Chapter MPS, Problem 3.1MPS

(a)

To determine

To Find: Equation that represents the given situation

Answer to Problem 3.1MPS

  $x\left(\$5.00\right)+\left(3400-x\right)\left(\$2.50\right)=\$12250$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Total people enter into the fair}=3400\\ \text{Total fees collected from admission fees}=\$12250\\ \text{Admission fee of adult}=\$5.00\\ \text{Admission fee of child}=\$2.50\end{array}$

Formula Used:

  $\text{number of adults}\left(\text{Admission fee of adult}\right)+\text{number of child}\left(\text{Admission fee of child}\right)=\text{Total fee collected}$

Calculation:

  $\begin{array}{l}\text{Total people enter into the fair}=3400\\ \text{→}\text{Let the number of adult enter into the fair}=x\\ \text{→}\text{Number of child enter into the fair}=3400-x\\ \text{⇒}\text{number of adults}\left(\text{Admission fee of adult}\right)+\text{number of child}\left(\text{Admission fee of child}\right)=\text{Total fee collected}\\ \text{⇒}x\left(\$5.00\right)+\left(3400-x\right)\left(\$2.50\right)=\$12250\end{array}$

Conclusion:

Thus the equation for the given situation is $x\left(\$5.00\right)+\left(3400-x\right)\left(\$2.50\right)=\$12250$

(b)

To determine

To Find: Number of adults and number of child enter into the fair.

Answer to Problem 3.1MPS

  $\begin{array}{l}\text{→}\text{Number of adult}=1500\\ \text{→}\text{Number of child}=1900\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Total people enter into the fair}=3400\\ \text{Total fees collected from admission fees}=\$12250\\ \text{Admission fee of adult}=\$5.00\\ \text{Admission fee of child}=\$2.50\end{array}$

Formula Used:

  $\text{number of adults}\left(\text{Admission fee of adult}\right)+\text{number of child}\left(\text{Admission fee of child}\right)=\text{Total fee collected}$

Calculation:

  $\begin{array}{l}\text{Total people enter into the fair}=3400\\ \text{→}\text{Let the number of adult enter into the fair}=x\\ \text{→}\text{Number of child enter into the fair}=3400-x\\ \text{⇒}\text{number of adults}\left(\text{Admission fee of adult}\right)+\text{number of child}\left(\text{Admission fee of child}\right)=\text{Total fee collected}\\ \text{⇒}x\left(\$5.00\right)+\left(3400-x\right)\left(\$2.50\right)=\$12250\\ \text{⇒}\$5.00x+\left(\$8500-\$2.50x\right)=\$12250\\ \text{⇒}\$5.00x+\$8500-\$2.50x=\$12250\\ \text{⇒}\$2.50x+\$8500-\$8500=\$12250-\$8500\left(\text{subtract both side by }\$8500\right)\\ \text{⇒}\$2.50x=\$3750\\ \text{⇒}x=\frac{\$3750}{\$2.50}\left(\text{divide both side by \$2}\text{.50}\right)\\ \text{⇒}x=1500\\ \text{→}\text{Number of child}=3400-1500=1900\\ \text{→}\text{Number of adult}=1500\end{array}$

Conclusion:

Thus the number of adults and child enter into the fair is $1500,1900$