A cyclist traveled for two days. On the second day the

Question

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

A. 5 mph
B. 10 mph
C. 20 mph
D. 30 mph
E. 40 mph

Bunuel · Accepted Answer

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

A. 5 mph
B. 10 mph
C. 20 mph
D. 30 mph
E. 40 mph

Approach 1 - Algebra: 
 
Given that on the second day the cyclist traveled for 4 hours more than on the first day and the total travel time over two days was 12 hours, we can set up the equation  t + (t + 4) = 12 . Solving for  t , we get  t = 4 . This means she traveled for 4 hours on the first day and 8 hours on the second day.
 
If we let the  rate on the second day  be  r  miles per hour, then the equation for the total distance covered on both days is:  4(r + 10) + 8r = 280 . Solving for  r , we find  r = 20  miles per hour.
 
 Approach 2 - Logic: 
 
The average rate of the cyclist is  \frac{	ext{total distance}}{	ext{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, we have  r \lt 23\frac{1}{3} \lt r+10 . Only answer choice B fits, as the rate from the 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

LalaB · Answer

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

thorinoakenshield · Answer

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

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

EMPOWERgmatRichC · Answer

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 12244 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:

Show Spoiler

C

GMAT assassins aren't born, they're made,
Rich

PareshGmat · Answer

............................. 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

Answer = C

sudhirmadaan · Answer

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

law258 · Answer

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

C.

Salsanousi · Answer

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

Total 280, answer choice C

Posted from my mobile device

Sa800 · Answer

I would like to see an answer to this....@Bunel....

Did you figure out why this does not work?

Sa800 · Answer

i think i found it

just because rate it slower does not mean that Day 2 had less travelling.....(remember day 2 travel longer than day 1 )...also you accounted for day 2 with 8 hours but did not account for day one with 4 hours...

KarishmaB · Answer

Simply use the weighted average concept. Her average speed over the two days is 280/12 = 70/3 mph She travelled for 4 hrs on Day 1 and 8 hrs on Day 2 (because she travelled 4 hrs longer on Day 2 and total she travelled for 12 hrs). So these will be the weights in our formula. \frac{w1}{w2} = \frac{S2- Savg}{Savg - S1} \frac{4}{8} = \frac{(S1- 10) - 70/3}{70/3 - S1} \frac{70}{3} - S1 = 2* (S1 - 10) - \frac{140}{3} S1 = 30 mph Then S2 = 20 mph Answer (C) The weighted average formula is discussed in this post: https://anaprep.com/arithmetic-weighted-averages/ and in this video: https://www.youtube.com/watch?v=_GOAU7moZ2Q Also check out these videos for other such TSD questions: https://youtu.be/qnhF9yLHNzU https://youtu.be/6Xjj5V2lxBg

Paras96 · Answer

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

Let v be speed on 1st day, therefore speed on 2nd day = v-10
Let t be time traveled on 1st day, therefore time traveled on 2nd day = t + 4

Given, t + t + 4 = 12
=> 2t = 8
=> t = 4

Now, vt + (v-10)(t+4) = 280
=> 4v + 8(v-10) = 280
=> 12v = 360
=> v = 30

Average speed on 2nd day, v-10=20

Hence C

bumpbot · Answer

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