GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Oct 2019, 14:41 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Sanjana jogged uphill for a while at an average speed of 3

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Intern  Joined: 20 Jan 2012
Posts: 20
Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

4
1
27 00:00

Difficulty:   85% (hard)

Question Stats: 60% (03:05) correct 40% (02:48) wrong based on 343 sessions

HideShow timer Statistics

Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

A. 1 mile
B. 1 1/3 miles
C. 1 1/2 miles
D. 1 3/5 miles
E. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2
Intern  Joined: 14 Aug 2013
Posts: 31
Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
GPA: 3.2
WE: Consulting (Consumer Electronics)
Re: Sanjana jogged uphill for a while  [#permalink]

Show Tags

12
4
piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

a. 1 mile
b. 1 1/3 miles
c. 1 1/2 miles
d. 1 3/5 miles
e. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2

Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr

D1/3+D2/8=40/60 hrs
=>8D1+3D2=16------Eq(1)

also given she travelled uphill and downhill at an avg of 4 miles/hr
=>total distance/Avg speed=40/60hrs
=>(D1+D2)/4=2/3
=>3D1+3D2=8-------Eq(2)
solving equations 1 and 2
D1=8/5 mile and D2=16/15 miles
=>D1=1 3/5 miles

the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case..

Hope it is helpful
General Discussion
Manager  B
Joined: 18 Dec 2012
Posts: 94
Location: India
Concentration: General Management, Strategy
GMAT 1: 660 Q49 V32 GMAT 2: 530 Q37 V25 GPA: 3.32
WE: Manufacturing and Production (Manufacturing)
Re: Sanjana jogged uphill for a while  [#permalink]

Show Tags

piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

a. 1 mile
b. 1 1/3 miles
c. 1 1/2 miles
d. 1 3/5 miles
e. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2

Here we can't use the S1*S2/ (S1+S2) formula because the uphill and downhill distances are not same.
_________________
I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?
SVP  Joined: 06 Sep 2013
Posts: 1573
Concentration: Finance
Re: Sanjana jogged uphill for a while  [#permalink]

Show Tags

1
kusena wrote:
piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

a. 1 mile
b. 1 1/3 miles
c. 1 1/2 miles
d. 1 3/5 miles
e. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2

Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr

D1/3+D2/8=40/60 hrs
=>8D1+3D2=16------Eq(1)

also given she travelled uphill and downhill at an avg of 4 miles/hr
=>total distance/Avg speed=40/60hrs
=>(D1+D2)/4=2/3
=>3D1+3D2=8-------Eq(2)
solving equations 1 and 2
D1=8/5 mile and D2=16/15 miles
=>D1=1 3/5 miles

the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case..

Hope it is helpful

But aren't the uphill and downhill distances actually the same?
Are we assuming is the same hill no?

Cheers
J Intern  Joined: 14 Aug 2013
Posts: 31
Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
GPA: 3.2
WE: Consulting (Consumer Electronics)
Re: Sanjana jogged uphill for a while  [#permalink]

Show Tags

jlgdr wrote:
kusena wrote:
piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

a. 1 mile
b. 1 1/3 miles
c. 1 1/2 miles
d. 1 3/5 miles
e. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2

Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr

D1/3+D2/8=40/60 hrs
=>8D1+3D2=16------Eq(1)

also given she travelled uphill and downhill at an avg of 4 miles/hr
=>total distance/Avg speed=40/60hrs
=>(D1+D2)/4=2/3
=>3D1+3D2=8-------Eq(2)
solving equations 1 and 2
D1=8/5 mile and D2=16/15 miles
=>D1=1 3/5 miles

the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case..

Hope it is helpful

But aren't the uphill and downhill distances actually the same?
Are we assuming is the same hill no?

Cheers
J Question says average speed of both the journeys is 4 miles/hr
Where as uphill speed 3miles/hr and downhill speed 8 miles/hr
If both uphill and down hill distances are same then the avg speed will be (v1*v2)/(v1+v2)=8*3/(8+3)=24/11 miles/hr
which is not equal to 4 miles/hr.hence the the distances traveled uphill and downhill aren't the same.
Hope its clear
SVP  Joined: 06 Sep 2013
Posts: 1573
Concentration: Finance
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

2
3
piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

A. 1 mile
B. 1 1/3 miles
C. 1 1/2 miles
D. 1 3/5 miles
E. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2

Easy question

3x + 8y / x+y = 4
x + y = 40

So solving y = 8
x = 32

So D is out correct answer choice

Hope it helps
Provide kudos!

Cheers
J SVP  Joined: 06 Sep 2013
Posts: 1573
Concentration: Finance
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

1
2
jlgdr wrote:
piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

A. 1 mile
B. 1 1/3 miles
C. 1 1/2 miles
D. 1 3/5 miles
E. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2

Easy question

3x + 8y / x+y = 4
x + y = 40

So solving y = 8
x = 32

So D is out correct answer choice

Hope it helps
Provide kudos!

Cheers
J Just found another way of doing it with weighted average.

Let's call x the uphill distance and y the downhill distance

Also we know that average speed is 4mph.

So x is at 3mph and y is at 8mph. By differentials

x = 4y. So if total time is 40 minutes then the time spent uphill must be 4 times that spend downhill

Therefore x = 32 min. Now, (3)(32/60)=1 3/5

Hope you enjoyed
Cheers
J
Manager  Joined: 24 Nov 2013
Posts: 56
Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

I used weighted avg to get uphill to downhill ratio. U/D = 4/1. i.e. U = 4D.
Uphill to total will be 4/5.

Does 4/5 ratio only apply to the total time OR does it also apply to the total distance.
I mean, can the total distance also be split in this ratio?

