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24 Dec 2024, 11:02
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We can have the following equation (assuming x bags of 8-pound bags, and y bags of 12-pound bags):
8x + 12y = 240,000,
which simplifies to:
2x + 3y = 60,000.
Looking at the options, we can easily eliminate 10,000 ; 14,000 ; 16,000 and 20,000 as values of x since these aren't multiple of 3 and thus wouldn't yield an integer value for y
So, let's use x = 12,000 to calculate y:
2(12,000) + 3y = 60,000
24,000 + 3y = 60,000
3y = 36,000
y = 12,000
Thus, x = 12,000 and y = 12,000.
8x + 12y = 240,000,
which simplifies to:
2x + 3y = 60,000.
Looking at the options, we can easily eliminate 10,000 ; 14,000 ; 16,000 and 20,000 as values of x since these aren't multiple of 3 and thus wouldn't yield an integer value for y
So, let's use x = 12,000 to calculate y:
2(12,000) + 3y = 60,000
24,000 + 3y = 60,000
3y = 36,000
y = 12,000
Thus, x = 12,000 and y = 12,000.
24 Dec 2024, 11:02
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Bunuel
Reducing all the values by 1000
The number of 8 pounds bag = x
The number of 12 pound bags = y
\(8x + 12y = 240\)
\(2x + 3y = 60\)
\(x = 30 - y - \frac{y}{2}\)
y = 2 ; x = 30 - 2 - 1 = 27
The other values of (x,y) = (24,4) ; (21,6) ; (18,8) ; (15,10) ; (12 , 12) ....
Hence,
8-pound bags = 12-pound bags = 12,000
Krunaal
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24 Dec 2024, 11:05
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Let x = no. of 8 pound bags, and y = no. of 12 pound bags.
8x + 12y = 240,000
2x + 3y = 60,000
Testing answer choices,
x = 10,000; then 3y = 40,000 => This is not possible as y won't be an integer.
x = 12,000; then 3y = 36,000 => y = 12,000 This is possible, and both bags are between 10,000 and 20,000. No need to test further.
Answer 12,000 and 12,000
8x + 12y = 240,000
2x + 3y = 60,000
Testing answer choices,
x = 10,000; then 3y = 40,000 => This is not possible as y won't be an integer.
x = 12,000; then 3y = 36,000 => y = 12,000 This is possible, and both bags are between 10,000 and 20,000. No need to test further.
Answer 12,000 and 12,000
