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

 It is currently 23 Oct 2019, 03:23 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.  On a particular trip, Jane drove x miles at 30 miles per hour and y

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

Hide Tags

Manager  Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 189
Location: India
MISSION : 800
WE: Design (Manufacturing)
On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

3
35 00:00

Difficulty:   85% (hard)

Question Stats: 56% (02:41) correct 44% (02:47) wrong based on 391 sessions

HideShow timer Statistics

On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x + 60
(2) y/x > 4/3

_________________
Thank you

+KUDOS

> I CAN, I WILL <

Originally posted by iamdp on 10 Jun 2015, 06:19.
Last edited by Bunuel on 10 Jun 2015, 07:11, edited 1 time in total.
Edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 58464
On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

19
3
dpo28 wrote:
On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x + 60
(2) y/x > 4/3

On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

$$average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}$$

Thus the question asks whether $$\frac{120(x+y)}{4x+3y}>35$$.
After simplifying the question becomes: is $$\frac{y}{x}>\frac{4}{3}$$.

(1) y > x + 60.

If x = 1 and y = 100, then y/x > 4/3.
If x = 1000 and y = 1200, then y/x < 4/3.

Not sufficient.

(2) y/x > 4/3. Directly gives an answer to our question. Sufficient.

Hope it's clear.

Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours):

x/30 < y/40.

Then continue as above.
_________________
Retired Moderator Joined: 06 Jul 2014
Posts: 1220
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

8
6
dpo28 wrote:
On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x + 60
(2) y/x > 4/3

35 miles this is average speed if both distance will take equal time.
For example x = 300 miles on the speed 30 mph - 10 hours
and y = 400 on the speed 40 mph - 10 hours

So for have average speed more than 35 miles Jane should spend more time on distance y than on distance x
time = distance / speed

$$\frac{y}{40}>\frac{x}{30}$$ --> $$\frac{y}{x}>\frac{4}{3}$$

1) this statement insufficient because we can't find ratio of distances x and y from given information

2) This statement gives us direct answer on the question.
Sufficient

_________________
General Discussion
Manager  Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 189
Location: India
MISSION : 800
WE: Design (Manufacturing)
On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Hi Bunuel

i found another explanation for this question

it says that avg speed ie 4 is one away from 3 and 4 away from 8. Therefore time spent travelling at 3 m/hr is 4/5 of total.

I know this explanation is in compliance with the weighted average formula.

Can you plz elaborate on this method how it is 4/5 and not 1/5
_________________
Thank you

+KUDOS

> I CAN, I WILL <
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

dpo28 wrote:
Hi Bunuel

i found another explanation for this question

it says that avg speed ie 4 is one away from 3 and 4 away from 8. Therefore time spent travelling at 3 m/hr is 4/5 of total.

I know this explanation is in compliance with the weighted average formula.

Can you plz elaborate on this method how it is 4/5 and not 1/5

Step 1: The distance between 3 and 8 is 5

Step 2: Realize that Weighted Average of 3 and 8 is 4

Step 3: Distance of 3 from Average 4 is 1 AND Distance of 8 from Average 4 is 4

Step 4: S1 / S2 = T2 / T1 by property because Speed S is Inversely proportional to Time T

Step 5: i.e. Speed 3 will take 4/5 of Total time and 8 will take 1/5 of Total time

I hope it clears!!!
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Originally posted by GMATinsight on 11 Jun 2015, 03:51.
Last edited by GMATinsight on 11 Jun 2015, 04:12, edited 1 time in total.
Retired Moderator Joined: 06 Jul 2014
Posts: 1220
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

1
GMATinsight wrote:

Step 1: The distance between 3 and 8 is 5

Step 2: Realize that Weighted Average of 3 and 8 is 4

Step 3: Distance of 3 from Average 4 is 1 AND Distance of 8 from Average 4 is 5

Step 4: S1 / S2 = T2 / T1 by property because Speed S is Inversely proportional to Time T

Step 5: i.e. Speed 3 will take 4/5 of Total time and 8 will take 1/5 of Total time

I hope it clears!!!

