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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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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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QZ
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:



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


Hi
Can you please explain what does the ratio '3/8' signify?
I understand on top row we have 12 and 4: %faulty bulb, then middle row gives aggregate %faulty bulb i.e. 7% ; now what does bottom row give? is it the total bulb in each consignment? if so, I don't get how?

Thanks in advance!
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EgmatQuantExpert
QZ
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:



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


Hi
Can you please explain what does the ratio '3/8' signify?
I understand on top row we have 12 and 4: %faulty bulb, then middle row gives aggregate %faulty bulb i.e. 7% ; now what does bottom row give? is it the total bulb in each consignment? if so, I don't get how?

Thanks in advance!

medha312, the bottom row simply gives the ratio of the 2 parties involved. Here the ratio of the 1st assignment to 2nd assignment = 3:5.
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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?

Let the first consignment have x bulbs, then the second consignments will have 4000-x bulbs

According to given question,
=>0.12x+0.04(4000-x)=0.07*4000
=>0.08x+16=28
=>0.08x=12
=>0.08x=12
=>x=1500

Faulty bulbs in 1st consignment => 0.12x=180
Hence D
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Lets use the teeter totter method:
4%. 7%. 12%

Now the avg.of (12+4)/2= 8%
So the 7% is between 4 & 8

Therefore consigment 2 has more bulbs than C1

The ratio will be (12-7)/(7-4)=
5:3
Therefore 3/8*4000= 1500 in C1

Now 1500*12% (faulty bulbs in C1) = 180

Ans D

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