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GMAT Club Olympics 2025 (Day 9): What was the average speed of a deliv

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lvillalon

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10 Jul 2025, 18:35

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E. Not enough information

With a simple fitting example you get to that:

(20*20+60*30)/80 = 220/80
(10*20+70*30)/80 = 230/80

Both fit the given statements. The problem here, is that the boxes may be loaded at different speeds over the shift.

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10 Jul 2025, 19:45

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Over a 4-hour shift, Mark loaded a total of 80 boxes into a truck. What was the average (arithmetic mean) weight of the boxes he loaded during the shift?

Average weight = Total boxes weight / number of boxes

(1) During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.

Avg weight during first 2 hours = 20; however, we don't know the number of boxes, so we can't calculate the total weight. Insufficient.

(2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.

Avg weight during last 3 hours = 30; however, we don't know the number of boxes can't calculate the total weight. Insufficient.

With both also we won't get the number of boxes. Hence, both together are insufficient.

Ans E.

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Statement 1: Insufficient Without knowing how many boxes loaded in the first 2 hours, we cannot calculate total weight in first 2 hours.

Without boxes count in last 2 hours, we cannot find total weight in last 2 hours.

So, overall total weight is unknown.

Statement 2: Insufficient
Last 2 hours average weight = 30 kg. But no info on boxes loaded last 2 hours or first 2 hours.

Combining statements 1 and 2,
x= boxes loaded in first 2 hours
80-x = boxes loaded in last 2 hours.
20x+30(80-x)=2400-10x
Avg weight of all boxes=30-x/8.

Insufficient.

Answer: (E) Neither statement is sufficient.

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10 Jul 2025, 21:06

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NOTES:
4 hour shift
Mark loads 80 boxes

ASK:
Whats average weight of boxes?

Solve:
1) During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.
NOT SUFFICIENT. We do not have any information for second half the shift
2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.
NOT SUFFICIENT. We do not have any information for first half the shift (when alone)

Combined) This works! we have first and second half.

ANSWER E - combined

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10 Jul 2025, 21:22

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$\text{Average weight}=\frac{\text{Total weight of boxes}}{80}$

Statement 1: During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.

Total weight of boxes in first 2 hours = No of boxes* Average weight=n*20

No of boxes are not known.

Not Sufficient

Statement 2: During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.

Total weight of boxes in last 2 hours = No of boxes* Average weight=m*20

No of boxes are not known.

Not Sufficient

Statement Combined:

Total weight of boxes = Weight of boxes in first 2 hours + Weight of boxes in last 2 hours=20n+30m

n & m are unknown

Not Sufficient

Answer: E

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10 Jul 2025, 21:29

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Given:
Total number of boxes = 80
Asked:
Average weight of the boxes he loaded during 4hrs shift
Statement (1):
Let the total weight of the boxes he loaded in first 2 hrs be $q$
Number of boxes he loaded = $x$
Average arthematic mean = $q/x=20$
==> $q=20x$-----eq1
Statement 1 alone not sufficient

Statement (2):
Let the total weight of the boxes he loaded during last 2 hrs be $p$

Number of boxes be loaded = $80-x$
Average arthematic mean = $p/(80-x)= 30$
==>$p=(80-x)30$----eq2
Statement 2 alone not sufficient

From Statement (1) and (2) :

Average arthematic mean = $Total weight/total boxes$ = $p+q/80$

from eq1 and eq2

==>$(20x+(80-x)30)/80$= $(2x+240-3x)/8$ = $(240-x)/8$

There is still a variable $x$ which cannot be determined...
Together also not sufficient

Option E correct

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For this we will need both statements as 1st one states avg of 1st 2 hours and 2nd one tells avg of last 2 hours which will amount to total 4 hours as given.
We can get avg as 25kg as time is same

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10 Jul 2025, 23:07

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Statement 1: During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.

let's suppose mark has loaded x boxes in first two hours

total weight of those x boxes = 20x

boxes left to load in last two hours = 80-x

let's suppose there AM weight is W

AM weight of the boxes he loaded during the shift = [20x + (80-x)W]/80

not sufficient.

