Mathematics Mcqs

A. 4.5 kmph
B. 4.8 kmph
C. 5 kmph
D. 5.1 kmph
Time taken for the forward journey
= 2/6 = (1/3) hrs Time taken for the return journey
= 2/4 = (1/2) hrs Total time = 5/6 hrs Average speed = 4/(5/6) = 24/5 = 4.8kmph

A. 1.8 km/hr
B. 3 km/hr
C. 3.6 km/hr
D. 4 km/hr
Let x be the speed of the river.
Ds = (6 + x) km/hr; Us = (6 – x) km/hr
If t hours is the time to row downstream then 4t hours is the time to row upstream.
(6 + x)*t = (6 – x)*4t
6 + x = 24 – 4x
x = 3.6 km/hr

A. 0.60 minutes
B. 0.36 minutes
C. 0.78 minutes
D. 0.42 minutes
Total distance = addition of length of the two trains = 140 + 120 = 260 metres
As the two trains are travelling in the same direction, their relative speed is:
v = | v1 – v2 | = | 40 – 60 | = 20 km/hr = 20*1000/60 = 1000/3 metres/min
t = 260/ 1000*3
t = 0.78 minutes

A. 2:5
B. 4:3
C. 1:2
D. 1:1
Rayan’s speed = 1000/300 = 3.33 m/s
Hence Ratio  =  1:1

A. 6:8
B. 3:7
C. 5:4
D. 4:3
If the ratio of the speeds of two objects is a:b, then the time taken by them to cover the same distance is
b:a Hence, the answer is 4:3

A. 400
B. 320
C. 240
D. 300
With a full tank, it can travel
360*4/3 = 480 km With 2/3 full tank, it can travel
480*2/3 = 320 km

A. 20 km
B. 40 km
C. 30 km
D. 10 km
x = 7 km/hr ; y = 3 km/hr
Ds = 10 km/hr ; Us = 4 km/hr
Distance (d) is same. Therefore, if time taken for downstream is t hours, the time for upstream is (t + 6) hours.
10*t = 4*(t + 6)
6t = 24 ; t = 4 hours
d = 10*4 = 40 km

A. 36
B. 24
C. 30
D. 18
Let t be his usual time to reach his office and v be his usual speed.
v = d/t ……….(d is the distance Afnan travels while going to his office)
vt = d
At v1 = 4v/5 ; t1 = t + 6
4v/5 = d/(t + 6)
4v/5* (t + 6) = d
4v/5* (t + 6) = vt
On solving we get,
t = 24 minutes

A. 17 hr
B. 14 hr
C. 12 hr
D. 19 hr
Relative speed = 5.5 – 5 = .5 kmph (because they walk in the same direction)
distance = 8.5 km
time = distance/speed=8.5/.5=17 hr

A. 30 kmph
B. 27 kmph
C. 25 kmph
D. 20 kmph
Distance = 24*50/60 = 20 km New Speed = 20/(40/60)) = 30 kmph