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Johnny travels a total of one hour to and from school. On the way ther

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Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 24 Aug 2015, 23:50
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Johnny travels a total of one hour to and from school. On the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour. How far is it to the school?

A. 2 miles
B. 4 miles
C. 4.8 miles
D. 8 miles
E. 10 miles

Kudos for a correct solution.

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Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post Updated on: 27 Aug 2015, 00:59
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Answer: B) 4 miles.

Average speed for round trip = 2*a*b/(a+b), where a,b are speeds
so, average speed was = 2*5*20/(5+20) = 8m/hr
the distance between school & home should be half of that. ie. 4 miles
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Originally posted by SOURH7WK on 25 Aug 2015, 00:08.
Last edited by SOURH7WK on 27 Aug 2015, 00:59, edited 1 time in total.
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 25 Aug 2015, 00:43
Bunuel wrote:
Johnny travels a total of one hour to and from school. On the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour. How far is it to the school?

A. 2 miles
B. 4 miles
C. 4.8 miles
D. 8 miles
E. 10 miles]


Ans: B

let the distance between Home and School is S
Solution: total time taken = time jog +time bus
= 1 = s/5 + s/20
= 5s/20=1
s=4 miles
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Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 25 Aug 2015, 04:44
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Backsolved, with a 5th of an hour is 12 minutes.
A. if the route took him \(24\) minutes running there, it will take less than that to return home, and therefore \(2\) miles is incorrect.
B. \(4*12=48\) and 4 miles going \(20\)mph takes \(12\) minutes, for \(4\)miles is \(\frac{1}{5}\)th of \(20\)miles, or it takes \(\frac{1}{5}\)hours to travel \(4\)miles going \(20\)mph. Anyhow, 48+12=60. Thus, B is correct.

IMO B.
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 25 Aug 2015, 11:19
t1+t2=1;
t1= d/5, t2=d/20; d= distance;
1= d/5+d/20 = 5d/20 =>d=4

Hence answer is B
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 25 Aug 2015, 12:04
Bunuel wrote:
Johnny travels a total of one hour to and from school. On the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour. How far is it to the school?

A. 2 miles
B. 4 miles
C. 4.8 miles
D. 8 miles
E. 10 miles

Kudos for a correct solution.


Let one way distance be =d
Now, d/5+d/20=1
5d/20=1
d=4 miles
Answer B
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 25 Aug 2015, 13:06
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Distance = d
Time taken while walking = t
=> d = 5 x t => t = d/5 ......(1)
For the return journey
Distance = d
Time taken = 1-t
=> d = 20 x (1-t) ....(2)
Calculating (1) & (2) for d
Distance d = 4 miles (B)
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 26 Aug 2015, 04:07
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Bunuel wrote:
Johnny travels a total of one hour to and from school. On the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour. How far is it to the school?

A. 2 miles
B. 4 miles
C. 4.8 miles
D. 8 miles
E. 10 miles

Kudos for a correct solution.


Spotting a the ratio between 2 speeds 20/5=4. The time should be divided by same ratio 20/5=48/12

d= 20 * 12/60= 4 or 5* 48/60 = 4mile
answer B
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 26 Aug 2015, 05:43
Bunuel wrote:
Johnny travels a total of one hour to and from school. On the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour. How far is it to the school?

Kudos for a correct solution.


IMO :: B

Given
\(S_1\) = 5
\(S_2\) = 20
Distance is constant

Since distance is constant we can say that speed is inversely proportional to time

Thus

\(\frac{S_1}{S_2}\) = \(\frac{T_2}{T_1}\)

Thus \(\frac{T_2}{T_1}\) = \(\frac{1}{4}\)

Thus \(T_1\) = 4/5 of total time
Distance = \(S_1\) * \(T_1\)
= 4 miles
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 26 Aug 2015, 09:06
SOURH7WK wrote:
Answer: B) 4 miles.

