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A man covered a certain distance at some spe

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A man covered a certain distance at some spe  [#permalink]

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New post 03 Nov 2018, 23:14
1
4
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

55% (03:09) correct 45% (02:17) wrong based on 54 sessions

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A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
A.35
B.36
C.37
D.40
E. 25
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New post 04 Nov 2018, 00:35
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dharam44 wrote:
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
A.35
B.36
C.37
D.40
E. 25


Let, Original speed = s
Distance = d
Time = t mins

Here we get three equations as per the given information

d = s*t

d = (s+3)*(t-40)

d = (s-2)*(t+40)


Solving equation 1 and 2

s*t = (s+3)*(t-40)
i.e. s*t = s*t +3t-40s-120

i.e. 3t-40s = 120

Solving equation 1 and 3

s*t = (s-2)*(t+40)
i.e. s*t = s*t -2t+40s-80

i.e. -2t+40s = 80

Solving the two blue and bold equations

t = 200 minutes
and s = 12 m/h

i.e. d = 12*(200/60) = 40 miles

Answer: Option D
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New post 04 Nov 2018, 01:34
\(\frac{d}{s} = t\) ; \(d = s*t\)

Given ,
\(d = (s-2)(t+\frac{2}{3})\) ------------------- 1

and ,

\(d = (s+3)(t-\frac{2}{3})\) -------------------- 2

Adding 1 & 2 ,

\(2d = (s-2)(t+\frac{2}{3})+(s+3)(t-\frac{2}{3})\)

\(2d = st + \frac{2}{3}s - 2t - \frac{4}{3} + st - \frac{2}{3}s +3t - 2\)

t = 3 hr 20 mins or 200 mins or \(\frac{10}{3}\)

If time is increased by 40 mins or \(\frac{40}{200}*100 = 20\)% , speed should be reduced by 16.67% or 1/6

\(\frac{5}{6}s = s - 2\)
s = 12 km/hr

\(d = \frac{10}{3} * 10\) = 40km
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A man covered a certain distance at some spe  [#permalink]

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New post 16 Nov 2018, 06:25
dharam44 wrote:
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
A.35
B.36
C.37
D.40
E. 25


OA:D

Let the initial speed of man be \(x\) Kmph
Let the initial time taken by man be \(t\) hours

\(xt=(x+3)(t-\frac{40}{60})\)

\(-\frac{2}{3}x+3t=2\) .....(1)

\(xt=(x-2)(t+\frac{40}{60})\)

\(\frac{2}{3}x-2t=\frac{4}{3}\) .....(2)

Adding \((1)\) and \((2)\), we get

\(t=2+\frac{4}{3}=\frac{10}{3}\) hours

Putting \(t =\frac{10}{3}\)hours into \((1)\), we get

\(-\frac{2}{3}x+3*\frac{10}{3}=2\)

\(x=\frac{2-3*\frac{10}{3}}{-\frac{2}{3}}=\frac{-8}{-\frac{2}{3}}=12\) Kmph

Distance travelled \(= xt = 12*\frac{10}{3}= 40\) Kms
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A man covered a certain distance at some spe   [#permalink] 16 Nov 2018, 06:25
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