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# How much did Tony pay for each game?

These are two math problems I need help with:
1. Tony paid the same price for a number of computer games totaling \$138.25. Willy bought the same number of games for \$117.25. Willy paid \$3 less for each game than Tony did. How much did Tony pay for each game?

2. Damon gave Shaquan a 20-yard head-start in a race. Damon can run 3 yards a second while Shaquan can run 2.5 yards a second. Damon crosses the finish line 2 seconds ahead of Shaquan. How many yards did Damon run?

### 4 Answers by Expert Tutors

Steve P. | Patient and Knowledgable Math and Chemistry TutorPatient and Knowledgable Math and Chemis...
4.8 4.8 (5 lesson ratings) (5)
0
1.  Let x be the number of games for Tony and Willy, let y be the price per game for Tony and y-3 the price per game for Willy.

Tony:

xy=138.25

Willy:

x(y-3)= 117.25

xy-3x=117.25

138.25-3y= 117.25

-3x=-21

x=7

Substitute into Tony's  equation:

7y=138.25

y=19.75

Tony paid \$19.75 for each game.

2.  Use the formula d=rt

Since this is a race to the finish line both runners run the same distance. Shaquan adds 2 seconds to his time because it takes him two seconds longer to run the same distance as Damon. Damon subtracts 8 seconds from his time because Shaquan had a 20 yard head start.

20yards/2.5yards/second=8 seconds

since d=d we can set rt=rt:

2.5(t+2)=3(t-8)

2.5t+5=3t-24

.5t=29

t=58s

d= 3(58-8) = 150 yards
Jason S. | My goal is the success of my students. Knowledge-Patience-HonestyMy goal is the success of my students. K...
4.9 4.9 (115 lesson ratings) (115)
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1.  Willy saved 21 dollars buying the same number of games.

138.25 - 117.25 = 21.00

Since the money he saved was \$3 per game,

Money saved
--------------                   =   number of games each bought.
Money saved per game

21/3 = 7.

They bought 7 games each.

Tony:

\$138.25/7 games = \$19.75/game.

2.

Distance = rate * time

They run the same distance D.

Let t be the time that it took Damon to run the distance.

Damon
D = 3yd/s * (t)

Shaquan took 2 seconds longer than Damon, so his time was t+2, and his distance was 20 yards shorter.

D - 20 = 2.5yd/s * (t+2)

Solve for D:

D = 2.5(t+2) + 20

Set them equal to each other:

3(t) = 2.5(t+2) + 20

3t = 2.5t + 5 + 20

3t - 2.5t = 25

.5t = 25

t = 25/.5 = 25*2 = 50s

That's the time for Damon.

The question wants the distance.

D = r*t
D = 3*50

D = 150 yards.

Double check:

If D = 150, then Shaquan only had to run 150-20=130 yards, because he had a headstart.

If Damon ran the race in 50 seconds, and Shaquan took 2 seconds longer, then it took Shaquan 50+2 = 52 seconds.

Does that check out?

D for Shaquan = (2.5 yds/s)(52s) = 130yds.  Check.

Michael H. | Tutor for Math and Computer SkillsTutor for Math and Computer Skills
4.9 4.9 (53 lesson ratings) (53)
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1. Lets think about what this problem is telling us.
• Tony bought a number of video games (n) at a fixed price (p)
• The total of these games was \$138.25
• We can write this as an equation:
• n*p = \$138.25

• Willy bought the same number of games as Tony (n)
• Willy paid \$3 less per game (p-3)
• Willy's total was \$117.25
• This can be written as: n*(p-3) = \$117.25

The problem asks us to find the price Tony paid per game (p)
• We have two equations and two variables in each.
• If we re-arrange the equations, we can get (n) on the left side of each
• n = 138.25 / p
• n = 117.25 / (p-3)

Now we have two different expressions for the value n, but we know that n = n
• So we can set the expressions for n equal to each other
• 138.25 / p = 117.25 / (p-3)
• At this point, we can solve the new equation for p
• 138.25 * (p-3) = 117.25 * p
• 138.25p - 414.75 = 117.25p
• 138.25p - 117.25p = 414.75
• 21p = 414.75
• p = 414.75 / 21 = 19.75

Tony paid \$19.75 for each game

An alternate method:
• We can take the difference of the total prices (\$138.25 - \$117.25 = \$21)
• Divide it by the price difference per game ( \$21 / \$3 = 7 )
• This is because the total difference divided by the per item difference will give us the count of items
• Therefore, each person bought 7 games.
• Now we can divide the total price paid by Tony by the number of games he bought (based on the equation from the first method):
• \$138.25 / 7 games = \$19.75 per game

You can use either of these methods to solve the second problem

You are correct, but I want to use algebra, choose a variable and convert English to algebraic statements.
Parviz, the first method I demonstrated converted English to algebraic statements. The first 8 steps explain the conversion. The variables I used are n and p. I demonstrated choosing the variable p, as the question asks about the price Tony paid.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
-1
nP = 138.25
n(p-3) = 117.25

138.25 = 117.25
p          p-3

P( 117.25) = p( 138.25)- 3( 138.25)

P( 138.25- 117.25) = 3( 138.25)

P = 3( 138.25 = 19.75
21
20/yd
S--------- D --------
2.5y/s       3.0Y/s

20 + 2.5(t+2) = 3 t
20 +5 = 3t-2.5t

25 = 0.5t         t = 50 second

50* 3yd/s = 150yd Damon Yards