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A man covered a certain distance at some spe

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Manager
Joined: 03 Nov 2018
Posts: 62
Location: India
Schools: LBS '21
GMAT 1: 580 Q44 V28
GMAT 2: 580 Q44 V28
GPA: 3.44
A man covered a certain distance at some spe  [#permalink]

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03 Nov 2018, 23:14
1
4
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Difficulty:

65% (hard)

Question Stats:

55% (03:13) correct 45% (02:19) wrong based on 56 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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A man covered a certain distance at some spe  [#permalink]

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04 Nov 2018, 00:35
3
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

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Joined: 14 Jun 2018
Posts: 212
A man covered a certain distance at some spe  [#permalink]

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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

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

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

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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Joined: 18 Jun 2018
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A man covered a certain distance at some spe  [#permalink]

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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
A man covered a certain distance at some spe   [#permalink] 16 Nov 2018, 06:25
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