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? Clear 0A. 440 B. 200 C. 250 D. 320 E. 500

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slicergiza slicergiza · Answer 1 · 2019-12-09T12:03:57+01:00

Answer:

The correct option is A. 440

Step-by-step explanation:

Let a, b and c represents the number of orders from forms A, B and C,

According to the question,

[tex]a = 40[/tex],

[tex]b = 4a + \frac{1}{2}c[/tex]

[tex]c = a + \frac{1}{2}c[/tex]

From these equations we get,

[tex]b = 160 + \frac{1}{2}c----(1)[/tex]

[tex]c = 40 + \frac{1}{2}b---(2)[/tex]

Substitute the value of b from (1) to (2),

[tex]c = 40 + \frac{1}{2}(160 + \frac{1}{2}c)[/tex]

[tex]c = 40 + 80 + \frac{1}{4}c[/tex]

[tex]c-\frac{1}{4}c = 120[/tex]

[tex]\frac{4c-c}{4}=120[/tex]

[tex]\frac{3c}{4}=120[/tex]

[tex]c=\frac{480}{3}=160[/tex]

From (1),

[tex]b = 160 + \frac{1}{2}(160)=160 + 80 = 240[/tex]

[tex]\because a + b + c = 40 + 240 + 160 = 440[/tex]

Hence, he used 440 forms total,

i.e. OPTION A is correct.