Five boxes together weigh 100 pounds. What is the median weight of the

Question

Five boxes together weigh 100 pounds. What is the median weight of the five boxes?
 
(1) One of the boxes weighs 20 pounds.
(2) When arranged in ascending order, the weights of the boxes form an arithmetic progression i.e., the difference between the weights of any two consecutive boxes is the same.

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agrasan · Answer

Five boxes together weigh 100 pounds. What is the median weight of the five boxes?

(1) One of the boxes weighs 20 pounds.
 Insufficient , we don't know what other weights are so we can't come up with a certain median value.

(2) When arranged in ascending order, the weights of the boxes form an arithmetic progression i.e., the difference between the weights of any two consecutive boxes is the same.
 Insufficient , we don't know what's the firm term and constant difference is, hence, this alone is insufficient.

Together (1) and (2),
Let's say the first term is a and difference is d
The sum is 5a + 15d = 100
 a + 3d = 20 
But the median term is a + 2d

We already know from S(1) that one of the values is 20 so even after combining (1) and (2), we don't have enough info to comment on median value.

(E)  is the answer.

akash9 · Answer

Statement-2 is sufficient as sum of AP is a+(a+d)+....+(a+4d)= 5a + 10d =100

--> a + 2d --> Median

JDAlonzo · Answer

Yes the median term is a + 2d, however, if we let a be the first term and the difference d, then, a + a+d + a+2d (median term) + a+3d + a+4d = 100
5a + 10d = 100 , a+2d = 20, The middle term is 20.

shivani1351 · Answer

Five boxes together weigh 100 pounds. What is the median weight of the five boxes?

(1) One of the boxes weighs 20 pounds.
(2) When arranged in ascending order, the weights of the boxes form an arithmetic progression i.e., the difference between the weights of any two consecutive boxes is the same.

Since there are an odd number of boxes, the weight of the box that comes 3rd in sequence when all of them are arranged in ascending/descending order (in terms of their weights) would give us the answer.

Now, let's look at each statement individually and then at both of them together (if required):

Statement 1: One of the boxes weighs 20 pounds.
Through this statement, we know that one of the boxes weighs 20 pounds, but we do not know whether this is the middle value. Hence, Statement 1 is insufficient alone.

Now, let's take a look at Statement 2 closely:
Statement 2: When arranged in ascending order, the weights of the boxes form an arithmetic progression i.e., the difference between the weights of any two consecutive boxes is the same.
Important: For an AP with odd number of terms, its Median = Average.
Since the total weight of the boxes is 100, and the total number of boxes is 5: Average = 100/5 = 20
Thus, the median is 20.

Hence, the correct answer is Option B - Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

Hope this helps!

SM15 · Answer

Statement A alone: Not sufficient 
Statement B alone: sufficient 
We need to find the median i.e. value of a.
(a-2n)+(a-n)+a+(a+n)+(a+2n)=100
On solving, we get 
5a=100
a=20.

Dereno · Answer

We need to find the median weight of the five boxes.

Sum of weights of the five boxes =100.

Statement 1:

One of the boxes weighs 20 pounds.

With one of the boxes as 20. We can find many combinations to find for the rest of 4 boxes. Either it can be same as 20. Or can hold values less than or more than 20. Hence,  Insufficient.

Statement 2:

When arranged in ascending order, the weights of the boxes form an arithmetic progression i.e., the difference between the weights of any two consecutive boxes is the same.

Let the weight of the boxes expressed in A.P be defined as

(a-2d) , (a-d) , a , (a+d), (a+2d)

sum of these values = 5a = 100.

Then, a = median of the values for the 5 boxes = 20.

Hence,  Sufficient

Option B

nemoexplicabo · Answer

This is a beautiful DS problem that combines median concepts with arithmetic progressions — a combination that tests whether you understand both topics deeply!

The question asks: Five boxes weigh 100 pounds total. What is the median weight?

With 5 boxes, the median will be the 3rd box when arranged in order from lightest to heaviest.

Statement (1): One of the boxes weighs 20 pounds

This tells us:
- One box = 20 pounds
- Remaining 4 boxes = 80 pounds total

But here's the problem: we don't know how those 80 pounds are distributed, and we don't know where the 20-pound box falls in the ordering.

Let me show you with examples:
- If weights are {5, 10, 20, 25, 40}: Median = 20
- If weights are {20, 20, 20, 20, 20}: Median = 20
- If weights are {10, 15, 20, 25, 30}: Median = 20
- But if weights are {1, 2, 20, 37, 40}: Median = 20
- And if weights are {1, 19, 30, 30, 20}: Ordered = {1, 19, 20, 30, 30}, Median = 20

Actually, wait! Let me reconsider with more diverse examples:
- If weights are {10, 15, 25, 25, 25}: Median = 25 (the 20-pound box isn't in this set!)
- If weights are {5, 15, 25, 30, 25}: Ordered = {5, 15, 25, 25, 30}, Median = 25

I realize I need to include the 20-pound box. Let me try again:
- If weights are {5, 10, 20, 30, 35}: Median = 20
- If weights are {2, 3, 4, 20, 71}: Median = 4

Different distributions give different medians! Statement (1) is INSUFFICIENT.

Statement (2): The weights form an arithmetic progression when arranged in ascending order

This is powerful! Let's call the five weights: a, a+d, a+2d, a+3d, a+4d (where d is the common difference).

Their sum equals 100:
a + (a+d) + (a+2d) + (a+3d) + (a+4d) = 100
5a + 10d = 100
a + 2d = 20

Now notice: the median is the middle term = a+2d = 20 pounds!

The beautiful thing here is that for any arithmetic sequence, the middle term always equals the average of all terms. And since the average of 5 boxes totaling 100 pounds is 20 pounds, the median must also be 20 pounds.

Statement (2) is SUFFICIENT.

Answer: B

Common traps to avoid:

The big trap with Statement 1 is assuming that knowing one specific value automatically gives you the median — it doesn't! You need to know either all values or enough structural information about how they're distributed. With Statement 2, some students might think "I need to find the value of a and d separately," but that's not necessary. Data Sufficiency only asks whether you CAN determine the answer, not whether you need to find every variable. Here, knowing that a+2d = 20 is enough because that expression IS the median.

Key takeaway: Arithmetic progressions have a special property: the median equals the mean (average). This happens because the values are symmetrically distributed around the middle value. Whenever you see "arithmetic progression" or "evenly spaced" in a DS problem about median or mean, remember they'll be equal! This property can turn an apparently insufficient statement into a sufficient one.

ExpertsGlobal5 · Answer

Explanation Video:

https://www.youtube.com/watch?v=PW5-ni0-JRY

Five boxes together weigh 100 pounds. What is the median weight of the

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GMAT Data Sufficiency (DS) Questions

Five boxes together weigh 100 pounds. What is the median weight of the

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