Latest TCS Aptitude Question SOLUTION: A and B run 1 k.m race.if A gives B a start of 50meter,A wins by 14 sec and if A gives B a start of 22 sec, B wins by 20 meter.find the time taken by A to run 1k.m...?

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A and B run 1 k.m race.if A gives B a start of 50meter,A wins by 14 sec and if A gives B a start of 22 sec, B wins by 20 meter.find the time taken by A to run 1k.m...?

Asked In TCS Akhilesh Agrawal (10 years ago)

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Yvaine  (10 years ago)

So here the only given quantity is 14 sec.So express equations in terms of time.
Let a and b be the speed of A and B respectively.
Given,
When A gives B a head start of 50m,
950/b-1000/a=14-----(1)
The second condition:
980/a+22=1000/b----(2)
Let 1/a=x 1/b=y
then,
950y-1000x=14
980x+22=1000y--->1000y-22=980x
solving we get..
a=10m/s
time taken by A to run 1000km=1000/10=100s

Is this solution Helpfull? Yes (40) | No (3)  

aman kesharwani  (10 years ago)

Let times taken by A and B to run a km, be x and y, respectively.
When B gets a start of 50 m, B runs
1000 – 50 = 950 m while A runs 1000 m.
(950/1000)y - x = 14,
i.e. 0.95y - x = 14
and, When B gets a start of 22 sec, A runs (y - 22) sec, while B runs for y sec.
1000-(1000/x)(y-22)=20
50y - 49x = 1100.
3.45y = 414
y = 120 sec.
x = 0.95 x 120 - 14 = 100 sec.
so the answer is 100 sec

Is this solution Helpfull? Yes (8) | No (7)  

priyanka kumari  (10 years ago)

could you plz inform me with the latest TCS 2013-2014 criteria?

Is this solution Helpfull? Yes (5) | No (11)  

BIBHUTI NARAYAN  (10 years ago)

in first case,
suppose time taken by A is 'T'.
After B is '50m' ahead in beginning ,still B takes '14sec' more than 'T' to reach destination.
That means,time taken by B to reach from '50m' to '1000m' is 'T+14'.
That means,time taken by B to travel 950m is 'T+14'.
So ,total time taken by B from '0m to 1000m' = 'T+14+(time taken to travel 50m );
which is ='T+14+{(T+14)*50)/950};
In second case ,
B is '22sec' ahead.
Now A reaches in 'T sec'.
In the same 'T sec' B reaches '20m' ahead.
So the time taken to travel '1000m' by B =[22+T-{time taken by B to travel '20m'

So the time taken to travel '1000m' by B={22+T-(20*(T+14)/950)};

[time taken by B to travel '20m' is achieved from case 1];;;

equating eqn in case 1 and 2 ,we get

T=102.5

Is this solution Helpfull? Yes (5) | No (6)  

rachel mathew  (10 years ago)

the answer is 100 sec
From the first part we calculate the speeds of A and B
if in t sec A covers 1000 m then B covers 950 m in (t+14) sec
thus A(speed)= 1000/t
B(speed)=950/(t+14)
It is further given that A gives B a head start of 22 sec. B covers 950 m in t+14 sec thus in 22 sec it will cover (950*22)/(t+14). To complete the race B has to cover 1000-(the distance covered in 22 sec as calculated earlier). and this time will be equal to the time taken by A to cover 980 m.

Is this solution Helpfull? Yes (3) | No (6)  

Akhilesh Agrawal  (10 years ago)

Plz explain me properly.bcoz here first is camparision between distance n time then in time and distance..i m totally blank to solve this

Is this solution Helpfull? Yes (2) | No   

raju  (7 years ago)

950B-1000A=14
980A+22=1000B

SOLVE ABOVE EQUATIONS AND YOU GET A=10;
SO A=1000/10= 100M

Is this solution Helpfull? Yes (2) | No   

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