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12 Days of Christmas GMAT Competition 2025 - Day 4: Each of the boxes

1 2 3 4

rahumangal

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18 Dec 2025, 12:10

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Bunuel

Each of the boxes in a shipment weighs either 5 pounds, , or . What is the average (arithmetic mean) weight per box in the shipment?

(1) The number of boxes that weigh 5 pounds is twice as many as the number that weigh 11 pounds.

(2) The ratio of the number of boxes weighing 5 pounds, 7 pounds, and 11 pounds is 4 to 2 to 2.

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Let no of box weighing 5pounds=a 7 pounds=b 11 pounds=c
we are asked avg= [5a+7b+11c][/a+b+c]
statement 1 says a=2c- in sufficient
statement 2 says - a:b:c =4:2:2
so let a= 4k,b=2k, c=2k
so avg = 56k/8k= 7
sufficient
ans= B

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18 Dec 2025, 12:21

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Acc. to ques 5 7 11 -> find avg -> need to now how many were each pounds.

A -> only ratio for 5 pounds and 11 pounds but nothing about 11 pounds => No
B -> ratio: 4:2:2 => avg = 4*5 + 2*7 + 2*11/8 = 7 => Yes is enough

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1 - If 11 pound boxes are 2x in number, 5 pound boxes are x in number.
This does not give the relation with 7 pound box which is necessary for answering.
Hence, statement 1 alone is not sufficient.

2 - 5, 7 and 11 pound boxes are in ratio 4:2:2
i.e. their total weights would be in the ratio 4 x 5 : 2 x 7 : 2 x 11
=> 20 : 14 : 22

Average weight = total weight / total units
=> 20+14+22 / 4+2+2
=> 56 / 8
=> 7 pounds

Hence statement 2 alone is sufficient but 1 alone is not sufficient.

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Bunuel

Each of the boxes in a shipment weighs either 5 pounds, 7 pounds, or 11 pounds. What is the average (arithmetic mean) weight per box in the shipment?

(1) The number of boxes that weigh 5 pounds is twice as many as the number that weigh 11 pounds.

(2) The ratio of the number of boxes weighing 5 pounds, 7 pounds, and 11 pounds is 4 to 2 to 2.

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There are boxes in a shipment where each weighs either 5 pounds, 7 pounds or 11 pounds.

We need to find the average weight per box in the shipment.

Statement 1:

The number of boxes that weigh 5 pounds is twice as many as the number that weigh 11 pounds.

Let the number of boxes weighing 5 pounds : 7 pounds : 11 pounds = a:b:c

Given that: a = 2c

Average =( 5a + 7b + 11 c )/ (a+b+c)

= (5*2c + 7b + 11c) / (2c+b+c)

= (21c + 7b) / (3c+b)

= 7*(3c+b) / (3c+b)

= 7

Hence, sufficient

Statement 2:

The ratio of the number of boxes weighing 5 pounds, 7 pounds and 11 pounds = 4:2:2

Average = (5*4x+ 7*2x + 11*2x) / (4x+2x+2x)

= (20x+ 14x+ 22x)/ 8x

= 56x/8x

= 7

Hence, Sufficient

Option D

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answer b

1) we dont know the number or ratio of 7 pound boxes. insufficent

2) 5 pounds x 4 + 7 pounds x2 + 11 pounds x 2 = 20 + 14 + 22 = 56 divided by 4+2+2=8 56/8 equals 7 sufficient

Bunuel

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Bunuel

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mean = n1xw1+n2w2+n3w3 / n1+n2+n3

lets look at options

1- n of (5 pounds) = 2 x y (n(11)) hence mean formula will look like - 5(2y) + 7(n) + 11 (y) / 3y+n

= 21y+7n/3y+n

taking 7 common

7(3y+n)/3y+n leaving us with 7 as the answer hence sufficient.