Statement 2: During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.

let's suppose mark has loaded y boxes in last 2 hours

total weight of those y boxes = 30y

boxes loaded in first two hours = (80-y)

let's suppose there AM weight is w

AM weight of the boxes he loaded during the shift = [30y + (80-y)w]/80

not sufficient

Taking both statements together

AM weight of the boxes he loaded during the shift = (20x + 30y)/80

not sufficient

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10 Jul 2025, 23:07

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Over a 4-hour shift, Mark loaded a total of 80 boxes into a truck.
Total Boxes N=80 Total weight= not known=W
What was the average (arithmetic mean) weight of the boxes he loaded during the shift?
We need to find avg weight= W/80

Statement 1

During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.
We don't know how many boxes have been loaded and no information on the remaining boxes weight.
Also no information on next 2 hours shift
So Avg weight can't be determined

(2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.
Next 2 hours avg weight of boxes= 30
We don't know how many nos of boxes and weight of other remaining boxes and also what happened in first 2 hours.

So Statement is not sufficient

Lets combine both statements
First 2 hours Avg weight= 20. Let X1 be the no. of boxes
remaining 2 hours avg weight=30. Let X2 be the no. of boxes
So avg weight= 20*X1+30*X2/(X1+X2)
X1+X2=80
As avg weight is not known so equation becomes insolvable.

Answer is E

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10 Jul 2025, 23:09

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Bunuel

Total Time = 4 Hours, Total Number of boxes = 80

Average weight = Total Weight if the boxes / Total Number of Boxes

1. Let the number of boxes mark loaded in first 2 hours be x.

Average weight for first two hours = 20 = total weight mark loaded in first 2 hours/ x
total weight mark loaded in first 2 hours = 20x -- total number of boxes is not give hence, Insufficient.

2. Let the number of boxes mark loaded in last 2 hours be y.

Average weight for last two hours = 20 = total weight mark loaded in last 2 hours/ y
total weight mark loaded in last 2 hours = 20y -- total number of boxes is not give hence, Insufficient.

Combining both ---- Average weight = Total Weight if the boxes / Total Number of Boxes

Average weight = (20x+20y)/80 ............. x & y not available hence still Insufficient. Answer is E

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10 Jul 2025, 23:21

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Over a 4-hour shift, Mark loaded a total of 80 boxes into a truck. What was the average (arithmetic mean) weight of the boxes he loaded during the shift?

(1) During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.
(2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.

To calculate the arithmetic mean of the boxes loaded, we need the total weight of the boxes and the number of boxes loaded. We know the number of boxes loaded to be 80. However, we do not know the total weight of the boxes.

Statement 1: The average weight of the boxes Mark loaded is 20kg. We do not know how many boxes were loaded so we cannot get the total weight of the boxes and thus infer the total weight of the rest boxes loaded.

Statement 2: Just like statement 1, it tells us the average weight of the boxes loaded but does not tell us how many boxes were loaded so we cannot get the total weight of theses boxes.

Combined: Statement 1 and 2 do not tell us how many boxes were loaded in either time sessions so we cannot get the weighted average nor the average of the total boxes loaded.

Therefore both statements are insufficient.

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11 Jul 2025, 00:37

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Bunuel

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Average : Total weight of boxes / 80

(1) During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.

Tells us the average weight for the first 2 hours, but not how many boxes he loaded in that time. So we can’t find the total weight for that period.
Not sufficient.

(2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.

Same as Statement 1. We get an average for the last 2 hours, but not the number of boxes. So we can’t find the total weight for the last two hpurs.
Also not sufficient.

(1) +(2)

Still no info on how many boxes were loaded in either half of the shift, so we can’t calculate total weight.

The statements combined is also not sufficient to find the average weight of the boxes.

Option E

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Till we don't know how many boxes he loaded in the 2-hour windows, we can't calculate the average weight of the boxes loaded in the 4-hour shift.

Statements 1 and 2 aren't useful without information on the quantity of boxes.

