Question

Extended Response
The table shows the annual fee at two gyms and G the fee each time members take an aerobics class.

1. In a year’s time, which gym is less expensive If you plan to take 4 aerobics classes each month? A \(540 \)3
2. Write and solve an inequality to determine the number B \(360 \)5

of aerobics classes for which the total cost for 1 year at gym A is less than that at gym B.

c. how many aerobics classes should you average each month so that

the total cost for I year at gym H Is less than that at gym A?

Short answer

(a) Gym A is more expensive.

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

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

Step-by-step solution

Step 1.  Apply the concept of Variable

In any expression or equation, the variable is an unknown quantity. It may be any alphabet or any special symbol.

Step 2.  Given information

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.

Step 3.  Explanation

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.

Thus, Gym A is more expensive.

Step 1.  Apply the concept of Variable

In any expression or equation, the variable is an unknown quantity. It may be any alphabet or any special symbol.

Step 2.  Given information

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.

Step 3.  Explanation

Let the number of Aerobics Classes in a year =.

Cost of Each Aerobics at Gym A =x\$540+3x5

Total Cost of a year at Gym B =

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.

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

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

Step 1.  Apply the concept of Variable

In any expression or equation, the variable is an unknown quantity. It may be any alphabet or any special symbol.

Step 2.  Given information

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.

Step 3.  Explanation

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

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

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

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

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

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