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# Word Problem - James started from his home and drove eastwards at a co

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Word Problem - James started from his home and drove eastwards at a co [#permalink]

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22 Jan 2017, 01:26
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95% (hard)

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45% (04:01) correct 55% (03:09) wrong based on 129 sessions

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Q.)
James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?

A. 25%
B. 33%
C. 50%
D. 67%
E. 75%

Thanks,
Saquib
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Word Problem - James started from his home and drove eastwards at a co [#permalink]

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Updated on: 27 Mar 2017, 05:51
The official solution has been posted. Looking forward to a healthy discussion..
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Originally posted by EgmatQuantExpert on 22 Jan 2017, 01:26.
Last edited by EgmatQuantExpert on 27 Mar 2017, 05:51, edited 1 time in total.
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Re: Word Problem - James started from his home and drove eastwards at a co [#permalink]

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22 Jan 2017, 03:21
Well i seriously think we dont expect such questions in GMAT. Although we should be prepared for the worst, but still one might not expect such questions on GMAT( my own perception)

As for the explanation. After James(J) travels for 90 min of 1.5 hours Patrick(P) starts and in 90 minutes he crosses J. So P covers a distance in 90 minutes what J covers in 180 minutes. So speed of P is double that of J. Let speed of J be x km/ph and speed of P be 2x km/hr. Then they continue to travel for 2 hours. So in this 2 hours P travels a distance of 4x( 2x x 2) and J travels a distance of 2x. Distance between them will be 2x(4x - 2x). Now in order to catch P in 8 hours, J must travel faster than P. Let new slower speed of P be "p". so the eqaution now becomes (2x/x-p) = 8.

x-p will be relative speed
2x distance between them
8 hours as given in the question that J must take.

So value of new speed of P will be 3/4x

so he has to reduce the speed from 2x to 3/4 x which is around 62.5 not in the options
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Word Problem - James started from his home and drove eastwards at a co [#permalink]

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22 Jan 2017, 03:58
4
EgmatQuantExpert wrote:
Q.)
James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?

A. 25%
B. 33%
C. 50%
D. 67%
E. 75%

Thanks,
Saquib
Quant Expert
e-GMAT
]

Hi
Since the distance covered by James in (90+90) minutes is same as Patrick in 90 minutes, speed of P is TWICE that of J..

Now it will be easier to give a number to each..
Let J's speed be 1kph, so P's speed will be 2kph....

We are looking for both to meet after 8 hour..
So in 8 HR , J will travel at constant rate @1kph= 8*1=8km..
P travels at 2kph for 2 hrs so travels 4km...
Remaining Kms=8-4=4...
So Phase to travel 4km in remaining 6hr...
So speed becomes 4/6=2/3kph...
Reduction=2-2/3=4/3
%=(4/3)/2*100=4/6*100=200/3=66.667%

EgmatQuantExpert pl correct your choices as you are asking for "meeting in EXACT 8 hrs"..
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Word Problem - James started from his home and drove eastwards at a co [#permalink]

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22 Jan 2017, 09:58
James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?
A. 25%
B. 33%
C. 50%
D. 67%
E. 75%

let p=P's speed
p/2=J's speed
x=fraction of p driven for last 6 hours
in the 8 hours after overtaking J,
P drives 2p+6xp miles
while J drives 8*(p/2) miles
2p+6xp=4p
x=1/3
P should reduce his speed by 2/3=67%
D
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Word Problem - James started from his home and drove eastwards at a co [#permalink]

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22 Jan 2017, 12:57
varundixitmro2512 wrote:
Well i seriously think we dont expect such questions in GMAT. Although we should be prepared for the worst, but still one might not expect such questions on GMAT( my own perception)

As for the explanation. After James(J) travels for 90 min of 1.5 hours Patrick(P) starts and in 90 minutes he crosses J. So P covers a distance in 90 minutes what J covers in 180 minutes. So speed of P is double that of J. Let speed of J be x km/ph and speed of P be 2x km/hr. Then they continue to travel for 2 hours. So in this 2 hours P travels a distance of 4x( 2x x 2) and J travels a distance of 2x. Distance between them will be 2x(4x - 2x). Now in order to catch P in 8 hours, J must travel faster than P. Let new slower speed of P be "p". so the eqaution now becomes (2x/x-p) = 8.

x-p will be relative speed
2x distance between them
8 hours as given in the question that J must take.

