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# Cole drove from home to work at an average speed of 75 kmh. He then

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Intern
Joined: 10 Sep 2015
Posts: 26
Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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15 Sep 2015, 13:44
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45% (medium)

Question Stats:

67% (02:33) correct 33% (03:09) wrong based on 167 sessions

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Cole drove from home to work at an average speed of 75 kmh. He then returned home at an average speed of 105 kmh. If the round trip took a total of 2 hours, how many minutes did it take Cole to drive to work?
A) 66
B) 70
C) 72
D) 75
E) 78

Source: GMAT Prep Now - http://www.gmatprepnow.com/module/gmat- ... /video/914
Intern
Joined: 20 Aug 2015
Posts: 4
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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16 Sep 2015, 01:36
3
3
[quote="skylimit"]Cole drove from home to work at an average speed of 75 kmh. He then returned home at an average speed of 105 kmh. If the round trip took a total of 2 hours, how many minutes did it take Cole to drive to work?
A) 66
B) 70
C) 72
D) 75
E) 78

Hi All,

Here is my first post...

I prefer to use the ratio approach when it is possible, in this case it is.

The ratios between the speeds Work:Home = 75:105 = 5:7
Given that the distance in both directions is the same, the ratios of times becomes Work:Home = 7:5 (inverse)

The total time is given 2 hours ==> , so: 2/12 *7 = 14/12 hours ==> Multiple by 5 to get minutes = 70/60 ==> Answer is 70 minutes.
##### General Discussion
Intern
Joined: 10 Sep 2015
Posts: 26
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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15 Sep 2015, 13:52
I tried solving the question by starting with this:
time to work + time home = 2
d/75 + d/105 = 2
105d/7875 + 75d/7875 = 15750
180d/7875 = 15750
180d = 7875*15750
I used my calculator for the rest and got d = 689062.5 which is obviously too big to work with the question.
So I stopped.
What went wrong?
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Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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15 Sep 2015, 14:55
1
Let the distance one way be x
Time from home to work = x/75
Time from work to home = x/105
Total time = 2 hrs
(x/75) + (x/105)= 2
Solving for x, we get x = 175/2

Time from home to work in minutes= (175/2)*60/75 = 70 minutes

Ans= B
Intern
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Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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15 Sep 2015, 14:57
2
skylimit wrote:
I tried solving the question by starting with this:
time to work + time home = 2
d/75 + d/105 = 2
105d/7875 + 75d/7875 = 15750
180d/7875 = 15750
180d = 7875*15750
I used my calculator for the rest and got d = 689062.5 which is obviously too big to work with the question.
So I stopped.
What went wrong?

The error lies in your 2nd step- 105d/7875 + 75d/7875 = 15750

It should be 105d/7875 + 75d/7875 = 2

A better approach would be to take the common factors out. This way the calculation becomes much simpler. In this case taking common factor "15" out we get:

(1/15)*[(d/5)+(d/7)]= 2 --> (d/5)+(d/7)=30; which is much easier to solve manually. Remember, if you think you need a calculator on any GMAT question then probably your approach is not correct. There is always a way around.
Intern
Joined: 10 Sep 2015
Posts: 26
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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15 Sep 2015, 15:12
sunnysid wrote:
skylimit wrote:
I tried solving the question by starting with this:
time to work + time home = 2
d/75 + d/105 = 2
105d/7875 + 75d/7875 = 15750
180d/7875 = 15750
180d = 7875*15750
I used my calculator for the rest and got d = 689062.5 which is obviously too big to work with the question.
So I stopped.
What went wrong?

The error lies in your 2nd step- 105d/7875 + 75d/7875 = 15750

It should be 105d/7875 + 75d/7875 = 2

A better approach would be to take the common factors out. This way the calculation becomes much simpler. In this case taking common factor "15" out we get:

(1/15)*[(d/5)+(d/7)]= 2 --> (d/5)+(d/7)=30; which is much easier to solve manually. Remember, if you think you need a calculator on any GMAT question then probably your approach is not correct. There is always a way around.

Shoot! Nice catch. Thanks
Intern
Joined: 10 Sep 2015
Posts: 26
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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16 Sep 2015, 09:07
zurvy wrote:
The total time is given 2 hours ==> , so: 2/12 *7 = 14/12 hours ==> Multiple by 5 to get minutes = 70/60 ==> Answer is 70 minutes.

