@krum : Similar to the qs on Co A and B that I posted above...using the same approach am not getting the rite ans ...see,A certain no of girls and 10 boys attended a party. No boy shakes hands with more than 4 grls.Out of all the grls, 4 of the grls shook hands with 4 boys each and each of the rem girls shook hands with 3 boys. the greatest possible no of grls?using the app above: 4+6C3=24 which is wrong!!
@krum : Similar to the qs on Co A and B that I posted above...using the same approach am not getting the rite ans ...see,
A certain no of girls and 10 boys attended a party. No boy shakes hands with more than 4 grls.Out of all the grls, 4 of the grls shook hands with 4 boys each and each of the rem girls shook hands with 3 boys. the greatest possible no of grls?
using the app above: 4+6C3=24 which is wrong!!
B1-B4 fix with G1-G4. now (x-4) girls r left. 3(x-4) = 24. x-4 = 8. x = 12. 12 girls.
There is a fam with father , mother and 6 children A thru F. They have to attend a party for which only one among the mother/father can attend along with any no of kids subject to the foll constraints:1) B or C won't go along with their mother2) F goes only if A or D goesIn how many ways can the fam attend the party?71677970100
the (x-4) girls have to compensate for the 6*4 handshakes (max possible) that the boys need to make....while te girls can make only 3 handshakes.hence, 3(x-4) = 6*4. hope it is clear.
aah..the max handshakes are 40(total)..out of which the 4 grls have made 16 so (g-4) have to make the rem 24, right?
In a quad PQRS, QR=16, RS=18, PS=20 and /_SPQ=60, /_PQR=120. Find PQ.
We get that PS || QR Now draw a line parallel to RS We get a triangle with2 sides 4 and 18 and one angle as 60' Uisng cosA=b^2+c^2-a^2 / 2bc, we get the PQ=19.7 Whats the OA?
We get that PS || QRNow draw a line parallel to RS We get a triangle with2 sides 4 and 18 and one angle as 60'Uisng cosA=b^2+c^2-a^2 / 2bc, we get the PQ=19.7Whats the OA?
There is a fam with father , mother and 6 children A thru F. They have to attend a party for which only one among the mother/father can attend along with any no of kids subject to the foll constraints:1) B or C won't go along with their mother2) F goes only if A or D goesIn how many ways can the fam attend the party?71677970100plz share the approach also..
Case 1) When mother goes , only options are A,D,E,F =>
4C0 + (4C1-1) + (4C2-1) +4C3 + 4C4 = 14 ( 4c1-1 coz only F not possible, 4C2-1 coz only EF not possible )
Case 2) When papa goes, options are A,B,C,D,E,F =>
6C0+ (6c1-1) + (6c2-3) + (6c3-3) + (6c4-1) + 6c5 + 6c6 = 56 ( only F not possible ; BF,CF,EF not possible; BCF,BEF,CEF not possible; BCEF not possible in the same order)
Case 1) When mother goes , only options are A,D,E,F =>
4C0 + (4C1-1) + (4C2-1) +4C3 + 4C4 = 14 ( 4c1-1 coz only F not possible, 4C2-1 coz only EF not possible )
Case 2) When papa goes, options are A,B,C,D,E,F =>
6C0+ (6c1-1) + (6c2-3) + (6c3-3) + (6c4-1) + 6c5 + 6c6 = 56 ( only F not possible ; BF,CF,EF not possible; BCF,BEF,CEF not possible; BCEF not possible in the same order)
= 14+56 = 70
Bhai in CASE 1 why cant we take - 6C4 as only 4 can go ... when the mother goes...
Bhai in CASE 1 why cant we take - 6C4 as only 4 can go ... when the mother goes...
6C4 would definitely mean that u are taking only 4 but which 4 is the question... how would you remove the part that B and C would not be a part of that 4.
6C4 gives you all combinations of 4 out of given 6 choices which doesnt eliminate cases like ABEF , ABCD etc......
Rajiv is a student in a business school. After every test he calculates his cumulative average.QT and OB were his last two tests.83 marks in QT increased his average by 2. 75 marks in OB further increased his average by 1. Reasoning is the next test, if he gets 51 in Reasoning, his average will be ___? A. 59 B. 60C. 61 D. 62E. 63