If 40 boxes were loaded in each of the 2-hour windows, we get a different answer. If 60 boxes were loaded in the first window and 20 in the second window, we get a different answer. Hence the two statements even together are insufficient.

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11 Jul 2025, 00:52

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1. We dont know the number of boxes
2. we don't know the number of boxes

Combined

we still don't know the number of boxesloaded in each shift NS

Ans E

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11 Jul 2025, 00:55

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Bunuel

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So, just from equation 1, it is difficult to find what was his total arithmetic mean for lifting the boxes as the weight and his speed of next 2 hours could be totally different. So alone equation 1 is not sufficient. Simillarly from equation 2, we just have information about last 2 hour. we dont know with what speed or weight he lifted the boxes in first 2 hr so equation 2 is also not sufficient.

But when we combine equation1 and 2, we know about his first 2 hour and last 2 hours. but when we try to find the average of his weight for both session we also need the number of boxes he lifted. Which ultimately we dont know still. So, E is correct.

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11 Jul 2025, 01:10

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Average weight of boxes = Total weight/Total boxes = Total weight/80
We need the total weight to find the average

(1) During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.

Total weight of boxes during first 2 hours = 20kgs * No of boxes loaded.
But we do not know number of boxes loaded in the first 2 hours, so statement 1 is not sufficient.

(2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.

Total weight of boxes during last 2 hours = 30kgs * No of boxes loaded.
But we do not know number of boxes loaded in the last 2 hours, so statement 2 is not sufficient.

(1) and (2) combined.

total weight = 20* Number of boxes loaded in first 2 hours + 30* Number of boxes loaded in last 2 hours

There could be several cases that satisify the constraints
for eg, 20 boxes loaded in first 2 hours, 60 boxes loaded in last 2 hours, then average weight =[20*(20) + 30*(60)]/ 80 = 27.5 kgs

for eg, 40 boxes loaded in first 2 hours, 40 boxes loaded in last 2 hours, then average weight =[20*(40) + 30*(40)]/ 80 = 25 kgs
Both would give different average weights

So statement 1 and 2 are both together also not sufficient
Answer is E

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11 Jul 2025, 01:17

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The Answer is E
The stem tells us that 80 boxes were loaded in 4 hours. But we can't say for sure the average boxes loaded in every hour. We need the total weight and total number of boxes to be loaded to find average weight.
Statement 1 gives us the average weight of boxes in 2 hours. Hence 20 =Total weight/x = 20x=Tw1. This is insufficient since we dont know about the next 2 hours
Statement 2 gives us the average weight of boxes in next 2 hours. Hence 30=Total Weight/80-x =2400-30x=TW2. This is insufficient since we dont know what happened in the former half
Combining the 2 statements, we know Average weight = Total weight/Total Boxes
Aw=20x+2400-30x/80
Even after combining we have 2 unknown variables. Hence E

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11 Jul 2025, 02:39

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4Hours=80 box
Avg of 80 boxes?
(1) During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.
-Avg of next two hours unknown
Insufficient
(2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.
Avg of first two hours unknown
Insufficient
1+2:
First 2 hours=x
next two hours=80-x
w of x=20x
w of 80-x=30(80-x)=2400-30x
total=20x+2400-30x=2400-10x
Avg=(2400-10x)/80
Since x is unknown combining both of them is also insufficient
IMO:E

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11 Jul 2025, 02:54

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Bunuel

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Time = 4 hours
Boxes = 80
Total weight = X
Mean = X/80
To find X?

(1) During the first 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 20 kilograms.
Time = First 2 hours
Avg Weight of Box = 20 kg
Number of boxes = A
Total weight = 20A for first 2 hours

Insufficient

(2) During the last 2 hours of the shift, the average (arithmetic mean) weight of the boxes Mark loaded was 30 kilograms.
Time = Last 2 hours
Avg Weight of Box = 30 kg
Number of boxes = B
Total weight = 30B for last 2 hours

Insufficient

Combining 1 and 2 we get

Total Boxes = A + B = 80
Total weight of boxes = 20A + 30B

Mean = {20A + 30B}/80

Insufficient

Option E

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