The inverse method is interesting and it seems pretty fast too.
I just don't understand the very last part above.
Why did you multiply 2/12 by 7 and then multiply by 5 to get minutes?
Intern
Joined: 20 Aug 2015
Posts: 4
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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16 Sep 2015, 23:34
1
skylimit wrote:
zurvy wrote:
The total time is given 2 hours ==> , so: 2/12 *7 = 14/12 hours ==> Multiple by 5 to get minutes = 70/60 ==> Answer is 70 minutes.

The inverse method is interesting and it seems pretty fast too.
I just don't understand the very last part above.
Why did you multiply 2/12 by 7 and then multiply by 5 to get minutes?

Hi there,

The ratio of time is as explained 5:7 during the entire trip , and it took a total of 2 hours ==> Imagine it like this, you have to split the 2 hours
with a ratio of 5:7 ==> The total parts would be in this case 5+7 = 12 ==> so The total time has to be divided first by 12, and then multiplied by 5.
If I told you to split 24 apples among two persons with a ratio of 5:7, you would first divide 24/12 (12=7+5) and than multiple by 5 and 7 respectively to get the answer.
24/12 * 5 = 10 for one person 24/12 * 7 = 14 for the other making a total of 24 apples.

Regarding the multiplication by 5, you need to convert 14/12 to fractions of 60 (so you get the minutes).
(5*14/5*12 = 70/60)
Or you could, 14/12 = 7/6 hour = 1 1/6 hour = 1 hour and 10 minutes = 70/60 hour (its just playing with fractions, and with time, having a denominator of 60, always
gives you the number of minutes at the top)
Intern
Joined: 10 Sep 2015
Posts: 26
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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17 Sep 2015, 10:06
zurvy wrote:
skylimit wrote:
zurvy wrote:
The total time is given 2 hours ==> , so: 2/12 *7 = 14/12 hours ==> Multiple by 5 to get minutes = 70/60 ==> Answer is 70 minutes.

The inverse method is interesting and it seems pretty fast too.
I just don't understand the very last part above.
Why did you multiply 2/12 by 7 and then multiply by 5 to get minutes?

Hi there,

The ratio of time is as explained 5:7 during the entire trip , and it took a total of 2 hours ==> Imagine it like this, you have to split the 2 hours
with a ratio of 5:7 ==> The total parts would be in this case 5+7 = 12 ==> so The total time has to be divided first by 12, and then multiplied by 5.
If I told you to split 24 apples among two persons with a ratio of 5:7, you would first divide 24/12 (12=7+5) and than multiple by 5 and 7 respectively to get the answer.
24/12 * 5 = 10 for one person 24/12 * 7 = 14 for the other making a total of 24 apples.

Regarding the multiplication by 5, you need to convert 14/12 to fractions of 60 (so you get the minutes).
(5*14/5*12 = 70/60)
Or you could, 14/12 = 7/6 hour = 1 1/6 hour = 1 hour and 10 minutes = 70/60 hour (its just playing with fractions, and with time, having a denominator of 60, always
gives you the number of minutes at the top)

Thanks zurvy!
I didn't realize you multiplied by 5 to get a denominator of 60.
Nice solution
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Joined: 07 Dec 2014
Posts: 1224
Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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17 Sep 2015, 17:00
let t=time to work
time to home (derived from inverse kmh ratio)=75t/105➡5t/7
t+5t/7=2 hours total time
t=7/6 hours➡70 minutes
Intern
Joined: 19 Mar 2014
Posts: 5
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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18 Sep 2015, 22:25
[quote="skylimit"]Cole drove from home to work at an average speed of 75 kmh. He then returned home at an average speed of 105 kmh. If the round trip took a total of 2 hours, how many minutes did it take Cole to drive to work?
A) 66
B) 70
C) 72
D) 75
E) 78

First round distance travelled (say) = d
Speed = 75 k/h
Time taken, T2 = d/75 hr

Second round distance traveled = d (same distance)
Speed = 105 k/h
Time taken, T2 = d/105 hr

Total time taken = 2 hrs
Therefore , 2 = d/75 + d/105
LCM of 75 and 105 = 525

2= d/75 + d/105
=> 2 = 7d/525 + 5d/525
=> d = 525 / 6 Km

Therefore, T1= d/75
=> T1 = 525 / (6 x 75)
=> T1 = (7 x 60) / 6 -- in minutes
=> T1 = 70 minutes.
Intern
Joined: 21 Feb 2016
Posts: 9
Location: United States (MA)
Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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30 Jun 2016, 11:53
Cole drove from home to work at an average speed of 75 kmh. He then returned home at an average speed of 105 kmh. If the round trip took a total of 2 hours, how many minutes did it take Cole to drive to work?
A) 66
B) 70
C) 72
D) 75
E) 78

Round-trip average speed: (2*75*105)/(75+105)=525/6
Round-trip time=2 hrs given
So round-trip distance=(526/6)*2=175 kms
Single-trip distance=175/2 kms...Time taken to cover this distance at 75kmph= (175/2)/75 hrs=70 mins.
Manager
Joined: 30 Apr 2013
Posts: 75
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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03 Oct 2017, 01:55
Time = distance /speed

So Lets consider D =100

Therefore , 100/75(60) +100/105(50)= 120 mins

Is the right way of thinking ?

