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# During a trip, Francine traveled x percent of the total distance at an

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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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24 Aug 2016, 11:08
2
Mariwa wrote:
Why can't I apply the simple average method here?

Average Speed=40*X/100+60*(100-X)/100

The average speed is always the total distance, divided by the total time.

When you multiplied 40*x and 60*(100-x), that actually doesn't make sense, mathematically. You're multiplying a rate (40 miles) by a percent of distance (x% of the total distance). That doesn't give you anything meaningful. Always check before you do the math: what am I taking a percent of?

Here's the version of your formula that would work:

total distance = d
total time for first half of trip = 40/(x/100 * d)
total time for second half of trip = 60/((100-x)/100*d)

average speed = total distance/(total time for first half + total time for second half)

If you try writing that out, though, you'll see why picking numbers is actually the right choice here!
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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24 Aug 2016, 11:27
1
Mariwa wrote:
ccooley wrote:
Mariwa wrote:
Why can't I apply the simple average method here?

Average Speed=40*X/100+60*(100-X)/100

The average speed is always the total distance, divided by the total time.

When you multiplied 40*x and 60*(100-x), that actually doesn't make sense, mathematically. You're multiplying a rate (40 miles) by a percent of distance (x% of the total distance). That doesn't give you anything meaningful. Always check before you do the math: what am I taking a percent of?

Here's the version of your formula that would work:

total distance = d
total time for first half of trip = 40/(x/100 * d)
total time for second half of trip = 60/((100-x)/100*d)

average speed = total distance/(total time for first half + total time for second half)

If you try writing that out, though, you'll see why picking numbers is actually the right choice here!

I got it thanks!

This is applicable to all problems concerning rates? No only the ones about speed right?

Any time a problem says 'average speed' or 'average rate', start by writing out 'total work' (or 'total distance') / 'total time' on your paper. Then, just fill in those two values as you work them out! That said, I can't remember ever seeing an official GMAT problem that asked for an average rate, rather than an average speed. But there's always a chance, and in that case, you'd do it exactly like this problem.
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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31 Oct 2016, 03:04
Can we solve it by assuming Francine travelled 100% of the distance at 40 - However, when i am plugging in the assumption, several options are yielding correct result. Where is it going wrong?

Thanks
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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31 Oct 2016, 06:12
WilDThiNg wrote:
Can we solve it by assuming Francine travelled 100% of the distance at 40 - However, when i am plugging in the assumption, several options are yielding correct result. Where is it going wrong?

Thanks

Yes, you can plug in values like that. If multiple options give you the correct answer, try some other values on the shortlisted options only.

For example, next you should try x = 0 (all distance travelled at 60 only and hence avg is 60). You get options (C) and (E).

Next try x = 50
Avg Speed = 2*40*60/(40+60) = 48
Out of options (C) and (E), only option (E) gives 48.
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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27 Feb 2017, 04:04
Shortcut: x =0 answer should be 60 and x =100 answer should be 40. After checking the option we can see that Option C and E could be the answer.
Now avg speed = 1/(d1/v1 + d2/v2). Since distance part is coming in denominator, x will be in the denominator. Hence Option E is the answer
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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29 May 2017, 01:26
vksunder wrote:
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)

1. Since the distance traveled is given in terms of percent, assume the total distance traveled as 100 units.
2. A distance of x at 40 miles per hour and a distance of 100-x at 60 miles per hour
3. Average speed = 100 / total time taken
4. Total time taken = x/40 + (100-x)/60
5. Substituting (4) in (3) we get, Average speed= 12,000/(x+200)
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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29 Jun 2017, 09:38
vksunder wrote:
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)

Though I tried to solve it by formula.. But really if we try to apply our mind, we can easily solve the problem without solving using complex x and y ...

So the solution goes like this..
x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour

x% with 40 miles/hr
(100-x) with 60 miles/hr

So, if we assume x as 0% i.e. whole distance is traveled with speed of 60 miles/hr.
Now start checking options .
A. 90
B. 15
C. 60 (May be correct).. Always check all options when we are solving through options and assuming the values.
D. 600/115
E. 60

So we can eliminate A,B,D.

So C or E is correct.

Lets assume x as 100% i.e. whole distance is traveled with 40 miles/hr.
Now start checking option C and E.
C. 200/5 = 40
E. 12000/300 = 40

So this was wrong assumption..

Lets assume x= 50% i.e. the average speed = 2x40x60/(40+60) = 48 miles/hr

Lets check the option C & E again
C. 250/5 = 50 miles /hr (It is wrong )
E. 12000/250 = 48 miles/hr (correct option )

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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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16 Jul 2017, 16:37
Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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17 Jul 2017, 05:07
1
suntorytea wrote:
Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.

How do you get 12000/(x + 2)?
You don't need to assume a value for d.

