Chapter 1.1, Problem 100E

(a)

To determine

To find: Equation of the line giving demand x in terms of the rent p .

Answer to Problem 100E

The equation is $x=50-\frac{2(p-600)}{25}$.

Explanation of Solution

Given data:

The relation between the rent and number of rooms occupied is a linear function.

Formula used:

Linear equation:

  $ax+by+c=0$

Calculation:

Let the rent of the house be p and number of rooms occupied be s.

If rent is 600$ then 50 rooms will be occupied and when rent is 625$ then 48 rooms will be occupied .

As it is a linear function and for every 50$ the rooms is decreased by 2. So,

  $x=50-\frac{2(p-600)}{25}$

Conclusion:

The equation is $x=50-\frac{2(p-600)}{25}$.

(b)

To determine

To find: Equation of the line giving demand x in terms of the rent p .

Answer to Problem 100E

44 rooms will be occupied when rent is 675$.

Explanation of Solution

Given data:

The relation between the rent and number of rooms occupied is a linear function.

Formula used:

Linear equation:

  $ax+by+c=0$

Calculation:

Graph for given equation will be:

x	50	48	46	44	42	40
p	600	625	650	675	700	725

  

No. Of room occupied at the rent of 675$

No. Of room occupied at the rent of 675$

  $\begin{array}{l}x=50-\frac{2(p-600)}{25}\\ x=50-\frac{2(675-600)}{25}\\ x=44\\ \end{array}$

44 rooms will be occupied.

Conclusion:

44 rooms will be occupied when rent is 675$.

(c)

To determine

To find: Equation of the line giving demand x in terms of the rent p .

Answer to Problem 100E

46 rooms will be occupied when rent is 650$.

Explanation of Solution

Given data:

The relation between the rent and number of rooms occupied is a linear function.

Formula used:

Linear equation:

  $ax+by+c=0$

Calculation:

No. Of rooms occupied at the rent of 650$

   $\begin{array}{l}x=50-\frac{2(p-600)}{25}\\ x=50-\frac{2(650-600)}{25}\\ x=46\\ \end{array}$

46 rooms will be occupied.

Conclusion:

46 rooms will be occupied when rent is 650$.