P Q R S are distinct integers numbered from 1 to 12 What is the possible smallest value for P

largest no. divided by smallest no.
1/12 + 2/11
10 years agoHelpfull: Yes(15) No(7)
inc deno to get lowest possible,possible-1/12,2/11 or 2/12,1/11
in both the case the denominator is 132...therfore the no with smaller numerator is smaller..which is 2nd case...
10 years agoHelpfull: Yes(10) No(1)
1/11+2/12
10 years agoHelpfull: Yes(5) No(2)
(1/12)+(2/11) = 35/132
But,
(2/12)+(1/11) = 34/132

So, the second one will have the least value.

Hence the answer is (2/12)+(1/11)
8 years agoHelpfull: Yes(4) No(1)
the integer can be negative value and they are takling about smallest value
which is(-1/12)+(-2/11) --- ans for sure
9 years agoHelpfull: Yes(3) No(6)
if the ans. should be an integer, then ANS = 1 .
10 years agoHelpfull: Yes(2) No(2)
smallest number is possible when 2/12 +1/11 = 0.25757576
not, 1/12 + 2/11 which is = 0.2651515152
10 years agoHelpfull: Yes(2) No(0)
the given options for the above question are, a.1/12+2/11 b.1/11+9/10 c.1/11+1/12 d.1/10+1/11
In mathematics, two things are called distinct if they are not equal.
so p,q,r,s are not equal.
Ans:- 1/12 + 2/11
8 years agoHelpfull: Yes(2) No(0)
Its 1/12+2/11
10 years agoHelpfull: Yes(0) No(2)
1/12+2/11 ans..
10 years agoHelpfull: Yes(0) No(5)
3/12 is the smallest possible value
10 years agoHelpfull: Yes(0) No(0)
It should be 1/11 + 1/12
9 years agoHelpfull: Yes(0) No(3)

P,Q,R,S are distinct integers numbered from 1 to 12.What is the possible smallest value for (P/Q)+(R/S)=?