total number of recommendations=15+17+20=52

let x be the number of students with 2 reco and y be that with 3 recos

then x+2y=52-30
or x+2y=22

now to minimize y we have to maximize x..since x and y both are integers max value of x can be 6 and that of y is 2

ans option B

But OA given is 0

Any opinions?

OA has to be 0 buddy..


Logical explanation: let say all 30 students are a1,a2,a3,...,a30

now a1,a2,a3,...a15 got the recommendation from the 1st prof.
now 2nd prof gives that to 17 student.so take a16,a17,....,a30 which will give u 15 and let say to a1,a2..so total 17.
till now a1,a2 gets both reco from 2 profs rite??
now 3rd prof gives 20 reco...take any 20 out of 28(except a1,a2 as they already got 2 before)..
so it is not necessary they there will be some ppl will get reco from all 3...so ans is 0 as we are looking for least no of ppl.

Now mathematical explanation:

let say the seven sections are a,b,c,d,e,f,g
a=only from prof 1
b=only from prof 2
c=only from prof 3
d=both 1,2
e=both 2,3
f=both 1,3
g=all 1,2,3

now a+b+c+...+f+g=30...
a+b+c+2(d+e+f)+3g=52
d+e+f+2g=22...eq 1
now d+g>=(15+17)-30=2
e+g>=(17+20)-30=7
f+g>=(20+15)-30=5
so d+e+f+3g>=14 and d+e+f+2g=22...2 equations
now 22+g>=14
g>=-8....mathematically only, and again logically not possible
so min value of g=0..
If there is any doubt,please post...i believe for this qn logical approach is quite easy...but for understanding we can have mathematical approach for homework pupose.

ern5231 wrote:In a university there are 30 applicants. These applicants are recommended by three teachers. 15 individuals received recommendation of the first teacher,17 by the second and 20 by the third. What is the least number of applicants who received recommendation from all three teachers?

A) 0 B) 2 C) 3 D) 5 E) 6

Let's solve this with a lot less math and a lot more intuition.

We know that there are 30 total applicants and at least some of them were recommended by more than 1 teacher (since the number of recommendations is greater than 30).

We have 30 students and 15+17+20 = 52 recommendations; so, we need to make up 22 extra recommendations.

We want to minimize the number of triples, so let's see if we can get all 22 people in double recommendations.

Let's say half of the 20 people recommended by the 3rd teacher were also recommended by exactly one of the first two teachers; that takes care of 20/22 duplicates.

Now, we still have 5 singles by the first teacher and 7 singles by the 2nd teacher; we can have 2 of those people recommended by both, accounting for all 22 of our duplicates. Accordingly, we don't need any triplicates at all: choose (A).

To illustrate the above solution, let's call our duplicates A through V:

Teacher 1: ABCDEFGHIJUV + 3 singles (15 students)
Teacher 2: KLMNOPQRSTUV + 5 singles (17 students)
Teacher 3: ABCDEFGHIJKLMNOPQRST (20 students)