Average Speed = Total Distance / Total Time

Total Distance = d

Time 1 $$= \frac{D1}{S1} = \frac{(\frac{x}{100})*d}{40} = \frac{xd}{4000}$$

Time 2 $$= \frac{D2}{S2} = \frac{(1 - x/100)*d}{60} = \frac{(100 - x)d}{6000}$$

Average Speed $$= \frac{d}{\frac{xd}{4000} + \frac{(100-x)d}{6000}}$$

We take d common from the denominator and it cancels out with the d in the numerator

Average Speed $$= \frac{d}{d * (\frac{x}{4000} + \frac{(100-x)}{6000})}$$

Average Speed $$= \frac{1}{(\frac{x}{4000} + \frac{(100-x)}{6000})}$$

Average Speed $$= \frac{12000}{x + 200}$$
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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24 Aug 2017, 12:17
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vksunder wrote:
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. $$\frac{(180-x)}{2}$$

B. $$\frac{(x+60)}{4}$$

C. $$\frac{(300-x)}{5}$$

D. $$\frac{600}{(115-x)}$$

E. $$\frac{12,000}{(x+200)}$$

I like to begin with a word equation:
Average speed = (total distance)/(total time)
For this question, let's let the total distance = D

Next, observe that: total time = (time spent driving 40 mph) + (time spent driving 60 mph)

time spent driving 40 mph = distance/speed
Aside: distance driven = (x/100)(D)
So, time spent driving 40 mph = (x/100)(D)/40

time spent driving 60 mph = distance/speed
Aside: if x% of the distance was driven at 40 mph, then the distance driven at 60 mph = [(100-x)/100](D)
So, time spent driving 60 mph = [(100-x)/100](D)/60

Here comes the awful algebra ...

Total time = (x/100)(D)/40 + [(100-x)/100](D)/60

Simplify ...
Total time = xD/4000 + [100D-xD]/6000
Total time = 3xD/12000 + [200D-2xD]/12000
Total time = (xD+200D)/12000

And finally,
Average speed = (total distance)/(total time)
= D/[(xD+200D)/12000]
= (12000D)/(xD+200D)
= (12000)/(x+200)

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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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09 Sep 2017, 18:56
1
vksunder wrote:
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. $$\frac{(180-x)}{2}$$

B. $$\frac{(x+60)}{4}$$

C. $$\frac{(300-x)}{5}$$

D. $$\frac{600}{(115-x)}$$

E. $$\frac{12,000}{(x+200)}$$

Check solution attached
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During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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28 Dec 2017, 06:24
1
Bunuel, niks18, amanvermagmat

Quote:
distance $$d$$ will cancel out from the equation (edited the earlier post to clear this). So we can assume distance to be some number. I chose 40 as it's easy for calculation.

I hope you meant d as TOTAL DISTANCE

Quote:
Timed needed for the rest of the trip: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}$$;

For above, I am getting:

$$t_2= \frac{distance_2}{speed_2}=\frac{d-xd}{60*100}$$;

To be more precise, see VeritasPrepKarishma solution above:
x% of d = xd/100

then complimentary distance will be
(1-x)d/100 or (d-xd)/100

Can you advise what mistake I am making?
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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28 Dec 2017, 11:17
Bunuel, niks18, amanvermagmat

Quote:
distance $$d$$ will cancel out from the equation (edited the earlier post to clear this). So we can assume distance to be some number. I chose 40 as it's easy for calculation.

I hope you meant d as TOTAL DISTANCE

Quote:
Timed needed for the rest of the trip: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}$$;

For above, I am getting:

$$t_2= \frac{distance_2}{speed_2}=\frac{d-xd}{60*100}$$;

To be more precise, see VeritasPrepKarishma solution above:
x% of d = xd/100

then complimentary distance will be
(1-x)d/100 or (d-xd)/100
Can you advise what mistake I am making?

i think you are making simple calculation errors. I would suggest write down your approach and try to get the answer. For the moment forget about other's method and focus on you approach and calculation steps.

d-dx/100 = (100d-dx)/100 =>(100-x)d/100. I am not sure how you got the highlighted portion
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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29 Dec 2017, 09:13
Bunuel wrote:
hardnstrong wrote:
Is there any clear way of getting correct answer without plugging in different numbers ????????????????

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?

A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

Algebraic approach:
$$Average \ speed=\frac{distance}{total \ time}$$, let's assume $$distance=40$$ (distance $$d$$ will cancel out from the equation, so we can assume distance to be some number.) so we should calculate total time.

Francine traveled $$x$$ percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: $$t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*40}{40}=\frac{x}{100}$$;

Timed needed for the rest of the trip: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}$$;

$$Total \ time=t_1+t_2=\frac{x}{100}+\frac{100-x}{150}=\frac{x+200}{300}$$;

$$Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}$$.

Hello Bunuel,

here is my solution:

let the first part of distance be X1
Speed = 40 mph
Time = t1
the first part distance done --> x1= 40*t1
----> T1 = x1/40

the rest distance let be x2
Spped = 60
Time = t2
the rest part of distance done -->x2= 60*t2
T2 = x2/60

now total distance / total time = average speed

40t+60t / x/40+x/60 = 12000t/5 <---- this is the answer i got

Can you please explain what is wrong in my solution ? Isnt my logic correct? I always have hard time solving such kind of questons with many variables.

Thank you!
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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29 Dec 2017, 10:57
dave13 wrote:
Bunuel wrote:
hardnstrong wrote:
Is there any clear way of getting correct answer without plugging in different numbers ????????????????

