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 October 17th, 2014, 04:08 PM #1 Newbie   Joined: Oct 2014 From: usa Posts: 10 Thanks: 0 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
 October 17th, 2014, 05:02 PM #2 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 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. Thanks from saha Last edited by Timios; October 17th, 2014 at 05:05 PM.
 October 17th, 2014, 05:09 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 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. Thanks from topsquark and saha
 October 17th, 2014, 05:28 PM #4 Newbie   Joined: Oct 2014 From: usa Posts: 10 Thanks: 0 Thanks for ur reply but please explain me step by step formation of the equation. 3(t+1/3) = 4(t-1/4)
 October 17th, 2014, 06:00 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 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. Thanks from topsquark and saha
 October 17th, 2014, 06:57 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 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) Thanks from topsquark and saha
 October 18th, 2014, 12:59 PM #7 Senior Member   Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 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 Thanks from saha
 October 18th, 2014, 01:32 PM #8 Senior Member   Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 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. Thanks from topsquark and saha

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# starting from my office i reach the house 20 minutes late if i walk at 3 kmph.instead if i walk at 4 kmph i reach 15 minutes early.how far is my house from my office

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