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Otto and Han are driving at constant speeds in opposite directions on

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Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post Updated on: 04 Aug 2015, 10:59
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Otto and Han are driving at constant speeds in opposite directions on a straight highway. At a certain time they are driving toward each other and are 60 miles apart. One and a half hours later, they are again 60 miles apart, driving away from each other. If Otto drives at a speed of x miles per hour, then, in terms of x, Han drives at a speed of how many miles per hour?

a) 80-x
b) 40-x
c) 80-2x
d) 120-x
e) 40-x/2

SOURCE: ECONOMIST GMAT

Originally posted by manpreetsingh86 on 08 Jan 2015, 08:51.
Last edited by reto on 04 Aug 2015, 10:59, edited 1 time in total.
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Re: Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post 08 Jan 2015, 11:27
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manpreetsingh86 wrote:
Otto and Han are driving at constant speeds in opposite directions on a straight highway. At a certain time they are driving toward each other and are 60 miles apart. One and a half hours later, they are again 60 miles apart, driving away from each other. If Otto drives at a speed of x miles per hour, then, in terms of x, Han drives at a speed of how many miles per hour?

a) 80-x
b) 40-x
c) 80-2x
d) 120-x
e) 40-x/2

Dear manpreetsingh86

I'm happy to respond. You may find the discussion of gaps in this blog article relevant:
http://magoosh.com/gmat/2014/gmat-pract ... on-motion/

Let's say the two cars have speeds V1 and V2. The fact that they are moving in opposite direction means that their relative speed is (V1 + V2). In other words, any gap between them will be changing in size at a rate of (V1 + V2). It doesn't matter whether they are moving toward each other or away from each other. If they are approaching each other, the gap between them is decreasing at a rate of (V1 + V2). If they are moving away from each other, the gap between them is increasing at a rate of (V1 + V2). Either way, the number for the rate of change remains the same.

Here, the two cars approach a distance 60 mi, then move away from each other another distance of 60 miles. That's a total distance of 120 miles in 1.5 hr, which gives a rate of:
R = (120 mi)/(1.5) = 80 mph
That's the rate of change of the gap, so it must equal the sum of the speeds of the two cars.

One of the speeds is x, and let's call the other y. We want y.
x + y = 80
y = 80 - x

Answer = (A)

Does all this make sense?
Mike :-)
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Re: Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post 08 Jan 2015, 20:44
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Answer = a) 80-x

Let the speed of Han = y mph

Speed of Otto = x mph

They are moving towards each other, so relative speed = (x+y) mph

Distance travelled = 60 + 60 = 120

Setting up the time equation:

\(\frac{Total distance}{Relative speed} = Time Elapsed\)

\(\frac{120}{x+y} = 1.5\)

x+y = 80

y = 80-x
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Re: Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post 09 Jan 2015, 03:12
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let speed of otto =x, and speed of han = h
Let's breakdown this question into three different segments, namely t0,t1, and t2. now, lets see what's happening in the each segment.

Image

As can be seen in the above pic, the initial distance between otto and han is 60 km. After time t=t0, they start moving towards other at their respective rate of x and h respectively.

now suppose at time t=t1, they meet each other at point z.
Image

Therefore total distance traveled by otto and han, when they met at point z is equal to 60.
thus, t1 = \(\frac{60}{(x+h)}\) ----------------------------------------1)

Now, from point z, they again start moving in their respective directions at their constant rates, until the distance between them becomes 60.

so, at time t=t2, distance between otto and han becomes 60.
Image

thus, t2 = \(\frac{60}{(x+h)}\) -------------------------------------------2)

also, as per the question t1+t2=(3/2) hours

now, put the value of t1 and t2 from 1 and 2, we have

\(\frac{60}{(x+h)}\) + \(\frac{60}{(x+h)}\) = 3/2

x+h =80
or h=80-x
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Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post 21 Nov 2015, 15:38
let h=Han's speed
120/(x+h)=3/2
h=80-x
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Re: Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post 06 Aug 2018, 22:43
Between them, Otto and Han cover 120 miles in 1.5 hours. Let 'd' be the distance covered by Otto so the distance covered by Han is (120-d). Let 'y' mph be Han's sped. Thus we have:

x=d/1.5 and y=(120-d)/1.5 or y=80 - (d/1.5) or y=80 -x. Ans. (A)
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Re: Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post 06 Aug 2018, 23:05
Let the speed of Han be y miles/hour

Speed of Otto is given as x miles/hour


They both are approaching towards each other with relative speed of (x+y) miles/hour

They covered 120 miles in 3/2 hours

Therefore ,

(120/(x+y))=(3/2)

Time = distance/speed

After solving we get y = 80-x

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Re: Otto and Han are driving at constant speeds in opposite directions on  [#permalink]

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New post 07 Aug 2018, 00:39
manpreetsingh86 wrote:
Otto and Han are driving at constant speeds in opposite directions on a straight highway. At a certain time they are driving toward each other and are 60 miles apart. One and a half hours later, they are again 60 miles apart, driving away from each other. If Otto drives at a speed of x miles per hour, then, in terms of x, Han drives at a speed of how many miles per hour?

a) 80-x
b) 40-x
c) 80-2x
d) 120-x
e) 40-x/2

SOURCE: ECONOMIST GMAT


t= 1.5 hours
distance = from 60 towards to 60 past = 120 miles
speed = 1200/15=80
which is the relative speed
since they both are travelling towards each other
80=O + H
where O and H are speeds of otto and Han

if otto = x
substitute the value
then Han = 80-x
Re: Otto and Han are driving at constant speeds in opposite directions on &nbs [#permalink] 07 Aug 2018, 00:39
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