Quantitative Aptitude Questions (Boats and Streams) for SBI Clerk 2018 Day-23:
Dear Readers, SBI is conducting Online preliminary Examination for the recruitment of Clerical Cadre. preliminary Examination of SBI Clerk was scheduled from March 2018. To enrich your preparation here we have providing new series of Boats and Streams – Quantitative Aptitude Questions. Candidates those who are appearing in SBI Clerk Prelims Exam can practice these Quantitative Aptitude average questions daily and make your preparation effective.
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- A man can row 40 km along the stream in 2.5 hours and against the stream in two times of the number of hours taken to along the stream. Find the speed of boat and speed of current
- 12 km/hr, 6 km/hr
- 12 km/hr, 4 km/hr
- 8 km/hr, 4 km/hr
- 9 km/hr, 15 km/hr
- 12 km/hr, 8 km/hr
- A man takes to row 120 km along the stream is one-third of the time taken to row against the stream. Find the speed of boat in still water if the speed of stream is 10 km/hr
- 12 km/hr
- 20 km/hr
- 24 km/hr
- 30 km/hr
- 25 km/hr
- A man can row 160 km downstream and 200 km upstream in 14 hours. If the speed ratio of downstream to upstream is 2:1, then how many hours takes to row 240 km along the stream?
- 12 hrs
- 8 hrs
- 4 hrs
- 6 hrs
- None of these
- On Tuesday, a man takes to row 80 km along the stream is half of the time taken to row 80 km against the stream and speed of the boat in still water is 15 km/hr. On Wednesday, he rows 120 km and return back in how many hours?
- 18 hrs
- 12 hrs
- 16 hrs
- 14 hrs
- 10 hrs
- A man can row 80 km along the stream and 48 km against the stream in 11 hours. What is the ratio of the speed of boat to speed of current, if the speed of downstream is 16 km/hr?
- 1:3
- 2:3
- 3:2
- 2:5
- None of these
- Speed of boat in still water is 12 km/hr and speed of current is 8 km/hr. Amar travels x km along the stream and (x+60) km against the stream in 21 hrs. How many hours required traveling 4x km along the stream?
- 2 hrs
- 3 hrs
- 6 hrs
- 4 hrs
- 5 hrs
- A boat covers certain distance along the stream in 8 sec and returns back 9/16 of the downstream distance against the current in 18 secs. What is the total distance if the speed of boat in still water is 50 m/s?
- 1000 m
- 10 km
- 100 m
- 1500 m
- 1200 m
- Boat A travels 120 km from P to Q downstream and boat B travels from Q to P upstream. If both the boats starts at 1.00 pm and meets at 2.30 pm and the speed of current is 10 km/hr, then find the sum of the speed of boat A and B
- 50 km/hr
- 60 km/hr
- 80 km/hr
- 40 km/hr
- Cannot be determined
- A swimmer swims certain distance downstream is 0.6 times of the same distance in upstream. Find the value of ‘x’ if the speed of boat in still water and speed of stream is x km/hr and 10 km/hr respectively.
- 20 km/hr
- 30 km/hr
- 50 km/hr
- 40 km/hr
- None of these
- A man travels certain distance (x km) along the stream in 3 hrs and against the stream in 5 hrs. Then find the total time taken to travels x+60 km downstream and upstream if the speed of stream is 10 km/hr
- 11 hrs 12 min
- 10 hrs 12 min
- 12 hrs 20 min
- 12 hrs 20 min
- 10 hrs 20 min
Answers:
- Answer b
Let us take speed of current and speed of boat be x and y
Given,
Upstream speed = 40/5=8 km/hr
Downstream speed = 40/2.5=16 km/hr
Speed of boat = (16+8)/2=12 km/hr
Speed of current = (16-8)/2=4 km/hr
- Answer b
Let us take speed of boat in still water be x km/hr
Given,
120/(x+10)=1/3*(120/(x-10))
3x-30=x+10
2x=40=>x=20 km/hr
- Answer d
Given,
160/2x+200/x=14
560/2x=14
X=20 km/hr
Required number of hours = 240/40=6 hrs
- Answer a
On Tuesday:
80/ (15+x) = ½*80/ (15-x)
30-15=3x=>15=3x
=>x=15/3 =5 km/hr
Required time = 120/20+120/10=18 hrs
- Answer e
Given,
Time taken to travel downstream
80/16=5 hrs
48/ (11-5) =48/6=8 km/hr
Speed of boat = (16+8)/2=24/2=12 km/hr
Speed of stream = (16-8)/2=8/2=4 km/hr
Required ratio = 12:4=3:1
- Answer d
Given,
x/20 + (x+60)/4=21
x+5x+300=420
6x=120=>x=20 km
Required time = (4*20)/20=4 hrs
- Answer a
Let us take speed of stream be x km
Given,
8 (50+x)*9/16=18*(50-x)
400+8x=32*50-32x
=>40x=120=>x=30 m/s
Required distance = 8 (50+30) +18*(50-30)
=>640+360=1000 m
- Answer c
Let us take speed of boat A as x and speed of boat B as y
120/ ((x+10) +(y-10)) =1 ½
120/(x+y) =3/2
=>x+y=80
- Answer d
Let us take distance as y
Given,
Y/(x+10)=0.6*(y/(x-10))
1/(x+10) =0.6*(1/(x-10))
X-10=0.6x+6
0.4x=16=>x=40 km/hr
- Answer a
Let Speed of boat in still water be ‘y’
3*(y+10) =5*(y-10)
3y+30=5y-50
=>2y=80=>y=40
Required distance = 3*50=150 km
Required time = 210/50+210/30=4 hr 12 min+7hr=11 hrs 12 min
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