2. Pretty straightforward option imo. Number of is given ..substitute the n as 4k,2k,2k.

4k x 5 + 7k x 2 + 11k x 2 / 4k+2k+2k ..k will get cancelled leaving the answer as 7.. hence sufficient

option D is the answer

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18 Dec 2025, 13:35

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Bunuel

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let quanity by x y z for 5, 7, 11 pound respectively

1) implies x = 2z
so avg =(10z + 7y + 11z ) / (3z + y)
can;t solve alone

2) implies, 2x = y = z
so avg = (10 z + 7z + 11z)/ (4z)
= 7pounds

hence answer is B, statment 2) is enough

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Let number of 5 pound boxes = n, 7 pound boxes = m, 11 pound boxes = r
(5n+7m+11r)/(n+m+r) = ?

1. n = 2r, m=? Insufficient

2. n:m:r = 4x : 2x : 2x
total = 8x

(5*4x + 7*2x + 11*2x)/8x = 56x/8x = 7
Sufficient

Answer B

Bunuel

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here we need to find out the average arithmetic mean weight per box?
1. no information about the no. of 7 pounds boxes so --insufficient
2.since the ratio of the no. of boxes of each type is given as 5 lbs:7 lbs:11 lbs=4:2:2 is clearly sufficient to find out the requisite as, (5*4x + 7*2x + 11*2x)/8x=7 pounds
so B

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There are 3 unknowns - quantity of each weight type boxes

(1) It only tells relationship between 2 of the 3. Can't compute avg from this. Not sufficient.

(2) It ratio of 3 weight type boxes is given. We can create equation from here.

Average weight per box = ((4x)*5+(2x)*7+(2x)*11) / (4x +2x +2x)

But cannot identify x. Hence, not sufficient.

Considering together, we will be able to identify x. Hence, (1) and (2) together is sufficient.

(C) is the answer

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1) No of boxes weighing 5 pounds to 11 pounds is 2:1, the average would be weighed more towards 5 pounds but we still dont know how many 7 pounds boxes are there. If they are the most, average will be closer to it. Not Sufficient

2) No of boxes of 5 pounds to 7 to 11 pounds = 4:2:2 = 2:1:1

let the no of boxes be 2x, x & x

then we have average = 10x + 7x + 11x / 4x = 28x / 4x = 7 pounds

Ans B

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18 Dec 2025, 15:14

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say num of 5 lb = x
num of 7 lb = y
num of 11 lb= a

S-1 x = 2a

10a+7y+11a / 3a +y
21a+7y / 3a+ y
7.
sufficient

S-2 given ratio= x:y:a = 4:2:2
20p + 14p + 22p / 8p
7
sufficient

ans D

Bunuel

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(1) We still need to know the number of boxes that weight 7 pounds so this is NOT SUFFICIENT.
(2) Whether we choose the number of boxes is {4,2,2} or {8,4,4}... The average weight is still (5x4+7x2+11x2)/(4+2+2). SUFFICIENT.

Answer: B

Bunuel

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Let no of boxes having weight 5,7,11 pounds be x,y,z respectively. We need to find average weight per box. let's look at statement 1:

(1) it says x=2z. so , Average= (5x+7y+11z)/(x+y+z)

= (10z+11z+7y)/(3z+y)
=7(3z+y)/(3z+y)
=7

So , 1st statement is sufficient.

(2) relationship between x,y,z is provided in the form of ratios. x:y:z =4:2:2 , so we can find the average weight. If you calculate it will come as 7. So Statement 2 is sufficient.

So answer is D.

Bunuel

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We are asked to find the weighted average weight per box
Let number of boxes for each of the weights be:
5 pounds - a
7 pounds- b
11 pounds - c

Using Weighted average formula: 5a+7b+11c/a+b+c

I) Only gives us a=2c
5(2c) + 7b+11c / 2c+b+2c= 21c+7b/4c+b => not sufficient

II)Let a= 4x, b=2x c=2x
Substituting in formula gives us = 5(4x) +7(2x) = 11 (2x)/ 4x+2x+2x= 56x/8x = 7
Giving a unique solution hence
ans. II alone is sufficient

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Formula for avg weight = 5x +7y + 11z/x+y+z
x , y and z are number of boxes weighing 5 , 7 and 11 pounds .

