@raopradeep said:if xy is a 2 digit number andd u,v,x,y are digits then find the number of solutions of the eqtn (xy)^2 =u!+v2305i am getting one of them is 11 plz find other 2
27^2 = 6! + 9
11^2 = 5! + 1
71^2 = 7! + 1
"In the parallelogram ABCD, AB=21, BC=13 and BD=14. Find the length of AC."
@techgeek2050 said:m getting 106. share ur approach
a=0
b and c can take the values from 4 to 26 and not 15 15: 22 cases
a=1
b and c can take the values from 4 to 25 : 22 cases
a=2
b and c can take the values from 4 to 24 and not 15 15: 20 cases
a=3
b and c can take the values from 4 to 23 : 20 cases
total 84 cases....i m not sure abt it...
@veertamizhan said:"In the parallelogram ABCD, AB=21, BC=13 and BD=14. Find the length of AC."
32? Used apollonius...
regards
scrabbler
regards
scrabbler
@Dexian said:a=0 b and c can take the values from 4 to 26 and not 15 15: 22 casesa=1b and c can take the values from 4 to 25 : 22 casesa=2b and c can take the values from 4 to 24 and not 15 15: 20 casesa=3b and c can take the values from 4 to 23 : 20 casestotal 84 cases....i m not sure abt it...
Since 3 diff var we should technically also do x 6 😞 To bahut bada answer aayega :(
regards
scrabbler
regards
scrabbler
@scrabbler said:Since 3 diff var we should technically also do x 6 To bahut bada answer aayega
regardsscrabbler
ye approach sahi hai na....
i thought of *6,
@veertamizhan said:"In the parallelogram ABCD, AB=21, BC=13 and BD=14. Find the length of AC."
32..
21^2 + 13^2 = 2(7^2 + a^2) (appollonius theorem of medians)
a= 16
so AC =32
@abhishek.2011 said: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
arre i m not taking time into picture. 😛
basically when A moves 2 parts, B moves 3.
basically when A moves 2 parts, B moves 3.
@raopradeep said:there are N men sitting around acircular table at N distinct points . every possible pair of men except the ones sitting adjacent to each other sings a 2 minute song one pair after other. of total time taken is 88 min. find N ?810911
11 ?
m(m - 3)/2*2 = 88
m = 11
zada has to distribute 15 choclates among her 5 kids(a,b,c,d,e) . such that "a" gets atleast 3 and at most 5 chocolates .in how many ways she can do this ?49577417435
@scrabbler bro i m doing it like this
first give 3 chocolates to a.. now the remaining 12 chocolates have to be distributed among 4 kids in 12c4
let a have 4 chocolates so you have to distribute 11 chocolates in 11c4 ways
Next you give 5 chocolates to a and then distribute remaining 10 chocolates to 4 people in 10c4 ways
So the total ways of distributing the chocolates is 12c4+11c4+10c4 = 445+330+210=985
What is the mistake i am committing here?
first give 3 chocolates to a.. now the remaining 12 chocolates have to be distributed among 4 kids in 12c4
let a have 4 chocolates so you have to distribute 11 chocolates in 11c4 ways
Next you give 5 chocolates to a and then distribute remaining 10 chocolates to 4 people in 10c4 ways
So the total ways of distributing the chocolates is 12c4+11c4+10c4 = 445+330+210=985
What is the mistake i am committing here?
@veertamizhan said:"In the parallelogram ABCD, AB=21, BC=13 and BD=14. Find the length of AC."
32 ?
21^2 + 13^2 = 2(7^2 + AC^2/4)
AC = 32
@raopradeep said: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
Whats the OA??
@iLoveTorres said:@scrabbler bro i m doing it like thisfirst give 3 chocolates to a.. now the remaining 12 chocolates have to be distributed among 4 kids in 12c4
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)
regards
scrabbler
@raopradeep said:if xy is a 2 digit number andd u,v,x,y are digits then find the number of solutions of the eqtn (xy)^2 =u!+v2305i am getting one of them is 11 plz find other 2
11^2 = 5! + 1
27^2 = 6! + 9
71^2 = 7! + 1 ??
@scrabbler
can u see that x,y.z sum = 30 wala sum.. plzzz