So value of new speed of P will be 3/4x

so he has to reduce the speed from 2x to 3/4 x which is around 62.5 not in the options

Hey,

I agree with your point that we should be prepared for the worst in GMAT and hence this question.

The main idea of this question was to see if the students can recognise the pitfall in the question. A few of them missed it and made a mistake.

Would you like to revisit your solution and focus on the last line of the question -

"By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?"

Is the solution correct or will we get a different answer?

Thanks,
Saquib
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Re: Word Problem - James started from his home and drove eastwards at a co [#permalink]

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22 Jan 2017, 13:03
chetan2u wrote:
EgmatQuantExpert wrote:
Q.)
James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?

A. 25%
B. 33%
C. 50%
D. 67%
E. 75%

Thanks,
Saquib
Quant Expert
e-GMAT

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Hi
Since the distance covered by James in (90+90) minutes is same as Patrick in 90 minutes, speed of P is TWICE that of J..

Now it will be easier to give a number to each..
Let J's speed be 1kph, so P's speed will be 2kph....

We are looking for both to meet after 2+8 hour..
So in 10 HR , J will travel at constant rate @1kph= 10*1=10km..
P travels at 2kph for 2 hrs so travels 4km...
Remaining Kms=10-4=6...
So Phase to travel 6km in remaining 10hr...
So speed becomes 6/8=3/4 kph...
Reduction=2-3/4=5/4
%=(5/4)/2*100=5/8 *100=500/8=62.5%

EgmatQuantExpert pl correct your choices as you are asking for "meeting in EXACT 8 hrs"..

Hey,

I agree they are meeting in exactly 8 hours, but then the last line clearly states - "in exactly 8 hours after Patrick overtook James?"

In the given solution presented here, you have considered 8+2 hours, which is incorrect.

The percentages have been rounded off in the options. But then if you find the correct solution, it will be easy to figure out the correct option.

Thanks,
Saquib
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Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line

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Re: Word Problem - James started from his home and drove eastwards at a co [#permalink]

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27 Mar 2017, 05:51
2

Solution

Given

• Patrick started 90 minutes = 1.5 hours after James started his journey
o So, James would have travelled for 1.5 hours when Patrick started his journey

• Patrick overtook James exactly after 1. 5 hours of Patrick starting his journey
o So, by the time Patrick overtook James, James would have travelled for 1.5 + 1.5 = 3 hours
• Patrick drove at the same speed for another 2 hours after overtaking James and then reduces his speed
o So, by that time James would have travelled a total of 3 + 2 = 5 hours
o Also, let Patrick reduce his speed by x%
o So, patrick’s reduced speed = P – x% of P kilometres per hour
• James has to catch up with Patrick exactly 8 hours after Patrick overtook James
o As Patrick overtook James when James had travelled for 3 hours, James would catch up with Patrick when James would have travelled for 3 + 8 = 11 hours

To Find: Value of x in percentage

Approach:

1. We need to find the reduced speed of Patrick in terms of the initial speed of Patrick.
2. Distance = Speed * Time. Since we know the time for which Patrick travelled, if we can find the distance travelled by Patrick, we would be able to find his speed.
3. Now, we can see from the diagram above that at certain points, Patrick and James are together. As they started from the same point, they would have travelled the same distance in reaching those points.
a. So, we would try to find the distance travelled by James and then equate it with distance travelled by Patrick to find the reduced speed of Patrick
4. We know that the speed of James is constant throughout the journey. Hence, the distance covered by him in time t will be J*t kilometres, where J is the constant speed of James. Therefore, the total distance covered by him in 11 hours is 11J.
5. We know that Partick starts 1.5 hrs later than James. He travels for 3.5 hrs with a speed P and then reduces his speed to P’ for the next 6 hours. Therefore, the distance covered by Patrick overall is 3.5P+6P’.
6. We know that there are 2 times in the journey when Patrick and James are together.
a. Time 1: When James has travelled for 3 hours and Patrick has travelled for 1.5 hrs.
i. 3J=1.5P
b. Time 2: At the end of 11 hours after James commenced his Journey.
i. 11J=3.5P+6P’.
7. Using these equations, we can find the relationship between P and P’. Once we find this relationship, we can then find out by how much percent did Patrick reduce his speed, by:
a. $$\mathrm x=\frac{\mathrm P-\mathrm P’}{\mathrm P}\ast100$$

