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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 09:00
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8-pound bags: 12,000 12-pound bags: 12,000
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Difficulty:

25% (medium)

Question Stats:

85% (02:19) correct 15% (02:22) wrong based on 460 sessions

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12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

 


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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.
8-pound bags 12-pound bags
10,000
12,000
14,000
16,000
20,000
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8-pound bags: 12,000 12-pound bags: 12,000
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:00
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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.
Show more


GMAT Club's Official Explanation:



Let x be the number of 8-pound bags and y be the number of 12-pound bags.

Equation: 8x + 12y = 240,000

Check each given option:

10,000 8-pound bags:
8(10,000) = 80,000 from 8-pound bags
Remaining = 160,000 pounds
Required 12-pound bags = 160,000 ÷ 12 = 13,333.33...
Not valid (requires non-whole number of bags)

12,000 8-pound bags:
8(12,000) = 96,000 from 8-pound bags
Remaining = 144,000 pounds
Required 12-pound bags = 144,000 ÷ 12 = 12,000
Valid solution! (12,000 is within range)

14,000 8-pound bags:
8(14,000) = 112,000 from 8-pound bags
Remaining = 128,000 pounds
Required 12-pound bags = 128,000 ÷ 12 = 10,666.67...
Not valid (requires non-whole number of bags)

16,000 8-pound bags:
8(16,000) = 128,000 from 8-pound bags
Remaining = 112,000 pounds
Required 12-pound bags = 112,000 ÷ 12 = 9,333.33...
Not valid (requires non-whole number of bags and below 10,000 minimum)

20,000 8-pound bags:
8(20,000) = 160,000 from 8-pound bags
Remaining = 80,000 pounds
Required 12-pound bags = 80,000 ÷ 12 = 6,666.67...
Not valid (requires non-whole number of bags and below 10,000 minimum)

Answer: Only the combination of 12,000 8-pound bags and 12,000 12-pound bags satisfies all requirements.
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 09:25
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We need to package 240,000 pounds of coffee using 8-pound and 12-pound bags. We are given that the number of bags used for each size is between 10,000 and 20,000.

Let:

x = number of 8-pound bags
y = number of 12-pound bags
The total weight equation is: 8x + 12y = 240,000

Simplify by dividing by 4: 2x + 3y = 60,000

Now, solve for different values of x:

If x = 10,000: 2(10,000) + 3y = 60,000 → 20,000 + 3y = 60,000 → 3y = 40,000 → y = 13,333.33 (not an integer)

If x = 12,000: 2(12,000) + 3y = 60,000 → 24,000 + 3y = 60,000 → 3y = 36,000 → y = 12,000 (valid solution)

If x = 14,000: 2(14,000) + 3y = 60,000 → 28,000 + 3y = 60,000 → 3y = 32,000 → y = 10,666.67 (not an integer)

If x = 16,000: 2(16,000) + 3y = 60,000 → 32,000 + 3y = 60,000 → 3y = 28,000 → y = 9,333.33 (not an integer)

If x = 20,000: 2(20,000) + 3y = 60,000 → 40,000 + 3y = 60,000 → 3y = 20,000 → y = 6,666.67 (not an integer)

Conclusion: The only valid solution is:

8-pound bags: 12,000
12-pound bags: 12,000
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 09:29
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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.
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8x + 12y = 240,000
2x + 3y = 60,000

12000 * 5 = 60000

So, x = y = 12000

Even, try with different option values to eliminate choices

if x = 10000, y = 40000/3 --- Not possible
if x = 14000, y = 32000/3 --- Not possible
if x = 16000, y = 28000/3 --- Not possible
if x = 20000, y = 10000/3 --- Not possible
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 09:53
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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.

