@raopradeep said:a box contains five sets of balls while there are three balls in each set . each set of balls has one colour which is different from every other set . what is the least number of balls need to be removed from the box so as to assure that a pair of same coloured balls has been removed67911
@iLoveTorres said:If p > 4 is a prime number, then 24 divides p2 - 1 without remainder (A) never (B) sometimes only (C) always (D) only if p = 5 (E) none of these
@raopradeep said:a box contains five sets of balls while there are three balls in each set . each set of balls has one colour which is different from every other set . what is the least number of balls need to be removed from the box so as to assure that a pair of same coloured balls has been removed67911
@Subhashdec2 said:5 all different have to be removed first 5then lets say we ahve 10 balls left lets make pairs of 2 haveing same colorso 5 different colors are there5+5+1=11Box 1=1 6 8 Box 2=2 7 9Box 3=3 8 10Box 4=4 7 9Box 5=5 6 10
Q:If both the roots of the quadratic equation ax^2+bx+c=0 lie in the interval (0,3) then a lies in
(1,3),(-1,-3),(-1,3),none of these
@raopradeep said:yaar need little more explanation samaj nahi aaraha hai
@Subhashdec2 said:let the balls 1 2 3 4 5 be of different colornow 10 balls are leftlet them be of 5 different colors and 2 of each color are therethese pairs of two are shared between different boxes (for me it is balls no 6 7 8 9 and 10)now pick up 1-5=5 ballsden pick up one each of 6-10=5 ballsthen the next one will ensure a pair11
@pakkapagal said:Q:If both the roots of the quadratic equation ax^2+bx+c=0 lie in the interval (0,3) then a lies in (1,3),(-1,-3),(-1,3),none of these
@Subhashdec2 i have a doubt.....question states that what is the least number of balls need to be removed from the box so as to assure that a pair of same coloured balls has been removed...
we have to stop when we have a pair of same colored balls...
so, even if the first 5 balls are of different color...then also from the 6th ball we will get a pair...
Please let me know if i missed sumthing here....
@raopradeep said:bhai ye clear nahi ho rahaa hai ki after picking of 1-5 balls =5 balls next any ball will generate pair from 6-10 balls i think plz correct me where i went wrong
@Sufi0469 said:@joyjitpal@raopradeepcan u please tell what does lie in interval (0,3) means..???
0 , 3 are excluded :)
@pakkapagal said:Q:If both the roots of the quadratic equation ax^2+bx+c=0 lie in the interval (0,3) then a lies in (1,3),(-1,-3),(-1,3),none of these
@Sufi0469 said:@Subhashdec2 i have a doubt.....question states that what is the least number of balls need to be removed from the box so as to assure that a pair of same coloured balls has been removed...we have to stop when we have a pair of same colored balls...so, even if the first 5 balls are of different color...then also from the 6th ball we will get a pair...Please let me know if i missed sumthing here....
@raopradeep said:(1,3) i guess
@cat_virus said:is it (1,3) as the eqn will be (x-r1)(x-r2)=0 where r1 and r2 lie in(0,3)and from this eqn , the coeff of x2 i.e a cant be a negative no..
@joyjitpal said:means interval between 0 and 30 , 3 are excluded

@Subhashdec2 let me explain....
suppose we have 15 balls for which we have 5 sets...
like
3 white
3 blue
3 red
3 green
3 pink
now in these knid of questions we have to find the worse case scenario...
there can b a case like his 1st & 2nd pick both comes out to be white..
so we have got a pair..
but as per worse case scenario.....we will assume that he picks all diff colors in his first 5 picks...
now then at his 6th pick any balls he picks it will always make a pair with any of the 5 balls he have already picked...
this is what i have understood from the question....
@Sufi0469 said:@Subhashdec2 let me explain....suppose we have 15 balls for which we have 5 sets...like3 white 3 blue3 red3 green3 pinknow in these knid of questions we have to find the worse case scenario...there can b a case like his 1st & 2nd pick both comes out to be white..so we have got a pair..but as per worse case scenario.....we will assume that he picks all diff colors in his first 5 picks...now then at his 6th pick any balls he picks it will always make a pair with any of the 5 balls he have already picked...this is what i have understood from the question....
oa is 11