Can someone please explain this with logic not formulas
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Joined: 22 May 2015
Posts: 126
Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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03 Oct 2017, 05:22
santro789 wrote:
Time = distance /speed

So Lets consider D =100

Therefore , 100/75(60) +100/105(50)= 120 mins

Is the right way of thinking ?

Can someone please explain this with logic not formulas

It is incorrect to assume the value of distance D as we have a constrain on time ( t1+t2 = 2 hours ).

We know the avg speed for first half is 75kmph => D = 75 * T1
We also know the avg speed for second half is 105kmph => D = 105*T2

Since the distance is same we can equate the above equations 75*T1 = 105*T2 => 5T1=7T2.
Now we know the total time taken is 2 hours => T1+T2 = 2
On solving we will get T1 = 7/6 hours which is 70 minutes.
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Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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05 Oct 2017, 10:12
1
skylimit wrote:
Cole drove from home to work at an average speed of 75 kmh. He then returned home at an average speed of 105 kmh. If the round trip took a total of 2 hours, how many minutes did it take Cole to drive to work?
A) 66
B) 70
C) 72
D) 75
E) 78

We use the formula distance = rate x time, or equivalently, time = distance/rate. We can let the distance either way = d; thus, Cole’s time from home to work is d/75 and his time from work to home is d/105. The total travel time is given as 2 hours, so we have:

d/75 + d/105 = 2

Multiplying the entire equation by 525, we have:

7d + 5d = 1050

12d = 1050

d = 1050/12 = 525/6 = 175/2 km

Thus, it took him (175/2)/75 = 175/150 = 7/6 hours = 7/6 x 60 = 70 minutes.

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Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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Updated on: 11 Feb 2018, 20:29
Thanks

Originally posted by Buttercup3 on 05 Oct 2017, 17:19.
Last edited by Buttercup3 on 11 Feb 2018, 20:29, edited 1 time in total.
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Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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05 Oct 2017, 19:03
1
skylimit wrote:
Cole drove from home to work at an average speed of 75 kmh. He then returned home at an average speed of 105 kmh. If the round trip took a total of 2 hours, how many minutes did it take Cole to drive to work?
A) 66
B) 70
C) 72
D) 75
E) 78

Source: GMAT Prep Now - http://www.gmatprepnow.com/module/gmat- ... /video/914

I avoid division if I can in RTD problems. I solved for time with multiplication. It's fast.

Total time = 2 hours

Let time from home to work = x
So time from work to home = (2-x)

Rate, home to work = 75 kmh
Rate, work to home = 105 kmh

$$r*t=D$$

D, home to work: (75*x)
D, work to home: (105)*(2-x)

$$D$$ is equal both ways, hence

$$75x = (105)(2-x)$$
$$75x = 210 - 105x$$
$$180x = 210$$

$$x = \frac{210}{180} = \frac{21}{18} =\frac{7}{6}hrs$$

Any fraction of an hour can be multiplied by 60 to obtain minutes.

$$x =\frac{7}{6}hrs * 60$$ = 70 minutes

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Re: Cole drove from home to work at an average speed of 75 kmh. He then  [#permalink]

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06 Mar 2019, 01:48
skylimit wrote:
Cole drove from home to work at an average speed of 75 kmh. He then returned home at an average speed of 105 kmh. If the round trip took a total of 2 hours, how many minutes did it take Cole to drive to work?
A) 66
B) 70
C) 72
D) 75
E) 78

Source: GMAT Prep Now - http://www.gmatprepnow.com/module/gmat- ... /video/914

Speed from home to work = 75 kmph

Say time spent = t hours

Therefore distance travelled = 75t kms

Speed from work to home = 105 kmph

Time spent = (2 - t) hours {as total time is 2 hours)

Distance travelled = 105 (t - 2) = 105t - 210 kms

As both distances are same, we can equate distances:

105t - 210 = 75t

t = 210/180 = 7/6

t = 7/6 * 60

t = 70 minutes

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Re: Cole drove from home to work at an average speed of 75 kmh. He then   [#permalink] 06 Mar 2019, 01:48
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# Cole drove from home to work at an average speed of 75 kmh. He then

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