I got total dist as 8/3 miles. (total time 40/60 * avg speed 4mph)
will the uphill distance be 4/5 of 8/3...It does not seem so...

Could somebody pls clarify whether 4/5 ratio applies only to the time and why should it not be applied to the total distance?

Thanks

PS - In the above post we have "if total time is 40 minutes then the time spent uphill must be 4 times that spend downhill".
I would like to know whether the same ratio applies to total distance also or just the time.
Director  G
Joined: 23 Jan 2013
Posts: 525
Schools: Cambridge'16
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

add times to get 2/3 hours (40min.) and consider full distance (2/3)*4=8/3:

x/3 + (8/3)-x)/8=2/3

5x=8

x=8/5 => 1+3/5

D
VP  P
Joined: 07 Dec 2014
Posts: 1223
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

1
let D=uphill distance
D/3 hours=uphill time
(4 mph avge speed)(2/3 hour total time)=8/3 miles total distance
8/3-D=downhill miles
(8/3-D)/8=downhill time
D/3+(8/3-D)/8=2/3
D=8/5➡1+3/5 miles
Board of Directors P
Joined: 17 Jul 2014
Posts: 2515
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

i used only 1 variable to solve this question...
uphill 3mph - time 2/3-x - distance 2-3x
downhill 8mph - time x - distance 8x

total distance 8/3 miles

so 2-3x+8x=8/3
5x=8/3 - 6/3
5x = 2/3
x=2/15

now, distance uphill is 2-3x
2-2/5 = 1 and 3/5

D
VP  P
Joined: 07 Dec 2014
Posts: 1223
Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

let t=uphill time
3t+8(2/3-t)=8/3 total miles
t=8/15 hours
(3 mph)(8/15 hours)=1 3/5 uphill miles
Manager  D
Joined: 17 May 2015
Posts: 247
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

3
Apply mixture and alligation method.
Please refer the attached diagram.
time spent at the rate of 3 miles/hr: time spent at the rate of 8 miles/hr = 4:1
So reqd time = 4/5*40 = 32 minutes

So distance traveled = 32/60 * 3 = 8/5 = 1 3/5 miles.
Attachments Jogg.jpeg [ 3.71 KiB | Viewed 6942 times ]

Manager  B
Joined: 31 Dec 2016
Posts: 68
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

4
1
D=Distance
R=Rate
T=Time
T1= Time 1 = Uphill
T2 = Time 2 = Downtown
Attachments GMAT 19.png [ 331.89 KiB | Viewed 6025 times ]

Manager  G
Joined: 23 Nov 2017
Posts: 113
Location: Singapore
Concentration: Strategy, Other
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

chetan2u, VeritasKarishma, Why can we not apply weighted average ratio of 4/5 on total distance traveled (ie-8/3 miles) directly instead of applying it on total time taken of 40 minutes and then multiplying it to the uphill speed?
CEO  V
Joined: 12 Sep 2015
Posts: 3997
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

Top Contributor
1
piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

A. 1 mile
B. 1 1/3 miles
C. 1 1/2 miles
D. 1 3/5 miles
E. 1 2/3 miles

(distance traveled uphill) + (distance traveled downhill) = TOTAL DISTANCE

Let t = the time spent jogging UPHILL (in hours)
The total travel time = 40 minutes = 2/3 HOURS
So, 2/3 - t = the time spent jogging DOWNHILL (in hours)

Distance = (speed)(time)
So, we can rewrite our word equation as: 3t + 8(2/3 - t) = (4)(2/3)
Expand to get: 3t + 16/3 - 8t = 8/3
Subtract 16/3 from both sides to get: 3t - 8t = -8/3
Simplify to get: -5t = -8/3
Solve: t = (-8/3)/(-5) = 8/15
So, Sanjana spent 8/15 hours traveling UPHILL

How far did Sanjana jog uphill?
Distance = (speed)(time)
So distance = (3)(8/15) = 8/5 = 1 3/5 miles

Cheers,
Brent
_________________
Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8043
Location: United States (CA)
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

A. 1 mile
B. 1 1/3 miles
C. 1 1/2 miles
D. 1 3/5 miles
E. 1 2/3 miles

We can let t = the number of hours Sanjana jogged uphill. Since 40 minutes = 2/3 of an hour, the time she jogged downhill was (2/3 - t) hours. We can create the equation:

3t + 8(2/3 - t) = 4(2/3)

3t + 16/3 - 8t = 8/3

-5t = -8/3

t = 8/15

Since she jogged 8/15 of an hour uphill at an average speed of 3 mph, she jogged 8/15 x 3 = 8/5 = 1 3/5 miles.

_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern  B
Joined: 26 Dec 2018
Posts: 2
Re: Sanjana jogged uphill for a while at an average speed of 3  [#permalink]

Show Tags

kusena wrote:
piyush272 wrote:
Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?

a. 1 mile
b. 1 1/3 miles
c. 1 1/2 miles
d. 1 3/5 miles
e. 1 2/3 miles

my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2

Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr

D1/3+D2/8=40/60 hrs
=>8D1+3D2=16------Eq(1)

also given she travelled uphill and downhill at an avg of 4 miles/hr
=>total distance/Avg speed=40/60hrs
=>(D1+D2)/4=2/3
=>3D1+3D2=8-------Eq(2)
solving equations 1 and 2
D1=8/5 mile and D2=16/15 miles
=>D1=1 3/5 miles

the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case..

Hope it is helpful

How did you get 8D1+2D2=16? Wouldn't 40/60 simplify just to 2/3? Re: Sanjana jogged uphill for a while at an average speed of 3   [#permalink] 10 Apr 2019, 00:37
Display posts from previous: Sort by

Sanjana jogged uphill for a while at an average speed of 3

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  