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605-655 Level|   Non-Math Related|               
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TargetMBA007
Bunuel & Others:

Would love your thoughts on this one. The Min response as per OG is 52. This response seems a bit strange, as there is nothing in the Q Stem to suggest that when the truck is carrying 52T, it is fully loaded. In fact, in choosing 53T as the Max value (and not 52), the OG Implies that the truck could very much be plying at below full capacity. So based on the same logic, I would have thought the minimum value would be 49. What am I missing?

A recommended load (PR here) is a load upto that the truck can carry the load safely. However, its not like that the truck can't carry more than that. It can carry more than the PR but that would be unsafe.

So, 53 is the maximum value possible for PR beyond which three loads two of 54 and one of 56 was hauled by truck. In that case truck carried four lesser weights. I hope ill this point things are clear as the conditions(exactly three more than the PR) laid in the Question stem is satisfied.

Lowest PR value possible is 52. Here the truck carried exactly two loads weighted exactly to PR and three more than that. Thus lesser weights carried are less compared to the earlier case. Again the condition is satisfied.

Initially, it seemed bit ambiguous though.

Hope this is clear.
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52 min
53 max

Posted from my mobile device
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TargetMBA007
Bunuel & Others:

Would love your thoughts on this one. The Min response as per OG is 52. This response seems a bit strange, as there is nothing in the Q Stem to suggest that when the truck is carrying 52T, it is fully loaded. In fact, in choosing 53T as the Max value (and not 52), the OG Implies that the truck could very much be plying at below full capacity. So based on the same logic, I would have thought the minimum value would be 49. What am I missing?

A recommended load (PR here) is a load upto that the truck can carry the load safely. However, its not like that the truck can't carry more than that. It can carry more than the PR but that would be unsafe.

So, 53 is the maximum value possible for PR beyond which three loads two of 54 and one of 56 was hauled by truck. In that case truck carried four lesser weights. I hope ill this point things are clear as the conditions(exactly three more than the PR) laid in the Question stem is satisfied.

Lowest PR value possible is 52. Here the truck carried exactly two loads weighted exactly to PR and three more than that. Thus lesser weights carried are less compared to the earlier case. Again the condition is satisfied.

Initially, it seemed bit ambiguous though.

Hope this is clear.
Can u explain this?
why can we take pr as 54 and other three weights be 56 each
also
for min why it cant be :
49 as then 3 can be 50 and all others can be 49
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lnm87
TargetMBA007
Bunuel & Others:

Would love your thoughts on this one. The Min response as per OG is 52. This response seems a bit strange, as there is nothing in the Q Stem to suggest that when the truck is carrying 52T, it is fully loaded. In fact, in choosing 53T as the Max value (and not 52), the OG Implies that the truck could very much be plying at below full capacity. So based on the same logic, I would have thought the minimum value would be 49. What am I missing?

A recommended load (PR here) is a load upto that the truck can carry the load safely. However, its not like that the truck can't carry more than that. It can carry more than the PR but that would be unsafe.

So, 53 is the maximum value possible for PR beyond which three loads two of 54 and one of 56 was hauled by truck. In that case truck carried four lesser weights. I hope ill this point things are clear as the conditions(exactly three more than the PR) laid in the Question stem is satisfied.

Lowest PR value possible is 52. Here the truck carried exactly two loads weighted exactly to PR and three more than that. Thus lesser weights carried are less compared to the earlier case. Again the condition is satisfied.

Initially, it seemed bit ambiguous though.

Hope this is clear.
Can u explain this?
why can we take pr as 54 and other three weights be 56 each
also
for min why it cant be :
49 as then 3 can be 50 and all others can be 49
If you read the question carefully you will see that exactly three weights must be more than PR so 53 is max. 54 would go against the question requirements.
For min PR see that the requirements would again be unfulfilled.
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robu1
if we take 53 max as PR then three loads greater than 53 are 54,54,56
if we take 52 min as PR then three loads greater than 52 are 54,54,56 in this case too.
if we take 49 min as PR then loads greater than 49 are 50, 51, 52, 52, 54, 54, 56. ( 07 nos) which conflict the given condition: The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR.


i hope this makes clear.

I am confused...For the maximum, the problem states that 3 loads are greater than the truck's PR which would mean 56, 54, and 52. We can't count a number twice...

That would mean 51 is the max. If you take 53 as the max, then that would violate the claim that there were 3 loads greater than the truck's PR, no?
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robu1
if we take 53 max as PR then three loads greater than 53 are 54,54,56
if we take 52 min as PR then three loads greater than 52 are 54,54,56 in this case too.
if we take 49 min as PR then loads greater than 49 are 50, 51, 52, 52, 54, 54, 56. ( 07 nos) which conflict the given condition: The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR.


i hope this makes clear.

I am confused...For the maximum, the problem states that 3 loads are greater than the truck's PR which would mean 56, 54, and 52. We can't count a number twice...

That would mean 51 is the max. If you take 53 as the max, then that would violate the claim that there were 3 loads greater than the truck's PR, no?