Hello GMATinsight

I think there is typo in 3 step: "Step 3: Distance of 3 from Average 4 is 1 AND Distance of 8 from Average 4 is 4"
_________________
Intern  Joined: 04 May 2014
Posts: 29
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Seriously, how can you simplify so fast to this;
Quote:
Thus the question asks whether 120(x+y)/(4x+3y)>35.
After simplifying the question becomes: is y/x>4/3.
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

2
1
tjerkrintjema wrote:
Seriously, how can you simplify so fast to this;
Quote:
Thus the question asks whether 120(x+y)/(4x+3y)>35.
After simplifying the question becomes: is y/x>4/3.

120(x+y)/(4x+3y)>35

Since x and y are Positive Numbers hence you can cross multiply

120x + 120 y > 35*(4x+3y)

i.e. 120x + 120 y > 140x + 105y

i.e. 15y > 20x

i.e. 3y > 4x

i.e. (y/x) > (4/3)

I hope it clears your doubt!
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Intern  Joined: 04 May 2014
Posts: 29
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

GMATinsight wrote:
tjerkrintjema wrote:
Seriously, how can you simplify so fast to this;
Quote:
Thus the question asks whether 120(x+y)/(4x+3y)>35.
After simplifying the question becomes: is y/x>4/3.

120(x+y)/(4x+3y)>35

Since x and y are Positive Numbers hence you can cross multiply

120x + 120 y > 35*(4x+3y)

i.e. 120x + 120 y > 140x + 105y

i.e. 15y > 20x

i.e. 3y > 4x

i.e. (y/x) > (4/3)

I hope it clears your doubt!

Wow I was going at it the whole wrong way. Thanks I guess a better strategy is always to get rid of the fraction first.
Manager  Joined: 24 Nov 2013
Posts: 56
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

1
Bunuel

you have mentioned "If x = 100 and y = 200, then y/x < 4/3."
In this case y/x = 2 and it is not less than 4/3.

we might need something like x = 240 and y = 320 to prove option 1 is not sufficient.
in this case y/x = 4/3 and thus not less than 4/3

Math Expert V
Joined: 02 Sep 2009
Posts: 58464
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Success2015 wrote:
Bunuel

you have mentioned "If x = 100 and y = 200, then y/x < 4/3."
In this case y/x = 2 and it is not less than 4/3.

we might need something like x = 240 and y = 320 to prove option 1 is not sufficient.
in this case y/x = 4/3 and thus not less than 4/3

Edited the typo. Thank you.
_________________
Manager  Joined: 06 Mar 2014
Posts: 220
Location: India
GMAT Date: 04-30-2015
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Bunuel wrote:
dpo28 wrote:
On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x + 60
(2) y/x > 4/3

On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

$$average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}$$

Thus the question asks whether $$\frac{120(x+y)}{4x+3y}>35$$.
After simplifying the question becomes: is $$\frac{y}{x}>\frac{4}{3}$$.

(1) y > x + 60.

If x = 1 and y = 100, then y/x > 4/3.
If x = 1000 and y = 1200, then y/x < 4/3.

Not sufficient.

(2) y/x > 4/3. Directly gives an answer to our question. Sufficient.

Hope it's clear.

Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours):
x/30 < y/40.

Then continue as above.

Could anyone please explain the logic behind the highlighted part. I m not able to get how more than average speed correlates to time taken to cover specific distances, x and y.
Manager  Joined: 06 Mar 2014
Posts: 220
Location: India
GMAT Date: 04-30-2015
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Harley1980 wrote:
dpo28 wrote:
On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x + 60
(2) y/x > 4/3

35 miles this is average speed if both distance will take equal time.
For example x = 300 miles on the speed 30 mph - 10 hours
and y = 400 on the speed 40 mph - 10 hours

So for have average speed more than 35 miles Jane should spend more time on distance y than on distance x
time = distance / speed

$$\frac{y}{40}>\frac{x}{30}$$ --> $$\frac{y}{x}>\frac{4}{3}$$

1) this statement insufficient because we can't find ratio of distances x and y from given information

