three variables x,y,z have a sum of 30. all three of them are non-negative integers . if any two integers dont have same value and exactly one variable has a value equal or less than three .find number of possible solutiins9828568294
M getting 504 ..way over all the options...
Approach :
a = 0 ; b+c =30 where b.c >= 4
total ways 22(subtracted the case where they are equal i.e. 15)
a = 1 ;total cases 22
a =2 ; total cases 20
a = 3 ; total cases 20
in all 84 cases..
since they are all different any value from the above values can be assigned..so it should be multiplied by 3!..
pata nai kya kiya hai...but answer toh nai aa raha hai esse....
12 chocs among 4 kids is not 12C4...! It is 15C3 (revise partitioning theory - identical objects into distinct groups...)However this would give 15C3 + 14C3 + 13C3 which doesn't match any of the options.If there is also a condition that no kid goes empty handed, then 11C3 + 10C3 +9C3...but alas this also doesn't seem to match any of the options (not sure, did it in head, could be wrong)regardsscrabbler12 chocs among 4 kids is not 12C4...! It is 15C3 (revise partitioning theory - identical objects into distinct groups...)However this would give 15C3 + 14C3 + 13C3 which doesn't match any of the options.If there is also a condition that no kid goes empty handed, then 11C3 + 10C3 +9C3...but alas this also doesn't seem to match any of the options (not sure, did it in head, could be wrong)regardsscrabbler
but is my approach right? like taking the cases and then proceeding or is there an easier way out?
the digits 1, 2 , 3, ....,9are written at random to form a Nine-digit number..... The probability that it is divisible by "11" is
number of terms at odd places will remain constant(5 in our case).but in ur answer we have 4 terms as well as 5 terms for Y(sum of terms at odd places ) how ? m i missing something ?
@jain4444the digits 1, 2 , 3, ....,9are written at random to form a Nine-digit number..... The probability that it is divisible by "11" isnumber of terms at odd places will remain constant(5 in our case).but in ur answer we have 4 terms as well as 5 terms for Y(sum of terms at odd places ) how ? m i missing something ?
that was when sum of odd places = 17 and when sum of even places = 17
@jain4444the digits 1, 2 , 3, ....,9are written at random to form a Nine-digit number..... The probability that it is divisible by "11" isnumber of terms at odd places will remain constant(5 in our case).but in ur answer we have 4 terms as well as 5 terms for Y(sum of terms at odd places ) how ? m i missing something ?
45 = 17 + 28
Now it could be the 4 terms adding to 17 and the 5 to 28...or the 5 adding to 17 and the 4 to 28...hence... regards scrabbler
A shopkeeper makes a profit of Q% by selling an object for rs.24/-. Had the cost price and selling price been interchanged ,it would have led to a loss of 62.5Q%.With the latter cost price , what would be the new selling price to get a profit of Q%?
actually u r doing a little mistaketotal time taken d/(p+q) + 6d/(p+q) = 7d/(p+q)distance travelled by a from p = 7dp/(p+q)now p:q = 2:3putting the ratio it gives distance travelled as 2.8dso 0.8d from p
bhai how is the total time taken = d/(p+q) +6d/(p+q) ???
A shopkeeper makes a profit of Q% by selling an object for rs.24/-. Had the cost price and selling price been interchanged ,it would have led to a loss of 62.5Q%.With the latter cost price , what would be the new selling price to get a profit of Q%?
i am getting an answer close to 36.. options would be helpful to decide the answer
