Practice Model questions set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A man can row three-quarters of a kilometer against the water stream in $11{1/4}$ minutes and along the stream in $7{1/2}$ minutes respectively. The speed in (km/hr) of the man in still water is
(a)
(b)
(c)
(d)
Q-2) The speed of a boat in still water is 8 km/hr. It can travel 20 km downstream at the same time as it can travel 12 km upstream, the rate of stream (in kmph) is
(a)
(b)
(c)
(d)
Let rate of stream = x kmph
∴ $20/{8+x}=12/{8-x}$
160 – 20x = 96 + 12x
64 = 32x
x = 2
∴ Rate of stream = 2 kmph
Q-3) The speed of a boat in still water is 15 km/h and the rate of stream is 5 km/h. The distance travelled downstream in 24 minutes is
(a)
(b)
(c)
(d)
Downstream speed = 15 + 5 = 20 km/h.
∴ Required distance $20×24/60$=8km
Q-4) A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is
(a)
(b)
(c)
(d)
Let speed in downstream = (x + y)
Speed in upstream = (x – y)
∴ (x + y) = 2 (x – y)
x = 3y
x : y = 3 : 1
Q-5) A man rows a distance downstream in 45 min and the same distance upstream in 75 min. What is the ratio of speed of the stream to the boat in still water ?
(a)
(b)
(c)
(d)
Let speed of boat in still water = x km/hr
Let speed of stream = y km/hr
Let distance covered = d km
∴ $d/{x+y}=45/60=3/4$ …(1)
$d/{x-y}=75/60=5/4$ …(2)
Form (1) & (2),
${x-y}/{x+y}=3/5 ⇒5x-5y=3x+3y$
⇒2x=8y ⇒ $y/x=1/4 $
∴ ratio of speed of the stream to boat in still water = 1 : 4
Q-6) A steamer goes downstream from one part to another in 4 hours. It covers the same distance upstream in 5 hours. If the speed of stream is 2 km/hr, the distance between the two ports is
(a)
(b)
(c)
(d)
Let the distance between the two parts = 'x' km
Let the speed of steamer in still water = 'y' km/hr
∴ $x/{y+2}$=4 ⇒x=4y+8 ….(1)
$x/{y-2}$=5⇒x=5y-10 …..(2)
From (1) and (2)
4y+8=5y-10
⇒ y = 18
∴ From (1)
x = 4 × 18 + 8 = 80 km.
Q-7) A boat takes half the time in moving a certain distance downstream than upstream. The ratio between rate in still water and rate of current is
(a)
(b)
(c)
(d)
Let speed of boat in still water = x km/h
speed of current = y km/h
∴ (x + y) × t = (x – y) × 2t
x = 3y
x : y = 3 : 1
Q-8) A man can row 5 kmph in the still water. If the river is running at 2 kmph, it takes him 5 hours to row up to a place and come down. How far is the place?
(a)
(b)
(c)
(d)
Let the distance = d km
Time taken to row upstream '$t_1$ ' = $d/{5-3}=d/2$ …(1)
Time taken to row downstream $'t_2 '$ = $d/{5+3}=d/8$ …(2)
$t_1 + t_2$ = 5 (Given)
∴ $d/2+ d/8 = 5$
⇒ ${4d+d}/8=5⇒ d=8km $
∴ Distance of the place = 8 km.
Q-9) A boat takes 19 hours for travelling downstream from point A to point B and coming back to point C, mid way between A and B. If the velocity of the stream is 4 km/hr and the speed of the boat in still water is 14 km/hr. then the distance between A & B is
(a)
(b)
(c)
(d)
Downstream speed = 14 + 4 = 18 km/hr
Upstream speed = 14 – 4 = 10 km/hr
Let the distance between A and B = 'x' km
∴ $x/18+{x/2}/10=19$
∴ $x/18+x/20=19$
${10x+9x}/180=19$
${19x}/180=19$ ⇒ x=180km
Q-10) A motor boat whose speed is 15 km/h in still water goes 30 km downstream and comes back in four and a half hours. The speed of the stream is :
(a)
(b)
(c)
(d)
Let the speed of the stream be x km/h.
Then, upstream speed = (15 – x) km/h.
and downstream speed = (15 + x) km/h.
Now, $30/{(15 + x)}+30/{ (15 - x)}$ = 4.5
Checking with options, we find that x = 5 km/h.