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• $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. A cyclist traveled for two days. On the second day the  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Hide Tags Manager Joined: 14 Mar 2010 Posts: 50 A cyclist traveled for two days. On the second day the [#permalink] Show Tags 13 May 2012, 23:00 9 00:00 Difficulty: 55% (hard) Question Stats: 72% (02:02) correct 28% (02:32) wrong based on 252 sessions HideShow timer Statistics A cyclist traveled for two days. On the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day. If during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling, what was her average speed on the second day? A. 5 mph B. 10 mph C. 20 mph D. 30 mph E. 40 mph _________________ MGMAT CAT MATH http://gmatclub.com/forum/mgmat-cat-math-144609.html MGMAT SC SUMMARY: http://gmatclub.com/forum/mgmat-sc-summary-144610.html Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: A cyclist traveled for two days. On the second day the [#permalink] Show Tags 13 May 2012, 23:34 1 4 A cyclist traveled for two days. On the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day. If during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling, what was her average speed on the second day? A. 5 mph B. 10 mph C. 20 mph D. 30 mph E. 40 mph Approach #1 - Algebra: Since on the second day the cyclist traveled 4 hours longer than she traveled on the first day and spent a total of 12 hours traveling then $$t+(t+4)=12$$ --> $$t=4$$. So, she traveled 4 hours on the first day and 8 hours on the second day; Let the rate on the second day be $$r$$ mile per hour, then: 4(r+10)+8r=280 --> $$r=20$$. Answer: C. Approach #2 - Logic: Average rate of the cyclist is $$\frac{total \ distance}{total \ time}=\frac{280}{12}=23\frac{1}{3}$$. Now, since the weighted average of 2 individual averages ($$r$$ and $$r+10$$), must lie between these individual averages then $$r<23\frac{1}{3}<r+10$$, only answer choice B fits (rate from correct answer choice must be less than $$23\frac{1}{3}$$ and that rate plus 10 must be more than $$23\frac{1}{3}$$). Answer: C. _________________ General Discussion Senior Manager Joined: 23 Oct 2010 Posts: 350 Location: Azerbaijan Concentration: Finance Schools: HEC '15 (A) GMAT 1: 690 Q47 V38 Re: A cyclist traveled for two days. On the second day the [#permalink] Show Tags 14 May 2012, 00:01 2 the first day speed=V time =t the 2nd day speed =V-10 time =t+4 total distance =280 total time=12 t+t+4=12 t=4 8(v-10)+4v=280 v=30 30-10=20 answ is C.20 _________________ Happy are those who dream dreams and are ready to pay the price to make them come true I am still on all gmat forums. msg me if you want to ask me smth Manager Joined: 03 Jan 2015 Posts: 61 Concentration: Strategy, Marketing WE: Research (Pharmaceuticals and Biotech) Re: A cyclist traveled for two days. On the second day the [#permalink] Show Tags 29 Jan 2015, 17:07 1 A cyclist traveled for two days. On the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day. If during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling, what was her average speed on the second day? A. 5 mph B. 10 mph C. 20 mph D. 30 mph E. 40 mph SOLUTION: D = 280 Mi T = 12 hrs Day 1 time = T1 Day 2 time = T2 T2 - T1 = 4 hrs ----- (I) T1 + T2 = 12 hrs ----- (II) Adding I and II, T2 = 8 hrs and T1 = 4 hrs Day 1 Rate = R1 Day 2 Rate = R2 R1 - R2 = 10 mph i.e. R1 = 10 + R2 280 = 8R2 + 4R1 i.e. 280 = 8R2 + 4 (10 + R2) i.e. R2 = 20 mph ANSWER: c EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13095 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A cyclist traveled for two days. On the second day the [#permalink] Show Tags 29 Jan 2015, 18:07 Hi All, These types of Average Speed questions require a certain amount of organization. While you have the freedom to approach this Test in any way you choose, having a "default" way to create tables will help you to set up questions quicker over the course of the entire Exam. Here's one way to organize the work in this question: Attachment: GCDrtBox.png [ 462.05 KiB | Viewed 2447 times ] Note that the question asks for the Average Speed on DAY 2, so we have one more step: R = 30 R - 10 = 20 Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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04 Feb 2015, 00:36
............................. Distance .................... Time ............................ Speed

Day I .................... d ................................. 4 ..................................... $$\frac{d}{4}$$ (Let "d" is the distance covered on Day I)

Day II .................. 280-d ........................... 8 ..................................... $$\frac{280-d}{8}$$

Given that $$\frac{d}{4} - \frac{280-d}{8} = 10$$

d = 120

Speed on day II $$= \frac{280-120}{8} = \frac{160}{8} = 20$$

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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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26 Oct 2015, 19:19
Bunuel, i understood the solution, but please tell me , what is the point i m in missing in my calculation

defiantly she trvelled 4 hour on first day and 8 hours on second day
total of 280 km
d1+ d2= 280

on second day speed is 10km/h is lesser than first and trevlled for 8 hours thus covers 80 km lesser in distance on second day
d1+ d1-80 = 280
d1 = 180
and d2 = 100
so speed is 100/8 = which is greater than 10 but reaching 20
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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09 Jan 2017, 14:07
Drawing a diagram is helpful for this problem!!!

Day 1 --> d = r x 4
Day 2 --> 280-d = (r-10)x(8)

Note: you can figure out that hours allocated to day 1 and day 2 by solving the following formula from the prompt: t+t+4=12 --> t=4 --> all you have to do then is plug the correct values for t into day 1 and day 2 based on the information provided in the prompt

Plug d from Day 1 into Day 2 and solve --> you'll find r = 30 --> making Day 2 r = 20

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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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28 Sep 2018, 00:20
Let

T = t1 + t2
t1 is hours for day 1
t2 is hours for day 2

We know t2 = 4+t1 so this means t2 = 8 and t1 = 4

we are also given that v2 = x - 10
v1 = x

D1 + D2 = Dtotal
4x + 8 (x - 10) = 280
4x + 8x - 80 = 280
12x = 360
x = 30

We want speed for second day so x - 10 = 20 answer choice C

We could also solve this by using the answer choices and backsolving.

If we let the second day avg speed be 30 then the first day speed is 40

D1= 40 * 4 = 160
D2 = 30 * 8 = 240

Adding these two together makes it more than the distance. So we know the speed should be less than the picked.

We try for 20

20 * 8 = 160
30 * 4 = 120

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Re: A cyclist traveled for two days. On the second day the &nbs [#permalink] 28 Sep 2018, 00:20
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