Problem Solving

<ACB is 60?
prolong AP for the length of PB and make a point O
triangles APB anf trg CPO are simalar
triangles ACP anf trg POB are simalar
and so on... I am stuck

4meonly wrote: triangles APB anf trg CPO are simalar
triangles ACP anf trg POB are simalar

I don't think this is true - you know that PO = PB, but you don't know that PA = PC.

sureshbala wrote:Folks, here is the next apple.....



In the above figure, PB/PC = 1/2 LCBA = 45° while LAPC = 60°. Find LACB.

This was difficult, and took a while, but I think I got it.

Since <APB + 60 = 180, <APB = 120, which means that <PAB = 15

PB:PC is 1:2... let's assign the value X to PB, meaning that PC = 2X

The ratio of any given angle in a triangle to 180 is equivalent to the ratio of the length that angle's opposite side to triangle's perimeter.

The ratio, <PAB:<PBA:<APB = 15:45:120, or 1:3:8. Following that ratio and the fact that PB=X and is opposite <PAB, AP=3X (opposite of <PBA), and AB=8X (opposite of <APB). (To further that, you can say that <PAB=X, <PBA=3X and <APB=8X, where X=15)

Since <CAP + <ACP + 60 = 180, <CAP + <ACP = 120.

Since the opposite side of <ACP is AP and AP=3X, <ACP=3X; and since the opposite side of <CAP is CP and CP=2X, <CAP=2X and the ratio of <ACP to <CAP is 3:2.

Taking that, you can rewrite the equation, <CAP + <ACP = 120, as 2X + 3X = 120:

2X + 3X = 120
5X = 120
X = 120/5 = 24

Since, <ACP = 3X, <ACP = 24*3 = 72

<ACB = <ACP = 72

utibay wrote: The ratio of any given angle in a triangle to 180 is equivalent to the ratio of the length that angle's opposite side to triangle's perimeter.

The ratio, <PAB:<PBA:<APB = 15:45:120, or 1:3:8. Following that ratio and the fact that PB=X and is opposite <PAB, AP=3X

Means that on 30-60-90 angles triangle ratio of angles is 1:2:3, BUT thr ratio of sides is 1:2:sqrt3
according to your reasoning this ration should be 1:2:3
What do you think?

sureshbala, where are you?

Folks, finally here is the solution....Thought of being more clear on this problem, so uploaded the image....Have a look at this....

Hi Sureshbala - I followed all of that to the point where you wrote:

AD = DB, so DC = AD

I don't follow the logic there - how do we know that DC = AD?

Love the rest of the puzzle though. It's kind of elegant. Have you got more?

We know that DPB=120 so DBP=PDB=30 each. Since ABP=45 and DBP=30 we get DBA=15 already BAD=15 hence ADB is isoceles triangle. Hence AD=DB. Its already proved that DB=DC hence AD=DC.

- Deepak

Guys,
I first got 45 degree angle but could not prove it. But then I remembered simple circle properties. Yep, the answer is right. Thanks.

Keep posting similar questions!

Folks, as a content developer I prepare these questions and I am sure you are going to like these problems...any feed back is welcome.....

Although your questions are challenging, I've never seen this type of problems on GMAT. IMHO, your questions are more about good (and pretty hard) geometry practice than the GMAT itself.

How to waste time 101

Come on guys, let's get back to Earth. We are not studying for Calculus in the college. This is a GMAT test and concepts are very VERY straightforward.

Keeping in mind the feedback(I agree to the point that these questions are a bit tough but at the same time much within the basic concepts) that I got, here is the next one...

Edusoft, a software firm has engaged certain number of men to finish a project in the stipulated time. But after working for 14 days, 1/4 of them have left and to complete the remaining project the remaining members have taken as many days as the initial number of men would have taken to complete the entire project. In how many days was the project completed?

A. 48 B. 60 C. 56 D. 40 E. None of these

IMHO this can be a good question for entrance exam of some engineeing test but not GMAT. GMAT is not for making you Aaryabhatta it is to judge whether you can easily understand the business calculations.

welcome wrote:IMHO this can be a good question for entrance exam of some engineeing test but not GMAT. GMAT is not for making you Aaryabhatta it is to judge whether you can easily understand the business calculations.

I agree that these questions are tough but as pointed out by sureshbala these are based only on the baisc concepts and not beyond that. I think Aaryabhatta math is much ahead of this. I seriously feel that if these questions are asked in terms of DS (where we don't need to find the answer but in order to crack these questions we need the concepts) some of us for sure would agree to the point that they may appear in the exam at least for a 780+ aspirant. sureshbala why don't you start a DS thread?