M4maths Previous Puzzles

30 Apr 2014 Time, Distance and Speed

Two train, Each 80 m long, passes each other on parallel lines. If they are going in same direction, the faster train takes one minute to pass the other completely. If they are going in the opposite directions, they over take each other in three second. Find the rate of each train in meter per second.

OPtion

1) 28 m/sec, 76/3 m/sec
2) 25 m/sec, 76/3 m/sec
3) 27 m/sec, 25 m/sec
4) 26 m/sec, 77/3 m/sec
5) 29 m/sec, 28 m/sec
6) 28 m/sec, 25 m/sec
7) 28 m/sec, 101/4 m/sec
8) 35 m/sec, 30 m/sec
9) 38 m/sec, 28 m/sec
10)None of these

Solution

Let the speed of faster train be x m/sec and other train be y m/sec
Since
T = D/S
Then we get
x - y = 8/3 (i)
x + y = 160/3 (ii)
Now solving (i) and (ii)
x=28 m/sec and y=76/3 m/sec

Option 1

Correct Option: 1 Best Solution (15)

Ann Theressa   11 years AGO

Let the speed of trains be x and y
distance d = 80+80 = 160

speed = distance/time
x-y = 160/60 ---(1)
x+y = 160/3 ---(2)

(1) + (2)
2x = 56
x = 28m/sec

y = 76/3 m/sec

Like? Yes (29) | No (10)  

nirmal kumar saini   11 years AGO

let first train speed=v1;
and second train speed=v2;
so when these are going in one direction the resultant speed is=v1+v2;
and when going in opposite direction the resultant speed=v1-v2;
so in first condition equation is:-
d/v1+v2=60; .............(1)
and in second condition equation is:-
d/v1-v2=3; .............(2)
solving both of equation we get
v1=28 m/sec; and
v2=76/3 m/sec;ans

Like? Yes (16) | No (8)  

Purosotam Gupta   11 years AGO

let
the speed of one train=x
speed of second train=y
then
the question
x-y=160/60........equ 1
x+y=160/3.......equ 2
then from 1 and 2
x=28 m/s
y=76/3 m/s

Like? Yes (6) | No (4)  

indrajeet yadav   11 years AGO

let both trains speed=u and v m/sec
according to question
160/u-v=60........(1)
160/u+v=3.........(2)
solve both equation
u=28 m/sec
v=76/3 m/sec

Like? Yes (5) | No (3)  

PAWAN KUMAR   11 years AGO

Let speed of trains be x,y m/sec
now, (x-y)=160/60=8/3 m/sec
and (x+y)=160/3 m/sec
hence, x=28 & y=76/3 m/sec.

Like? Yes (4) | No (3)  

MD SOHAIL AKHTER   11 years AGO

let the speed of two trains be x and y (m/sec)

A/Q
160/(y-x)=60........... (1)
and
160/(y+x)=3...............(2)

on solving (1) & (2), we get
x=76/3m/sec
y=28 m/sec
ans- option 1

Like? Yes (4) | No (5)  

AJAY KUMAR   11 years AGO

3x-3y=8
3x+3y=160
x=28m/s
y=(76/3)m/s

Like? Yes (2) | No (3)  

Indira   11 years AGO

say s1 n s2 r speeds of two trains..
s1-s2 = 80+80/60 ..
s1+s2 = 80+80/3 ..
solve and
s1= 28m/sec and s2= 76/3 m/sec

Like? Yes (2) | No (3)  

gaurav arora   11 years AGO

ans is 1st option

Like? Yes (2) | No (3)  

Udhaya kumar   11 years AGO

160/(x-y)=60
160/(x+y)=3
by solving this we get x=28 , y=76/3.....

Like? Yes (4) | No (5)  

Rashmi bharti   11 years AGO

28 m/sec,76/3 m/sec

Like? Yes (3) | No (5)  

ASHISH MITTAL   11 years AGO

60x-60y=160
x+y=160
after solving we get
x=28
and y=76/3
so option (1)

Like? Yes (3) | No (5)  

KALIA MUDULI   11 years AGO

Let speed of first and second train are S1 and S2.
Given that:lenght of train are 80 m each
1 min=60 sec

according to the first condition:
60=(80+80)/(S1-S2)
S1-S2=8/3 ...........(1)

and according to second condtion:
3=(80+80)/(S1+S2)
S1+S2=160/3 ............(2)

by solving both eque (1)&(2)
we get....

S1=28 m/sec, S2=76/3 m/sec.

Like? Yes (4) | No (6)  

Aakash Jain   11 years AGO

after solving partially these 2 equations are formed,
r1+r2=160/3 and r1-r2=8/3
after solving these two equation, ans is 28m/s and 76/3 m/s

Like? Yes (2) | No (5)  

harshal agrawal   11 years AGO

160/(v-u)=60
160/(v+u)=3
v=28 u=76/3

Like? Yes (2) | No (5)