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

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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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Question Stats:

66% (02:08) correct 34% (02:32) wrong based on 111 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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In the first consignment, 12% bulbs were faulty. In the second  [#permalink]

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New post 12 May 2018, 18:31
1
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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Re: In the first consignment, 12% bulbs were faulty. In the second  [#permalink]

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New post 27 Jun 2018, 23:50
Hi,

Can you explain how to solve this by using weighted average method taking percents into consideration.

I mean W1/W2 = 12 - 7/7-4

I am getting 175.

QZ
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Re: In the first consignment, 12% bulbs were faulty. In the second  [#permalink]

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New post 28 Jun 2018, 01:03
QZ wrote:
Hi,

Can you explain how to solve this by using weighted average method taking percents into consideration.

I mean W1/W2 = 12 - 7/7-4

I am getting 175.

QZ


Solution



Given:
    • In the first consignment, 12% bulbs were faulty
    • In the second consignment, 4% bulbs were faulty
    • In the two consignments, 7% bulbs were faulty
    • The two consignments combined had 4000 bulbs

To find:
    • The number of faulty bulbs in the first consignment

Approach and Working:
We can use allegation to find the ratio of bulbs in each of the consignment, as follows:

Image

Therefore, the total number of bulbs in the first consignment = \(4000 * \frac{3}{8} = 1500\)
    • Faulty bulbs in first consignment = 12% of 1500 = \(1500 * \frac{12}{100} = 180\)

Hence, the correct answer is option D.

Answer: D
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Re: In the first consignment, 12% bulbs were faulty. In the second  [#permalink]

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New post 02 Jul 2018, 09:47
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


We can let the number of bulbs in the first consignment = a and the number of bulbs in the second consignment = b and create the equations:

0.12a + 0.04b = 0.07(a + b)

12a + 4b = 7a + 7b

5a = 3b

and

a + b = 4000

b = 4000 - a

Substituting, we have:

5a = 3(4000 - a)

5a = 12000 - 3a

8a = 12000

a = 1,500

The first consignment had 0.12 x 1500 = 180 faulty bulbs.

Answer: D
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Re: In the first consignment, 12% bulbs were faulty. In the second &nbs [#permalink] 02 Jul 2018, 09:47
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