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Speed Problem - Speed Problem

For COMPETITION
Number of Total Problems: 10.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: None

Section:Speed Problem

Theme:None
Adjustment# :

Difficulty: 1

Category Speed Problem

Analysis

Solution/Answer

Problem Num : 2

From : AMC10

Type:

Section:Speed Problem

Theme:
Adjustment# : 0

Difficulty: 1

'
Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$ext{(A)} 45 qquad ext{(B)} 48 qquad ext{(C)} 50 qquad ext{(D)} 55 qquad ext{(E)} 58$
'

Category Speed Problem

Analysis

Solution/Answer
Solution 1

Let the time he needs to get there in be t and the distance he travels be d. From the given equations, we know that $d=left(t+frac{1}{20} ight)40$ and $d=left(t-frac{1}{20} ight)60$ . Setting the two equal, we have $40t+2=60t-3$ and we find $t=frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$ . From $d=rt$ , we find that r, and our answer, is $oxed{ ext{(B)} 48 }$ .

Solution 2

Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonic mean of a and b is $frac{2}{frac{1}{a}+frac{1}{b}}=frac{2ab}{a+b}$ . In this case, a and b are 40 and 60, so our answer is $frac{4800}{100}=48$ , so $oxed{ ext{(B)} 48}$ .

Solution 3

A more general form of the argument in Solution 2, with proof:
Let $d$ be the distance to work, and let $s$ be the correct average speed. Then the time needed to get to work is $t=frac ds$ .
We know that $t+frac 3{60} = fracd{40}$ and $t-frac 3{60} = frac d{60}$ . Summing these two equations, we get: $2t = frac d{40} + frac d{60}$ .
Substituting $t=frac ds$ and dividing both sides by $d$ , we get $frac 2s = frac 1{40} + frac 1{60}$ , hence $s=oxed{48}$ .
(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighed sum in step two, and hence obtain a weighed harmonic mean in step three.)

Answer:

Problem Num : 3

From : AMC10

Type:

Section:Speed Problem

Theme:
Adjustment# : 0

Difficulty: 1

'
Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$mathrm{(A) } 250 qquad mathrm{(B) } 300 qquad mathrm{(C) } 350 qquad mathrm{(D) } 400qquad mathrm{(E) } 5...$
'

Category Speed Problem

Analysis

Solution/Answer
Solution 1

Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion: $frac{100}{x- 150} = frac{frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350 Longrightarrow mathrm{(C)}$ .

Solution 2

The total distance the girls run between the start and the first meeting is one half of the track length.
The total distance they run between the two meetings is the track length.
As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting.
Thus between the meetings Brenda will run $2cdot 100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $Rightarrow mathrm{(C) }$

Answer:

Problem Num : 4

From : NCTM

Type: Application

Section:Speed Problem

Theme:Speed
Adjustment# : 0

Difficulty: 2

Category Speed Problem

Analysis ????????? ??? ?????????? ???????? ?????? ???. ??? ?????? ?????? ???????? ?????? ???? ????? ???????.

Solution/Answer

Problem Num : 5

From : AMC10

Type:

Section:Speed Problem

Theme:
Adjustment# : 0

Difficulty: 1

'
Doug can paint a room in $5$ hours. Dave can paint the same room in $7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $t$ be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by $t$ ?
$mathrm{(A)} left(frac{1}{5}+frac{1}{7} ight)left(t+1 ight)=1qquadmathrm{(B)} left(frac{1}{5}+frac{1}{7} ight)t...$
'

Category Speed Problem

Analysis

Solution/Answer
Solution 1

Doug can paint $frac{1}{5}$ of a room per hour, Dave can paint $frac{1}{7}$ of a room in an hour, and the time they spend working together is $t-1$ .
Since rate times time gives output, $left(frac{1}{5}+frac{1}{7} ight)left(t-1 ight)=1 Rightarrow mathrm{(D)}$

Solution 2

If one person does a job in $a$ hours and another person does a job in $b$ hours, the time it takes to do the job together is $frac{ab}{a+b}$ hours.
Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in $frac{5*7}{5+7} = frac{35}{12}$ hours. They also take 1 hour for lunch, so the total time $t = frac{35}{12} + 1$ hours.
Looking at the answer choices, $(D)$ is the only one satisfied by $t = frac{35}{12} + 1$ .

