@ScareCrow28 why multiply by 2 3 and 4 in step 2??
@ScareCrow28 said:Let N2, N3, N4 be exactly 2, exactly 3 and exactly 4Now, N2+N3+N4= 525Also, 2(N2) + 3(N3) + 4(N4) = 375+275+375+375 = 1400Put N4 = 375..It won't satisfy..Put N4= 275=> 2(N2)+3(N3)=400 and N2+N3= 250 (But this is also not possible) Since 2(N2)+2(N3) =500So N4=225@chandrakant.k
if N4 is 225
N2+N3 = 300
2*N2+2*N3 = 600....(1)
and put N4 in 2*N2+3*N3+4*N4 = 1400
2*N2+3*N3 = 500....(2)
(1)-(2)
we get -ve value of N3 .....
so our assumption must be wrong..
so 225 cant be ans 
@ScareCrow28 said:Let N2, N3, N4 be exactly 2, exactly 3 and exactly 4Now, N2+N3+N4= 525Also, 2(N2) + 3(N3) + 4(N4) = 375+275+375+375 = 1400Put N4 = 375..It won't satisfy..Put N4= 275=> 2(N2)+3(N3)=400 and N2+N3= 250 (But this is also not possible) Since 2(N2)+2(N3) =500So N4=225
if N4=275 , then 2(N2)+3(N3)=300, right?!
and i made a typo in the qs, my bad
....
Usme bhi ans jugaad se nikaal diya :P
This is the right one:
In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 375 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?
@FoolNFinal , @chandrakant.k
and i made a typo in the qs, my bad
....Usme bhi ans jugaad se nikaal diya :P
This is the right one:
In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 375 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?
@FoolNFinal , @chandrakant.k
coding
BREAD --> CZDQA
BUTTER --> QDSSTA
SALAD ?
OPTIONS :- RZKZC , CBKBR , CZKZR , BZKZR
@FoolNFinal said:if N4 is 225 N2+N3 = 3002*N2+2*N3 = 600....(1)and put N4 in 2*N2+3*N3+4*N4 = 14002*N2+3*N3 = 500....(2)(1)-(2) we get -ve value of N3 .....so our assumption must be wrong..so 225 cant be ans
Damn!! I did this for 275 but not for 225 thinking nothing else is left in options 
You are absolutely right!! 
Wrong options?
You are absolutely right!! Wrong options?@sonamaries7 said:if N4=275 , then 2(N2)+3(N3)=300, right?!and i made a typo in the qs, my bad ....Usme bhi ans jugaad se nikaal diya This is the right one: In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 375 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?
Now it all makes sense :)
@rkshtsurana said:codingBREAD --> CZDQABUTTER --> QDSSTASALAD ?OPTIONS :- RZKZC , CBKBR , CZKZR , BZKZR
CZKZR
Help in this problem
Calvin and David play a game of dice. They both throw fair 6-sided dice. The probability that the number on the David's dice is bigger than the number on Calvin's dice is ab, where a and b are positive coprime integers. What is the value of a+b?
My Solution : such cases (2,1)(3,1)(3,2)(4,1)...(6,5) = 15 such cases
total cases = 36
now since the probability is given as ab, should i take it as a %???
in that case 15/36*100 = 41.6 %
all are co prime but it is not of the form ab 😞
@rkshtsurana said:codingBREAD --> CZDQABUTTER --> QDSSTASALAD ?OPTIONS :- RZKZC , CBKBR , CZKZR , BZKZR
CZKZR eazzzzzzzzzy
A pt P is 3 units away from the line 3x+4y+k=0(L1) and 4 units away from the line 4x-3y+l=0 (L2). The reflection of the pt on L1 is P1, the reflection of P1 on L2 is P2, the ref of P2 on L1 is P3, the ref of P3 on L2 is P4. Find the area of the fig formed by joining P1 ,P2, P3 and P4.
48
96
24
Infinite
48
96
24
Infinite
@sonamaries7 said:if N4=275 , then 2(N2)+3(N3)=300, right?!and i made a typo in the qs, my bad ....Usme bhi ans jugaad se nikaal diya This is the right one: In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 375 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?@FoolNFinal , @chandrakant.k
II + III + IV = 525
2II + 3III + 4IV = 1500
III + 2IV = 450
max IV = 225 when III=0.
..............................................
@sonamaries7 said:A pt P is 3 units away from the line 3x+4y+k=0(L1) and 4 units away from the line 4x-3y+l=0 (L2). The reflection of the pt on L1 is P1, the reflection of P1 on L2 is P2, the ref of P2 on L1 is P3, the ref of P3 on L2 is P4. Find the area of the fig formed by joining P1 ,P2, P3 and P4.489624Infinite
48
Perpendicular lines... Area of figure made by point and lines is 4*3=12
Since reflection is taken 4 times.. total area= 4*12=48
@sonamaries7 said:A pt P is 3 units away from the line 3x+4y+k=0(L1) and 4 units away from the line 4x-3y+l=0 (L2). The reflection of the pt on L1 is P1, the reflection of P1 on L2 is P2, the ref of P2 on L1 is P3, the ref of P3 on L2 is P4. Find the area of the fig formed by joining P1 ,P2, P3 and P4.489624Infinite
48
@Brooklyn said:kaise??
The lines given r perpendicular to each other
Assume any 2 perpendicular lines say co-ordinate axes and plot the points
A rectangle is formed
Assume any 2 perpendicular lines say co-ordinate axes and plot the points
A rectangle is formed
@deedeedudu said:The lines given r perpendicular to each otherAssume any 2 perpendicular lines say co-ordinate axes and plot the pointsA rectangle is formed
points kaise nikale??
@Brooklyn said:points kaise nikale??
Bhai koi b point assume kar le since the ques offers such leeway
say a point (3,4) its reflection on x axis (3,-4) its reflection on y axis (-3,-4) and finally (-3,4)
say a point (3,4) its reflection on x axis (3,-4) its reflection on y axis (-3,-4) and finally (-3,4)
@Brooklyn said:points kaise nikale??
Reflections will have same area as original figure.. 4 like figures are made since 4 points are made by tking reflections. Just multiply area of one such figure by 4. no need for taking any point