Question

Please help me answer the following questions with showed work because that's what my math teacher wants! :)

1) At one store, a trophy costs $12.50. Engraving is then $0.40 per letter. At another store, the same trophy costs $14.75 with engraving being $0.25 per letter. How many letters can you add to the trophy to make them cost the same?

2) Lenny makes $55,000 and is getting annual raises of 2,500 each year. Karl makes $62,000 with a raise of $2,000 per year. After how many years will Karl and Lenny be making the same amount of money?

3) Container A has 200L of water and is being filled at a rate of 6L per minute. Container B has 500L of water and is being drained at a rate of 6L per minute. After how many minutes will the two containers be holding the same amount of water?

This is due tmr! :) :)

Answer 1

1) 15 Letters

2) 14 years

3) 25 minutes

Explanation:

1)

Store 1:

Trophy = $12.50

(Engraving) One letter = $0.40

Store 2:

Trophy = $14.75

(Engraving) One letter = $0.25

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Let's make an equation. L will equal the amount of letters added.

12.5 + 0.4L = 14.75 + 0.25L

-12.5 -12.5

0.4L = 2.25 + 0.25L

-0.25 -0.25

0.15L = 2.25

÷ 0.15 ÷ 0.15

L = 15

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You can engrave 15 letters on each trophy

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2)

Lenny:

-Starts with: $55,000

-Receives Annually: $2,500

Karl:

-Starts with: $62,000

-Receives Annually: $2,000

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Let's have the variable be r for the amount of raises needed.

55,000 + 2,500r = 62,000 + 2,000r

-2,000r -2,000r

55,000 + 500r = 62,000

-55,000 -55,000

500r = 7,000

÷ 500 ÷ 500

r = 14

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They will have the same pay in 14 years

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3)

Container A:

-Starts with: 200L

-Added (Minutes): 6L

Container B:

-Starts with: 500L

-Drained (Minutes): 6L

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Variable: M equals the amount of minutes that go by

200 + 6M = 500 - 6M

-200 -200

6M = 300 - 6M

+6M +6M

12M = 300

÷ 12 ÷ 12

M = 25

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The containers will be holding the same amount in 25 minutes.