Mgmat cat qn: Intresting question

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dinesh19aug: Senior | Next Rank: 100 Posts; Posts: 78; Joined: Thu Feb 21, 2008 9:05 am; Thanked: 7 times

by dinesh19aug » Thu Jul 29, 2010 5:59 pm

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Source: Beat The GMAT — Data Sufficiency | Thu Jul 29, 2010 5:59 pm

selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Thu Jul 29, 2010 7:51 pm

Dinesh,

I think this is PS problem.

Can you please post the answer options?

--Anand--

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Thu Jul 29, 2010 7:55 pm

y=(3y + 4)^1/2

y^2=3y+4

y^2-3y-4=0

(y-4)(y+1)=0

y=4,y=-1

Product of all possible solutions=4*(-1)=-4

--Anand--

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dinesh19aug: Senior | Next Rank: 100 Posts; Posts: 78; Joined: Thu Feb 21, 2008 9:05 am; Thanked: 7 times

by dinesh19aug » Fri Jul 30, 2010 10:31 am

Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Fri Jul 30, 2010 5:41 pm

Yes I thought about this root.

I have doubt here.

y=(3y + 4)^1/2

y=4,y=-1

y=-1

-1=sqrt(1)

sqrt(1)=1 or -1.

y=4

4=sqrt(16)

sqrt(16)=4 or -4.

See these above values.y=-1 satisfy the above equation.

Am I missing something?

--Anand--

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Fri Jul 30, 2010 9:01 pm

dinesh19aug wrote:Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.

If the only valid root is 4, then how can you take the product of all possible roots? You can't take the product of 1 number (that's non-sensical).

When you say "MCAT 1", do you mean MGMAT CAT 1? Or is this from another source? MGMAT doesn't usually create invalid questions.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Fri Jul 30, 2010 9:36 pm

dinesh19aug wrote:Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.

Did you copy the question exactly? Or did you add the "^1/2" because you couldn't copy the square root sign properly?

If the right side were "sqrt(3y + 4)", then it would only include the positive root. However, if it's reproduced correctly, then it should include both the positive and negative root.

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dinesh19aug: Senior | Next Rank: 100 Posts; Posts: 78; Joined: Thu Feb 21, 2008 9:05 am; Thanked: 7 times

by dinesh19aug » Sat Jul 31, 2010 8:47 am

Stuart Kovinsky wrote:
dinesh19aug wrote:Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.
If the only valid root is 4, then how can you take the product of all possible roots? You can't take the product of 1 number (that's non-sensical).

When you say "MCAT 1", do you mean MGMAT CAT 1? Or is this from another source? MGMAT doesn't usually create invalid questions.

Yes this MGMAT cat question and I copied the question exactly as it was in the CAT.

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Sat Jul 31, 2010 7:23 pm

I am not sure about this.

sqrt(1)=1 or -1

If you say -1 not satisfy the equation.lets try with y=4

4=sqrt(16)

sqrt(16)=4 or -4.Why should v take only 4 as its root?-4 also is valid.

--Anand--

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Sat Jul 31, 2010 7:23 pm

Can you please post the answer options?

--Anand--

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haidgmat: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Mon Jul 12, 2010 6:32 am

by haidgmat » Tue Aug 03, 2010 11:58 am

What's the OA!!!!!!!!!!! PILIZZ GIVE IT !!

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Tue Aug 03, 2010 12:20 pm

hi guys
to me the answer is 4
my reasoning
for sure the given equation has two roots after squaring
y1=4, and y2=-1

but now insert y2=-1 in both parts of the equation and check equality
(3*(-1)+4)^1/2=(-3+4)^1/2=1
so right part of the equation=1
but left part still y=-1
-1=1 no! y=-1 is not valid root here ( to me that is the trap)
so we are left with y=4

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Tue Aug 03, 2010 7:55 pm

I am not still convinced...sqrt(16) can be 4 or -4

--Anand--

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dinesh19aug: Senior | Next Rank: 100 Posts; Posts: 78; Joined: Thu Feb 21, 2008 9:05 am; Thanked: 7 times

by dinesh19aug » Tue Aug 03, 2010 10:13 pm

Selango: I have already posted the answer as 4.
Still here are the choices
A) -4
B) 1
C) -1
D) 4
E) 8

The question does not say that there HAS to be two roots. The question says what is product of possible root(s). There could be 1 or 2 or more than 2 roots.

Clock60 very well explained. There can be no detailed explanation than this.

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neonaan: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Fri Jul 09, 2010 3:55 am

by neonaan » Thu Aug 05, 2010 10:16 pm

Q: If y=(3y + 4)^1/2 then the product of all possible solution(s) for y is?

I got this question in MCAT 1 and got it wrong. I read the solution and was able to understand the reasoning. I DO NOT have any doubts on this question. Just wanted to share the question with everyone on the forums.

Solution: the 1/2 on the R.H.S (Right Hand side) of the = sign when transferred to the L.H.S makes it

yÂ² = 3Y+4 => yÂ²-3Y-4 =0 (Product is 4 and the Sum is -3 so -4 &1 are your best options)

yÂ²-4Y+Y-4=0 => y(Y-4) +1(Y-4)=0 => (Y-4) =0 ; (Y+1)=0 => Y= 4 or Y=-1

Substitute the values in the equation yÂ²-3Y-4=0 => First take 4 => 4Â² -3*4 -4 =0 ; 16-12-4 =>16-16 =0

Take the second value "-1" => -1Â² -3*(-1) -4 =0 +> 1 + 3 -4 => 4-4 =0.

From the list of Answer option it should be either 4 or -1

Feel free to comment if the approach is incorrect.

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