At a certain fast food franchise with 120 locations,

Question

At a certain fast food franchise with 120 locations, 90 locations introduced item X, 85 locations introduced item Y, and 72 locations introduced item Z.
If a and b are respectively the greatest and smallest possible number of locations that introduced all three items, what is the value of a-b?

A. 7
B. 18
C. 36
D. 58
E. 65

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A. 7
B. 18
C. 36
D. 58
E. 65

demawtf · Accepted Answer

At a certain fast food franchise with 120 locations, 90 locations introduced item X, 85 locations introduced item Y, and 72 locations introduced item Z. If a and b are respectively the greatest and smallest possible number of locations that introduced all three items, what is the value of a-b?

Applied the method seen here https://gmatclub.com/forum/in-a-village ... ml#p937898

max value = 72 (smallest number between the 3 locations)

min value: sum of the max of each location which doesn't have the product
120-90 =30 no X
120-85=35 no Y
120 -72=48 no Z
so total of 113 (30+35+48) combined who might not have some combination of the 3 products. Subtract from 120 --> 120-113 = 7 so the minimum of locations that could have all 3.

72-7 = 65 so E

Archit3110 · Answer

all three locations max
120=90+85+72- all three
all three = 127

overlaps such that all three items are introduced will be 
90+72 ; 162
90+85 ; 175
85+72 ; 157
max will be 162 and min 157
all three 127
∆ 162-127 ; 35
∆ 157-127 ; 30
total 65
option E

MissionAdmit · Answer

Max no of stores with all three items = Min ( 90,85,72) = 72

Min no of stores with all three items 
Here no of stores = 120 
No of Item-store combination = 90+85+72= 247

Lets spread these 247 over 120 stores to have min overlap of all three= (90 + 30)+(55+65)+7

1st fill 90+30
2nd fill - Two in a store = 55+65 
34d fill - Three in a store = 7

Min value = 3

Thus a-b = 72-7 = 65

Credit : GMAT Ninja

prachisaraf · Answer

Can someone provide a technique for these kind of problems ?

Kinshook · Answer

@Laila12618Given: At a certain fast food franchise with 120 locations, 90 locations introduced item X, 85 locations introduced item Y, and 72 locations introduced item Z.
Asked: If a and b are respectively the greatest and smallest possible number of locations that introduced all three items, what is the value of a-b?

For a: - 
----------------------------------120-----------------------------------
---------------90------X----------------------
-------------------85-----Y----------------
-----72-------Z-------------------

a = 72

For b: - 
----------------------------------120-----------------------------------
---------------90------X------------------------
                    --------------------------85----------Y----------------
-----35--Z------                           --------------37------Z--------

b = 7

a - b = 72 - 7 = 65

IMO E

Ggt1234 · Answer

There are two (relatively) easy ways to answer this type of question.
firstly, the max number for all three is the minimum among the major groups X, Y and Z, here Z = 72. So our a= 72. Trouble is with finding the minimum. For that, we need to maximise the two-group overlap. Here are two ways to do that:

1) Pick any two of the three groups. Say, X (90) and Y (85). Added, they give 175. Imagine they are the only two groups. So the overlap becomes 175-120 = 55. Now treat this 55 as one group and Z as another group that overlaps with it. The overlap now becomes (55+72) - 120 = 7. That is the minimum three-group overlap. hence a-b = 72-7 = 65

2) X+Y+Z = 247. We have 120 locations. so have a total overlap of 247-120 =127. Remember, we want to minimise three-group overlap and for that, maximise two-group overlap. So coming back to this 127. We have a total of 120 locations. Imagine that all 120 locations have two products, that is, all 120 are in 2-group overlap. that leaves 127-120 = 7 locations remaining from the total overlap, which has to be the 7 with three-group overlap. hence, a-b = 72-7 = 65.

Similar question : https://gmatclub.com/forum/in-a-class-o ... NT6Apc1bpM

prachisaraf

Can someone provide a technique for these kind of problems ?

Ggt1234 · Answer

There are two (relatively) easy ways to answer this type of question.
firstly, the max number for all three is the minimum among the major groups X, Y and Z, here Z = 72. So our a= 72. Trouble is with finding the minimum. For that, we need to maximise the two-group overlap. Here are two ways to do that:

1) Pick any two of the three groups. Say, X (90) and Y (85). Added, they give 175. Imagine they are the only two groups. So the overlap becomes 175-120 = 55. Now treat this 55 as one group and Z as another group that overlaps with it. The overlap now becomes (55+72) - 120 = 7. That is the minimum three-group overlap. hence a-b = 72-7 = 65

2) X+Y+Z = 247. We have 120 locations. so have a total overlap of 247-120 =127. Remember, we want to minimise three-group overlap and for that, maximise two-group overlap. So coming back to this 127. We have a total of 120 locations. Imagine that all 120 locations have two products, that is, all 120 are in 2-group overlap. that leaves 127-120 = 7 locations remaining from the total overlap, which has to be the 7 with three-group overlap. hence, a-b = 72-7 = 65.

Similar question : https://gmatclub.com/forum/in-a-class-o ... NT6Apc1bpM

prachisaraf

Can someone provide a technique for these kind of problems ?

JuniqueLid · Answer

Can somone please clarify, how can the max be 72 and min be 7? Let's say there are 72 stores with all 3 items XYZ, then there are 18 stores (90-72) with X but not all three, and 13 stores (85-72) with Y but not all three. If there is no overlap between the 18 and 13, there are in total 72+18+13 = 103 stores which is less than the 120 given in the stem - incorrect; if there is any overlep between the 18 and 13, then the total number of stores would be less than 103, also incorrect. Therefore, the max cannot be 72.

To find the max, Set A(stores with X) + Set B(stores with Y) + Set C(stores with Z) - Set D(stores with exactly 2 items) - 2*Set E(stores with exactly 3 items) should equal to 120, i.e. 90+85+72-D-2E = 120 => D+2E = 127. For E to be max, D = 0, such that 2E = 127, E = 63.5 (noting number of stores cannot be non-integer, but the math based on the information given in the stem happens to give this result - we could leave as is or could say max(E) is 63 and D has to be at least 1). This satisfies that E needs to be no larger than min(90,85,72) = 72.

For E to be min, D needs to be as large as possible. max(D) has to be no greater than 120. Knowing there are 90+85+72-120=127 stores with at least 2 items, for E to be the smallest, we maximise the number of stores with 2 items, which is 120. So 7 stores have all 3 items => min(E) is 7 while D=113. So answer is 63-7=56.

None of the options have 56! Would appreciate if someone can help explain.

Krunaal · Answer

17 locations introduced  neither  of the three items; it is  nowhere mentioned  in the question stem that each of the 120 locations introduced atleast one item.

Set A(stores with X) + Set B(stores with Y) + Set C(stores with Z) - Set D(stores with exactly 2 items) - 2*Set E(stores with exactly 3 items) = Total  - Neither

JuniqueLid · Answer

Aha. Thanks Krunaal! Brilliant pick-up.

At a certain fast food franchise with 120 locations,

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