# AMCAT Quantitative Ability Previous Papers-4

Ques. In one hour, a boat goes 11km along the stream and 5 km against it. Find the speed of the boat in still water
6
7
8
9

Explanation:

We know we can calculate it by 1/2(a+b)

=> 1/2(11+5) = 1/2(16) = 8 km/hr

Ques. If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the stream is
5 km/hr
4 km/hr
2 km/hr
1 km/hr

Explanation:

Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 – 5)kmph = 1 kmph

Ques. A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water is
4 kmph
5 kmph
6 kmph
7 kmph

Explanation:

Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18)*(18/5) kmph
= 5 kmph

If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 3 kmph, then the speed of the boat in still water is
12 kmph
13 kmph
14 kmph
15 kmph

Explanation:

Rate upstream = (7/42)*60 kmh = 10 kmph.
Speed of stream = 3 kmph.
Let speed in sttil water is x km/hr
Then, speed upstream = (x —3) km/hr.
x-3 = 10 or x = 13 kmph

Ques. 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 stream is
3:1
1:3
2:4
4:2

Explanation:

Let speed downstream = x kmph
Then Speed upstream = 2x kmph

So ratio will be,
(2x+x)/2 : (2x-x)/2
=> 3x/2 : x/2 => 3:1

A man’s speed with the current is 20 kmph and speed of the current is 3 kmph. The Man’s speed against the current will be
11 kmph
12 kmph
14 kmph
17 kmph

Explanation:

If you solved this question yourself, then trust me you have a all very clear with the basics of this chapter.

If not then lets solve this together.
Speed with current is 20,
speed of the man + It is speed of the current
Speed in still water = 20 – 3 = 17

Now speed against the current will be
speed of the man – speed of the current
= 17 – 3 = 14 kmph