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(#M40164655) ELITMUS Numerical Ability question Keep an EYE

A, B, C walk 1km in 5 min ,8 min and 10 min respectively .C starts walking from a point at a certain time,B starts from the same point 1 min later and A starts from the same point 2 min later than C .Then A meets B and C at times

Asked In Elitmus Sniper (7 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Siddharth Verma  (7 years ago)

A covers 1km in 5 min
Speed of A is 1km/5min=1000m/300sec=10/3 in m/sec

Similarly
Speed of B is 1km/8min=1000m/480sec=25/12 in m/sec

Speed of C is 1km/10min=1000m/600sec=10/6 in m/sec

As A starts after 1 min of B and 2 min of C:
Distance covered by B in 1 min=25*60/12=125m
Distance Covered by C in 2 min=10*120/6=200m

Time taken by A to meet B assuming to be x can be calculated by the below mentioned equation:
x*10/3=125+(x*25/12)
solving for x we get [x=100sec]Ans

Time taken by A to meet C assuming to be x can be calculated by the below mentioned equation:
x*10/3=200+(x*10/6)
solving for x we get [x=120sec]Ans

Is this solution Helpfull? Yes (12) | No (2)  

  (7 years ago)

Speed of A is 1km/5min
Speed of B is 1km/8min
Speed of C is 1km/10min

A nd C cover same distance when they meet, but A takes 2 min less => t/10 = (t - 2)/5
A meets C at time => t = 4 min

A nd B cover same distance when they meet, but A takes 1 min less => (t - 1)/8 = (t - 2)/5
A meets B at time => t = 3.66 min

Is this solution Helpfull? Yes (3) | No (1)  

Narayan yadav Bhosale  (6 years ago)

1km=1000m
s=d/t
A:1000/5=200m/min
B:1000/8=125m/min
c:1000/10=100m/min

When A started B cover 1min*125s=125m And C cover 2min*100s=200m
Difference between A and C speed:
200-125=75
A have to cover 125m
In 125m A covers 75 m in 1min
remaining 50 m covers in 40 sec=(50/75*60)convert min to sec
=100sec

Difference between A and C speed:
200-10=100
A have to cover 200m
In 1min A cover 100m
Means 200 m covers in 2mi
=120 sec

Is this solution Helpfull? Yes (2) | No   

Devendra Marghade  (7 years ago)

A meets B at 2 min 40 sec and C at 4 min.

Speed of A=1/5 , B=1/8 , C=1/10 km/min.
Here, A starts 1 min later than B and 2 min later than C.
Let A meets B after t1 min, then Bt1 = A(t1 - 1) => (1/8)t1 = (1/5)(t1 -1)
=> t1 = 8/3 min= 2 min 40 sec.

Similarly let A meets C after t2 min, then Ct2 = A(t2 - 2) => (1/10)t2 = (1/5)(t2 - 2)
=> t2=4

Is this solution Helpfull? Yes (1) | No (2)  

neha kumari  (7 years ago)

A walks 1 km in 5 min.
Then, distance covered by A in 1 min = 1000/5 = 200 m

B walks 1 km in 8 min.
Then, distance covered by B in 1 min = 1000/8 = 125 m

C walks 1 km in 10 min.
Then, distance covered by C in 1 min = 1000/101 = 100 m

Let A meets B and C in x and y min, respectively.

Then, according to the question,
Distance covered by C in(x + 2) min = Distance covered by A in x min
⇒ 100(x + 2) = 200 x
⇒100 x + 200 = 200 x
⇒ 200 = 100 x
∴ x = 200/100 = 2 min

Now, for A and B
Distance covered by B in (y + 1) min = Distance covered by A in y min
⇒ 125(y + 1) = 200y
⇒ 125y + 125 = 200y
⇒ 125 = 200y - 125y
⇒ 125 = 75y
∴ y = 125/75 = 5/3 min

Is this solution Helpfull? Yes (1) | No (1)  

shekhar verma  (7 years ago)

A meet B after 112.5 sec and C after 2 min.

As given in problem A=1km/5min
that is speed of A=200m/min

that of B=1km/8min
speed of B=125m/min
speed of C=1km/10min=100m/min
C B A
at 2 min 100*2=200m 125*1=125m not started yet (0 m)
distance cover after 1 min B
by each of them started

after 3 min 300m 250m 200m
after 4 min 400m 375m 400m


so clearly it saw that A meet C at 2 min after when A get started.
and A meet B at before 2 min.....
so ....A has a speed of 200m/120sec
so time need to cover 375m is
(120*375)/400==112.5sec.

Is this solution Helpfull? Yes | No (1)  

mahesh  (7 years ago)

speed of a,b,c respectively 1/5,1/8 and 1/10 in km/min.
a is late by 2 min tan c-->so,(1/5)(t-2)=(1/10)*t-->t=4min--> in 4 min c travels 400m.
similarly for b

Is this solution Helpfull? Yes | No (4)  

Matches  (3 years ago)

no way 1*8

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