# Higher Level Maths ( also useful for PGDCM )

94 Posts  ·  54 Users
Okay peeps, We are yet to have a thread on higher level maths. In this thread we'll discuss : 1)differential and Integral calculus. 2)probability & Statistics. 3)Limits, Continuity and Discontinuity. 4)Convergence and Divergence of ...
Page 1 of 10    can anyone send me attachment shortcut method of train,mixture,alligations etc

Commenting on this post has been disabled by the moderator. what will be the answer to this question :
If you have 3 tickets to a lottery for which 10 tickets were sold and 5 prizes are to be given, the probability that you will win at least one prize is:
a. 7/12
b. 9/12
c. 1/12
d. 11/12

i think 11/12 is right

if the 3 tickets are such that none of them has any prize on them, then these three can be selected from 5 in 5C3 ways = 10 ways

total ways of selecting 3 tickets from 10 is 10C3 = 120 ways

so probability that none wins a prize = 10/120

so atleast one = 1 - 1/12 = 11/12
Commenting on this post has been disabled by the moderator. what will be the answer to this question :
If you have 3 tickets to a lottery for which 10 tickets were sold and 5 prizes are to be given, the probability that you will win at least one prize is:
a. 7/12
b. 9/12
c. 1/12
d. 11/12

i got the answer! its 11/12.. correct me if i m wrong..
my logic : Probability that you will win at least one prize = 1 - probability that you will not win any prize. therefore 1-(7C5/10C5) i.e. 1-(1/12) = 11/12
7C5 coz all 5 prize will b from the remaining 7 tickets and 10C5 coz from 10 tickets, 5 prizes will be picked.
Commenting on this post has been disabled by the moderator. what will be the answer to this question :
If you have 3 tickets to a lottery for which 10 tickets were sold and 5 prizes are to be given, the probability that you will win at least one prize is:
a. 7/12
b. 9/12
c. 1/12
d. 11/12

Commenting on this post has been disabled by the moderator. any idea how to solve questions like this.
x^2 - 2x + y^2 -4y +5 = 0 on the XY plane represents
1.a point
2.a circle
3.an ellipse
4.a hyperbola

x^2 -2x + 1 + y^2 -2*2y + 4 = 0

(x-1)^2 + (y-2)^2 = 0

so its a point !
Commenting on this post has been disabled by the moderator. any idea how to solve questions like this.
x^2 - 2x + y^2 -4y +5 = 0 on the XY plane represents
1.a point
2.a circle
3.an ellipse
4.a hyperbola

Commenting on this post has been disabled by the moderator. Derrida Says
And I ask why to do it? I'm sorry if you find my response stupid, but wouldn't there be other easier pickings in the paper?

It looks scary coz writing a matrix as such is not possible for me on computer. if omeobdy can solve it, it will be great help, otherwise coz i dont know how 2 do it, i would b leaving it. IMS has given such questions in JMET sample papers. dats y i wanted 2 know how 2 do it.
Commenting on this post has been disabled by the moderator. Ques. A= {(1,-1,0),(0,0,-1),(0,1,0)} is the matrix of linear transformation T with respect to basis {(1,0,0),(0,1,0),(0,0,1)}. Find the matrix of T with respect to the basis {(0,1,-1),(-1,0,1),(0,0,-1)}

i) A= {(1,0,-1),(1,-1,0),(2,2,-1)}
ii) A= {(0,1,0),(-1,0,0),(-1,-1,-1)}
iii) A= {(-1,1,-1),(1,1,0),(1,-2,-2)}
iv) A= {(1,-1,1),(1,1,0),(-1,2,-1)}

How to do this? And I ask why to do it? I'm sorry if you find my response stupid, but wouldn't there be other easier pickings in the paper?
Commenting on this post has been disabled by the moderator. Ques. A= {(1,-1,0),(0,0,-1),(0,1,0)} is the matrix of linear transformation T with respect to basis {(1,0,0),(0,1,0),(0,0,1)}. Find the matrix of T with respect to the basis {(0,1,-1),(-1,0,1),(0,0,-1)}

i) A= {(1,0,-1),(1,-1,0),(2,2,-1)}
ii) A= {(0,1,0),(-1,0,0),(-1,-1,-1)}
iii) A= {(-1,1,-1),(1,1,0),(1,-2,-2)}
iv) A= {(1,-1,1),(1,1,0),(-1,2,-1)}

How to do this? Commenting on this post has been disabled by the moderator. z +6 = +/- 3

z = -3 or -9

substitute

largest magnitude = 6

You see you've reduced the whole argand plane to the real number line....Nice .

BTW, anyone has any sample ATM tests?
Commenting on this post has been disabled by the moderator.