8-pound bags100001200014000160002000015000900060000
12-pound bags1333312000106679333666710000140001600020000

Since only options of 12000, 12000 is provided and there is no option to select 15000,9000,6000 or 0, 8-pound bags = 12000 & 12-pounds bags = 12000

8-pound bags12-pound bags
1200012000
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 09:53
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8 pounds and 12 pounds of bags
For both number of bags used between 10,000 and 20,000

Now, total pounds = 2,40,000

8*x + 12*y = 2,40,000

Satisfying from the options we get
x = y = 12,000

Answer: (12,000; 12,000)
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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.
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4x + 8x = 240000
12x = 240000
x= 12000

so, 8 pound bags = 8*12000 = 96000
12 pound bags = 12*12000 = 144000

Total = 240000

IMO 12000 nos for both 8 & 12 pound bags
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:07
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We need to package 240,000 pounds of coffee using a combination of 8-pound and 12-pound bags, and the number of each type of bag must be between 10,000 and 20,000. Let’s denote:

x
x: Number of 8-pound bags.
y
y: Number of 12-pound bags.
The total weight equation is:

8
x
+
12
y
=
240
,
000
8x+12y=240,000
Simplifying the equation:
Divide through by 4:

2
x
+
3
y
=
60
,
000
2x+3y=60,000
Constraints:
10
,
000
≤
x
≤
20
,
000
10,000≤x≤20,000
10
,
000
≤
y
≤
20
,
000
10,000≤y≤20,000
Solving for
y
y in terms of
x
x:
Rearrange
2
x
+
3
y
=
60
,
000
2x+3y=60,000 to solve for
y
y:

3
y
=
60
,
000
−
2
x
3y=60,000−2x
y
=
60
,
000
−
2
x
3
y=
3
60,000−2x
​

For
y
y to be an integer,
60
,
000
−
2
x
60,000−2x must be divisible by 3.

Testing integer solutions:
Let’s test possible values of
x and check if y
y falls within the range
10,
10,000≤y≤20,000.

Case
x=10,000:
y=(Not an integer)
.
y= 60,000−2(10,000) = 60,000−20,000]/3
​40,000/3 =13,333.33(Not an integer).
Case
x=12,000:
y= [60,000−2(12,000) ]/3
​
= 36,000/3 =12,000.
This works!
y=12,000.

Case
x = 666.67
(Not an integer)
.
y= 60,000−2(14,000)
60,000−28,000
​
= 32,000/3
​
=10,666.67(Not an integer).
Case
x=16,000:
y = 28
Namol000
3 = 9,333.33
(Not valid as y = 60,000−2(16,000)
60,000−32,000 = 28,000/3 = 9,333.33(Not valid as y<10,000).

Case
x=20,000:

(Not valid as (60,000−40,000) / 3
=> 20,000/3 = 6,666.67(Not valid as y<10,000).
Solution:
The only valid combination is:
x=12,000(8-pound bags)and y=12,000(12-pound bags).

Selections:
8-pound bags: 12,000
12-pound bags: 12,000
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:10
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If we used equal number of bags, say x, to pack 8oz and 12oz coffee, then we have

8x +12x = 240000
20x = 240000
x = 12000
given this is in the range, we choose 12000 bags for both
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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.
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:11
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Total pounds = Positive difference(Total pounds in 12-pound bags, Total pounds in 8-pound bags)
Total pounds = 240000;
Assume the number of 8 pound bag = 12000;
Total pounds in 8-pound bags = 96000;
Therefore, Total pounds in 12-pound bags = 144000

So the number of 12-pound bags = 144000/12 = 12000;

Option B for both the columns
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:13
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The correct answer is 12,000 8-pound bags and 12,000 12-pound bags

1. If we take 12,000 8-pound bags, 96000 (12000*8) pounds of coffee can be packed. Remaining coffee (240000-96000 = 144000 pounds) can be packed in 144000/12 = 12000 12-pound bags.

Similarly, on trying other options, we get

2. If we take 10,000 8-pound bags, 80000 (10000*8) pounds of coffee can be packed. Remaining coffee (240000-80000 = 160000 pounds) can be packed in 160000/12 bags - Not getting a round number - Out

3. If we take 14,000 8-pound bags, 112000 (14000*8) pounds of coffee can be packed. Remaining coffee (240000-112000 = 128000 pounds) can be packed in 128000/12 bags = 10666.67 - Not getting a round number - Out
We will not get round numbers using 16000 and 20000 8-pound bags.


