Three clients need a rough estimate of what their houses will cost them with similar
fittings. To develop a property, the (rand per square meter) usually shows that
the larger the house the cheaper the cost per square meter that a customer pays. The
floor area of client A’s house is three times larger than client B’s and the floor area of
client is half the size of client . The total floor area of all the houses is .
a) Calculate the floor area of each house.

Ask by Burgess Washington. in South Africa

Nov 19,2024

Question

Three clients need a rough estimate of what their houses will cost them with similar
fittings. To develop a property, the (rand per square meter) usually shows that
the larger the house the cheaper the cost per square meter that a customer pays. The
floor area of client A’s house is three times larger than client B’s and the floor area of
client is half the size of client . The total floor area of all the houses is .
a) Calculate the floor area of each house.

Ask by Burgess Washington. in South Africa

Nov 19,2024

UpStudy AI · Accepted Answer

Let's denote the floor area of client A's house as $$ A $$, the floor area of client B's house as $$ B $$, and the floor area of client C's house as $$ C $$.

Given:
- The floor area of client A's house is three times larger than client B's house: $$ A = 3B $$
- The floor area of client C's house is half the size of client A's house: $$ C = \frac{1}{2}A $$
- The total floor area of all the houses is $$ 682 \, 	ext{m}^2 $$: $$ A + B + C = 682 $$

We can solve this system of equations to find the floor area of each house.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}A=3B\C=\frac{1}{2}A\A+B+C=682\end{array}ight.$$
- step1: Substitute the value of $$A:$$
  $$\left\{ \begin{array}{l}C=\frac{1}{2}	imes 3B\3B+B+C=682\end{array}ight.$$
- step2: Multiply the numbers:
  $$\left\{ \begin{array}{l}C=\frac{3}{2}B\4B+C=682\end{array}ight.$$
- step3: Substitute the value of $$C:$$
  $$4B+\frac{3}{2}B=682$$
- step4: Simplify:
  $$\frac{11}{2}B=682$$
- step5: Multiply by the reciprocal:
  $$\frac{11}{2}B	imes \frac{2}{11}=682	imes \frac{2}{11}$$
- step6: Multiply:
  $$B=124$$
- step7: Substitute the value of $$B:$$
  $$C=\frac{3}{2}	imes 124$$
- step8: Simplify:
  $$C=186$$
- step9: Substitute the value of $$B:$$
  $$A=3	imes 124$$
- step10: Simplify:
  $$A=372$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}A=372\B=124\C=186\end{array}ight.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}A=372\B=124\C=186\end{array}ight.$$
- step13: Rewrite:
  $$\left(A,B,Cight) = \left(372,124,186ight)$$
The solution to the system of equations is:
- $$ A = 372 \, 	ext{m}^2 $$
- $$ B = 124 \, 	ext{m}^2 $$
- $$ C = 186 \, 	ext{m}^2 $$

Therefore, the floor area of each house is:
- Client A's house: $$ 372 \, 	ext{m}^2 $$
- Client B's house: $$ 124 \, 	ext{m}^2 $$
- Client C's house: $$ 186 \, 	ext{m}^2 $$
- Client A's house: $$ 372 \, 	ext{m}^2 $$ - Client B's house: $$ 124 \, 	ext{m}^2 $$ - Client C's house: $$ 186 \, 	ext{m}^2 $$

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