A firm decides to inspect 48 of its distribution centers.They have a

Question

A firm decides to inspect 48 of its distribution centers.They have a staff of 23 employees available for the inspection.If each person will inspect at least 1 center and at most 4 centers, what is the maximum and the minimum number of people who will inspect 4 centers, if all the 48 centers will be inspected?

A. (8,2)
B. (8,0)
C. (9,0)
D. (9,2)
E. (9,1)

A. (8,2)
B. (8,0)
C. (9,0)
D. (9,2)
E. (9,1)

Bunuel · Answer

Please name topics properly. Read rule 3 in our  RULES OF POSTING . Thank you.

deepudiscover · Answer

Lets start with the easier calculation to find min employees. 
-If each employees goes to one one center..we cover 23 center 
- for each employee to 2 center ... we cover 46 center 
- with employee to 3 shop...>48 center 
so min is 0 employee can go to 4 center

This leaves with option B and C
- Lets pick 9,0. if we have 9 people going to 4 center we have covered 9*4=36 shops. 
employee left are 23-9=14 and center left are 48-36=12.

Hence not a options as 14-12=2 will still be left to visit any center

Ramyanag · Answer

For maximum number of people, 48-23= 25
After each person inspects one center 25 centers will be left. In order to inspect 4 centers some people has to inspect 3 more centers. We know 3*8= 24. So, out of 23 people 8 will inspect 4 centers, 1 will inspect 2 centers (remaining one after 24 out of 25 centers are inspected) and remaining people will inspect 1 center each. 
Maximum: 8

For minimum number of people, 23*2= 46 
If each person inspects 2 centers, a total of 46 centers' inspection is done. We are left with 2 centers which we can divide among any 2 persons. Thus, 2 of the 23 people will inspect 3 centers and rest will inspect 2 centers. So, there is a possibility where none of the 23 people investigates 4 centers.
Minimum: 0

Answer: B. (8,0)

yashmatai · Answer

chetan2u, Bunuel, VeritasKarishma, generis

Experts can you please help me with this ?

GMATinsight · Answer

yashmatai

because every employee has to inspect at least 1 so let's exclude 23 centers to be inspected by 23 employees (1 center each employee)

Now we have 48-23 = 25 centers to be allotted
which have to be divided in group of 3 (the employee who has inspected one center can inspect maximum3 more centers
25/3 = 8
i.e. 8 people can inspect 3 centers each and one left center will be inspected by some other employee who will inspect total 2 centers

i.e. Total Maximum People who inspect 4 centers each = 8

Now, in order to minimize the employees inspecting 4 centers we will distribute centers such that maximum people get 3 or less centers to inspect

23*2 = 46 and 2 centers left

i.e. 21 employees can inspect 2 centers each and remaining 2 will inspect 3 centers each 
21*2+2*3 = 42+6 = 48
i.e. minimum employees inspecting 4 centers = 0

Answer: (8, 0)

Answer: Option B

Kinshook · Answer

Given: A firm decides to inspect 48 of its distribution centers.They have a staff of 23 employees available for the inspection.

Asked: If each person will inspect at least 1 center and at most 4 centers, what is the maximum and the minimum number of people who will inspect 4 centers, if all the 48 centers will be inspected?

Maximum number of people inspecting 4 centres

Let the maximum number of people inspecting 4 centers be x.
4x + y = 48
x + y <= 23
3x <= 25
x <= 25/3 = 8 1/3
x = 8

Maximum number of people inspecting 4 centres

Since there are 23 people and 48 centers, if 21 people inspect 2 centers and 2 people inspect 3 centers, there is no requirement for a person to inspect 4 centers.

Minimum number of people inspecting 4 centers = 0

Answer = (8,0)

IMO B

egmat · Answer

Looking at this problem, I need to find the maximum and minimum number of people who will inspect exactly  4  centers.

Key Constraints: 
-  48  centers total
-  23  employees available
- Each person inspects  at least 1  and  at most 4  centers

Finding Maximum: 
Let  x  = number of people inspecting  4  centers

To maximize  x , I want the remaining ( 23 - x ) people to inspect as few centers as possible. Since each person must inspect  at least 1  center:

4x  +  1 ·( 23 - x ) =  48 
 4x + 23 - x = 48 
 3x = 25 
 x = 8.33...

Since  x  must be a whole number, maximum  x = 8 .

Verification:8  people ×  4  centers =  32  centers. The remaining  15  people must cover  16  centers. This works if  14  people inspect  1  center each and  1  person inspects  2  centers.

Common mistake:   Thinking the answer could be  9 . If  9  people inspect  4  centers =  36  centers, then  14  people need to cover only  12  centers. But since each person must inspect  at least 1  center,  14  people would cover  at least 14  centers, which is too many!

Finding Minimum: 
To minimize  x , I want others to inspect as many centers as possible (up to  3  each, since we're minimizing those who inspect  4 ).

Can  x = 0 ? This means all  23  people inspect  1 ,  2 , or  3  centers.
Average needed:  48 ÷ 23 ≈ 2.09  centers per person

This is achievable! For example:
-  2  people inspect  3  centers =  6  centers 
-  21  people inspect  2  centers =  42  centers
- Total:  6 + 42 = 48  centers ✓

Therefore, minimum  x = 0 .

Answer: (8, 0) - Choice B

Key principle:   In optimization problems with constraints, push the non-optimized variables to their extreme values (minimum when maximizing, maximum when minimizing) to find the bounds of your target variable.

A firm decides to inspect 48 of its distribution centers.They have a

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A firm decides to inspect 48 of its distribution centers.They have a

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards