 10/60
4x/60 = 1/6
x=2.5km
This satisfies the given condition.
From Option (b)
x/4 – x/10 = 10/60
6x/40=1/6
X=40/36=10/9
This not satisfies the given condition.
From Option (c)
x/5 – x/10 = 10/60
5x/50=1/6
X=10/6
X=5/3
This not satisfies the given condition.
From Option (d)
x/3 – x/10=10/60
7x/30=1/6
X=5/7
This not satisfies the given condition.
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