official quant thread for cat08

[selebratinglife]

Originally Posted by srikar2097

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...

May b th same method, not sure..
(y+y^3)^5 * (1+y^4)^5..In this expression take y common so it doesn't make any difference to th number of terms
(1+y^2)^5 * (1+y^4)^5..So ors in th expansion would be..
1 2 4 6 8 10
1 4 8 12 16 20
So the powers that can be obtain through the combination is
2,3,5,6,7,8,9,10,11,13,14,16,17,18,20,21,22,24,26, 28,30

Actual powers ar obtain by adding 5 to th above powers

So total 21 terms

[harkat]

it is 16.
@UDITMOHANls confirm.

[nsitaditya]

(1) If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^2 t^3 s^4, then the number of ordered pair (p, q) is


(A) 256 (B) 360 (C) 315 (D) 324 (E) 323


(2) In triangle ABC, side AB = 20, AC = 11, and BC = 13. What is the diameter of the semicircle inscribed in ABC, whose diameter lies on AB, and that is tangent to AC and BC?


(A) 9 (B) 10 (D) 19/2 (D) 23/2 (E) none of the foregoing

[selebratinglife]

Originally Posted by srikar2097

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.

passes are of the form n(n-1)/2 + 1..Now equating this to 50k+1(To come back to the first person..)..So n=25 instead of 24

[selebratinglife]

Originally Posted by nsitaditya

(2) In triangle ABC, side AB = 20, AC = 11, and BC = 13. What is the diameter of the semicircle inscribed in ABC, whose diameter lies on AB, and that is tangent to AC and BC?


(A) 9 (B) 10 (D) 19/2 (D) 23/2 (E) none of the foregoing

2. Say 'o' is th center of the semicircle..An th semicircle touchs AC an Dc at say p,q respectively..So OP = OQ =r ..we can fin AO,OB in terms of r which come out to b 5r/3,65r/3..So 2r=d=11..option(E)

[themystique]

Originally Posted by selebratinglife

2. Say 'o' is th center of the semicircle..An th semicircle touchs AC an Dc at say p,q respectively..So OP = OQ =r ..we can fin AO,OB in terms of r which come out to b 5r/3,65r/3..So 2r=d=11..option(E)

i did not understand this,

how did you get in terms of r?

ap = (11 - x) and bq = ( 13 - x)

oa = r + some value
ob = r + some value

how did you proceed next?


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[selebratinglife]

Originally Posted by themystique

i did not understand this,

how did you get in terms of r?

ap = (11 - x) and bq = ( 13 - x)

oa = r + some value
ob = r + some value

how did you proceed next?


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we can find CosA using cosine rule,,
An in th right angled triangle OPA, OP=r,So OA=OP/SinA

[themystique]

(1) If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^2 t^3 s^4, then the number of ordered pair (p, q) is


(A) 256 (B) 360 (C) 315 (D) 324 (E) 323


puys,

i thought of doing it like this,

initially, we find the number of ways the lcm can be expressed as a product.
then we take out common terms one at a time and repeat the process.
the common terms can be taken out in more than 50 ways,
this turns out to be extremely laborious,

any short cut i am missing out on?

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[shivam_01]

Originally Posted by ravijanjanam

sol: |x| can be expressed as +/- x

therfore |x+y|= +/-(x+y)
|x-y|= +/-(x-y)

=> case1: +(x+y)+(x-y)-4 = Z(suppose)

=> 2x-4=Z------------eq 1
=> case 2: -(x+y)-(x-y)-4 =Z1 (suppose)

=> -2x-4=Z1-----------eq 2

plotting eqns 1 and 2 on graph provieds us triangle with points (-2,0),(0,-4),(2,0)

therfore finding the area of this triangle is the solution

well i think this is the equation of a square not a triangle
as u can see from the fact that the point given by u (0,-4) does not satisfy the equation
the square will be form of the sides (0,-2) (0,2) (2,0) (-2,0)

hence the area of the square will be 16 square units

correct me if i am wrong

[themystique]

please solve this puys
from CAT 06-

1. the number of employees in a company is a prime number and is less than 300.
ratio of employees who are graduates and above to those who are not can be:

(1)101:88 (2) 87:100 (3) 110:111 (4) 85:98 (5) 97:84


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