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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.
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:15
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Given 8x + 12y =240000
=> 2x + 3y = 60000
Substituting each of the values in the options for x, we arrived at x = 12000, y = 12000 is the only valid combination.
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:19
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8x+12y=240 (ignoring 1000 as makes life much more simple)

2x+3y=60

now x should be a multiple of 3 then only y is an integer. {my quick way of checking RHS is divisible by 3 so LHS should also be divisible by 3, for which x must be a multiple of 3}. Only option present is 12.

So 2*12+3*y=60 hence y=12 as well. So 12000 for both

PS: i did end up double checking my calculations because its not often that you see same option being ans for both
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:28
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Between means 10,000 and 20,000 are not included so ignore the extreme limits in the option choices
10,000< number of 8 pd/12 pd bags < 20,000

putting in the values for 8pd bags we get
96,000 we get 144,000 - divides 12, and we get 12,000
112,000 minus this from 240,000 and we get 128,000 not a multiple of 12 (rejected)
128,000 minus this from 240,000 and we get 112,000 not a multiple of 12 (rejected)
Answer is 12,000 and 12,000
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:31
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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.
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Since the smallest bag size from the column is 10k, let's start with it:

8*10k + 12*x = 240k
x won't be integer hence 10k not possible.

Next let's try 12k
8*12k + 12*x = 240k
96k + 12x = 240k
12x = 144k
x = 12k

Hence 12k, 12k bags of 8pound and 12pound bags
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:42
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With the first line If they used between 10,000 and 20,000 bags of each size, eliminate options 10K and 20K.
Now pick 14000-- 12-P bags-- 168000 Pounds- So we need 72000/8 ---> 8P bags= 9000 which is <10000.

We can now directly pick 12000 - 12P bags as taking 16000 - 12P bags would further reduce the 8P bags from 9000.
So answer is 12000- 12P bags and 12000-8P bags.

12000*12=144000
240000-144000= 96000--->96000/8 = 12000. This satisfies the condition.
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:50
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number of 8 pound bags + number of 12 pound bags = 240000
8x + 12y = 240000
2x + 3y = 60000
10000<= x,y <= 20000

From the options taking x,y = 12000 will satisfy the equation

2(12000) + 3(12000) = 60000

Hence number of 8 pound bags are 12000
and the number of 12 pound bags are 12000
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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.
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:53
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we have to divide 240,000 pounds into packets of 8 pound and 16 pound

we can form an equation in 2 variable - 8x + 16y = 240000
and use trial and error for each option

8 pound12 pound
10,00080,000120,000
12,00096,000144,000
14,000112,000178,000
16,000128,000198,000
20,000160,000240,000

when both 8 and 12 pounds are in 12,000 quantity each do wet get the total amount




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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.
Show more
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:54
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let no. of 8pound bags be x
let no. of 12 pound bags be y
8x + 12y = 240000

Putting all the values in options in the equation
it only satisfys for x=y=12,000

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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.
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nishantswaft
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12 Days of Christmas GMAT Competition - Day 7: A coffee distributor 24 Dec 2024, 10:58
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In TPA questions its always better to form a strategy first before solving.

The stem tells us that;
  • A coffee distributor must package their entire annual production of 240,000 pounds, using only a combination of 8-pound and 12-pound bags.
  • they used between 10,000 and 20,000 bags of each size

And we have to select;
  • for 8-pound bags the number of 8-pound bags to be used (a)
  • for 12-pound bags the number of 12-pound bags to be used (b)

Let's form an equation;

8a+12b=240000
Or, 2a+3b= 60000

Since 10000<a or b<20000

Let's manipulate the equation;

2a=60000-3b
OR, 2a=3(20000-b)

Since a and b will have integer value only and 2 is not a multiple of 3 so definitely a is a multiple of 3.

Looking at the options
Only possible value for a =12000
Substituting this in the equation;
Value for b= 12000

Hence we can fill these respective options.
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