Answer:

Problem Num : 6

From : AMC10B

Type:

Section:Speed Problem

Theme:
Adjustment# : 0

Difficulty: 1

'
Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?
$extbf{(A)} 18 qquad extbf{(B)} 21 qquad extbf{(C)} 24 qquad extbf{(D)} 27 qquad extbf{(E)} 30$
'

Category Speed Problem

Analysis

Solution/Answer
We know that $d = vt$
Since we know that she drove both when it was raining and when it was not and that her total distance traveled is $16$ miles.
We also know that she drove a total of $40$ minutes which is $dfrac{2}{3}$ of an hour.
We get the following system of equations, where $x$ is the time traveled when it was not raining and $y$ is the time traveled when it was raining:
$left{egin{array}{ccc} 30x + 20y & = & 16 \x + y & = & dfrac{2}{3} end{array} ight.$
Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:
$-10y = -4 Leftrightarrow y = dfrac{2}{5}$
We know now that the time traveled in rain was $dfrac{2}{5}$ of an hour, which is $dfrac{2}{5}*60 = 24$ minutes
So, our answer is $oxed{ extbf{(C)} 24}$

Answer:

Problem Num : 7

From : AMC10

Type:

Section:Speed Problem

Theme:
Adjustment# : 0

Difficulty: 1

'
Eric plans to compete in a triathlon. He can average $2$ miles per hour in the $frac{1}{4}$ -mile swim and $6$ miles per hour in the $3$ -mile run. His goal is to finish the triathlon in $2$ hours. To accomplish his goal what must his average speed in miles per hour, be for the $15$ -mile bicycle ride?
$mathrm{(A)} frac{120}{11}qquadmathrm{(B)} 11qquadmathrm{(C)} frac{56}{5}qquadmathrm{(D)} frac{45}{4}qquadmath...$
'

Category Speed Problem

Analysis

Solution/Answer
Since $d=rt$ , Eric takes $frac{frac{1}{4}}{2}=frac{1}{8}$ hours for the swim. Then, he takes $frac{3}{6}=frac{1}{2}$ hours for the run. So he needs to take $2-frac{5}{8}=frac{11}{8}$ hours for the $15$ mile run. This is $frac{15}{frac{11}{8}}=frac{120}{11} frac{ ext{miles}}{ ext{hour}}$
$longrightarrow fbox{A}$

Answer:

Problem Num : 8

From : AMC10

Type:

Section:Speed Problem

Theme:
Adjustment# : 0

Difficulty: 1

'
Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
$mathrm{(A)} 20qquadmathrm{(B)} 30qquadmathrm{(C)} 55qquadmathrm{(D)} 65qquadmathrm{(E)} 80$
'

Category Speed Problem

Analysis

Solution/Answer
Let their speeds in kilometers per hour be $v_A$ and $v_L$ . We know that $v_A=3v_L$ and that $v_A+v_L=60$ . (The second equation follows from the fact that $1,unit{ m km/min} = 60,unit{ m km/h}$ .) This solves to $v_A=45$ and $v_L=15$ .
As the distance decreases at a rate of $1$ kilometer per minute, after $5$ minutes the distance between them will be $20-5=15$ kilometers.
From this point on, only Lauren will be riding her bike. As there are $15$ kilometers remaining and $v_L=15$ , she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $5+60 = oxed{65}$ .

Answer:

Problem Num : 9

From : AMC10

Type:

Section:Speed Problem

Theme:
Adjustment# : 0

Difficulty: 1

'
Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$extbf{(A)} 10qquad extbf{(B)} 15qquad extbf{(C)} 20qquad extbf{(D)} 25qquad extbf{(E)} 30$
'

Category Speed Problem

Analysis

Solution/Answer
Cagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in $frac{20*30}{20+30} = frac{600}{50} = 12$ seconds. In 300 seconds (5 minutes), they can frost $oxed{ extbf{(D)} 25 ext{ cupcakes }}$ .

Answer:

Problem Num : 10

From : NCTM

Type: Application

Section:Speed Problem

Theme:None
Adjustment# : 0

Difficulty: 1

Category Speed Problem

Analysis

Solution/Answer

Array ( [0] => 5959 [1] => 7759 [2] => 7800 [3] => 3005 [4] => 7873 [5] => 8132 [6] => 7885 [7] => 7898 [8] => 7921 [9] => 3069 ) 10