ST 1 : x= 2z , substituting this 5(2z) + 7 y +11z /2z+y+z , we still don't know about y . ❌
ST 2 : Exact Ratios given , 4:2:2 .
5(4) +7(2) +11(2)/4+2+2= 56/8 = 7 pounds. Sufficient.✅

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(1) This might look like it is not suffiicient, but if we use scale method(weighted averages) and put the numbers on the number line and see, 7 is 2 units away from 5 and 4 units away from 11, which means that if 7 is overall average, then the number of boxes in 5 pounds category will be twice the number of boxes in 11 pounds category (the weights here are the number of boxes in each weight category, and the weights ratio is the inverse ratio of the distances from the average).
Now, this statement gives exactly that. So irrespective of the number of boxes in 7 pounds category, this alone is sufficient to say the average is 7 pounds.
To understand better:
= (5*2x + 11*x + 7*_) / (2x + x + _), here for 7, if you take x, you get 7 as the average. Even if you take 2x, you would still get 7 as the answer.

P.S. If it had been the case that the number of boxes in 5 pounds category is thrice the number of boxes in 11 pounds category, then the overall avg will be greater than 7 but still less than 11. In this case, we would need the number of boxes in 7 pounds category to get the actual average value.

(2) This alone is sufficient, as in the weighted averages formula, the multiplier gets cancelled from the numerator and denominator.
As in (5*4x + 7*2x + 11*2x) / 8x, x gets cancelled, giving 7 as the average. Sufficient.

Option D.

Bunuel

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Statement (1)
Gives us that number of 5lbs boxes = 2 x (Number of 11lbs boxes)
Even if we have 1 variable for 2 types of boxes, i.e, x and 2x there is still 7lbs boxes pending for which we do not know the number of boxes.
Not sufficient.
Statement (2)
Ratio of weight--> 5:7:11
Ratio of boxes--> 4:2:2

So, [ (5 x 4) + (7 x 2) + (7 x 2) ] / (4+2+2) = 56/8 = 7 (Average arithmetic mean)

Sufficient !!

Ans choice B

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19 Dec 2025, 01:28

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Let 5 pound box be x, 7 pound box be y and 11 pound box be z

To find

Avg=(5x+7y+11z)/x+y+z

St-1 x=2z

then Num 5x+7y+11z=10z+7y+11z=21z+7y=7(3z+y)

and Denom x+y+z=2z+y+z=3z+y

So Avg=7
Sufficient

St-2 x:y:z=4:2:2

x=4a,y=2a, z=2a

Inserting these value we again get Avg=7, Sufficient

Option D

Bunuel

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19 Dec 2025, 01:57

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Okay, we know that we have 5 pound, 7 pound and 11 pound boxes as choices. We'll need to find the average. This is a weighted average question. For instance, if there are two 5-pounders, three 7-pounders and one 11-pounder, that's 5*2 + 7*3 + 11*1 = 42kgs / 6 = 7kgs as the average. We know this value will vary depending on different conditions and we need to find the exact answer.

With that in mind, we look at the statements:

Statement I: Okay, cool. We now know that there are exactly twice the number of 5-pound boxes as there are 11-pounders. But will this be enough? For instance, if there is one 11-pounder, we'll get two 5-pounders. If, say, there are three 7-pounders, that'll take the average to 5*2 + 7*3 + 11*1 = 42kgs / 6 = 7kgs. But if there were ten 11-pounders, then there'd be twenty 5-pounders, thus making the average 5*20 + 3*7 + 11*10 = 221 / 37 = ~5.9kgs. Different values. hence, NOT SUFFICIENT.

Statement II: Okay, so, easy. We now know the ratios. This will be enough, because as we increase the count in the same ratio, the average will stay consistent. For instance, as per the 4:2:2 ratio here, we'd get 5*4 + 7*2 + 11*2 = 56 / 8 = 7kgs average, we take the ratios directly.

But if it's 5*40 + 7*20 + 11*20 = 560 / 80 = 7kgs average, we see with the ratio adapting to different values, the answer is still consistent.

Hence, STATEMENT 2 IS SUFFICIENT. B.

Bunuel

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