2) This statement gives us direct answer on the question.
Sufficient

Harley1980
Could you please elaborate on how does one say that " average speed more than 35 miles Jane should spend more time on distance y than on distance x"
Retired Moderator Joined: 06 Jul 2014
Posts: 1220
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

2
earnit wrote:
Harley1980
Could you please elaborate on how does one say that " average speed more than 35 miles Jane should spend more time on distance y than on distance x"

Hello earnit

This is weighted average question
We can imagine scales:
on one side --------------------- on another side
speed 40 mph ----------------- 30mph

if both side have equal weight (in our case weight is time) when speed will be in middle:
speed 40 mph--------------------- 30mph
10 hours --------------------------10 hours

Average = 35 mph
===============================

if we shift our weight to left: more time with speed 40 mph then average will grow

speed 40 mph---------------------30mph
20 hours---------------------------0 hours

Average = 40 mph
===============================

if we shift our weight to right: more time with speed 30 mph then average will decline

speed 40 mph--------------------- 30mph
0 hours --------------------------- 20 hours

Average = 30 mph

So if we want average to be more than 35 mph we should put more weight on 40 mph scale (distance y in this question)

Try to read Karishma's articles on this topic. She like this topic and write really good posts about it.
At first weighted average was very difficult for me but when I read her articles and solve like 50 tasks this topic became very intuitive:

http://www.veritasprep.com/blog/2014/12 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2014/09 ... difficult/
http://www.veritasprep.com/blog/2013/12 ... -question/

There is much more articles I just post first that I find. You can use search by words "weighted average"
_________________
Manager  Joined: 06 Mar 2014
Posts: 220
Location: India
GMAT Date: 04-30-2015
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Harley1980 wrote:
earnit wrote:
Harley1980
Could you please elaborate on how does one say that " average speed more than 35 miles Jane should spend more time on distance y than on distance x"

Hello earnit

This is weighted average question
We can imagine scales:
on one side --------------------- on another side
speed 40 mph ----------------- 30mph

if both side have equal weight (in our case weight is time) when speed will be in middle:
speed 40 mph--------------------- 30mph
10 hours --------------------------10 hours

Average = 35 mph
===============================

if we shift our weight to left: more time with speed 40 mph then average will grow

speed 40 mph---------------------30mph
20 hours---------------------------0 hours

Average = 40 mph
===============================

if we shift our weight to right: more time with speed 30 mph then average will decline

speed 40 mph--------------------- 30mph
0 hours --------------------------- 20 hours

Average = 30 mph

So if we want average to be more than 35 mph we should put more weight on 40 mph scale (distance y in this question)

Try to read Karishma's articles on this topic. She like this topic and write really good posts about it.
At first weighted average was very difficult for me but when I read her articles and solve like 50 tasks this topic became very intuitive:

http://www.veritasprep.com/blog/2014/12 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2014/09 ... difficult/
http://www.veritasprep.com/blog/2013/12 ... -question/

There is much more articles I just post first that I find. You can use search by words "weighted average"

Harley1980 +1 explanation. Terrific stuff. I am aware of this concept but just needed that missing link that connects weighted average with this qs.

($$c1w1+c2w2)/ (w1+w2)$$ = $$Cavg$$

If w1>w2 then Cavg tilts towards c1 and vice versa.

Thank you so much for clearing the clutter and making me see the plain logic behind.
Intern  B
Joined: 25 Feb 2016
Posts: 1
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Bunuel,

If y=41miles and x=40 miles
the average speed will be 34,7 (less than 35!)

41 miles = ~1 hour
40 miles = ~1 hour and 20 minutes

total distance: 81 miles/ 2hours and 20 minutes = 37,7 miles.

Am I doing something wrong?

Bunuel wrote:
dpo28 wrote:
On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x + 60
(2) y/x > 4/3

On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

$$average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}$$

Thus the question asks whether $$\frac{120(x+y)}{4x+3y}>35$$.
After simplifying the question becomes: is $$\frac{y}{x}>\frac{4}{3}$$.

(1) y > x + 60.

If x = 1 and y = 100, then y/x > 4/3.
If x = 1000 and y = 1200, then y/x < 4/3.

