CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[implex]

Originally Posted by Aarav

Plug-in the value from the options. For x = 32 we have sin(76-32).sin(72) = sin44.sin(76+32).

Let's see who comes up with pure geometrical solution

what's wrong with a trigonometrical solution . geometry will make it easier, but the chances of it striking are lesser, but I know trigo will always work for me !!

[Aarav]

Originally Posted by implex

what's wrong with a trigonometrical solution . geometry will make it easier, but the chances of it striking are lesser, but I know trigo will always work for me !!

Nothing wrong - even I follow the same approach and as you said thanks to the genius Loney. But purists may argue for geometrical solution and most of these solutions teach a thing or two more about geometry

[Daniel Ocean]

Let M be the point where the perp. bis. cuts DA

we can prove that MDX and AMX are equilateral => DX = AX (SAS axiom)
also let <DXA = 2x;

now, using sine law in ABX,
AB/sin(<AXB) = AX/sin(44)
or AB/sin(76-x) = AX/sin(44) ...(1)

using sine law in CDX,
CD/sin(104-x) = DX/sin(72)
use sin(180-x)=sin(x)

we get CD/sin(76+x) = DX/sin(72) ...(2)

now in 1 & 2 AB = CD and AX=DX so eliminate them

we have sin(76-x)/sin(76+x) = sin(44)/sin(72)

use componendo dividendo, then use identities: sin(90-x)=cos(90-x) and for tan(90-x)=cot(x)

upon simplification we get cot(x)=tan(58=cot(32)

so x=32
hence ans is (3) 64

[implex]

Originally Posted by Daniel Ocean

Let M be the point where the perp. bis. cuts DA

we can prove that MDX and AMX are equilateral => DX = AX (SAS axiom)
also let <DXA = 2x;

now, using sine law in ABX,
AB/sin(<AXB) = AX/sin(44)
or AB/sin(76-x) = AX/sin(44) ...(1)

using sine law in CDX,
CD/sin(104-x) = DX/sin(72)
use sin(180-x)=sin(x)

we get CD/sin(76+x) = DX/sin(72) ...(2)

now in 1 & 2 AB = CD and AX=DX so eliminate them

we have sin(76-x)/sin(76+x) = sin(44)/sin(72)

use componendo dividendo, then use identities: sin(90-x)=cos(90-x) and for tan(90-x)=cot(x)

upon simplification we get cot(x)=tan(58=cot(32)

so x=32
hence ans is (3) 64

I know its a typo
still to not confuse others
Sin(90-x)=cosx

[dewan_iitr]

Originally Posted by Aarav

The problem has been solved correctly by most of you in the 1st attempt. Very well done

Here is the official solution to 001 - we might change this if by end of the day someone comes with a better approach

Solution:

Given a(a+1) + b(b+1) + c(c+1) + d(d+1) = 312 => (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2 = 312 - 3S + 4 = 256. Let a-1 = A, b-1 = B, c-1 = C, d-1 = D => we have non-negative integers A, B, C, D such that A^2 + B^2 + C^2 + D^2 = (A+B+C+D)^2 (the LHS <= RHS always) and can only be true if all but one number is zero => three among a, b, c, d are 1 and fourth number is 17.

Thus (3) holds true - (2) can guarantee the values for c and d. For a=b we have exactly 2 determinable values - thus (1) is not true.

=> choice (5) is the right answer

Hi Arav!
Late joinee
i tried the problem and i feel answer should be (3)
as we all have got 17,1,1,1 is the only sol. and if a>b then all of them are uniquely determinable. a-17, b=c=d=1

plz clear my doubt

-Ashish

[dewan_iitr]

Originally Posted by dewan_iitr

Hi Arav!
Late joinee
i tried the problem and i feel answer should be (3)
as we all have got 17,1,1,1 is the only sol. and if a>b then all of them are uniquely determinable. a-17, b=c=d=1

plz clear my doubt

-Ashish

Gott it man
(2) is also correct so

(5) is the answer

Thanks anyways

[dewan_iitr]

Originally Posted by Aarav

Hello All,

Kindly report us in case you aren't receiving the NLs. I hope all of you got the announcement QQAD 2008 mailer?

Quite a number of PGites PMed me today as they didn't get QQAD # 001 and I have forwarded their email id to PG technical team.
For any future concerns please write to estranged_gnrs or me. Writing directly to Rohit (estranged_gnrs) will invoke quicker response on technical issues.

Hi Arav

I am not receiving the QQAD mails this year.
I am registered user from past two years - email ID <snipped>
I also registered with another ID - <snipped>
currently i am looking questions from the post and solving them.

Thnax
Ashish

[vineet.nitd]

Originally Posted by dewan_iitr

Hi Arav

I am not receiving the QQAD mails this year.
I am registered user from past two years - email ID #######
I also registered with another ID - XXXXXX

currently i am looking questions from the post and solving them.

Thnax
Ashish

Please remove your email ID info by editing your post. It is against the forum rules.

You need to subscribe again in order to receive the QQAD problems of this season. Find the link for the same on the home page of this forum.

If you still face problems, PM(personal message) Aarav for that.

Thank You!

[Aarav]

Originally Posted by dewan_iitr

Hi Arav

I am not receiving the QQAD mails this year.
I am registered user from past two years - email ID I also registered with another ID

currently i am looking questions from the post and solving them.

Thnax
Ashish

I have been informed by PG technical team that users who subscribed for text format are facing issues. Did you subscribe earlier in text format? If not, request you to unsubscribe and subscribe again for HTML format. If the issue still persists then please PM estranged_gnrs or me.

[dewan_iitr]

Originally Posted by Aarav

I have been informed by PG technical team that users who subscribed for text format are facing issues. Did you subscribe earlier in text format? If not, request you to unsubscribe and subscribe again for HTML format. If the issue still persists then please PM estranged_gnrs or me.

ya man i did it in text format
i ll do that in HTML

thanx