55934.A goods train leaves a station at a certain time and at a fixed speed. After 8 hours, an express train leaves the same station and moves in the same direction at a uniform speed of 120 kmph, this train catches up the goods train in 7 hours. Find the speed of the goods train.
50
48
56
60
Explanation:
Let the speed of the goods train be x kmph.
Distance covered by goods train in 15 hrs = Distance covered by express train in 7 hrs
So, 15x = 7 × 120 or x = 56.
So, Speed of goods train = 56 kmph
Let the speed of the goods train be x kmph.
Distance covered by goods train in 15 hrs = Distance covered by express train in 7 hrs
So, 15x = 7 × 120 or x = 56.
So, Speed of goods train = 56 kmph
55935.A train travelled at a speed of 15 km/hr and reached destination 20 minutes late. Had the speed increased to 20 km/hr, it would reach the destination 20 minutes early. Find the distance travelled by train.
30 km
28 km
40 km
45 km
Explanation:
Let the distance travelled be ‘D’ km.
Time taken to travel D km at 15 kmph = D/15 hours
Time taken travel D km at 20 kmph = D/20 hours
Given the difference between time = 40 minutes =40/60 =2/3.
=>$\dfrac{D}{15}$-$\dfrac{D}{20}$=$\dfrac{2}{3}$.
=>$\dfrac{3D-2D}{60}$=$\dfrac{2}{3}$
=>$\dfrac{D}{60}$=$\dfrac{2}{3}$
=>D=40 km
Let the distance travelled be ‘D’ km.
Time taken to travel D km at 15 kmph = D/15 hours
Time taken travel D km at 20 kmph = D/20 hours
Given the difference between time = 40 minutes =40/60 =2/3.
=>$\dfrac{D}{15}$-$\dfrac{D}{20}$=$\dfrac{2}{3}$.
=>$\dfrac{3D-2D}{60}$=$\dfrac{2}{3}$
=>$\dfrac{D}{60}$=$\dfrac{2}{3}$
=>D=40 km
55936.Two trains is moving in opposite directions at 70 kmph and at 80 kmph. Their lengths are 1.3 km and 0.7 km respectively. The time taken by the slower train to cross the faster train in seconds is:
24 sec
48 sec
76 sec
102 sec
Explanation:
Relative speed = (70 + 80) = 150 km/hr.
=>(150 $\times \dfrac{5}{18})$ m/sec
=>(125/sec)m/sec
Distance covered = (1.30 + 0.7)km = 2 km = 2000 m
Reqd. time = ($\dfrac{3}{125} \times 2000)$ sec
=48 sec
Relative speed = (70 + 80) = 150 km/hr.
=>(150 $\times \dfrac{5}{18})$ m/sec
=>(125/sec)m/sec
Distance covered = (1.30 + 0.7)km = 2 km = 2000 m
Reqd. time = ($\dfrac{3}{125} \times 2000)$ sec
=48 sec
55937.A train started from point A at a speed of 60 km/hr and after 2 hours another train of same length started from A at a speed of 80 km/hr in the same direction as the first one. After how much time the second train will meet the first train?
5 hours
3 hours
6 hours
8 hours
Explanation:
Let after x hours the second train will meet the first train.
Because distance is same,
S1 t1 = S2 t2
60 (x + 2) = 80 × x
60x + 120 = 80x
80x – 60x = 120
20x = 120
x = 6 hours
Let after x hours the second train will meet the first train.
Because distance is same,
S1 t1 = S2 t2
60 (x + 2) = 80 × x
60x + 120 = 80x
80x – 60x = 120
20x = 120
x = 6 hours
55938.Two trains are running on parallel lines in the same direction at speeds of 60 km/h and 35 km/h respectively. The faster train crosses a man in the slower train in 54 second. If the length of the slower train is 4/5th of the faster train, find the length of the slower
300 m
320 m
430 m
340 m
Explanation:
According to the question,
(60–35)×5/18 =D/54
D = 375 m
Length of the faster train = 375m,
Length of the slower train =375×4/5
= 300 m
According to the question,
(60–35)×5/18 =D/54
D = 375 m
Length of the faster train = 375m,
Length of the slower train =375×4/5
= 300 m
55939.Two trains start at the same time, P from A to B and Q from B to A.If they arrive at B and A,respectively, hours and 10 hours after they passed each other, and the speed of P is 90 km/hr, then the speed of Q in km/hr is?
80
75
45
60
Explanation:
Let the distance between A and B be ‘d’
speed of Q be q
‘t’ be the time taken for them to meet
d/(90+q) =t
d/(t+10)=q
d/(t+(5/2))=90
d=90t+225
d=90t+qt
d=qt+10q
qt=225
q=9t
t^2 =25
t=5hours
q=9*5=45 km/hr
Let the distance between A and B be ‘d’
speed of Q be q
‘t’ be the time taken for them to meet
d/(90+q) =t
d/(t+10)=q
d/(t+(5/2))=90
d=90t+225
d=90t+qt
d=qt+10q
qt=225
q=9t
t^2 =25
t=5hours
q=9*5=45 km/hr
55940.A train passes a platform in 40 sec and a woman standing on the platform in 30 sec. If the speed of the train is 108 km/hr, what is the length of the platform ?.
100 m
300 m
900 m
1020 m
Explanation:
Speed of the train =(108 * 5/18)=30 m/sec
Length of the train = Distance travelled to cross the woman x Time taken = (30 × 30) m = 900m.
Let the length of the platform be x meters.
Then, $\dfrac{(x+900)}{40}$=30
=>x+900=1200
=>x=300 m.