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?

A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

Algebraic approach:
$$Average \ speed=\frac{distance}{total \ time}$$, let's assume $$distance=40$$ (distance $$d$$ will cancel out from the equation, so we can assume distance to be some number.) so we should calculate total time.

Francine traveled $$x$$ percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: $$t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*40}{40}=\frac{x}{100}$$;

Timed needed for the rest of the trip: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}$$;

$$Total \ time=t_1+t_2=\frac{x}{100}+\frac{100-x}{150}=\frac{x+200}{300}$$;

$$Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}$$.

Hello Bunuel,

here is my solution:

let the first part of distance be X1
Speed = 40 mph
Time = t1
the first part distance done --> x1= 40*t1
----> T1 = x1/40

the rest distance let be x2
Spped = 60
Time = t2
the rest part of distance done -->x2= 60*t2
T2 = x2/60

now total distance / total time = average speed

40t+60t / x/40+x/60 = 12000t/5 <---- this is the answer i got

Can you please explain what is wrong in my solution ? Isnt my logic correct? I always have hard time solving such kind of questons with many variables.

Thank you!

hi dave13

you have used 4 different variable in you solution t1, t2, x1 & x2. Finally while calculating you are using only 2 variable t & x ( see the highlighted part) which is wrong.
you need to have another equation connecting all your variables.
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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29 Dec 2017, 11:25
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2
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During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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29 Dec 2017, 11:33
dave13 wrote:
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2

Hi dave13

what you are assuming here is t1=t2=t. is this really the case?

do you know the values of t1, t2 or t. no

as per your equation 40t1+60t2/x1/40+x2/60, if you solve this you should get

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2)

how are you going to proceed further from here?
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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30 Dec 2017, 07:40
niks18 wrote:
dave13 wrote:
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2

Hi dave13

what you are assuming here is t1=t2=t. is this really the case?

do you know the values of t1, t2 or t. no

as per your equation 40t1+60t2/x1/40+x2/60, if you solve this you should get

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2)

how are you going to proceed further from here?

Hi niks18, and Bunuel

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2) a minor correcton

I solved denominator seperatly but the point is that i didnt indicate x as x1 and x2 and t as t1 and t2 - was it a must ?

so i did this 3x+2x/120 = 5x/120 is total total time

than numerotor calculated 40t+60t = 100t is a total distance

average speed = total distance / total time

100t / (5x/120) = 100t/1 *120/5x = 12000t/5x this is how i solved it

you are saying i need additional equation for linking variables, but i seperatly expressed time and distance for one part and seperatly expressed time and distance for second part

so i i linked all these variables through formula of finding average speed = total distance divided by total time.

if so which additional equation am i missing and why ?

many thanks!
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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30 Dec 2017, 12:02
1
dave13 wrote:
niks18 wrote:
dave13 wrote:
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2

Hi dave13

what you are assuming here is t1=t2=t. is this really the case?

do you know the values of t1, t2 or t. no

as per your equation 40t1+60t2/x1/40+x2/60, if you solve this you should get

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2)

how are you going to proceed further from here?

Hi niks18, and Bunuel

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2) a minor correcton

I solved denominator seperatly but the point is that i didnt indicate x as x1 and x2 and t as t1 and t2 - was it a must ?

so i did this 3x+2x/120 = 5x/120 is total total time

than numerotor calculated 40t+60t = 100t is a total distance

average speed = total distance / total time

100t / (5x/120) = 100t/1 *120/5x = 12000t/5x this is how i solved it

you are saying i need additional equation for linking variables, but i seperatly expressed time and distance for one part and seperatly expressed time and distance for second part

so i i linked all these variables through formula of finding average speed = total distance divided by total time.

if so which additional equation am i missing and why ?

many thanks!

hi dave13
what you are doing is really confusing . Nonetheless i'll take a final shot in explaining the calculation to you using the variables you used.

distance traveled in the first part $$=x_1$$

so time taken to travel first part $$t_1=\frac{x_1}{40}$$

distance traveled in the second part $$=x_2$$

so time taken to travel second part $$t_2=\frac{x_2}{60}$$

Hence total time taken to complete the journey $$= t_1+t_2=\frac{x_1}{40}+\frac{x_2}{60}=\frac{3x_1+2x_2}{120}$$------------(1)

Now let the total distance traveled be $$d$$.

given distance traveled in first part $$x_1= x$$% of $$d=\frac{xd}{100}$$

so distance traveled during the second part $$x_2= d-\frac{xd}{100}=\frac{d(100-x)}{100}$$

Now substitute the value of $$x_1$$ & $$x_2$$ in equation (1) to get total time $$= \frac{\frac{3xd}{100}+\frac{2d(100-x)}{100}}{120}$$

=> total time $$= \frac{d(x+200)}{12000}$$

therefore average speed $$= \frac{total distance}{total time}$$ $$=\frac{d}{\frac{d(x+200)}{12000}}=\frac{12000}{(x+200)}$$
Re: During a trip, Francine traveled x percent of the total distance at an &nbs [#permalink] 30 Dec 2017, 12:02

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