Reggie S.

asked • 03/02/13# A math word problem

A theater group made appearances in two cities. The hotel charge before tax in the second city was $500 lower than in the first. The tax in the first city was 3%, and the tax in the second city was 4.5%. The total hotel tax paid for the two cities was $258.75. How much was the hotel charge in each city before tax?

## 2 Answers By Expert Tutors

**The second hotel is $500 less than the first hotel:**

Hotel_{1} = x

Hotel_{2} = x-500

**Tax rates for each hotel:**

Rate_{1} = .03

Rate_{2}= .045

**Solve for x:**

Total Tax = Hotel_{1}Rate_{1} + Hotel_{2}Rate_{2}

258.75 = (x)(.03) + (x-500)(.045)

258.75 = .03x +.045x - 22.5

281.25 = .075x

281.75/.075 = x

x=3750

**Solution:**

3750 = x = Hotel_{1}

Hotel_{2} = x-500 = 3750 - 500 = 3250

This is a word problem and let's admit this that word problems are the most convoluted part of math because they apply our knowledge to real life.

Let's make this easy:

Step 1: List all the numerical info, ORGANIZE

*1st City (A):

Tax = 3%

*2nd City (B):

hotel charge (minus tax) = $500 less than A (note here is we get an equation; We have our left and right side)

Tax = 4.5%

-----------------

Total tax for A and B = $258.75 (note here is where we get an equation; we have our left side and right side of equation)

Step 2: Put info from Step 1 into math terms

* Tax for A = 0.03A

* Tax for B = 0.045B

Hence: 0.03A + 0.045B = 258.75

Prior tax info (in orange before this) can be expressed as ---> B = A-500 (says that 2nd city before tax price is $500 less than first city)

Step 3: Solve the two equations you've come up with

0.03A + 0.045B = 258.75

B = A-500

How?

* Isolate either A or B in first equation and substitute in second.

Ex) B = (258.75 - 0.03A)/0.045

(258.75 - 0.03A)/0.045 = A -500 (plug in B value to left side of 2nd eqn in place of A)

A = $3750 (remember that A and B represent hotel charge prior to tax)

B = (3750) - 500 (can substitute value A in either first or second eqn to get B)

B = $3250

Organization is key here. In problems where there are 2 variables (like A and B), find 2 equations and work with them like this.

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Rizul N.

I am not sure why my explanation is not showing up but here is what I did:

This is a word problem and let's admit this that word problems are the most convoluted part of math because they apply our knowledge to real life.

Let's make this easy:

Step 1: List all the numerical info, ORGANIZE

*1st City (A):

Tax = 3%

*2nd City (B):

hotel charge (minus tax) = $500 less than A (note here is we get an equation; We have our left and right side)

Tax = 4.5%

-----------------

Total tax for A and B = $258.75 (note here is where we get an equation; we have our left side and right side of equation)

Step 2: Put info from Step 1 into math terms

* Tax for A = 0.03A

* Tax for B = 0.045B

Hence: 0.03A + 0.045B = 258.75

Prior tax info (in orange before this) can be expressed as ---> B = A-500 (says that 2nd city before tax price is $500 less than first city)

Step 3: Solve the two equations you've come up with

0.03A + 0.045B = 258.75

B = A-500

How?

* Isolate either A or B in first equation and substitute in second.

Ex) B = (258.75 - 0.03A)/0.045

(258.75 - 0.03A)/0.045 = A -500 (plug in B value to left side of 2nd eqn in place of A)

A = $3750 (remember that A and B represent hotel charge prior to tax)

B = (3750) - 500 (can substitute value A in either first or second eqn to get B)

B = $3250

Organization is key here. In problems where there are 2 variables (like A and B), find 2 equations and work with them like this.

03/02/13