I initially made that same mistake thinking i cannot count 52 Twice or 54 Twice

But 52 CAN BE COUNTED twice.

The following suggests that

If 52 and 54 CANNOT BE counted twice as you mention above

-- Min value becomes : 51 [as you 52 (albeit 2 isntances of 52) , 54 (albeit 2 instances of 54) and 56 (one instance only)]

-- Max value becomes -- there is no max value you can come up with as you need 3 different payloads (irrespective of number of occurences)

i) lets say you try 52 -- only 54 (two instances) and 56 (one instance) -- only two payloads are seen - 54 and 56.
ii) lets say you try 53 -- only 54 (two instances) and 56 (one instance) -- only two payloads are seen - 54 and 56.
iii) lets say you try 54 -- 56 (one instance) -- only one instance of 56 is seen

But per the question stem - you should be able to get 3 payloads (irrespective of number of occurences) when choosing a max value too


But there
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Official Explanation

According to the information provided, the PR is a whole number of tonnes, and among the seven loads (weighing 50, 51, 52, 52, 54, 54, and 56 tonnes), exactly three exceeded the truck’s PR. Given this, the PR must be a whole number of tonnes less than 54 and no less than 52, so the truck’s PR can be either 52 or 53 tonnes. Therefore, the least possible value for the truck’s PR is 52 tonnes.

The correct answer for Least is 52.

Based on the explanation above, the greatest possible value for the truck’s PR is 53 tonnes.

The correct answer for Greatest is 53.
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Explanation:
A: Apply

According to the information provided, the PR is a whole number of tonnes, and among the seven loads (weighing 50, 51, 52, 52, 54, 54, and 56 tonnes), exactly three exceeded the truck's PR. Given this, the PR must be a whole number of tonnes less than 54 and no less than 52, so the truck's PR can be either 52 or 53 tonnes. Therefore, the least possible value for the truck's PR is 52 tonnes.

The correct answer is D, 52.

B: Apply

Based on the explanation for A, the greatest possible value for the truck's PR is 53 tonnes.

The correct answer is E, 53.
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The payload rating (PR) of a truck is the truck's recommended load weight, which is specified as a number of tonnes (t), where 1 tonne = 1,000 kilograms.

The statement presents a definition of payload rating (PR) of a truck. PR = the truck’s recommended load weight.

(It is important to note that this is a recommended weight, not the actual weight that the truck carries nor the maximum weight that a truck can carry. Perhaps, a truck can carry more weight, but it may not be safe to do so. Also, the recommended weight may be the best balance of economy and safety or may be best for safety but not for economy. We don’t know how the recommended weight is arrived at. It’s a part of comprehension to know what we don’t know. Many people otherwise make assumptions.)

A certain truck's PR is a whole number of tonnes.

PR is a ‘whole’ number of tonnes. Thus, PR will not be a decimal number, such 2.3, 16.7 etc.

The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR.

We’re talking about a specific truck now. It has carried 7 loads, out of which 3 had a greater weight than the truck’s PR. So, the remaining 4 loads were equal to or less than the truck’s PR.

The weights of these 7 loads, in tonnes, are as follows:

50, 51, 52, 52, 54, 54, 56.

Since 3 loads are greater than the truck’s PR, we can deduce that the truck’s PR has to be less than 54, i.e., 53 or lower. Also, if the truck’s PR goes below 52, the number of loads greater than the truck’s PR will go above 3. Thus, the truck’s PR has to be at least 52.

Thus, the truck’s PR has to be either 52 or 53.

Select the Least and Greatest possible values for the truck's PR, in tonnes. Make only two selections, one in each column.

As discussed above, the least possible value is 52, and the greatest possible value is 53.­
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I'd like your assistance Bunuel GMATCoachBen avigutman GMATNinja

I calculated the average of the weights (369/7 = 52.71) and since it has to be an integer I thought that it has to be either 52 or 53. We also know that 3 of the weights are more than the PR so they should exceed the average too, which means that these 3 are 54,54,56 and I confirmed the least and greatest values.

Is my reasoning correct or is just a coincidence that I found the correct answer?

Thank you in advance
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Gmatguy007
I'd like your assistance Bunuel GMATCoachBen avigutman GMATNinja

I calculated the average of the weights (369/7 = 52.71) and since it has to be an integer I thought that it has to be either 52 or 53. We also know that 3 of the weights are more than the PR so they should exceed the average too, which means that these 3 are 54,54,56 and I confirmed the least and greatest values.

Is my reasoning correct or is just a coincidence that I found the correct answer?

Thank you in advance
­@Gmatguy007  I'm curious, what prompted you to calculate the average? The wording of the question does not ask for this.

The key is reading carefully, especially the bolded portion below:

"The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR. The weights of these 7 loads, in tonnes, are as follows:

50, 51, 52, 52, 54, 54, 56."


