Chapter 3, Problem 23PTTS

a.

To determine

To find out which gym is less expensive if one plan to take 4 aerobics classes each month.

Answer to Problem 23PTTS

Gym A is more expensive.

Explanation of Solution

Given:

The annual fee of Gym A is given = $540.

The annual fee of Gym B is given = $360.

The Class fee of Aerobics Each time of Gym A = $3.

The Class fee of Aerobics Each time of Gym B = $5.

Concept Used:

The concept of forming and solving inequality is used.

Calculation:

Number of Classes of Aerobics one takes each month at = 4

Number of Classes in a year = 4 x 12 =48

Cost of Each Aerobics at Gym A = $3 x 48 = $144

Total Cost of a year at Gym A = $144 + $540 = $684.

Cost of Each Aerobics at Gym B = $5 x 48 = $240

Total Cost of a year at Gym B = $360 + $240 = $600

Hence, cost of a year at Gym A is more than cost of a year at Gym B.

Conclusion:

Thus, Gym A is more expensive.

b.

To determine

To write and solve an inequality to determine the number of aerobics class 1 year at Gym A is less than that at Gym B.

Answer to Problem 23PTTS

The inequality is x<90. So, 91 Aerobics Classes at Gym A is less than that at Gym B.

Explanation of Solution

Given:

The annual fee of Gym A is given = $540.

The annual fee of Gym B is given = $360.

The Class fee of Aerobics Each time of Gym A = $3.

The Class fee of Aerobics Each time of Gym B = $5.

The cost of 1 year at Gym A is less than Gym B.

Concept Used:

The concept of forming and solving inequality is used.

Calculation:

Let the number of Aerobics Classes in a year = $x$ .

Cost of Each Aerobics at Gym A = $3 $x$ .

Total Cost of a year at Gym A = $\$540+3x$

Cost of Each Aerobics at Gym B = $5 $x$

Total Cost of a year at Gym B = $\$360+5x$

The cost of 1 year at Gym A is less than Gym B.

  $\Rightarrow 540+3x<360+5x$

Take like Terms on the same side and solve.

  $\begin{array}{l}\Rightarrow 540-360<5x-3x\\ \Rightarrow 180<2x\\ \Rightarrow x>90\end{array}$

Conclusion:

Thus, for 91 aerobics classes, Gym A is cheaper than Gym B.

c.

To determine

To find how many aerobics classes one should average each month so that the total cost for 1 year at gym B is less than that at gym A.

Answer to Problem 23PTTS

For 22 aerobics classes each month, Gym B is cheaper than Gym A.

Explanation of Solution

Given:

The annual fee of Gym A is given = $540.

The annual fee of Gym B is given = $360.

The Class fee of Aerobics Each time of Gym A = $3.

The Class fee of Aerobics Each time of Gym B = $5.

The cost of 1 year at Gym B is less than Gym A.

Concept Used:

The concept of forming and solving inequality is used.

Calculation:

Let the number of Aerobics Classes in a year = $x$ .

Cost of Each Aerobics at Gym A = $3 $x$ .

Total Cost of a year at Gym A = $\$540+3x$

Cost of Each Aerobics at Gym B = $5 $x$

Total Cost of a year at Gym B = $\$360+5x$

The cost of 1 year at Gym B is less than Gym A.

  $\Rightarrow 540+3x>360+5x$

Take like Terms on the same side and solve.

  $\begin{array}{l}\Rightarrow 540-360>5x-3x\\ \Rightarrow 180>2x\\ \Rightarrow x<90\end{array}$

Average classes each month = $90/4$ = 22.5

Conclusion:

Thus, for 22 aerobics classes each month, Gym B is cheaper than Gym A.