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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor

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BillyButcher0956

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24 Dec 2024, 22:52

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8x + 12y = 240,000

As all options are in '000, lets assume X = x/1000 and Y = y/1000
ie 8X + 12Y = 240
Simplify, 2X + 3Y = 60

A) If Y=10 => X= 15
B) If Y=12 =>X=12 CORRECT

Hence answer would be 12,000 & 12,000

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Bunuel

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Let x be the number of 8 pound bags
Let y be the number of 12 pound bags
10,000<x/y<20,000

8x + 12y = 240,000
2x + 3y = 60,000

By try and error, if x = 12,000, y = 12,000 within the range of 10,000 to 20,000

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25 Dec 2024, 00:04

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Hi All,

given

8x+12y=240000

rearraging the equation

x=(240000-12y)/8

the only option value for which this will solve is when y is 12,000 (given the constraint that x an y values are between 10,000 and 12,000)

Therefore x is also 12,000

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Let the number of 8-pound and 12-pound bags be a and b. We have:

8a + 12b = 240000
b = 20000 - 2a/3

So a must be divisible by 3. The only solution is a = 12000. Then b = 12000.

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I would say (12000,12000)

8x+12y=240,000
2x+3y=60,000
x=12000 and y= 12000 is the only solution

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25 Dec 2024, 01:46

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Ans: 12,000 and 12,000 bags

Bunuel

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given 8*(x,000) + 12*(y,000) = 240,000 => 8x+12y =240

x = (240 -12y)/8
x = 30 - (3/2)*y = now put values of y and get the x
when y = 12 , x = 12

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25 Dec 2024, 01:53

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Hi everyone

Straight forward.

240,000 pounds = 8*x+12*y
60,000 = 2x + 3y
10,000 <= x/y <= 20,000

x=12
y=12

24,000+36,000 = 60,000

Our two answers are 12,000.

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25 Dec 2024, 02:26

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8m+12b =240000
Only 12000,12000 fit from answer choices

Ans 12000,12000

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HIT AND TRIAL:

total bags = 240,000

To be divided into 2 types of bags, 8 pounds and 12 pounds.

So, 240,000 should be divided into 2 multiples of 8 and 12 each such that the result comes out to be between 10,000 and 20,000.

If we take 10,000 bags of 8 pounds each, that would mean 80,000 pounds. That leaves 160,000 pounds which is not divisible by 12. Hence, incorrect.
If we take 12,000 bags of 8 pounds each, that would mean 96,000 pounds. That leaves 144,000 pounds which would be 12,000 bags of 12 pounds each.

Therefore, 12,000 and 12,000.

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8-pound * 8 + 12-pound * 12 = 240,000

12-pound = 20,000 - 2*8-pound/3

To be positive integers, 8-pound must have 3 as a factor.
12,000 has 3 as a factor.

Answers:

8-pound bags=12,000
12-pound bags=12,000

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25 Dec 2024, 04:04

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Easy one
just frame the equation as
8x+12y=240,
now out of the options check the values which can satisfy
one such solution is if x=12 and y=12
as 8*12+12*12= 96+144=240
hence 12,12 is the answer.

Bunuel

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Given that 8*x + 12*y = 240,000

10,000 =< x,y =< 20,000

Checking for different cases of x, for which will satisfy a value given

If x = 10,000

y = 160,000/12 - which is not possible

If x = 12,000

y = 144,000/12 = 12,000 - possible

If x = 14,000

y = 128,000/12 - not possible

If x = 16,000

y = 112,000/12 - not possible

If x = 20,000

y = 80,000/12 - not possible

Therefore x = y = 12,000 is the correct answer

Bunuel

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Bunuel

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240000 pounds of coffee
8-pound and 12-pound bags (8 + 12 = 20)
240000/20 = 12000
It has also been mentioned that they have used between 10000 and 20000 bags of each size.

12000 * 8 = 96000 and 12000 * 12 = 144000 (since 144000 + 96000 = 240000)

96000/8 = 12000 (8-pound bags)
144000/12 = 12000 (12-pound bags)

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Bunuel

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Let x be the number of 8-pound bags and y the number of 12-pound bags.
The total weight of coffee is: 8x+12y=240,000
The constraints on the number of bags are: 10,000≤x≤20,000 and 10,000≤y≤20,000
Solving 8x+12y=240,000
2x + 3y = 60000

Trying various values for x
If x = 10000, y = 40000/3, Not possible (y should be an integer)
If x = 12000, y = 36000/3 = 12000, Possible
If x = 14000, y = 32000/3, Not possible (y should be an integer)
If x = 16000, y = 28000/3, Not possible (y should be an integer)
If x = 20000, y = 20000/3, Not possible (y should be an integer)

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25 Dec 2024, 04:57

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Let x be the number of 8-pound bags & y be the number of 12 pound bags
8x + 12 y = 240000
10,000 > or equal to x,y > or equal to 20,000

After substituting the options in the equation, with the present constraint only
x = 12000 & y = 12000 works

IMO 12000 & 12000

Bunuel

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25 Dec 2024, 05:05

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A coffee distributor must package their entire annual production of 240,000 pounds, using only a combination of 8-pound and 12-pound bags.

If they used between 10,000 and 20,000 bags of each size, select for 8-pound bags the number of 8-pound bags to be used, and select for 12-pound bags the number of 12-pound bags to be used. Make only two selections, one in each column.

Let’s assume, there are 10000 bags of 8 pound bags. So, 10000x8=80000. Then, 12 pounds bag would be, 240000-80000/12000=160000/12= is not an integer.

Let’s assume, there are 12000 bags of 8 pound bags. So, 12000x8=96000. Then, 12 pounds bag would be, 240000-96000/12000=144000/12= 12.

Answer: there are 12000 bags of 8 pound bags. So, 10000x8=80000. Then, 12 pounds bag would be 12000.

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25 Dec 2024, 06:06

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8-pound bags	12-pound bags
		10,000
		12,000
		14,000
		16,000
		20,000

say, 8-pound bags = x
12-pound bags = y

then,

8x + 12y = 240000
2x + 3y = 60000........(1)

Only, the corresponding pairs with x = 12000, y = 12000 (from amongst the given prompts) fits the Equation (1)

So, 8-pound bags = 12000
12-pound bags = 12000 is the CORRECT answer

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Let no. of 8-pound bags= a
Let no. of 12- pound bags= b

Total weight of packaged coffee= 240000
therefor, 8a+12b=240000=> 2a+3b=60000

Given,
10,000<=a<=20,000, 10,000<=b<=20,000.

So, we need to find the value of a and b that satisfies the equation- 2a+3b=60000 as well as lies in the above given range.

So, by doing hit and trial from the given opts. for a, we can find the value of b.

Ans. a= 12000 and b=12000 only satisfies

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25 Dec 2024, 07:06

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Quote:

I just set up a formula and start testing

Let's call number of 8lb bags a, 12lb bags b
8a + 12b = 240k lbs
a + b between 20k and 30k inclusive

- a = 10k -> 12b = 160k. b not integer. Eliminate
- b = 12k -> 12b = 144k. b is 12k. Make 12,000 choice for both a and b in table

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