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All the widgets manufactured by Company X are stored in two warehouses, A and B. Warehouse A...

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All the widgets manufactured by Company X are stored in two warehouses, A and B. Warehouse A contains three times as many widgets as does warehouse B. If 1/15 of the widgets in warehouse A and 1/20 of the widgets in warehouse B are defective, what fraction of all the widgets are NOT defective?

A. 1/5
B. 1/16
C. 15/16
D. 34/35
E. 74/75

OA C
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Re: All the widgets manufactured by Company X are stored in two warehouses, A and B. Warehouse A...

by Targt780 » Sat Aug 22, 2020 3:07 am
Let the number of widgets in Warehouse A be x and
Let the number of widgets in Warehouse B be y

Given that x=3y (Number of widgets in Warehouse A is 3 times the number of widgets in Warehouse B)

Total number of widgets manufactured by the company = x+y= 3y+y = 4y

Warehouse A:

Number of widgets in the warehouse = 3y

Fraction of defective widgets = \(\frac{1}{15}\)

Fraction of working widgets or NOT defective widgets = 1 - \(\frac{1}{15}\) = \(\frac{14}{15}\)

Number of working widgets in Warehouse A = 3y* \(\frac{14}{15}\) = \(\frac{14y}{5}\)

Warehouse B:

Number of widgets in the warehouse = y

Fraction of defective widgets = \(\frac{1}{20}\)

Fraction of working widgets or NOT defective widgets = 1 - \(\frac{1}{20}\) = \(\frac{19}{20}\)

Number of working widgets in Warehouse B = y* \(\frac{19}{20}\) = \(\frac{19y}{20}\)

Total number of working widgets = Working widgets of Warehouse A + Working widgets of Warehouse B
= \(\frac{14y}{5}\) +\(\frac{19y}{20}\) = \(\frac{56y\ +\ 19y}{20}\) = \(\frac{15y}{4}\)


Now to determine what fraction of the widgets are NOT defective we simply need to divide this by the total number of widgets by the company

\(\frac{\frac{15y}{4}}{4y}\) = \(\frac{15}{16}\)
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