Latest CTS Aptitude Question SOLUTION: A person starts from A to B. After 1hr his tyre gets punctured and he repairs it for 10 mins and travels the rest of the distant with 30km/hr and arrives 30 min late. had the punct

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(#M40007426) CTS Numerical Ability question Analytical problem Keep an EYE

A person starts from A to B. After 1hr his tyre gets punctured and
he repairs it for 10 mins and travels the rest of the distant with
30km/hr and arrives 30 min late. had the puncture occured 30 Kms earlier he would been only 15 mins late. Find the ditance betn. A & B and his initial speed.

Asked In CTS (12 years ago)

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Read Solution (1) :

Vignesh  (12 years ago)

Distance = 80kms.
let the velocity be 'V' kmph, time be 't' hrs and distance between A & B is 'd' km.
now t = (d/V)
in first case,
t + (1/3) = (d1/v) + {(d-d1)/30} assuming after travelling 'd1' tyre got punctured
second case,
t+ (1/12) = {(d1+30)/v} + {(d-d1-30)/30} asuming after travelling 'd1+30' tyre got punctured
now replacing t with (d/v),
firstcase, (d-d1)/v +(1/3) = (d-d1)/30
using this in second case we get V = 40 kmph
now, d1 = v * 1hr = 40km
using first case,
(d/40) + (1/3) = 1 + {(d-40)/30)}
==> d = 80kms

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