official quant thread for cat08

[themystique]

*** How many terms are there in the expression : (y+y^3+y^5+y^7)^5


we can factorize it as:

(y* (1 + y^2)*(1 + y^4) )^5

the single y term can be neglected as it will not produce any new sum term.

for [(1 + y^2)(1 + y^4)]^5,
we get all terms with exponents from 0 to 30.

hence number of terms must be 31.

please correct me if i am wrong.



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[prince_chittu]

@themystique

for [(1 + y^2)(1 + y^4)]^5,
we will have all the terms of y having power ranging from 0 to 30

thus,
31 terms

@UDITMOHAN
confirm the answer please......

thank u

[phasmophobe007]

Originally Posted by Fuzon

I have doubt regarding the case where A picks 4 balls , not the 5 balls one ...

plz refer the bold part of my post

u ARE DOIN A MISTAKE
KEEP IN MIND A can pick a minimum of 1 and a maximum of 3 balls.So how can he pick 4??

[phasmophobe007]

Originally Posted by themystique

@phasmo,

could you please tell me if the formula that you provided is true in all cases?
and in which cases it could be applied and in which it shouldn't be?

thanks </div>

This formula is valid only when the minimum no. of balls picked is 1,in all other cases try a different approach

[srikar2097]

Originally Posted by 80199

In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x -y| - 4 is

Choose one answer.
a. 8 b. 12 c. 16 d. 20

Remove the modulus. We get x=2; x=-2; y=2; y=-2.
So area = 16. Hence (C)

[srikar2097]

Originally Posted by Udit Mohan

hi folks ,
pls solve these questions ...
*** How many terms are there in the expression : (y+y^3+y^5+y^7)^5

simplify we get [(y+y^3)(1+y^4)]^5
i.e. (y+y^3)^5 * (1+y^4)^5
Expand each term using binomial theorem and group together terms with same powers we get 15 terms.

Is there any shorter way. Mine seems to be lengthy...

[harkat]

@phasmophobe007
how u reached to the solution,is it by expansion...

[srikar2097]

Originally Posted by Udit Mohan

hi folks ,
pls solve these questions ...
*** Find the no of positive integral solution of A*B*C*D*E=1050??

Are the values unique? Is the needed set ordered?

1) If the values of (a,b,c,d,e) are unique and unordered, then (a,b,c,d,e) takes 1 set of solutions i.e. (2,1,25,3,7)

2) if values of (a,b,c,d,e) are not unique and unordered, then the no. of integral sols. are factors of 1050 i.e. 24 solutions.

Please post the answers…

[srikar2097]

Originally Posted by harkat

how u reached to the solution,is it by expansion...

Yup! Expansion it is. Since there were bound to be common terms (i.e. term with same powers) I had no other way of figuring that out without expansion..

[srikar2097]

Originally Posted by Udit Mohan

hi folks ,
pls solve these questions ...
***50 kids are standing in a cirular arrangement and one of them is having a ball .....the ball is passed counter-clock wise skipping 0,1,2,3,4 & so on kids respectively i.e. the 1st kid gives it to 2 nd kid, then 2nd kid gives it to 4th kid, 4th to 7 th & so on .....after how many hands change will the ball back to the first kid ???

Since we come back to the first kid. so the number of passes has to be a multiple of the total kids. i.e. 50. Also since the passes are in the form of n(n+1)/2. So we need that value of n which is a multiple of 50 and in the first one in that sequence. By trial and error, I got n=24.