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30 Apr 2025, 21:51
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Cost of Producing
one white case: \(B+2\) Maximum Number
of each colour case: \(\frac{200,000}{2B+2}\)
one white case: \(B+2\) Maximum Number
of each colour case: \(\frac{200,000}{2B+2}\)
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (02:32) correct
15%
(02:59)
wrong
based on 314
sessions
History
Date
Time
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A major manufacturer of MP3 players wants to produce both black and white plastic cases for its new 32 GB MP3 line. The manufacturer wants to spend a maximum of $200,000 dollars on case production. White plastic cases cost $2 more per case to produce than black plastic cases.
Using W to represent the cost of producing each white case and B to represent the cost of producing each black case, select the expression that reflects the cost of producing one white case, and select the expression that reflects the maximum number of each type of case that can be produced within the company’s $200,000 budget if the same number of black and white MP3 cases are made. Make only two selections, one in each column.
Using W to represent the cost of producing each white case and B to represent the cost of producing each black case, select the expression that reflects the cost of producing one white case, and select the expression that reflects the maximum number of each type of case that can be produced within the company’s $200,000 budget if the same number of black and white MP3 cases are made. Make only two selections, one in each column.
|
Cost of Producing one white case |
Maximum Number of each colour case |
Expression-------------- |
| \(B+2\) | ||
| \(2W+B\) | ||
| \(\frac{200,000}{2B+2}\) | ||
| \(200,000(W+B)\) | ||
| \(\frac{200,000B}{2W}\) |
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Official Answer
Cost of Producing
one white case: \(B+2\) Maximum Number
of each colour case: \(\frac{200,000}{2B+2}\)
one white case: \(B+2\) Maximum Number
of each colour case: \(\frac{200,000}{2B+2}\)
01 May 2025, 03:35
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Solution:
Given in the question stem:
1. The manufacturer can spend $200,000 on total case production.
2. The cost of producing each white case = W
3. The cost of producing each black case = B
4. White plastic cases cost $2 more per case to produce than black plastic cases. W = B + 2
By using the formula Average = \(\frac{Total number of observations}{No of observations}\), we can solve the question.
1. The cost of producing one white case = W = B + 2.
2. The maximum number of each type of case that can be produced within the company’s $200,000 budget if the same number of black and white MP3 cases are made = B + 2 + B = 200,000 = \(\frac{200,000}{2B + 2}\)
Given in the question stem:
1. The manufacturer can spend $200,000 on total case production.
2. The cost of producing each white case = W
3. The cost of producing each black case = B
4. White plastic cases cost $2 more per case to produce than black plastic cases. W = B + 2
By using the formula Average = \(\frac{Total number of observations}{No of observations}\), we can solve the question.
1. The cost of producing one white case = W = B + 2.
2. The maximum number of each type of case that can be produced within the company’s $200,000 budget if the same number of black and white MP3 cases are made = B + 2 + B = 200,000 = \(\frac{200,000}{2B + 2}\)
Bismuth83
16 Jun 2025, 08:31
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Let's plug in numbers.
8 for each Black (B)
Cost of one white case.
8 + 2 = 10
Option 1 -
B + 2. (Correct)
Second part:
Let's say the manufacturer would produce 20 in total, then 10 black and 10 white.
Apply the formula Cost x Quantitity = Total Cost considering the two products.
8x10 + 10x10 = 180 (too small, try 22 in total, 11 each case)
8x11 + 10x11 - 198 (we can't go further, otherwise it would be higher than 200.)
Checking in the answers:
Option 2 -
2W + B
2x10 + 8 = 28 (not eleven)
Option 3 -
200.000/(2B + 2)
200.000/18 = 11111.11 (Correct)
Since we reduced three 0's in the original number, this answer is actually giving us 11.11 (consider 200 in the numerador, the only difference is the 0's)
Moreover, we can't have 11.11 as maximum, we only can produce 11 or 12, hence we have 11 as answer here.
8 for each Black (B)
Cost of one white case.
8 + 2 = 10
Option 1 -
B + 2. (Correct)
Second part:
Let's say the manufacturer would produce 20 in total, then 10 black and 10 white.
Apply the formula Cost x Quantitity = Total Cost considering the two products.
8x10 + 10x10 = 180 (too small, try 22 in total, 11 each case)
8x11 + 10x11 - 198 (we can't go further, otherwise it would be higher than 200.)
Checking in the answers:
Option 2 -
2W + B
2x10 + 8 = 28 (not eleven)
Option 3 -
200.000/(2B + 2)
200.000/18 = 11111.11 (Correct)
Since we reduced three 0's in the original number, this answer is actually giving us 11.11 (consider 200 in the numerador, the only difference is the 0's)
Moreover, we can't have 11.11 as maximum, we only can produce 11 or 12, hence we have 11 as answer here.
