You could obviously test numbers and solve it, which is actually reasonably fast or, you could deduce that 1188 = 2*2*3*3*3*11

Obviously one of the numbers has to be divisible by 11. So if you look at the choices and see for x = 84, y = 156 and neither are divisible by 11, then you could look at x = 96, y = 144, and neither are divisible by 11. So then, you've killed three choices already. You look at 108 and 132. 132 is divisible by 11. Hence 108 is the answer. Or of course you could use the fact that HCF * LCM = x * y

or in this case 14256 = x(240 - x)

you can use the following formula; ((-b (+ or -))sqrt(b^2 - 4ac))/2a

from where you can get both values of x and the smaller one is the answer.

sureshbala wrote:

sureshbala wrote:Folks, sorry for the delay (I am on a vacation).

Anyway, here is the next one.....

There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there?

A. 4

B. 6

C. 8

D. 10

E. Cannot be determined

Folks, this can be answered very quickly if you can conclude the following two results.

Result 1: Since the number of chairs occupied by the girls in both the cases is same there is no need to consider the number of girls.

Result 2: In the second case we are able to seat 2 boys per chair and no boy is left unseated, hence the number of boys must be even.

Let the number of boys be 2x.

Seats occupied in the first case = 2x-1

Seats occupied in the second case = x

Given that 2x-1-x = 3

So x = 4

Hence the number of boys 2x = 8.

That's one solution and again you could just test the numbers. If you try 4 boys and 5 girls, then according to statement 1, 8 chairs are there.

Then 2 boys and 5 girls = 7 chairs are there with 3 unoccupied so no. of chairs = 10. So these two statements don't agree. Using this, if you go on, you will see that 8 is the best answer.

Dear sureshbala,

I would like to thank you for these great problems. I did them just for fun (I wont take the GMAT in the next 3 years) and it was great pleasure to solve your questions!

Hope you continue with problems of this (or even higher ) difficulty!

Best regards,
Antoni

My 1st Post ....!!!

A room has the following dimensionz..

Length : 6 metres
Breadth : 6 metres
Height : 6 metres

an Ant which is @ one end of the longest diagonal needs to reach the other end of the diagonal in the shortest possible path ..!!

whats the length of the shortest path ?

Assuming that the Ant cannot fly

The shortest path will be one side and one smaller diagonal.

6 + 6 * root(2)
i.e. approx 14.49

Dear Sureshbala

Really good questions .. can we have the next question plsssss ??

The answer is not 6+6sqrt(2). The answer is 6sqrt(5). The trick is to unfold the cube and then draw a line from the two points and calculate that distance.

@prindaroy

I think the ans will be 6sqrt(3).

Qn:
An Ant which is @ one end of the longest diagonal needs to reach the other endof the diagonal in the shortest possible path ..!!

The shortest possible path will be a straight line. Hence, the qn asks for length of the longest diagonal in Cube (Room with L=B=W).

Formula: sqrt(L^2+W^2+B^2)
Here, L=B=W=6
sqrt(6^2(3)) = 6sqrt(3).

Please explain, how the ans is 6sqrt(5)?

@abhi27
The Ant can fly, but it has fly on a straight line to reach the other end of the diagonal in shorterst path. :!:

regards,
hhk.

[/b]

length of line = sqrt(12^2 + 6 ^2)

= 6sqrt(5)

You can't just square all sides. Look at the attachment for a clearer picture.

@prindaroy

UnFolding the cube is fine. But, the question is this the Shortest path for the Ant to reach the other end of the longest diagonal?

Here, we don't need to unfold the Cube (Room) to calculate the length.

First, find out the longest diagonal.
The Longest diagonal on cube is left end corner of bottom of the side to the right end corner of the top.

The shortest path to reach this longest diagonal will be a straight line....
(Ant has to fly in a Straight line...)

Join the LEC@bottom and REC@top, it is the longest straight line and consitute a diagonal.

See the attached for detail.

I have read in formums here abt the formula for 3 dimensional figure with lognest straight line.
Formula: sqrt(L^2+W^2+B^2)
Here, L=B=W=6
sqrt(6^2(3)) = 6sqrt(3).

Please correct me if I am wrong...

rgrds,
hhk.

Your answer is correct if the ant can fly. Ants cannot fly, which is why the word Ant was in bold in the op's post. I was assuming the ant cannot fly, and therefore it would have to crawl to the other end in which case the answer is 6sqrt(5).

Oops...Thanks for explanation prindaroy. I was comparing this question with other question in Cube. Hence, I have not given much imp. to Ant.....
Hence, the flaw...

Anyways, now I understand the prob correctly..

regrds,
hhk

Hi,

It is my first post. I wanted to know whether there is a thread for 780+ aspirants for DS also. I really liked the 780+ PS (sticky) thread started by sureshbala. Is there any such threads available for DS questions too?

@ Sureshbala: Thanks a lot for those tough question. Have you started any such thread for DS questions too?

Thanks,
Syed

Guys,

I got it in a different way.
I have constructed a 3-4-5 right triangle and extended it to a prism, where the height is 5. (since hypo- becomes the side of the square.)
Then, if I draw a plane, which passes through points B and O, it must divide the right triangle into two equal angles.

Hence answer is 45.

please see the attached image.

Thanks,
Vivek

Folks, I am back after a long break.....

Here is the next question.....

In a certain school, out of every 7 students playing Rugby, 3 play Football as well; for every student playing at least one of these two sports, there are 3 students who take up neither. It is known that 10% of the students play only Rugby. Find the percentage of students who play Football.

A. 10%

B. 15%

C. 20%

D. 18%

E. None of these