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A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.
During one week, from Monday to Sunday, the number of requests received on each day was as follows:


810, 790, 830, 823, 780, 795, 820

First, let's sort the nos. 780, 790, 795, 810, 820, 823, 830

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

Now, the RT values must be such that four days have at least one request that has not been addressed, or three days requests must be fully addressed.

For 780,790,795 must be fully addressed. Minimum RT can be 795.

For 810, 820, 823, 830, at least one request must not be addressed. Maximum RT can be 810-1 = 809;

On the following Monday, the site received 812 requests.

Minimum Not Addressed: 812 - 809 = 3

Maximum Not Addressed: 812 - 795 = 17
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Let x be the limit, then the interval is:
795>=x>810

Minimum = 812-809 = 3
Maximum = 812-795 = 17
Bunuel
 


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A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

During one week, from Monday to Sunday, the number of requests received on each day was as follows:


810, 790, 830, 823, 780, 795, 820

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

On the following Monday, the site received 812 requests.

Based on this information, select the Minimum and the Maximum number of requests that could have been not addressed on that Monday. Make only two selections, one in each column.
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A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

During one week, from Monday to Sunday, the number of requests received on each day was as follows:

810, 790, 830, 823, 780, 795, 820

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

1st case only one request was not addressed
RT=823
So 830 was discarded

2nd case all 4 requests have been discarded
RT=795
So 810, 820, 823 & 830 requests have been discarded

3rd case RT =810 three requests 820, 823 & 830 discarded
Now On the following Monday, the site received 812 requests.

For Min requests that have not been addressed RT should be max =810 (taking 823 will lead to zero rejections)

So min requests that have been discarded =2

For max requests to be discarded
RT should be min = 795
So max requests discarded =812-795=17
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"During one week, from Monday to Sunday, the number of requests received on each day was as follows:

810, 790, 830, 823, 780, 795, 820"

Ordering by request amount:

780, 790, 795, 810, 820, 823, 830

"On exactly 4 out of the 7 days that week, there was at least one request that was not addressed."


So: 795 <= RT < 810

"On the following Monday, the site received 812 requests.


Based on this information, select the Minimum and the Maximum number of requests that could have been not addressed on that Monday. Make only two selections, one in each column."

To minimize the possible number of dropped requests, we need to maximize the RT:
Max RT = 809
Minimum drops: 812 - 809 = 3

To maximize the possible number of dropped requests, we need to minimize the RT:
Min RT = 795
Maximum drops: 812 - 795 = 17

Answer:
Minimum: 3
Maximum: 17
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NOTES:
Website threshold (RT) - max it can handle in a single day
If requests is >RT it can only process RT. All more are discarded.
Number of Requests Mon-Sunday
Monday - 810
Tuesday - 790
Wednesday -830
Thursday - 823
Friday - 780
Saturday - 795
Sunday - 820

On 4 of the 7 days at least 1 request was not addressed ( must be the 4 largest)

On the following Monday - 812 requests.

ASK:
Min and max requests that could have been not addressed following Monday.

SOLVE:

Monday - 810
Tuesday - 790
Wednesday -830
Thursday - 823

Friday - 780
Saturday - 795

Sunday - 820

BOLD DAYS - are 4 highest so those days have at least 1 request that was not addressed.
UNDERLINED DAYS - All requests were answered

That means 810 is at least 1 request too high
and 795 is at least 1 request in the clear.
The unknown failure range then is 796-809

If the following Monday they had 812 asks. That means that at minimum 3 were discarded (812-809) and max 17 (812-795).

Min - 3
Max - 17
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arranging the given values in ascending order, we get:

780
790
795
.......... since 4 days the requests were more than RT, the minimum RT is 796, maximum RT will be 809
810
820
823
830


so when we have 812 requests on next Monday,
maximum discarded requests = 812-796 = 16
minimum discarded requests = 812 - 809= 3
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Bunuel
 


This question was provided by GMAT Club
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A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

During one week, from Monday to Sunday, the number of requests received on each day was as follows:


810, 790, 830, 823, 780, 795, 820

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

On the following Monday, the site received 812 requests.

