AMCAT Quantitative Ability Previous Papers-6

Ques. A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?
16 km
14 km
12 km
10 km
Answer And Explanation
Answer: Option A

 

Explanation:

 

Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 – x) hr

 

Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4

 

So distance traveled on foot = 4(4) = 16 km

 

A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. Find the average speed for first 320 km of tour.
70.11 km/hr
71.11 km/hr
72.11 km/hr
73.11 km/hr
Answer And Explanation
Answer: Option B

 

Explanation:

 

We know Time = Distance/speed

 

So total time taken =
(160/64+160/80)=92hours
Time taken for 320 Km
= 320∗2/9
=71.11km/hr
Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
9 km/hour
10 km/hour
11 km/hour
12 km/hour
Answer And Explanation
Answer: Option D

 

Explanation:

 

We need to calculate the distance, then we can calculate the time and finally our answer.
Lets solve this,

 

Let the distance travelled by x km
Time = Distance/Speed

 

x/10−x/15=2
[because, 2 pm – 12 noon = 2 hours]

 

3x−2x=60
x=60
Time=Distance/Speed
Time@10km/hr=60/10=6hours
So 2 P.M. – 6 = 8 A.M
Robert starts at 8 A.M.

 

He have to reach at 1 P.M. i.e, in 5 hours

 

So, Speed = 60/5 = 12 km/hr

 

A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ?
44 km/hour
46 km/hour
48 km/hour
50 km/hour
Answer And Explanation
Answer: Option C

 

Explanation:

 

Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr

 

Average speed =
2xy/(x+y)
=2∗40∗60/(40+60)
=4800/100
=48Km/hr