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This Week's Manhattan challenge problem

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by enjoylife1788 » Sun Oct 16, 2011 11:34 pm
Dear members,

I came across this week's manhattan question and have no clue how to go about solving it. So can someone make me understand please?

Line m and line n intersect, forming 4 angles. Does any of these angles measure greater than 120°?

(1) The product of the measures, in degrees, of the four angles is less than 2^10 3^4 5^4.
(2) The product of the measures, in degrees, of the four angles is greater than 2^14 5^4.

I dont know what OA is.
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Source: — Data Sufficiency |
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by shankar.ashwin » Mon Oct 17, 2011 2:21 am
For 2 intersecting lines we know opposite sides are equal.

Let the 2 angles be x & y.
We know x + y = 180.

From (1) we know;

x^2 * y^2 = 2^10 * 3^4 * 5^4
Taking root, we have
xy = 2^5 * 3^2 * 5^2 = 7200.
We know, x+y = 180.
y = 180 -x

x(180-x) < 7200.

Angles can take values from (0-60). If x < 60, then Y will be greater than 120 (or) vice-versa. Sufficient.
Sufficient.

From (2);

xy= 2^7 * 5^2 = 3200.
And, similarly, x(180-x) > 3200.

The angles can take values from 20-160. We can't say if it will be less than 120 or greater. Insufficient.

A IMO.
Last edited by shankar.ashwin on Mon Oct 17, 2011 4:09 am, edited 1 time in total.
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by enjoylife1788 » Mon Oct 17, 2011 3:51 am
I guess you got the inequalities wrong.

Statement 1 says its less than 7200 and Statement 2 says its greater. :)
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by enjoylife1788 » Mon Oct 17, 2011 4:08 am
Also, I got a bit confused there. So did it my way. A bit length at first. So please correct me if m wrong :)

Statement 1 says,

x(180-x)<7200
180x-x^2<7200
-x^2+180x-7200<0

x^2-180x+7200>0 (Multiplying both sides by -1)
(x-120) (x-60) > 0

x> 120 and x>60

so when x is greater than 120, y is less than 60 and when x is greater than 60, y is less than 120.

so one of them is greater, hence sufficient

Statement 2 says,

xy>3200
x(180-x)>3200
similarly as 1, x^2 - 180x + 3200 < 0

(x-160) (x-20) < 0

x< 160 and x< 20.

Here, I don't understand, is it sufficient. It says its less than 160, so when x is less than 160, y is greater than 20. Accordingly, when x is less than 20, y is greater than 160.


First of all, is this explanation correct. And even if it is, m sure there is a better explanation than i have given. Please GMAT gurus, explain this problem. :)
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by shankar.ashwin » Mon Oct 17, 2011 4:10 am
Thanks for pointing out the mistake enjoylife1788 :) Edited the post. I was wondering where I went wrong :)
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by enjoylife1788 » Mon Oct 17, 2011 4:19 am
shankar.ashwin wrote:Thanks for pointing out the mistake enjoylife1788 :) Edited the post. I was wondering where I went wrong :)
Your Welcome :)

But i need your help to understand how you reached to those range of values. I know its too stupid. But I am just not getting it right now. ;)
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by shankar.ashwin » Mon Oct 17, 2011 4:29 am
enjoylife1788 wrote:
shankar.ashwin wrote:Thanks for pointing out the mistake enjoylife1788 :) Edited the post. I was wondering where I went wrong :)
Your solution is correct. Solving the quadratic equation gives you limits for the equation. You know roots of the equation are 20,160.

Roots are when the equations change sign usually. So test the inequality for 3 segments.

(0-20) (20-160) and (160-180)

You can see the condition x(180-x)>3200 holds true only for the range (20-160).

From this we cannot say if one angle would be greater than 120.
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by enjoylife1788 » Mon Oct 17, 2011 4:49 am
shankar.ashwin wrote:
enjoylife1788 wrote:
shankar.ashwin wrote:Thanks for pointing out the mistake enjoylife1788 :) Edited the post. I was wondering where I went wrong :)
Your solution is correct. Solving the quadratic equation gives you limits for the equation. You know roots of the equation are 20,160.

Roots are when the equations change sign usually. So test the inequality for 3 segments.

(0-20) (20-160) and (160-180)

You can see the condition x(180-x)>3200 holds true only for the range (20-160).

From this we cannot say if one angle would be greater than 120.
Ok I get it now. Thank You :)

So, The answer should be A.
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by enjoylife1788 » Mon Oct 17, 2011 4:52 am
Confirmed. OA answer is also A
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by thestartupguy » Mon Oct 17, 2011 8:51 am
Good question :)
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