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Minimum: 35
Maximum: 100
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
74% (02:11) correct
26%
(02:35)
wrong
based on 614
sessions
History
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Time
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Not Attempted Yet
12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes
At a regional food festival, there were 240 attendees in total.
Select for Minimum the minimum number of attendees who could have chosen both pasta and grilled dishes, and select for Maximum the maximum number of attendees who could have chosen both. Make only two selections, one in each column.
|
At a regional food festival, there were 240 attendees in total.
- 150 attendees chose pasta dishes.
- 100 attendees chose grilled dishes.
- At least 25 attendees chose other food options—neither pasta nor grilled dishes.
Select for Minimum the minimum number of attendees who could have chosen both pasta and grilled dishes, and select for Maximum the maximum number of attendees who could have chosen both. Make only two selections, one in each column.
| Minimum | Maximum | |
| 35 | ||
| 45 | ||
| 90 | ||
| 100 | ||
| 150 |
ShowHide Answer
Official Answer
Minimum: 35
Maximum: 100
25 Dec 2024, 10:00
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Bunuel
GMAT Club's Official Explanation:
Total = Pasta + Grilled - Both + Neither
240 = 150 + 100 - Both + Neither
Both = Neither + 10
Since Neither is at least 25, Both is at least 35, making the minimum for Both 35.
For the maximum, Both cannot exceed the smallest group, so it cannot be more than 100. Can it be exactly 100? Yes, when Neither = 90:
240 = 150 + 100 - 100 + 90
General Discussion
25 Dec 2024, 09:22
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imo
min-35
max-100
To determine the minimum and maximum number of attendees who could have chosen both pasta and grilled dishes, we need to use the principle of inclusion-exclusion.
Given:
Total attendees: 240
Attendees who chose pasta dishes: 150
Attendees who chose grilled dishes: 100
At least 25 attendees chose other food options (neither pasta nor grilled dishes).
First, let's denote:
A
A as the number of attendees who chose pasta.
B
B as the number of attendees who chose grilled dishes.
C
C as the number of attendees who chose both pasta and grilled dishes.
D
D as the number of attendees who chose neither pasta nor grilled dishes.
We are given:
A
=
150
A=150
B
=
100
B=100
D
≥
25
D≥25
The total number of attendees is:
A
+
B
−
C
+
D
=
240
A+B−C+D=240
Since
D
≥
25
D≥25, we can substitute
D
D with 25 for the minimum case:
150
+
100
−
C
+
25
=
240
150+100−C+25=240
275
−
C
=
240
275−C=240
C
=
275
−
240
C=275−240
C
=
35
C=35
So, the minimum number of attendees who could have chosen both pasta and grilled dishes is 35.
For the maximum number of attendees who could have chosen both pasta and grilled dishes, we need to consider the case where the number of attendees who chose neither pasta nor grilled dishes is minimized. Since
D
≥
25
D≥25, the minimum value for
D
D is 25.
A
+
B
−
C
+
D
=
240
A+B−C+D=240
150
+
100
−
C
+
25
=
240
150+100−C+25=240
275
−
C
=
240
275−C=240
C
=
275
−
240
C=275−240
C
=
35
C=35
However, to find the maximum, we need to consider the situation where the overlap is maximized. The maximum overlap occurs when the sum of the individual groups minus the overlap is equal to the total number of attendees.
A
+
B
−
C
≤
240
A+B−C≤240
150
+
100
−
C
≤
240
150+100−C≤240
250
−
C
≤
240
250−C≤240
C
≥
10
C≥10
But since we are given that at least 25 attendees chose neither pasta nor grilled dishes, we need to adjust:
A
+
B
−
C
+
D
=
240
A+B−C+D=240
150
+
100
−
C
+
25
=
240
150+100−C+25=240
250
−
C
+
25
=
240
250−C+25=240
275
−
C
=
240
275−C=240
C
=
35
C=35
Thus, the minimum number of attendees who could have chosen both pasta and grilled dishes is 35, and the maximum number of attendees who could have chosen both pasta and grilled dishes is:
Maximum
C
=
min
(
A
,
B
)
=
min
(
150
,
100
)
=
100
Maximum C=min(A,B)=min(150,100)=100
Therefore, the correct selections are:
Minimum: 35 Maximum: 100
min-35
max-100
To determine the minimum and maximum number of attendees who could have chosen both pasta and grilled dishes, we need to use the principle of inclusion-exclusion.
Given:
Total attendees: 240
Attendees who chose pasta dishes: 150
Attendees who chose grilled dishes: 100
At least 25 attendees chose other food options (neither pasta nor grilled dishes).
First, let's denote:
A
A as the number of attendees who chose pasta.
B
B as the number of attendees who chose grilled dishes.
C
C as the number of attendees who chose both pasta and grilled dishes.
D
D as the number of attendees who chose neither pasta nor grilled dishes.
We are given:
A
=
150
A=150
B
=
100
B=100
D
≥
25
D≥25
The total number of attendees is:
A
+
B
−
C
+
D
=
240
A+B−C+D=240
Since
D
≥
25
D≥25, we can substitute
D
D with 25 for the minimum case:
150
+
100
−
C
+
25
=
240
150+100−C+25=240
275
−
C
=
240
275−C=240
C
=
275
−
240
C=275−240
C
=
35
C=35
So, the minimum number of attendees who could have chosen both pasta and grilled dishes is 35.
For the maximum number of attendees who could have chosen both pasta and grilled dishes, we need to consider the case where the number of attendees who chose neither pasta nor grilled dishes is minimized. Since
D
≥
25
D≥25, the minimum value for
D
D is 25.
A
+
B
−
C
+
D
=
240
A+B−C+D=240
150
+
100
−
C
+
25
=
240
150+100−C+25=240
275
−
C
=
240
275−C=240
C
=
275
−
240
C=275−240
C
=
35
C=35
However, to find the maximum, we need to consider the situation where the overlap is maximized. The maximum overlap occurs when the sum of the individual groups minus the overlap is equal to the total number of attendees.
A
+
B
−
C
≤
240
A+B−C≤240
150
+
100
−
C
≤
240
150+100−C≤240
250
−
C
≤
240
250−C≤240
C
≥
10
C≥10
But since we are given that at least 25 attendees chose neither pasta nor grilled dishes, we need to adjust:
A
+
B
−
C
+
D
=
240
A+B−C+D=240
150
+
100
−
C
+
25
=
240
150+100−C+25=240
250
−
C
+
25
=
240
250−C+25=240
275
−
C
=
240
275−C=240
C
=
35
C=35
Thus, the minimum number of attendees who could have chosen both pasta and grilled dishes is 35, and the maximum number of attendees who could have chosen both pasta and grilled dishes is:
Maximum
C
=
min
(
A
,
B
)
=
min
(
150
,
100
)
=
100
Maximum C=min(A,B)=min(150,100)=100
Therefore, the correct selections are:
Minimum: 35 Maximum: 100