Speed of the train =(108 * 5/18)=30 m/sec
Length of the train = Distance travelled to cross the woman x Time taken = (30 × 30) m = 900m.
Let the length of the platform be x meters.
Then, $\dfrac{(x+900)}{40}$=30
=>x+900=1200
=>x=300 m.
55941.Two trains A and B start from Town X and Y going towards Y and X respectively. Towns X and Y are 900kms apart and it takes A 9 hours to travel between X and Y. A starts at 6:00 am while B starts at 8:00 am. The ratio of speeds of A and B is 5:4. The trains cross each other at point P and then stay at Y and X respectively. The next day, B starts from X at 6:00 am while A starts from Y at 8:00 am.The trains cross each other at point Q. Find the distance between points P and Q.
110
125
100
150
Explanation:
Speed of A,a=900/9=100 km/hr
Speed of B,b=4/5 * 100 =80 km/hr
On day 1 ,distance travelled by A before B starts=2 * 100=200 km
When B starts moving,the relative speed between A and B ,r=100 + 80 =180 km/hr
When B starts moving,the distance between A and B =900-200=700 km
Thus ,time taken for the 2 trains to meet =700/180=35/9 hours
Location of point P=35/9 * 80 from Y=2800/9 km from Y.
On day 2,distance travelled by B before A starts =2 * 80 =160 km
When B starts moving ,the relative speed between A and B,r=100+80=180 km/hr
When B starts moving ,the distance between A and B=900-160=740 km.
Thus time taken for the 2 trains to meet =740/180 =37/9 hours.
Location of point Q =37/9 * 100 from Y=3700/9 km from Y
Therefore ,distance between P and Q =3700 /9 -2800/9 =100 km.
Speed of A,a=900/9=100 km/hr
Speed of B,b=4/5 * 100 =80 km/hr
On day 1 ,distance travelled by A before B starts=2 * 100=200 km
When B starts moving,the relative speed between A and B ,r=100 + 80 =180 km/hr
When B starts moving,the distance between A and B =900-200=700 km
Thus ,time taken for the 2 trains to meet =700/180=35/9 hours
Location of point P=35/9 * 80 from Y=2800/9 km from Y.
On day 2,distance travelled by B before A starts =2 * 80 =160 km
When B starts moving ,the relative speed between A and B,r=100+80=180 km/hr
When B starts moving ,the distance between A and B=900-160=740 km.
Thus time taken for the 2 trains to meet =740/180 =37/9 hours.
Location of point Q =37/9 * 100 from Y=3700/9 km from Y
Therefore ,distance between P and Q =3700 /9 -2800/9 =100 km.
55942.The ratio of the length of two trains X and Y is 4 : 7 and the ratio of the time taken by both trains to cross a man standing on a platform is 2 : 3. If the speed of the train X is 36 km/h find the speed of the train Y in m/s.
35 m/s
30 m/s
20 m/s
15 m/s
Explanation:
Let the length of train X = 4a , time = 2b
Speed = 4a/2b
Let the length of train Y = 7a , time = 3b
Speed = 7a/3b
Ratio of the speed =$\dfrac{4a/2b}{7a/3b}$=6:7
Speed of train Y = 36/6 * 7=42 km/h
in m/s =42 * 5/18 =35/3 m/s
Let the length of train X = 4a , time = 2b
Speed = 4a/2b
Let the length of train Y = 7a , time = 3b
Speed = 7a/3b
Ratio of the speed =$\dfrac{4a/2b}{7a/3b}$=6:7
Speed of train Y = 36/6 * 7=42 km/h
in m/s =42 * 5/18 =35/3 m/s
55943.A train travels a distance of 480 km at uniform speed. Due to breakdown, its speed is reduced by 20 km/hr and hence it travels the destination 8 hours late. Find the initial speed of the train.
30 km/hr
60 km/hr
25 km/hr
35 km/hr
Explanation:
Given that the distance=480km
Let the speed of the train= x km/hr
Then, time in which it can travel 480 km = 480/x hr
Now, Speed is reduced by 20 km/hr
Reduced speed =(x-20) km/hr
Time in which it can travel 480 km at reduced speed = $\dfrac{480}{(x-20)}$
Difference in time =4 hours
=> $\dfrac{480}{(x-20)}$-$\dfrac{480}{x}=4$
=>$\dfrac{480(x-x+20)}{X^2 - 20x}$=4
=>9600=$4(x^{2}-20x)$
=>$x^{2}-20x-2400=0$
=>$x^{2}-60x+40x-2400$=0
=>x(x-60)+40(x-60)=0
(x+40)(x-60)=0
x+40=0|x-60=0
x=-40|x=60
Speeds cannot be negative
Hence,initial speed =60km/hr
Given that the distance=480km
Let the speed of the train= x km/hr
Then, time in which it can travel 480 km = 480/x hr
Now, Speed is reduced by 20 km/hr
Reduced speed =(x-20) km/hr
Time in which it can travel 480 km at reduced speed = $\dfrac{480}{(x-20)}$
Difference in time =4 hours
=> $\dfrac{480}{(x-20)}$-$\dfrac{480}{x}=4$
=>$\dfrac{480(x-x+20)}{X^2 - 20x}$=4
=>9600=$4(x^{2}-20x)$
=>$x^{2}-20x-2400=0$
=>$x^{2}-60x+40x-2400$=0
=>x(x-60)+40(x-60)=0
(x+40)(x-60)=0
x+40=0|x-60=0
x=-40|x=60
Speeds cannot be negative
Hence,initial speed =60km/hr