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In the first consignment, 12% bulbs were faulty. In the second

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Joined: 26 Feb 2016
Posts: 2587
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In the first consignment, 12% bulbs were faulty. In the second [#permalink]

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New post 12 May 2018, 14:14
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Difficulty:

  45% (medium)

Question Stats:

71% (01:02) correct 29% (02:28) wrong based on 24 sessions

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In the first consignment, 12% bulbs were faulty. In the second consignment, 4% bulbs were faulty.
In the two consignments, 7% bulbs were faulty. If the two consignments combined had 4000 bulbs,
how many faulty bulbs did the first consignment have?

A. 60
B. 100
C. 175
D. 180
E. 300

Source: Experts Global

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Joined: 22 May 2016
Posts: 1680
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In the first consignment, 12% bulbs were faulty. In the second [#permalink]

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New post 12 May 2018, 18:31
pushpitkc wrote:
In the first consignment, 12% bulbs were faulty. In the second consignment, 4% bulbs were faulty.
In the two consignments, 7% bulbs were faulty. If the two consignments combined had 4000 bulbs,
how many faulty bulbs did the first consignment have?

A. 60
B. 100
C. 175
D. 180
E. 300

Source: Experts Global

Weighted average, in which
\(x\) = number of bulbs in first consignment
\(4000 - x\) = number of bulbs in second consignment

\(.12(x) + .04(4000-x) = .07(4000)\)
\(.12x + 160 -.04x = 280\)
\(.08x = 120\)
\(x = \frac{120}{.08}=\frac{12000}{8}=1500\)

\(x\) is the number of bulbs in first consignment.
\(12\) percent are faulty
\(12\) percent of \(1500\) is
180, Answer D



The formula I used.
(%) = percent faulty
A = consignment #1
B = consignment #2

(% A)(# of A) + (% B)(# of B) = (% of A+B)(# of A+B)

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In the first consignment, 12% bulbs were faulty. In the second   [#permalink] 12 May 2018, 18:31
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In the first consignment, 12% bulbs were faulty. In the second

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