DeeptiM wrote:
Three interviewers, A, B, and C are interviewing 40 applicants. Only with three interviewers' admission can an applicant be admitted. If interviewer A admitted 15 applicants, B admitted 17 applicants, and C admitted 20 applicants, at least how many applicants get the admission?
(A) 0
(B) 2
(C) 6
(D) 8
(E) 12
VeritasKarishma Bunuel Could you please tell me what's wrong with my approach? I do understand Karishma's explanation here:
https://gmatclub.com/forum/three-interv ... l#p1456069. I noticed that this Q is similar to 2 other Q (one of them is quoted below). Even though they are similar, approach isn't the same for both since my derived formula works for one but not the other. Could you please help me understand the difference and how to tackle such problems? Is there any common strategy/formula? Thanks!
Min (A & B & C) = (A + B + C) % Total + None
No constraint on None => Take None = 0
(15+17+20) % 40 = 52 % 40 = 12
Do you think my approach (notes posted below) make sense?
My approach
• To minimize the all item overlapping group, distribute the items as fairly as possible. Think of distributing chocolates to kids. You want to give first to everyone and then the second to everyone and so on. Take None=0 unless there's a min constraint on it.
• (Divisor: Interviewers; Dividend: Students)
https://gmatclub.com/forum/three-interv ... 18918.html ○ We don't have L1 sets so can't use the formula mentioned below.
• (Divisor: Households; Dividend: Electronics)
https://gmatclub.com/forum/in-a-village ... 57-40.html ○ Let "at least 1" => L1
○ 3 Overlapping Sets: Min (L1A & L1B & L1C) = (L1A + L1B + L1C) % Total Divisor + None
○
3 Overlapping Sets: Min (A & B & C) = (A + B + C) % Total + None • 2 Overlapping Sets: Min (A & B) = (A + B) % Total + None
https://gmatclub.com/forum/a-group-of-p ... 04442.htmlSimilar Q with 3 Sets:
https://gmatclub.com/forum/in-a-village ... l#p2312717dabaobao wrote:
Hussain15 wrote:
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:
A. 65
B. 55
C. 45
D. 35
E. 25
x - y = ? => Max - Min = ?
Max (A&B&C) = Min (A, B, C) = Min (75, 80, 55) = 55
To minimize the all item overlapping group, distribute the items as fairly as possible. Think of distributing chocolates to kids. You want to give first to everyone and then the second to everyone and so on.
3 Overlapping Sets: Min (A & B & C) = (A + B + C)%Total
2 Overlapping Sets: Min (A & B) = (A + B)%Total
Total Items = 75 + 80 + 55 = 210
210%100 = 10 => y = 10
x - y = 55 - 10 = 45
Similar Q with 2 Sets:
https://gmatclub.com/forum/a-group-of-p ... l#p2365267dabaobao wrote:
Bunuel wrote:
A group of people were given 2 puzzles. 79% people solved puzzle X and 89% people solved puzzle Y. What is the maximum and minimum percentage of people who could have solved both the puzzles?
(A) 11%, 0%
(B) 49%, 33%
(C) 68%, 57%
(D) 79%, 68%
(E) 89%, 79%
Kudos for a correct solution.
Method: Direct Formula (Better)
Max (A&B) = Min(A, B) = 79
Min (A&B) = A + B - Total + None (To minimize Both, minimize None by taking None=0 except when there's a min constraint on None.)
Min (X&Y) = A + Y - 100 + None = 79 + 89 - 100 + 0 = 68 (None has no constraint)
Since there’s only one choice with max=79, no need to calculate min. => D
Alternate Method for Min (A&B):
Min (A & B) = (A + B) % Total + None
Since None has no constraint, minimize None by taking it equal to 0 => Min (A & B) = (79 + 89)%100 + 0 = 168%100 = 68
ANSWER: D