Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 31)
31.
Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:
Answer: Option
Explanation:
Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.
Discussion:
108 comments Page 5 of 11.
Sundar said:
1 decade ago
@Pawan :
Detailed explanation:
---------------------
Formula: Distance = (Speed x Time).
Speed = 90 km/hr
Convert km/hr into m/sec.
Therefore, Speed = (90 x 1000/3600) = 90000/3600 = 25 m/s
Now, Time = 4 hr 40 min
Convert it into secs.
Therefore = (4 x 3600) + (40 x 60 )= 14400+2400 = 16800 secs.
Now, Distance = 25 m/sec x 16800 sec = 25 x 16800 m
= 4,20,000 m [or]
= 420 km. [1000 m = 1 km]
Some short-cut tricks :
-----------------------
Distance = (Speed x Time)
We can write 4 hr 40 mins = (4 + 40/60) hrs.
So, Distance = 90 x (4 + 2/3)
= (90 x 4) + (90 x 2/3)
= 360 + 60
= 420 km. [or] 420000 m.
Detailed explanation:
---------------------
Formula: Distance = (Speed x Time).
Speed = 90 km/hr
Convert km/hr into m/sec.
Therefore, Speed = (90 x 1000/3600) = 90000/3600 = 25 m/s
Now, Time = 4 hr 40 min
Convert it into secs.
Therefore = (4 x 3600) + (40 x 60 )= 14400+2400 = 16800 secs.
Now, Distance = 25 m/sec x 16800 sec = 25 x 16800 m
= 4,20,000 m [or]
= 420 km. [1000 m = 1 km]
Some short-cut tricks :
-----------------------
Distance = (Speed x Time)
We can write 4 hr 40 mins = (4 + 40/60) hrs.
So, Distance = 90 x (4 + 2/3)
= (90 x 4) + (90 x 2/3)
= 360 + 60
= 420 km. [or] 420000 m.
Dheeraj said:
1 decade ago
(A's Speed):(B's Speed)= sq.under root a: sq.under root b
Sudipto Maity said:
1 decade ago
Good Problem !!!Just memorize the FORMULA .
Ishaq Ahamed said:
1 decade ago
d1+d2= D;(total distance)
v1=d1/t1; =>v1=(D-d2)/t1----> EQ1
v2=d2/t2; => v2=(D-d1)/t2---->EQ2
d1 is distance traveled by first train to meet the 2nd, similarly d2 is distance traveled by second train to meet first train;
t1 is time taken by train 1 to meet the other
t2 is time taken by train 2 to meet first train
now d1+d2= D;(total distance)
according to given data
v1=(D-d1)/9;
v2=(D-d2)/16;
v1/v2=(16/9) *((D-d1)/(D-d2));
=(16/9) *((v2/t2)/(v1/t1))----->from EQ1 and EQ2
(v1/v2)^2 =(16/9)(t1/t2)
t1 and t2 are equal as trains start at same time and both meet after equal amounts of time so, memorize the above formula
v1=d1/t1; =>v1=(D-d2)/t1----> EQ1
v2=d2/t2; => v2=(D-d1)/t2---->EQ2
d1 is distance traveled by first train to meet the 2nd, similarly d2 is distance traveled by second train to meet first train;
t1 is time taken by train 1 to meet the other
t2 is time taken by train 2 to meet first train
now d1+d2= D;(total distance)
according to given data
v1=(D-d1)/9;
v2=(D-d2)/16;
v1/v2=(16/9) *((D-d1)/(D-d2));
=(16/9) *((v2/t2)/(v1/t1))----->from EQ1 and EQ2
(v1/v2)^2 =(16/9)(t1/t2)
t1 and t2 are equal as trains start at same time and both meet after equal amounts of time so, memorize the above formula
Sanchita said:
1 decade ago
Its an simple formula nd you all are making it so much complicated.
Srihari said:
1 decade ago
For this question the direct formulas are:
(A's speed) : (B's speed) = root(b) : root(a)
Time at which they meet= root(a*b)
Derivation to the formulas:
Let X be the total distance, t be the time at which they meet
a,b are times at which they reach the destination after they meet.
s1, s2 are speeds of train1, train2 respectively.
then X = t(s1+s2) ...(1)
X = (t+a)s1 ...(2)
X = (t+b)s2 ...(3)
after equating (1) and (2)
t.s2 = a.s1 ...(4)
after equating (1) and (3)
t.s1 = b.s2 ...(5)
multiply (4) and (5)
t.t.a.b = s1.s2.a.b
" t = root(a*b) "
substitute t value in either (4) or (5)
then " s1/s2 = root(b)/root(a) "
(A's speed) : (B's speed) = root(b) : root(a)
Time at which they meet= root(a*b)
Derivation to the formulas:
Let X be the total distance, t be the time at which they meet
a,b are times at which they reach the destination after they meet.
s1, s2 are speeds of train1, train2 respectively.
then X = t(s1+s2) ...(1)
X = (t+a)s1 ...(2)
X = (t+b)s2 ...(3)
after equating (1) and (2)
t.s2 = a.s1 ...(4)
after equating (1) and (3)
t.s1 = b.s2 ...(5)
multiply (4) and (5)
t.t.a.b = s1.s2.a.b
" t = root(a*b) "
substitute t value in either (4) or (5)
then " s1/s2 = root(b)/root(a) "
Sandeep said:
1 decade ago
Its ok the formula is working but can anyone proove that formula!
Praveen said:
1 decade ago
A train travelled from Delhi to Patna and back in a certain time at the rate of 60kmph. But if the train had travelled from Delhi to Patna at rate of rate 80Kmph. And back from Patna to Delhi at the rate of 40Kmph. It would take two hours Longer. Find the distance between Delhi and Patna?
Detailed solution please.
Detailed solution please.
Parvatraj said:
1 decade ago
L.C.M of 60,80 and 40 is 240..
so let be 240 is the distace .
then train with 60kmh will take 8 hrs to complete.
and trains with 80 kmh and 40 kmh will take 9 hrs to complete.
so the time difference is (9-8) =1.. so what is required is 4 hrs lap.
so multiply with 4 we will get 240*4 = 960 km ..that is the answer.
so let be 240 is the distace .
then train with 60kmh will take 8 hrs to complete.
and trains with 80 kmh and 40 kmh will take 9 hrs to complete.
so the time difference is (9-8) =1.. so what is required is 4 hrs lap.
so multiply with 4 we will get 240*4 = 960 km ..that is the answer.
Zubi said:
1 decade ago
If distance between delhi and patna is x then, time taken in two conditions are :
1. 2x/60 and x/80+x/40
as eq. x/80+x/40= 2x/60+2
Simplified x=480.
1. 2x/60 and x/80+x/40
as eq. x/80+x/40= 2x/60+2
Simplified x=480.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers