It is currently 23 Oct 2017, 14:14

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If a motorist had driven 1 hour longer on a certain day and

Author Message
TAGS:

### Hide Tags

Manager
Joined: 05 Sep 2007
Posts: 144

Kudos [?]: 51 [2], given: 0

Location: New York
If a motorist had driven 1 hour longer on a certain day and [#permalink]

### Show Tags

24 Mar 2008, 13:37
2
KUDOS
13
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

69% (02:14) correct 31% (02:14) wrong based on 467 sessions

### HideShow timer Statistics

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100
(B) 120
(C) 140
(D) 150
(E) 160
[Reveal] Spoiler: OA

Kudos [?]: 51 [2], given: 0

Senior Manager
Joined: 15 Aug 2007
Posts: 252

Kudos [?]: 71 [2], given: 0

Schools: Chicago Booth

### Show Tags

24 Mar 2008, 14:48
2
KUDOS
D-150

assuming t is the original travel time in hrs, and x is the original speed in miles per hour

from the first half - 5t+(x+5)=70 => 5t+x=65
Now need to find out 10t+2(x+10) = 10t+2x+20 = 2(5t+x)+20 = 150

Kudos [?]: 71 [2], given: 0

SVP
Joined: 28 Dec 2005
Posts: 1545

Kudos [?]: 179 [10], given: 2

### Show Tags

24 Mar 2008, 19:32
10
KUDOS
1
This post was
BOOKMARKED
(t+1)(s+5) = d+70

expand to get ts+5t+s+5 = d+70, and recognize that ts = d (from standard equation speed = distance/time)

simplify down to 5t+s=65

now, (t+2)(s+10) = ts+10t+2s+20 = ts+2(5t+s)+20 = d+150

Kudos [?]: 179 [10], given: 2

Senior Manager
Joined: 15 Aug 2007
Posts: 252

Kudos [?]: 71 [0], given: 0

Schools: Chicago Booth

### Show Tags

24 Mar 2008, 19:41
el1981 wrote:
i don't understand. how did you get 5t and x+5?

70 additional miles came from 2 sources - 5 miles additional distance traveled for t hours (5t) + distance traveled in the additional hour. The later one is x+5 where in x is distance traveled in an hour with the original speed and 5 is the distance came from the additional speed (making total speed x+5 mph)

Kudos [?]: 71 [0], given: 0

CEO
Joined: 29 Mar 2007
Posts: 2554

Kudos [?]: 518 [2], given: 0

### Show Tags

25 Mar 2008, 13:37
2
KUDOS
el1981 wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

Interesting problem.

first we have (r+5)(t+1)=d+70 ---> rt+r+5t+5=d+70. Also notice that rt=d

thus we now have r+5t=65.

Second equation. (r+10)(t+2)=d+x

rt+2r+10t+20=d+x --> plug in rt and 2(r+5t)=2(65) ---->130 ---> rt+150=rt+x ---> x=150.

D

Kudos [?]: 518 [2], given: 0

SVP
Joined: 04 May 2006
Posts: 1881

Kudos [?]: 1408 [0], given: 1

Schools: CBS, Kellogg

### Show Tags

25 Mar 2008, 19:19
el1981 wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

D1=100, V1=50, T1=2
T2=3, V2=55 -->D2= 165
T3=5, V3=65 --> D3=325
D3-D2=150
D
_________________

Kudos [?]: 1408 [0], given: 1

Senior Manager
Joined: 16 Aug 2004
Posts: 320

Kudos [?]: 52 [12], given: 0

Location: India

### Show Tags

25 Mar 2008, 23:47
12
KUDOS
3
This post was
BOOKMARKED
Hey I have a quick approach for this.

The guy covered 70 miles more in an hour by driving 5 miles/hour faster. Therefore his speed current is 70 Miles/hr and his original speed was 65 Miles /Hr (since he is traveling 5Miles/Hr faster). Therefore his new speed is 75Miles/Hr (65+10) and he will travel 150 (75x2) Miles more in 2 Hours.

Hence D.

Kudos [?]: 52 [12], given: 0

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 320

Kudos [?]: 875 [3], given: 28

### Show Tags

07 Jun 2010, 13:16
3
KUDOS
my approach was similar too than the posts above ... but i'll share it anyway

avg rate of 5mph faster means that the motorist drove 5miles more in each hour.
since he drove 5miles in the last (extra) hour too, so he drove 70-5=65 miles more in each of the hour earlier
65/5=13hrs
this is the actual time he drove

avg rate of 10mph more means that he drove 10*13=130miles more in the first 13 hrs
he covered 2*10=20miles more in the last 2 hours
130+20=150 more miles covered than he actually did

so D it is

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Kudos [?]: 875 [3], given: 28

SVP
Joined: 04 May 2006
Posts: 1881

Kudos [?]: 1408 [0], given: 1

Schools: CBS, Kellogg
Re: Problem related to time and distance [#permalink]

### Show Tags

08 Jun 2010, 02:37
20. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

D.150, but I dont like my approach

1. (t+1)(v+5)=d+70
2. 5t+v =65 (taken from 1)
3. (t+2)(v+10)=[t+1 +1][v+5 + 5] =[(t+1)(v+5)+5(t+1)+(v+5) +5]=d+70+80=d+150
_________________

Kudos [?]: 1408 [0], given: 1

Manager
Joined: 02 May 2010
Posts: 57

Kudos [?]: 44 [0], given: 3

Schools: IU, UT Dallas, Univ of Georgia, Univ of Arkansas, Miami University
WE 1: 5.5 Yrs IT
Re: Problem related to time and distance [#permalink]

### Show Tags

08 Jun 2010, 03:33
Hi sondenso,
My approach was very similar to yours except for a minor change in the 3rd step;

"...How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?" can be rephrased as;

(t+2)(v+10) = ?
tv+10t+2v+20 = ?
10t+2v+20 = 2(5t+v)+20 = 2*65 (from 2) +20 = 150.

Kudos [?]: 44 [0], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129505 [3], given: 12201

Re: Problem related to time and distance [#permalink]

### Show Tags

08 Jun 2010, 07:54
3
KUDOS
Expert's post
3
This post was
BOOKMARKED
20. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

Let $$t$$ be the actual time and $$r$$ be the actual rate.

"If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did" --> $$(t+1)(r+5)-70=tr$$ --> $$tr+5t+r+5-70=tr$$ --> $$5t+r=65$$;

"How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?" --> $$(t+2)(r+10)-x=tr$$ --> $$tr+10t+2r+20-x=tr$$ --> $$2(5t+r)+20=x$$ --> as from above $$5t+r=65$$, then $$2(5t+r)+20=2*65+20=150=x$$ --> so $$x=150$$.

OR another way:

70 miles of surplus in distance is composed of driving at 5 miles per hour faster for $$t$$ hours plus driving for $$r+5$$ miles per hour for additional 1 hour --> $$70=5t+(r+5)*1$$ --> $$5t+r=65$$;

With the same logic, surplus in distance generated by driving at 10 miles per hour faster for 2 hours longer will be composed of driving at 10 miles per hour faster for $$t$$ hours plus driving for $$r+10$$ miles per hour for additional 2 hour --> $$surplus=x=10t+(r+10)*2$$ --> $$x=2(5t+r)+20$$ --> as from above $$5t+r=65$$, then $$x=2(5t+r)+20=150$$.

Note that the solutions proposed by dushver and dimitri92 are not correct (though correct answer was obtained). For this question we can not calculate neither $$t$$ not $$r$$ of the motorist.

Hope it helps.
_________________

Kudos [?]: 129505 [3], given: 12201

Senior Manager
Joined: 03 Mar 2010
Posts: 422

Kudos [?]: 361 [1], given: 22

Schools: Simon '16 (M)
Re: motorist - work problem [#permalink]

### Show Tags

26 Aug 2011, 11:16
1
KUDOS
Let original speed be s, distance be d, and time t
d=st
Also, d+70=(s+5)(t+1)
d+70=st+s+5t+5
st+70=st+s+5t+5
s+5t=65

And, d1=(s+10)(t+2)
d1=st+2s+10t+20
so, d1-d=2s+10t+20
=2(s+5t)+20
=2(65)+20
=130+20=150

OA D.
_________________

My dad once said to me: Son, nothing succeeds like success.

Kudos [?]: 361 [1], given: 22

VP
Joined: 24 Jul 2011
Posts: 1347

Kudos [?]: 644 [1], given: 20

GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
Re: motorist - work problem [#permalink]

### Show Tags

26 Aug 2011, 11:30
1
KUDOS
The motorist covers 70 miles in an hour when he drives 5 miles/hr faster
=> He is currently driving at 65 miles/hr

If he drives 10 miles/hr faster for 2 more hours he would cover 75*2 = 150 miles

(D) it is.
_________________

GyanOne | Top MBA Rankings and MBA Admissions Blog

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738

Kudos [?]: 644 [1], given: 20

Manager
Joined: 21 Aug 2010
Posts: 186

Kudos [?]: 129 [1], given: 141

Location: United States
GMAT 1: 700 Q49 V35
Re: motorist - work problem [#permalink]

### Show Tags

04 Oct 2011, 12:14
1
KUDOS
Lets take x as time and y as avg speed
If he has driven 1 hr longer and 5 km/hr faster the distance covered Is 70

(X+1)*(Y+5) – XY = 70
XY+Y+5+5X – XY = 70

Y+5X = 65
Scenario 2

(X+2)(Y+10)-XY = D
2x+10Y+20 = D
2(X+5Y)+20 = D
2*65 +20 = D
D = 150
_________________

-------------------------------------

Kudos [?]: 129 [1], given: 141

Intern
Joined: 02 Oct 2011
Posts: 11

Kudos [?]: 1 [0], given: 0

Re: motorist - work problem [#permalink]

### Show Tags

04 Oct 2011, 13:27
GyanOne wrote:
The motorist covers 70 miles in an hour when he drives 5 miles/hr faster
=> He is currently driving at 65 miles/hr

If he drives 10 miles/hr faster for 2 more hours he would cover 75*2 = 150 miles

(D) it is.

despite the correct answer, the logic here is flawed, isn't it ?

Quote:
The motorist covers 70 miles in an hour when he drives 5 miles/hr faster

M drives less than 70miles in one hour. Because M drove more than the original distance in the time before the extra hour starts. So that delta + another hour's worth of driving (at the elevated speed) is 70miles.

Kudos [?]: 1 [0], given: 0

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17405 [2], given: 232

Location: Pune, India
Re: motorist - work problem [#permalink]

### Show Tags

06 Oct 2011, 04:47
2
KUDOS
Expert's post
igemonster wrote:
despite the correct answer, the logic here is flawed, isn't it ?

Quote:
The motorist covers 70 miles in an hour when he drives 5 miles/hr faster

M drives less than 70miles in one hour. Because M drove more than the original distance in the time before the extra hour starts. So that delta + another hour's worth of driving (at the elevated speed) is 70miles.

Yes, 70 miles is the sum of extra distance covered by increasing speed + extra distance covered in an hour.

But remember, this is a PS question. You do have the correct answer in the options and there is only one correct option.
I will just assume any case and whatever I get will be my answer.

Say the original speed was 60 miles/hr. The speed increased to 65.
So the motorist would have traveled 5 miles + 65 miles (= 70) extra.
(He travels for only one hour initially.)
If he instead increases speed by 10 miles/hr, his speed becomes 70. In the initial 1 hr, he will travel an extra 10 miles and then in additional 2 hrs he will travel an extra 70*2 = 140 miles.
So he will travel an extra 150 miles.

I could have just as well assumed the original speed to be 55 miles/hr. If the speed increases to 60 and extra distance covered is 70, it means that the motorist travels for 2 hrs initially.
70 = 5 + 5 + 60
If instead, the speed increases by 10 miles/hr, the speed becomes 65.
Extra distance covered = 10+10+2*65 = 150

In any case, the answer has to be the same since it is a PS question.

I could also have assumed the original speed to be 65 miles/hr (as done above) If the speed increases to 70 and I cover 70 miles extra, it means I didn't travel at all before.
If the speed instead becomes 75, I travel 2*75 = 150 miles extra.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17405 [2], given: 232

Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 172

Kudos [?]: 98 [0], given: 1

Location: India
WE: Information Technology (Investment Banking)
Re: motorist - work problem [#permalink]

### Show Tags

15 Oct 2011, 23:31
We can make use of the below 3 equations
(s+5)(t+1) = 70+d
(s+10)(t+2) = x
and st=d

On solving we will get x=150 (D)

Kudos [?]: 98 [0], given: 1

Intern
Status: Stay Hungry, Stay Foolish.
Joined: 05 Sep 2011
Posts: 40

Kudos [?]: 9 [0], given: 6

Location: India
Concentration: Marketing, Social Entrepreneurship
Re: motorist - work problem [#permalink]

### Show Tags

16 Oct 2011, 09:58
Solution :
s-speed.
t-time.
d-distance.

Two equations will be formed.
As per the conditions given well get,

(s+5)(t+1) - st = 70.....(1)
on simplifying this,we get
s + 5t = 65....(a)

Another one will be,
(s+10)(t+2) - st = d
on simplifying this well get,

2s + 10t +20 = d...(b)
substituting (a) in (b)

we get 65*2 + 20 = 150.

Kudos [?]: 9 [0], given: 6

Manager
Joined: 18 Jun 2010
Posts: 142

Kudos [?]: 38 [0], given: 2

Re: motorist - work problem [#permalink]

### Show Tags

18 Oct 2011, 20:11
Let original speed be x

Extra distance travelled in 1 hr:
70 = (x + 5)*1 or x+5 = 70 ---- (1)

Now, question asks us to find:
d = (x+10)*2
d = (x+5+5)*2

From (1), d = (70+5)*2 = 150. Hence (D).

Kudos [?]: 38 [0], given: 2

Intern
Joined: 29 Jan 2012
Posts: 6

Kudos [?]: 2 [0], given: 1

GMAT 1: 650 Q49 V28
Re: motorist - work problem [#permalink]

### Show Tags

10 Mar 2012, 11:55
tm=(1+t)
dm=70+d
rm=(r+5)

Being r,t and d the actual rate, time and distance.

If you realize that the motorist has actually covered a distance of 70 miles in 1 hour, going at (r+5)speed, it is much easier...

70= (r+5)*1 then r=65mph

If the new rate is 10 mph more then it is r+10=75mph

Therefore in two hours the distance covered at this speed is 150miles

Kudos [?]: 2 [0], given: 1

Re: motorist - work problem   [#permalink] 10 Mar 2012, 11:55

Go to page    1   2    Next  [ 35 posts ]

Display posts from previous: Sort by