Latest TCS Aptitude Question SOLUTION: 1.A passenger train takes 5hrs less for a journey of 396 kmif its speed is increased by 55kmph from its normal speed.The normal speed (to the nearest kmph) is a)None b)44 c)55 d)66

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(#M40154829) TCS Numerical Ability question Apti Keep an EYE

1.A passenger train takes 5hrs less for a journey of 396 kmif its speed is increased by 55kmph from its normal speed.The normal speed (to the nearest kmph) is
a)None b)44 c)55 d)66

Asked In TCS dhritiman (8 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

dee hack kr  (8 years ago)

let speed=x
396/x=t ===>396=xt.....1
396/x+55=t-5===>396=(x+55)(t-5)...2
equating eq1 and eq2
we get
11t-55=x
heat and trail method (xt=396)
x=44 then t=9
xt=396 satisfied
so
ANS=: 44

Is this solution Helpfull? Yes (7) | No   

ayesha chand  (8 years ago)

396/x- 396/x+5 = 5
since time differecnce is 5 hour
check by option

Is this solution Helpfull? Yes (3) | No (2)  

sindan   (8 years ago)

distance=(product of speed/diff of speed )* time difference
so, 396=[x(let)*(x+55)/55]*5
396*11=x*(x+55)
now either you will solve it in a proper way or just put the option int it.
suppose i have chosen option 44 as answer because 44*(44+55) unit digit is 6 so choose option 44 first if you multiply it you will get L.H.S=R.H.S
so correct answer is : 44

Is this solution Helpfull? Yes (2) | No   

komal kumari  (8 years ago)

let normal speed of train is = xkm/hr
so 396/x=T
and 396/(x+55)=T-5
=>396/x-396/(x+55)=5
=>(396x+396*55-396x)/x(x+55)=5
=>x^2+55x-4356=0
=>(x+99)(x-44)=0
so x=44km/hr

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

Anika  (8 years ago)

pls anyone provide the clear solution to this problem asap.

Is this solution Helpfull? Yes | No (2)  

priyanka rajas  (8 years ago)

you can go through option u will get the ans

Is this solution Helpfull? Yes | No (1)  

Prashanti  (8 years ago)

Ans is 55...

Is this solution Helpfull? Yes | No (5)  

Raghvendra Dubey  (3 years ago)

ans is 36
None

Is this solution Helpfull? Yes | No   

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