All the widgets manufactured by Company X are stored in two warehouses

Question

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

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

IdiomSavant · Accepted Answer

Assigning values is less of a headache.

Warehouse A 
Total = 60 widgets
Defective =  \frac{1}{15}  or 4 of 60

Warehouse B 
Total = 20 widgets 
Defective =  \frac{1}{20}  1 of 20

Total defective  \frac{5}{80} 
Total not defective  \frac{75}{80}  or  \frac{15}{16}

akleiman · Answer

Let be A and B the number of widgets in warehouse A and B respectively. Let be X the total number of widgets for Company X.

We know that: A + B = X.

" Warehouse A contains three times as many widgets as does warehouse B " => A = 3B => 4B = X

"1/15 of the widgets in warehouse A and 1/20 of the widgets in warehouse B are defective" =>

14/15 of the widgets in warehouse A and 19/20 of the widgets in warehouse B are non-defective.

(14/15)*A + (19/20)*B = (15/4)*B (Because A = 3B)

4B is the total of the widgets in Company X, therefore (15/4)*B is the (15/16) fraction of Company X. So the correct answer is C.

KarishmaB · Answer

Use weighted average:

\frac{w1}{w2} = \frac{3}{1} = \frac{(1/20 - x)}{(x - 1/15)}

3 * (x - \frac{1}{15}) = (\frac{1}{20} - x)

x = \frac{1}{16}

The fraction of NOT defective widgets =  \frac{15}{16}

Answer (C)

gracie · Answer

let total widgets=4x
1/15*3x+1/20*x=x/4
(x/4)/4x=1/16=fraction of defective widgets
1-1/16=15/16=fraction of non-defective widgets
C

nikithaiyer5 · Answer

A - 3x 
B - 1x
Fraction of A - 3x/4x = 3/4
Fraction of B - x/4x = 1/4

Defective in A = 1/15 
Non defective A = 1-1/15 = 14/15
==> 14/15 * (fraction of A )= (14/15) * (3/4)

= 7/10

Now B,
Defective = 1/20
Non defective = 19/20

==> (19/20) * (1/4) = 19/80

Total non defective = (7/10)+(19/80) = 75/80 = 15/16

Therefore, Answer C .

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ScottTargetTestPrep · Answer

We can let the number of widgets in warehouse B = w, and since warehouse A has 3 times as many widgets as warehouse B, the number of widgets in warehouse A is 3w.

Since 1/15 of the widgets in warehouse A are defective, 1 - 1/15 = 14/15 are not defective. Thus, (14/15)(3w) = (14/5)w widgets in warehouse A are not defective.

Since 1/20 of the widgets in warehouse B are defective, 1 - 1/20 = 19/20 are not defective. Thus, (19/20)w widgets in warehouse B are not defective.

Thus, the fraction of all widgets that are not defective is:

/(3w + w)

Let’s multiply this expression by 20/20:

(56w + 19w)/(60w + 20w)

(75w)/(80w)

15/16

Answer: C

MvArrow · Answer

VeritasPrepKarishma Can you please tell me where I can learn about this method? It would be widely helpful! Thanks a lot

nimaki · Answer

Be calm. You got this.
Substitute x for a number 80
x = 20
3x = 60
Defective A: 60 * 1/15 = 4
Defective B: 20 * 1/20=1
Total Defective = 5

Ratio of Defective to Total : 5/80
Good widgets = 1 - 5/80 = 75/80 = 15/16

BrettonJames · Answer

But arent 1/15 defect in warehouse A?

Krunaal · Answer

It must be a typo. 4 of 60 is 4/60 = 1/15

All the widgets manufactured by Company X are stored in two warehouses

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