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Two people walk the same distance of 15miles, in which A spends 15 min

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COolguy101
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Two people walk the same distance of 15miles, in which A spends 15 min [#permalink] New post   12 Aug 2021, 19:34
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Two people walk the same distance of 15 miles, in which A spends 15 minutes less than B, and the speed of B is 10 miles/h slower than the speed of A. What is the speed of B?


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Re: Two people walk the same distance of 15miles, in which A spends 15 min [#permalink] New post   13 Aug 2021, 08:05
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COolguy101 wrote:
COolguy101 wrote:
Two people walk the same distance of 15 miles, in which A spends 15 minutes less than B, and the speed of B is 10 miles/h slower than the speed of A. What is the speed of B?


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GreenlightTestPrep KarunMendiratta rx10 sir, please help me .


Let,
D be the distance
T be the time
and S be speed

Given:
\(D_A = D_B = 15\)
\(T_A = T_B - \frac{15}{60}\)
\(S_A = S_B + 10\)

\(T_A = T_B - \frac{15}{60}\)

\(\frac{15}{S_B + 10} = \frac{15}{S_B} - \frac{15}{60}\)

\(\frac{15}{S_B} - \frac{15}{S_B + 10} = \frac{15}{60}\)

\(\frac{15(S_B + 10) - 15(S_B)}{S_B(S_B + 10)} = \frac{1}{4}\)

\((15S_B + 150 - 15S_B)(4) = S_B^2 + 10S_B\)

\(S_B^2 + 10S_B - 600 = 0\)

Solve for \(S_B = 20\)
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Re: Two people walk the same distance of 15miles, in which A spends 15 min [#permalink] New post   13 Aug 2021, 06:52
COolguy101 wrote:
Two people walk the same distance of 15 miles, in which A spends 15 minutes less than B, and the speed of B is 10 miles/h slower than the speed of A. What is the speed of B?


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GreenlightTestPrep KarunMendiratta rx10 sir, please help me .
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Re: Two people walk the same distance of 15miles, in which A spends 15 min [#permalink] New post   13 Aug 2021, 07:12
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COolguy101 wrote:
Two people walk the same distance of 15 miles, in which A spends 15 minutes less than B, and the speed of B is 10 miles/h slower than the speed of A. What is the speed of B?
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The speed of B is 10 miles/h slower than the speed of A
Let x = B's speed (in miles per hour)
So, x + 10 = A's speed (in miles per hour)

A spends 15 minutes less than B
Time = distance/rate
So, B's travel time = 15/x
And A's travel time = 15/(x + 10)

Since we're dealing with rates measured in miles per HOUR, will convert 15 minutes to 0.25 HOURS
We can write: (A's travel time) = (B's travel time) - 0.25
Substitute to get: 15/(x + 10) = 15/x - 0.25

To solve for x, will first multiply both sides of the equation by (x + 10) to get: 15 = 15(x + 10)/x - 0.25(x + 10)
Then we'll multiply both sides of the equation by x to get: 15x = 15(x + 10) - 0.25(x + 10)(x)
Expand: 15x = 15x + 150 - 0.25x² + 2.5x
Subtract 15x from both sides: 0 = 150 - 0.25x² - 2.5x
To make matters easier, let's multiply both sides of the equation by 4 to get: 0 = 600 - x² - 10x
Rearrange to get: x² + 10x - 600 = 0
Factor: (x - 20)(x + 30) = 0

So, x = 20 or x = -30
Since B's cannot be negative, the correct answer is 20
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Two people walk the same distance of 15miles, in which A spends 15 min [#permalink] New post   13 Aug 2021, 07:22
GreenlightTestPrep wrote:
The speed of B is 10 miles/h slower than the speed of A
Let x = B's speed (in miles per hour)
So, x + 10 = A's speed (in miles per hour)

A spends 15 minutes less than B
Time = distance/rate
So, B's travel time = 15/x
And A's travel time = 15/(x + 10)

Since we're dealing with rates measured in miles per HOUR, will convert 15 minutes to 0.25 HOURS
We can write: (A's travel time) = (B's travel time) - 0.25
Substitute to get: 15/(x + 10) = 15/x - 0.25

To solve for x, will first multiply both sides of the equation by (x + 10) to get: 15 = 15(x + 10)/x - 0.25(x + 10)
Then we'll multiply both sides of the equation by x to get: 15x = 15(x + 10) - 0.25(x + 10)(x)
Expand: 15x = 15x + 150 - 0.25x² + 2.5x
Subtract 15x from both sides: 0 = 150 - 0.25x² - 2.5x
To make matters easier, let's multiply both sides of the equation by 4 to get: 0 = 600 - x² - 10x
Rearrange to get: x² + 10x - 600 = 0
Factor: (x - 20)(x + 30) = 0

So, x = 20 or x = -30
Since B's cannot be negative, the correct answer is 20


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