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# A car travelled the first quarter of a certain distance at

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Director
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A car travelled the first quarter of a certain distance at  [#permalink]

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18 Mar 2013, 07:52
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8
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Difficulty:

45% (medium)

Question Stats:

66% (02:02) correct 34% (02:00) wrong based on 277 sessions

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A car travelled the first quarter of a certain distance at twice the speed it traveled the remaining distance. What proportion of the total time travelled, was the time taken to travel the first quarter of the distance?

A. 1/4
B. 1/5
C. 1/6
D. 1/7
E. 1/8

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Re: A car travelled the first quarter of a certain distance at  [#permalink]

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18 Mar 2013, 08:26
SravnaTestPrep wrote:
A car travelled the first quarter of a certain distance at twice the speed it traveled the remaining distance. What proportion of the total time travelled, was the time taken to travel the first quarter of the distance?

A. 1/4
B. 1/5
C. 1/6
D. 1/7
E. 1/8

These problems can be solved through algebra or sly number picking. Being a big fan of solving problems with numbers, let's pick a total distance divisible by 4 (say 40) so we can break it up into quarters, and a speed that can easily be halfed, say 10. Each quarter is thus 10 kilometers (or miles or feet or angstroms for all it matters), and the runner's speed is 10 km/h for the first quarter and 5 km/h for the remaining quarters.

He'll take 1 hour to do the first quarter and then 2 hours for the second quarter, 2 hours for the third and 2 hours for the fourth. On total he will take 7 hours to complete this race, of which 1 hour was spent on the first quarter. So 1/7. Answer D.

Some people prefer algebra, but picking the correct numbers can not only get you the right answer but make the problem come alive instead of simply plugging into a formula. Ideally, you should be fairly comfortable with either approach.

Hope this helps!
-Ron
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Re: A car travelled the first quarter of a certain distance at  [#permalink]

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18 Mar 2013, 08:59
We can use some "real numbers" to solve this.

Tot Distance = 4 km.
Speed = 1 km/h

First Quarter = 1 km.
2xSpeed = 2 km/h.

Total Time = $$(1km)/(2 km/h) + (3km)/(1km/h) = 3.5 h$$
First Quarter Time = $$(1km)/(2 km/h) = 0.5 h$$

$$FQT/TOT = 0.5/3.5 = 1/7$$
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Re: A car travelled the first quarter of a certain distance at  [#permalink]

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19 Mar 2013, 03:08
SravnaTestPrep wrote:
A car travelled the first quarter of a certain distance at twice the speed it traveled the remaining distance. What proportion of the total time travelled, was the time taken to travel the first quarter of the distance?

A. 1/4
B. 1/5
C. 1/6
D. 1/7
E. 1/8

Solution:

1.$$S1, T1, D1$$ be the speed, time of travel and the distance in the first quarter. $$S2, T2$$ be the speed and time of travel in the remaining distance and$$D2$$ be the remaining distance.

2. We need to find $$T1/ (T1 + T2)$$ = $$(D1/S1) / (D1/S1 + D2/S2)$$

3. $$S1= 2S2$$, $$D2=3D1$$

4. Substituting (3) in RHS of (2), the RHS becomes: $$(D1/2S2) / (D1/2S2 + 3D1/S2)$$

$$= (D1/2S2) / (7D1/ 2S2)$$
$$=1/7$$

Therefore the answer is choice D.
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Re: A car travelled the first quarter of a certain distance at  [#permalink]

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19 Mar 2013, 04:02
An excellent rates problem. Thanks SravnaTestPrep.

Alright, let's solve it

The first notion you've got to keep in mind is : distance (d) = speed (S) x time (t)

The total distance can be represented as follows :
Attachment:

PS1.jpg [ 26.06 KiB | Viewed 2756 times ]

So taking into account the fact that the car travelled d1 at twice the speed it travelled d2, ergo S1 = 2 x S2 we get :

The speed used to travel d1 is : $$S1 = \frac{d1}{t1} = \frac{d}{(4*t1)}$$
The speed used to travel d2 is : $$S2 = \frac{d2}{t2}= \frac{(3*d)}{(4*t2)}$$

Therefore, by using S1 = 2 x S2 we get :
$$\frac{d}{(4*t1)} = 2*\frac{(3*d)}{(4*t2)}$$

Simplifying the "d"s and dividing 4 by 2 yields :
$$\frac{1}{(4*t1)} = \frac{3}{(2*t2)}$$

Which gives us :
$$2*t2 = 12*t1$$ which is equivalent to $$t2 = 6*t1$$

And since we're interested in the proportion of the total time travelled necessary to travel the first quarter of the distance then :

Knowing that the total time is : $$t = t1 + t2$$

We substitute t2 in the expression which gives us : $$t = t1 + 6*t1$$

Meaning that : $$t = 7*t1$$ or $$t1 =\frac{t}{7}$$ which is answer choice D.

Hope that helped
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Re: A car travelled the first quarter of a certain distance at  [#permalink]

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30 Sep 2014, 21:47
1
Lets say 1st quarter distance = $$\frac{1}{4}$$ & speed = 2s

So, remaining distance = $$\frac{3}{4}$$ & speed = s

Time required in First Quarter$$= \frac{\frac{1}{4}}{2s} = \frac{1}{8s}$$

Total Time required for travelling complete distance$$= \frac{1}{8s} + \frac{3}{4s} = \frac{7}{8s}$$

$$Fraction = \frac{\frac{1}{8s}}{\frac{7}{8s}} = \frac{1}{7}$$

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Re: A car travelled the first quarter of a certain distance at  [#permalink]

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24 Oct 2014, 10:54
Substitution method-

let distance be 120 . So, quarter of 120= 30 km.

Let speed at which 120-30 = 90 km traveled be 10 kmph.. So, speed at which quarter portion traveled be twice of 10 = 20 kmph .

Now,
time taken to travel quarter distance = 30/20 = 3/2 hours.

time taken to travel rest of distance = 90/10 = 9 hours.

Hence, total time = 9 + 3/2 = 21/2

therefore, time required to travel quarter distance / total time

= 3/2 / 21/2

= 1/7

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Re: A car travelled the first quarter of a certain distance at  [#permalink]

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28 Feb 2016, 11:06
distance first quarter = 0.25d
speed first quarter = 2x.

distance remaining part = 0.75d
speed remaining part = x.

time first quarter is d/8x. time for the remaining part 3d/4x

now, total time: 7d/8x.

since we are asked the ratio of the first portion to the total time: d/8x * 8x/7d => 1/7
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28 Feb 2016, 17:59
let d=first quarter distance
2s=first quarter speed
(d/2s)+(3d/s)=7d/2s total time
(d/2s)/(7d/2s)=1/7 ratio of first quarter time to total time
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28 Jul 2018, 04:03
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