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

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All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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06 Apr 2017, 10:08
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Question Stats:

74% (02:09) correct 26% (01:13) wrong based on 74 sessions

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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
[Reveal] Spoiler: OA

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Re: All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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06 Apr 2017, 20:35
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KUDOS
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.
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Re: All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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07 Apr 2017, 01:49
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KUDOS
Expert's post
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

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 = $$\frac{15}{16}$$

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Re: All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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07 Apr 2017, 06:34
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

Assigning values is less of a headache.

Warehouse A
Total = 60 widgets
Defective = $$\frac{1}{20}$$ 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 $$\frac{15}{16}$$
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Posts: 906
Re: All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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07 Apr 2017, 18:59
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

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
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All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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08 Apr 2017, 01:16
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

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Re: All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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14 Apr 2017, 04:42
1
KUDOS
Expert's post
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

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:

[(14/5)w + (19/20)w]/(3w + w)

Let’s multiply this expression by 20/20:

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

(75w)/(80w)

15/16

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Re: All the widgets manufactured by Company X are stored in two warehouses [#permalink]

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08 Feb 2018, 07:01
VeritasPrepKarishma wrote:
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

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 = $$\frac{15}{16}$$

Re: All the widgets manufactured by Company X are stored in two warehouses   [#permalink] 08 Feb 2018, 07:01
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