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October 17th, 2014, 05:08 PM   #1
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Time & Distance!!

Please help me out in this question!!

QUE-- Starting from my office, I reach the house 20 min late if I walk at 3kmph. Instead, if I walk at 4kmph, I reach the house 15 min early. How far is my house from my office?

A.4 km

B.5 km

C.7 km

D.6 km
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October 17th, 2014, 06:02 PM   #2
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I don't think any of the options are correct.
For if you travel 4 km/h and arrive EARLY, that means that the distance to your house is LESS than 4 km from your office. But none of the options are less than 4 km.
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Last edited by Timios; October 17th, 2014 at 06:05 PM.
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October 17th, 2014, 06:09 PM   #3
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Distance equals speed times time, so

$\displaystyle 3\left(t+\frac13\right)=4\left(t-\frac14\right)\Rightarrow t=2$

$\displaystyle 3\left(2+\frac13\right)=7$ so the distance from your house to your office is 7 km.
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October 17th, 2014, 06:28 PM   #4
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Thanks for ur reply but please explain me step by step formation of the equation.
3(t+1/3) = 4(t-1/4)
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October 17th, 2014, 07:00 PM   #5
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Distance equals speed times time, so we have 3(t + 1/3) for some time t. All this means is
that at 3 km/h, you took t + 20/60 = t + 1/3 hours to cover the distance between your house
and your office. Also, you took t - 1/4 hours at 4 km/h to cover the distance between your
house and your office. This gives us an equation which we can solve for t and then we can
substitute that value into our equation(s) for distance to find the total distance traveled.
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October 17th, 2014, 07:57 PM   #6
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Anutter way:
@4 takes 35 minutes less than @3; so (time @3 = t):
3t = 4(t-35) : t = 140 min = 2 1/3 hr

So distance = 3(2 1/3) = 7 (as per Greg)
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October 18th, 2014, 01:59 PM   #7
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Let the distance between the office and the house be 1km
Time required at 3kmph =$\displaystyle \frac{1}{3}$ hr = 20 min.
Time required at 4kmph =$\displaystyle \frac{1}{4}$ hr = 15 min.
Difference =20−15=5 min.

Actual difference in timings =20+15=35 min
If difference is 5 min, distance is 1 km
⇒ If difference is 35 min, distance is 7 km
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October 18th, 2014, 02:32 PM   #8
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It seems that this is a Multiple Choice question.
One Tip : Sometimes, instead of solving the entire question, one can solve MCQ's faster just by checking the options one by one.
In this question, it must be observed that difference in timings is 35 min $\displaystyle \approx$ 0.583333 hr , or , $\displaystyle 0.58 \bar{3}$ hr.
So, in your mind, just calculate different times for speed = 3 km/h and speed = 4 km/h.
For option A, difference in timing is $\displaystyle 0. \bar{3}$
Only for option C, difference in timing $\displaystyle 0.58 \bar{3}$
So, C is correct.
This way, this MCQ can be solved faster.
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