Based on this information, select the Minimum and the Maximum number of requests that could have been not addressed on that Monday. Make only two selections, one in each column.
So we need to find certain value of RT so that the site has maximum and minimum number of requests that will not be addressed. So, let's firsst see top 4 requests which were, 823, 820, 830 and 810. Now we nneed such value of RT so that their is some un addressed requests.

This can only happen if RT is less than or equal to 810. So for minimum value of RT so that their is atleast 1 or more un answered request will be 810 with 2 un answered request.

Now as we lower the value of RT below we have a limit at 795 because in that case more than 4 days will have unanswered requests. So maximum value for RT exist when RT is 795. which creates a total of 17 more such unanswered requests.

So, now
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Bunuel
 


This question was provided by GMAT Club
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A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

During one week, from Monday to Sunday, the number of requests received on each day was as follows:


810, 790, 830, 823, 780, 795, 820

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

On the following Monday, the site received 812 requests.

Based on this information, select the Minimum and the Maximum number of requests that could have been not addressed on that Monday. Make only two selections, one in each column.

A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

Maximum number of requests it can handle in a single day = RT

Monday to Sunday, the number of requests received

810, 790, 830, 823, 780, 795, 820

RT breached on 4 days. Sorting the requests we get
780, 790, 795, 810, 820, 823, 830

Hence, RT breached on
810, 820, 823, 830

Monday requests = 812

For minimum we need to make the RT as close to the lowest value as possible (810). Since at least one RT breach was committed, the lowest value of RT denotes by \(RT_{min}\) can be 809 as 810 is the breach number

Minimum requests that would not been handled on Monday = 812 - \(RT_{min}\) = 812 - 809 = 3
Minimum = 3

For maximum we need to make the RT as distant from lowest value as possible (810) but not beyond the previous data (795). Since at least one RT breach was committed, the highest value of RT denoted by denotes by \(RT_{max}\) can be 795 as no breach happened for that day

Maximum requests that would not been handled on Monday = 812 - \(RT_{max}\) = 812 - 795 = 17
Maximum = 17


Hence, we get final values as
Minimum = 3
Maximum = 17
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Exactly 4 days had requests exceeding RT, so RT must be between 796 and 810.

If RT = 810, Monday’s discarded = 812 - 810 = 2 (minimum discarded).

If RT = 796, Monday’s discarded = 812 - 796 = 16 (maximum discarded).

RT values lower than 796 or higher than 810 don’t fit the 4-day discard condition.

Therefore, the minimum and maximum discarded requests on Monday are 2 and 16.

Answer: minimum is 2
Maximum is 16.
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Step 1: Identify RT Range
Let’s sort the weekly requests:

780, 790, 795, 810, 820, 823, 830
If exactly 4 days exceeded RT, then RT must lie between the 4th and 5th largest values, because:

If RT = 810 -> only 3 days (820, 823, 830) exceed RT -> too few
If RT = 809 -> 4 days (810, 820, 823, 830) exceed RT -> just right
If RT = 820 -> only 2 days exceed it -> too few
If RT = 789 -> 6 days exceed it -> too many
So, possible values of RT that make exactly 4 days exceed it:
Try RT = 809 or RT = 810 or RT = 811
Let’s test them:

RT = 809 -> Days above: 810, 820, 823, 830 => 4 days
RT = 810 -> Days above: 820, 823, 830 => 3 days
RT = 811 -> Days above: 820, 823, 830 => still 3
So the only feasible RT is 809

Step 2: Requests on next Monday = 812
So unaddressed requests = 812 − RT = 812 − 809 = 3
That’s when RT is maximized (RT = 809)

Step 3: Try lower values of RT
We can’t go below RT = 809 and still have exactly 4 overflow days.
But if we suppose RT = 795, then:
Days with requests -> 795: 810, 820, 823, 830 => 4 days
Unaddressed requests on Monday = 812 − 795 = 17
If RT = 794 -> 5 days exceeded -> too many
So minimum RT that still keeps 4 days over threshold is 795

Final Answer:

Minimum unaddressed requests = 812 − 809 = 3
Maximum unaddressed requests = 812 − 795 = 17

Bunuel
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 



A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

During one week, from Monday to Sunday, the number of requests received on each day was as follows:


810, 790, 830, 823, 780, 795, 820

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

On the following Monday, the site received 812 requests.

