Took AIMCAT 0611 today morning after taking the q paper from my friend:
VA/RC: 23.66
PS: 15.66
DI: 17.33
Total: 56.66
Hi All,
All the best for AIMCAT12..
Job Life has been going really tough
Still found out some time to give a mock
My Scores
QA : 10/6/8 
DI :19/4/17.6
VA: 28/5/26.33
Total : 52 ==98.6%
I know my weak areas. Hope i get some time to improve upon them
Currently, the strengths (QA) is not being used to its fullest. It sucked big time
take 3 boys and 4 girls.
each boy is friend of all the 4 girls and each girl is friend of all 3 boys. this satisfies the given condition.
the next number satisfying this condition will be 14, next 21 and so on..
you cannot take those two conditions in isolation and say that the number will be divisible by 4 and 5. assuming yours is right, then 20 is a number that satisfies this condition. can you say the number of boys and girls from that ? no. it is wrong..
Try finding number of boys or girls. you'll realise your fault. if total = 60, x =12 => no. of girls = 48... but u get a different number for y.
your solution.. when you say one boy is friends with four girls , this does not mean there are four times as many girls, because each girl has three boys as friends. not one.
thanks .such silly mistakes :satisfie:
ok let us take this approach...
suppose we think there are x boys then there must be 4x girls.so total strength of the class shud be a multiple of 5 as(strength x+4x)
and also if u take the case of every gal is exactly friends with 3 guys...
we can as earlier get to know tht total strength of class shud be a multiple of 4(y+3y)
so the number shud be diviisible by both 4 n 5
so only number between 59 and 68 divisible by 4 n 5 is 60
can u say where i have faulted??
Try finding number of boys or girls. you'll realise your fault. if total = 60, x =12 => no. of girls = 48... but u get a different number for y.
your solution.. when you say one boy is friends with four girls , this does not mean there are four times as many girls, because each girl has three boys as friends. not one. X boys would NOT mean 4x girls. Instead it would mean 4x pairs. similarly with girls, there are 3y pairs.
now 4x =3y.
=> x:y =3:4
=> x+y is a multiple of seven.
only 17 desks in the classroom ,each of which can hold more than four students => max of 68 students
there are 59 students in the class who study german => min of 59 students
every boy is friends with exactly 4 girls and every girl is friends with exactly three boys : this is possible only when boys+girls is a multiple of 7.
there is only one number (63) between 59 and 68 which is multiple of 7. that is the answer.
ok let us take this approach...
suppose we think there are x boys then there must be 4x girls.so total strength of the class shud be a multiple of 5 as(strength x+4x)
and also if u take the case of every gal is exactly friends with 3 guys...
we can as earlier get to know tht total strength of class shud be a multiple of 4(y+3y)
so the number shud be diviisible by both 4 n 5
so only number between 59 and 68 divisible by 4 n 5 is 60
can u say where i have faulted??
only 17 desks in the classroom ,each of which can hold more than four students => max of 68 students
there are 59 students in the class who study german => min of 59 students
every boy is friends with exactly 4 girls and every girl is friends with exactly three boys : this is possible only when boys+girls is a multiple of 7.
there is only one number (63) between 59 and 68 which is multiple of 7. that is the answer.
hi guys,
This is my first post in this forum.
Can u please help me out in analysin the problem given in aimcat11
Q31.In a class, every boy is friends with exactly 4 girls and every girl is friends with exactly three boys.it is known tht there r only 17 desks in the classroom ,each of which can hold more than four students.if there are 59 students in the class who study german,find the number of students in the class.
1.60 2.63 3. 66 4.cnbd
hoping ot see ur replies and analysis
orca SaysDont the sectional scores posted add upto 58.67
Yup ..u r right...I didnt expect such a silly mistake from a guy gettin 58 in 0611.
All the best 2 everybdy for 0612....
