Question

A salesman used three types of order forms: A, B, C. He used 40 of Form A. Of Form B he used four times as many as Form A plus 1/2 as many as Form C. He used as many of Form C as Form A plus 1/2 as many as Form B. Altogether, how many forms did he use? a. 440 b. 200 c. 250 d. 320 e. 500

Answer 1

Answer:

Given that

A = 40

B = 4A + 0.5C = 160 + 0.5C  .......... (i)

C =  A +  0.5B =  40 + 0.5 B  .......... (ii)

multiply (ii) with 2

B = 2C -40              .......... (iii)

B = 0.5C + 160        ...........(i)

subtract (i) from (iii)

0 = 1.5C -200

C = 400/3

put in (i)

B = 160 + 0.5(400/3) = 680/3

total forms = A+B+C = 40 + 680/3 + 400/3 = 487 approx

So e.500 is the better option

Answer 2

Answer:

a. 440

Step-by-step explanation:

The information provided allows us to set up the following system of linear equations:

$A = 40\\B=4A+0.5C\\C=A+0.5B$

We already have the value for A, solving the system gives us the values for B and C:

$A = 40\\B=(4*40)+0.5C\\C=40+0.5B\\\\B -B-80 = (4*40)+0.5C-2C\\C=\frac{160+80}{1.5}\\C=160\\B= (4*40) +(0.5*160)\\B=240$

The total number of forms used is:

$A+B+C = 40+240+160 = 440$

He used 440 forms.