Average speed for round trip = 2*a*b/(a+b), where a,b are speeds
so, average speed was = 2*5*20/(5+20) = 8Km/hr
the distance between school & home should be half of that. ie. 4 miles


HI, whats the logic here? Why is the distance half the average speed?
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 26 Aug 2015, 21:30
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i solved using Rate * Time = Distance

Going: Rate (5 mph) * Time (x) = Distance

Coming Rate (20mph) * Time (1-x) = Same Distance

5x=20-20x;
25x=20
x=4/5

5 * 4/5 = 4
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Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 27 Aug 2015, 00:58
noTh1ng wrote:
SOURH7WK wrote:
Answer: B) 4 miles.

Average speed for round trip = 2*a*b/(a+b), where a,b are speeds
so, average speed was = 2*5*20/(5+20) = 8m/hr
the distance between school & home should be half of that. ie. 4 miles


HI, whats the logic here? Why is the distance half the average speed?


First I calculated the avg speed ie 8m/hr. In the question it was mentioned that he travelled for 1 hr to and fro. That means he travelled 8m to and fro. So the distance one way should be 4m.
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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 30 Aug 2015, 09:53
Bunuel wrote:
Johnny travels a total of one hour to and from school. On the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour. How far is it to the school?

A. 2 miles
B. 4 miles
C. 4.8 miles
D. 8 miles
E. 10 miles

Kudos for a correct solution.


PRINCETON REVIEW OFFICIAL SOLUTION:

What is the question asking?

The question is looking for a distance that Johnny travels. Good news for us. All the answer choices are in miles.

Where are you starting?

I have two rates of travel and a total time to apportion to and from the school. Let me suggest that there are many ways to attack this one. I will offer just a few to demonstrate that multiple methods lead to the same spot.

A) The Algebra Student says: I will work with the rates and I can see that one is four times the other in terms of time, because D=R*T (Distance = Rate X Time). I know they are inversely related. So 5T= 1 hour. Therefore T=.2 hours. I can take either the bus speed or the bicycle speed and plug in: 20 miles an hour multiplied by .2 hours results in 4 miles. We check the work and we are right.

B) The Plug-In Master says, let me try each of the distances and see what I get. So he starts at 4.8 miles and divides that by 4 miles per hour and recognizes that he is already over an hour time, tries 4 miles and the math works.

C) The Class Math Wiz tells us that there is an elegant way to take the reciprocal of the rates, add them together, flip that result over and voilá we get 4 miles. While our Wiz is absolutely correct, we all shake our heads and look for something that we all can use.

D) Finally, our Class Ball Parker says that all of this was unnecessary, because we only had an hour available to us for both jogging and the bus. Since this is true, we exclude answers C, D, and E. They are too big. We look at 2 miles and realize it is too small and won’t even come close to totaling one hour and without any math, arrive at the exact same answer.

I guarantee that there are even more ways to attack this one problem. Which method is the best? There really is no answer to that. As a student taking the GMAT you really have to determine for yourself which method works for you. However, the first step is to understand where you are starting and where you are going. You need to know exactly what you are looking for, and what tools and information you have available to help you get to the answer.

This is why it is crucial to work as many problems in your practice tests as possible. The more familiar you are, both with the math and the way in which the questions are presented, the better the chances for success.
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Johnny travels a total of one hour to and from school. On the way ther  [#permalink]

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New post 16 Feb 2019, 11:50
Johnny travels a total of one hour to and from school. On the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour. How far is it to the school?

A. 2 miles
B. 4 miles
C. 4.8 miles
D. 8 miles
E. 10 miles

total time=1 hour

total time=time from home to school + time from school to home.

we know that distance is same in both cases so we assume distance to be D

D/5+D/20=1

or 5D/20=1


or D=20/5=4.

Option B:4 miles is correct.
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Re: Johnny travels a total of one hour to and from school. On the way ther   [#permalink] 16 Feb 2019, 11:50
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