So, "54, 54, 56" are greater than the PR, and 52 is not greater than the PR. Therefore, 52 \(\leq\) PR < 54

You correctly noticed that the PR must be an integer. So, the max that the PR could be is 53, and the minimum is 52.­
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Gmatguy007
I'd like your assistance Bunuel GMATCoachBen avigutman GMATNinja

I calculated the average of the weights (369/7 = 52.71) and since it has to be an integer I thought that it has to be either 52 or 53. We also know that 3 of the weights are more than the PR so they should exceed the average too, which means that these 3 are 54,54,56 and I confirmed the least and greatest values.

Is my reasoning correct or is just a coincidence that I found the correct answer?

Thank you in advance
­@Gmatguy007  I'm curious, what prompted you to calculate the average? The wording of the question does not ask for this.

The key is reading carefully, especially the bolded portion below:

"The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR. The weights of these 7 loads, in tonnes, are as follows:

50, 51, 52, 52, 54, 54, 56."


So, "54, 54, 56" are greater than the PR, and 52 is not greater than the PR. Therefore, 52 \(\leq\) PR < 54

You correctly noticed that the PR must be an integer. So, the max that the PR could be is 53, and the minimum is 52.­
­I thought that these 3 loads being greater than the PR, they should exceed the average (as an outlier) and consequently it would help me to find the min and max.

How did you reach the conclusion that these 3 values are the biggest ones? I mean that it could be 52, 54, 54 since the stem says only that the PR is the truck's recommended load weight.

Thank you for your help GMATCoachBen
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Gmatguy007
GMATCoachBen
Gmatguy007
I'd like your assistance Bunuel GMATCoachBen avigutman GMATNinja

I calculated the average of the weights (369/7 = 52.71) and since it has to be an integer I thought that it has to be either 52 or 53. We also know that 3 of the weights are more than the PR so they should exceed the average too, which means that these 3 are 54,54,56 and I confirmed the least and greatest values.

Is my reasoning correct or is just a coincidence that I found the correct answer?

Thank you in advance
­@Gmatguy007  I'm curious, what prompted you to calculate the average? The wording of the question does not ask for this.

The key is reading carefully, especially the bolded portion below:

"The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR. The weights of these 7 loads, in tonnes, are as follows:

50, 51, 52, 52, 54, 54, 56."


So, "54, 54, 56" are greater than the PR, and 52 is not greater than the PR. Therefore, 52 \(\leq\) PR < 54

You correctly noticed that the PR must be an integer. So, the max that the PR could be is 53, and the minimum is 52.­
­I thought that these 3 loads being greater than the PR, they should exceed the average (as an outlier) and consequently it would help me to find the min and max.

How did you reach the conclusion that these 3 values are the biggest ones? I mean that it could be 52, 54, 54 since the stem says only that the PR is the truck's recommended load weight.

Thank you for your help GMATCoachBen
­@Gmatguy007 Regarding the idea of using the "average", make sure to not bring your own ideas and or interpretation, just stick to what the question says literally. 

The question tells us the values for the 7 loads, in order from least to greatest, so "54, 54, 56" are the biggest. It could not be 52, 54, 54.

"The weights of these 7 loads, in tonnes, are as follows:

50, 51, 52, 52, 54, 54, 56."


 
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­@Gmatguy007 Regarding the idea of using the "average", make sure to not bring your own ideas and or interpretation, just stick to what the question says literally. 

The question tells us the values for the 7 loads, in order from least to greatest, so "54, 54, 56" are the biggest. It could not be 52, 54, 54.

"The weights of these 7 loads, in tonnes, are as follows:

50, 51, 52, 52, 54, 54, 56."



 
­Thank you!
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Gmatguy007
I'd like your assistance Bunuel GMATCoachBen avigutman GMATNinja

I calculated the average of the weights (369/7 = 52.71) and since it has to be an integer I thought that it has to be either 52 or 53. We also know that 3 of the weights are more than the PR so they should exceed the average too, which means that these 3 are 54,54,56 and I confirmed the least and greatest values.

Is my reasoning correct or is just a coincidence that I found the correct answer?

Thank you in advance
­@Gmatguy007  I'm curious, what prompted you to calculate the average? The wording of the question does not ask for this.

The key is reading carefully, especially the bolded portion below:

"The truck has hauled exactly 7 loads, exactly 3 of which had a greater weight than the truck's PR. The weights of these 7 loads, in tonnes, are as follows:

50, 51, 52, 52, 54, 54, 56."


So, "54, 54, 56" are greater than the PR, and 52 is not greater than the PR. Therefore, 52 \(\leq\) PR < 54

You correctly noticed that the PR must be an integer. So, the max that the PR could be is 53, and the minimum is 52.­
­Thanks . Out of all solutions posted , your's breaks the problem into the most comprehensible form. 
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When PR is 52, greater weights are 54,54,56 i.e. exactly 3

When PR is 53, still the 3 greater weights are 54,54,56 i.e. exactly 3

So either its 52 or 53, the greater truck loads would be 54,54,56. Now we can choose least possible (52) and maximum possible (53) for this to satisfy. More of a logical question than math.
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If we take a mean of all 7 weights we get 52.7.
The question stem says the PR has to be a whole number.
Then the least could be 52 and greatest is 53.

Is this logic flawed?

Asking as this seemed the easiest solution but no one has mentioned.
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