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Traveling at a constant rate of 15 miles per hour, it took Bob x hours

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Traveling at a constant rate of 15 miles per hour, it took Bob x hours  [#permalink]

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New post 24 Jan 2019, 04:05
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

87% (01:11) correct 13% (01:05) wrong based on 70 sessions

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Traveling at a constant rate of 15 miles per hour, it took Bob x hours  [#permalink]

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New post 24 Jan 2019, 04:11
Bunuel wrote:
Traveling at a constant rate of 15 miles per hour, it took Bob x hours to go from his home to his school. Traveling at a constant rate of 20 miles per hour, it took Bob x − 1 hours to travel the same route. What is the value of x?


A 2
B 3
C 4
D 5
E 6


15x = 20(x-1)

15x = 20x -20

x = 4.

*** Distance = time * speed.
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Re: Traveling at a constant rate of 15 miles per hour, it took Bob x hours  [#permalink]

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New post 24 Jan 2019, 05:30

Solution


Given:
    • \(s_1 = 15\) miles per hour, and \(t_1 = x\) hours
    • \(s_2 = 20\) miles per hour, and \(t_2 = x – 1\) hours
    • Both distances are same

To find:
    • The value of x

Approach and Working:
We know that distance = speed * time
    • Implies, \(s_1 * t_1 = s_2 * t_2\)
    • Thus, 15x = 20(x - 1)

Therefore, \(x = \frac{20}{5} = 4\)

Hence, the correct answer is Option C

Answer: C

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Re: Traveling at a constant rate of 15 miles per hour, it took Bob x hours  [#permalink]

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New post 24 Jan 2019, 06:07
Bunuel wrote:
Traveling at a constant rate of 15 miles per hour, it took Bob x hours to go from his home to his school. Traveling at a constant rate of 20 miles per hour, it took Bob x − 1 hours to travel the same route. What is the value of x?


A 2
B 3
C 4
D 5
E 6


Using logic, we know that in second case, Bob travelled 1 hr less because he travelled 5 miles extra for each of the (x - 1) hrs. This is the distance he would have covered in the last hour in the first case.

5*(x - 1) = 15
x = 4

Answer (C)
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Re: Traveling at a constant rate of 15 miles per hour, it took Bob x hours  [#permalink]

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New post 24 Jan 2019, 08:07
Bunuel wrote:
Traveling at a constant rate of 15 miles per hour, it took Bob x hours to go from his home to his school. Traveling at a constant rate of 20 miles per hour, it took Bob x − 1 hours to travel the same route. What is the value of x?


A 2
B 3
C 4
D 5
E 6


Let the distance travelled be 60 miles (LCM of 15 & 20)

So, Travelling at 15 miles/hour time taken is 4 hours & Travelling at 20 miles/hour time taken is 3 hours

Thus, the value of x is 4, Answer must be (C) 4
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Re: Traveling at a constant rate of 15 miles per hour, it took Bob x hours  [#permalink]

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New post 27 Jan 2019, 20:58
When I first saw these types of questions I could answer the simple ones, but was still confused by the units. I'm putting the below for others in similar situations.

1st trip \(\frac{15(miles)}{1(hour)}·x(hours)=distance\)

2nd trip \(\frac{20(miles)}{1(hour)}·(x-1)(hours)=distance\)

The distance is the same in both so we can have the equations equal each other

\(\frac{15(miles)}{1(hour)}·x(hours)=\frac{20(miles)}{1(hour)}·(x-1)(hours)\)

Simplification: Each side has hours in the denominator and numerator so they can be cancelled from each side. Just remember that we're trying to find x and it uses the unit hours

\(\frac{15(miles)}{1}·x=\frac{20(miles)}{1}·(x-1)\)

Simplification: Both sides of the equation use the unit miles, so it can be cancelled from both sides of the equation.

\(\frac{15}{1}·x=\frac{20}{1}·(x-1)\)

15x=20x-20
0=5x-20
0=x-4
4=x
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Re: Traveling at a constant rate of 15 miles per hour, it took Bob x hours   [#permalink] 27 Jan 2019, 20:58
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