Working Out

1. Total Distance travelled by James = 11J….(1)
2. Total Distance travelled by Patrick = 3.5 P + 6P’, where P’ = P – x% of P….(2)
3. Since the total Distance covered by James = total distance covered by Patrick, we can equate (1) and (2) to get
a. 11J = 3.5 P + 6P’…..(3)
4. Also, as Patrick overtook James at t =3 hours, distance covered by James (till t = 3 hours) = Distance covered by Patrick (till t= 3 hours). So, we have
a. 3J = 1.5 P
b. J = 0.5 P….(4)
5. Substituting (4) in (3), we have
a. 5.5P = 3.5P + 6P’
b. 0.33*P = P’
c. P – 67% of P = P – x% of P
d. So, x = 67%

Correct Answer = 67% = Option D
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Re: Word Problem - James started from his home and drove eastwards at a co [#permalink]

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10 Apr 2017, 07:25
the answer should be D,here is my solution-
let's take J's speed=10 km/h so he has traveled for 3 hrs initially that is 10*3=30km.
now take P's speed =20km/h as we get to know that the time at which they have met is 90 minutes after J has traveled so p has also traveled
1.5*20=30kms.

now P has traveled for the same constant rate for 2 hours after meeting so he has traveled in 3.5 hours = 30+20+20=70 kms. and as J has to catch him in 8hours after first meeting with P, so in reaming 6hrs . P must have reduced his speed, and he has traveled in remaining 6hrs that will be 6*20=80 so total distance covered by P is 70+80=150kms.

now j's speed has been constant during the whole journey so he must have traveled 8*10+30=110kms .
so the difference btw those distance is 40 , now look for option suppose if p had to reduce his speed with 50% then he must have traveled in remaining 6hrs for 60+70=130kms that is still more so take 67%which almost 2/3 so his current speed for 6 hours will be 20*1/3 and he has traveled 20*1/3*6=40 and total distance covered will be 110(70+40) . hence D is the answer

thanks
please kudos if you it helped .
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Re: Word Problem - James started from his home and drove eastwards at a co [#permalink]

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07 Jun 2018, 17:14
EgmatQuantExpert wrote:
Q.)
James started from his home and drove eastwards at a constant speed. Exactly 90 minutes after James stated from his home, his brother Patrick started from the same point and drove in the same direction as James did at a different constant speed. Patrick overtook James exactly 90 minutes after Patrick started his journey and then continued driving at the same speed for another 2 hours. By what percentage should Patrick reduce his speed so that James could catch up with Patrick in exactly 8 hours after Patrick overtook James?

A. 25%
B. 33%
C. 50%
D. 67%
E. 75%

We can let James’ speed be 60 mph. Thus, when James drove 180 minutes (or 3 hours), Patrick drove only 90 minutes. So James drove 60 x 3 = 180 miles. Although Patrick only drove 90 minutes (or 1.5 hours), he drove the same distance as James (since he overtook James exactly in 90 minutes), so Patrick drove at a speed of 180/1.5 = 120 mph. He continued at this speed for another 2 hours, which means he drove another 2 x 120 = 240 miles. He will drive another 6 hours, however, at a different speed, so that James could catch up with him exactly 8 hours after overtaking James. We can let this new speed be x. So the total distance Patrick travels is 180 + 240 + 6x = 420 + 6x.

Now let’s look at the distance James traveled. Recall that he had to catch up with Patrick exactly 8 hours after Patrick overtook him. When Patrick overtook him, each had driven 180 miles. Since James’ speed was 60 mph (and he continued to drive at that speed), then, in another 8 hours, he will have driven 60 x 8 = 480 miles. Thus the total distance James will have traveled is 180 + 480 = 660.

Now we can equate the distances traveled by the two brothers as follows:

420 + 6x = 660

6x = 240

x = 40

Since Patrick’s original speed was 120 mph and his new speed is 40 mph, he must have reduced his original speed by 2/3, or 67%.

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Re: Word Problem - James started from his home and drove eastwards at a co   [#permalink] 07 Jun 2018, 17:14
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