Not sufficient.

(2) y/x > 4/3. Directly gives an answer to our question. Sufficient.

Hope it's clear.

Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours):

x/30 < y/40.

Then continue as above.
Math Expert V
Joined: 02 Sep 2009
Posts: 58464
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

latinamerica2016 wrote:
Bunuel,

If y=41miles and x=40 miles
the average speed will be 34,7 (less than 35!)

41 miles = ~1 hour
40 miles = ~1 hour and 20 minutes

total distance: 81 miles/ 2hours and 20 minutes = 37,7 miles.

Am I doing something wrong?

Bunuel wrote:
dpo28 wrote:
On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

(1) y > x + 60
(2) y/x > 4/3

On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

$$average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}$$

Thus the question asks whether $$\frac{120(x+y)}{4x+3y}>35$$.
After simplifying the question becomes: is $$\frac{y}{x}>\frac{4}{3}$$.

(1) y > x + 60.

If x = 1 and y = 100, then y/x > 4/3.
If x = 1000 and y = 1200, then y/x < 4/3.

Not sufficient.

(2) y/x > 4/3. Directly gives an answer to our question. Sufficient.

Hope it's clear.

Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours):

x/30 < y/40.

Then continue as above.

Are you talking about the second statement? If yes, then you cannot use y = 41 miles and x = 40 miles because these values do not satisfy the condition y/x > 4/3.
_________________
Manager  S
Joined: 14 Jul 2014
Posts: 86
Location: India
Concentration: Social Entrepreneurship, Strategy
GMAT 1: 620 Q41 V34 WE: Information Technology (Computer Software)
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Explanation: For average speed problems, you always need to remember that the average speed is total distance divided by total time, not a simple average of the speeds (unless the times are the same). Or in weighted average terms: it is the time that affects the weighting of the two speeds when you average them together.

To answer this question, you need to determine at which speed more time was spent. If you can prove more time was spent at one speed, then you can answer the question or in other words, this question is really: is time for y greater than the time for x.

Algebraically this can be written as: is y/40>x/30 or is y/x>4/3

From this you realize clearly that statement (2) is sufficient.
For statement (1) you cannot prove this, because you do not know how big x and y are.
If x and y were really large distances, then the average might not be greater than 35, and if x and y were really small distances then it would be.

Manager  G
Joined: 21 Jun 2017
Posts: 234
Concentration: Finance, Economics
WE: Corporate Finance (Commercial Banking)
On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Hi chetan2u

While analyzing statement 1, is there any algebraic method of solving these two equation y/x>3/4 and y>x+60, instead of picking numbers or using weighted average concept ?
_________________
Even if it takes me 30 attempts, I am determined enough to score 740+ in my 31st attempt. This is it, this is what I have been waiting for, now is the time to get up and fight, for my life is 100% my responsibility.

Dil ye Ziddi hai !!!

GMAT 1 - 620 .... Disappointed for 6 months. Im back Im back. Bhai dera tera COMEBACK !!!
Senior Manager  P
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 435
Re: On a particular trip, Jane drove x miles at 30 miles per hour and y  [#permalink]

Show Tags

Stem:
rate*time=distance
30 * t1 = x
40 * t2 = y

Question:
(x+y)/(t1+t2) > 35, expressing time as Work/rate, (x+y)/(x/40+y/30) > 35
The question is asking whether the distance traveled at 40 mph is more than that at traveled at 30 mph.

1) y>x+60
if y = 80 and x=0, then obviously Yes since y>x
if y = 400 and x = 330 then 400/40 = 10 and 330/30 = 11, then No since x>y
Depends on values of y and x, NS

2) 3y > 4x, so y is more than x in this ratio, which is another way of saying that whatever the distance is, the part traveled at 40 mph is more than the part traveled at 30 mph, which means the Avg Speed will be >35. Sufficient. Re: On a particular trip, Jane drove x miles at 30 miles per hour and y   [#permalink] 08 Apr 2019, 09:35
Display posts from previous: Sort by

On a particular trip, Jane drove x miles at 30 miles per hour and y

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

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