Based on this information, select the Minimum and the Maximum number of requests that could have been not addressed on that Monday. Make only two selections, one in each column.
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Let's first sort the daily requests in ascending order. \(780, 790, 795, 810, 820, 823, 830\)

4 days out of these 7, there was at least one request that was not addressed. So these 4 days must be having the 4 largest number of requests received.

Some of the requests were not addressed on those days having more than equal to 810.

Minimum:

If we consider 1 request was not addressed out of \(810\), \(RT\) will be \(809\).

\(812-809=3\)

Maximum:

We know that 795 requests were addressed on a single day. If we consider \(RT=795\), requests more than 795 will not be addressed

\(812-795=17\)

Minimum: 3

Maximum: 17
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On exactly 4 out of the 7 days that week, there was at least one request that was not addressed. --> that means RT >= 795 and RT <810

So min RT range = 795 to 809.

So Min # of requests NOT addressed on Monday = 812 - 809 = 3.
Max # of requests NOT addressed on Monday = 812 - 795 = 17.
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Bunuel
 


This question was provided by GMAT Club
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A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

During one week, from Monday to Sunday, the number of requests received on each day was as follows:


810, 790, 830, 823, 780, 795, 820

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

On the following Monday, the site received 812 requests.

Based on this information, select the Minimum and the Maximum number of requests that could have been not addressed on that Monday. Make only two selections, one in each column.
Let's arrange the number of request for given week in ascending order = 780, 790, 795, 810, 820, 823, 830

The four days for which at least one request was not addressed must have received 810, 820, 823, 830 requests

On the day when 795 request were received, every request was addressed, hence minimum numbers of request that can be handled can be 795.
on the other hand at least one request was not addressed when 810 requests were received, hence maximum number of request that can be handled on a day can be 809.

Minimum number of requests that could have been not addressed on that Monday = no. of requests received - maximum number of request that can be handled on a day = 812- 809 = 3

Maximum number of requests that could have been not addressed on that Monday = no. of requests received - minimum number of request that can be handled on a day = 812- 795 = 17
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Ans: (Max 17) (Min 3)

4 out of 7 days at least 1 request was not addressed, meaning 4 days had more requests (at least 1 more) than RT.
4 higher values from the given numbers are: 830, 823, 820, and 810
So, from the set, maximum 795 requests were addressed without leaving any requests.

From the above statements, we can say that RT >= 795 and RT<=809
Next Monday's requests = 812
For Max RT has to minimum: if RT = 795, then 812-795 = 17 requests were not addressed (Max 17)
For Min RT has to Max: if RT <= 809, then 812-809 = 3 requests were not addressed. (Min 3)
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RT must be set so that exactly 4 days had requests above the threshold and 3 days at or below it.
Ascending order: 780, 790, 795, 810, 820, 823, 830
So, RT should be greater than the bottom three (780, 790, 795)
And RT should be lower than the top three (810, 820, 823, 830)
795<=RT<809
so for monday 812 requests.
Min. = 812-809 = 3
Max. = 812-795 = 17
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The number of unaddressed requests = 812 – RT if RT < 812
If RT ≥ 812, then no unaddressed requests

to find:
Minimum unaddressed, use highest possible RT < 812
Maximum unaddressed, use lowest possible RT

Highest possible RT=809
thus min unaddressed= 812-809=3

Lowest possible RT=795
Thus max unaddressed=812-795=17

Ans 3,17
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Bunuel
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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A certain website has a request threshold (RT), defined as the maximum number of requests it can handle in a single day. If the number of requests on a given day exceeds the RT, the site processes only the first RT requests. The remaining requests are not addressed and are permanently discarded.

During one week, from Monday to Sunday, the number of requests received on each day was as follows:


810, 790, 830, 823, 780, 795, 820

On exactly 4 out of the 7 days that week, there was at least one request that was not addressed.

On the following Monday, the site received 812 requests.

Based on this information, select the Minimum and the Maximum number of requests that could have been not addressed on that Monday. Make only two selections, one in each column.

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