Problem Solving

sureshbala wrote:Here is the next question...

On Sunday John left the place A for B at 3:30 am and on the same day Carsten started from B towards A at 8:30 am. They met each other at 12:30 pm. After meeting each other they took equal amount of time to reach their respective destinations. At what time did they reach their respective destinations?

A. 7:00 pm
B. 7:30 pm
C. 6:00 pm
D. 8:00 pm
E. None of these

OA:E

My answer is (E).

Assume the speed of John is v1, the speed of Carsten is v2, the time from the meeting point to their destinatino is t:

v1*9hours = v2*t..................... (1)
v1*t = v2*4hours......................(2)

(1)/(2) => 9/t=t/4

So t=6 hours

And the time John reachs home should be 6:30pm.

Well done folk, this is absolutely the solution that I am looking for....In fact if one can use the point that their speeds ratio is constant this can be done fast.




As shown in the above figure the time taken by John and Carsten to cover the distance AM is 9 hrs and x hrs respectively. So the time ratio is 9/x

Similarly when we consider the distance MB, the time ratio is x/4.

Since 9/x = x/4, we get x = 6 hrs.

Hence after meeting each other they take 6 hrs to reach their respective destinations i.e they reach by 6:30 pm

Folks, here is the next question

If a,b and c are real numbers such that a^2 + 2b =7, b^2 + 4c = -7 and c^2 + 6a = -14, find the value of a^2 + b^2 + c^2

A. 14
B. 16
C. 21
D. 28
E. 35

OA:A

are suresh bala p***a gaya hai kya, kis lihaj se ye 780+ gmat q hai...kaha padha raha hai gmat faculty ke roop me...kya score hai?

sureshbala wrote:Folks, here is the next question

If a,b and c are real numbers such that a^2 + 2b =7, b^2 + 4c = -7 and c^2 + 6a = -14, find the value of a^2 + b^2 + c^2

A. 14
B. 16
C. 21
D. 28
E. 35

OA:A

Add them up and complete the square:

a^2+6a+b^2+2b+c^2+4c=-14
(a^2+6a+9)+(b^2+2b+1)+(c^2+4c+4)=-14+9+1+4=0
(a+3)^2+(b+1)^2+(c+2)^2=0
a=-3,b=-1,c=-2
a^2+b^2+c^2=14

maihuna wrote:are suresh bala p***a gaya hai kya, kis lihaj se ye 780+ gmat q hai...kaha padha raha hai gmat faculty ke roop me...kya score hai?

Dear maihuna, please go through the above solution given by griscomtestprep. I am sure it is not as tough as you feel and definitely this is much within the standard of the GMAT questions. I have worked as a faculty in institutes like TIME and CL and I am sure they are the better people to judge my ability.

sureshbala wrote:

maihuna wrote:are suresh bala p***a gaya hai kya, kis lihaj se ye 780+ gmat q hai...kaha padha raha hai gmat faculty ke roop me...kya score hai?

Dear maihuna, please go through the above solution given by griscomtestprep. I am sure it is not as tough as you feel and definitely this is much within the standard of the GMAT questions. I have worked as a faculty in institutes like TIME and CL and I am sure they are the better people to judge my ability.

sureshbala, after going through this question i really thought it is no way a GMAT question. But as you said after going through the above solution this seems to be simple and of course this is all about GMAT.

Dear griscomtestprep thanks a lot for the simple solution provided

Also maihuna, having more than 350 posts to your credit (if you are not able to solve a question) doesn't mean that you are good at judging the standard of GMAT or the capability of sureshbala. I think the moderators have recognized what he is capable of and hence the result...this thread is made sticky.

sureshbala, waiting for the next question.....

Here is the next question....

Let f(x) be a real-valued function such that f(x) + 2f(2002/x) = 3x for all x>0. Find the value of f(1/2).

A. 8005.5
B. 2002.5
C. 8007.5
D. 4006.5
E. 4004.5

OA: C

sureshbala wrote:Here is the next question....

Let f(x) be a real-valued function such that f(x) + 2f(2002/x) = 3x for all x>0. Find the value of f(1/2).

A. 8005.5
B. 2002.5
C. 8007.5
D. 4006.5
E. 4004.5

OA: C

f(x) = 3x - 2f(2002/x) …(eqn 1)

If x = 2002/x, then:
f(2002/x) = 3(2002/x) - 2f(x*2002/2002)
f(2002/x) = 6006/x - 2f(x) …. (eqn 2)

Inserting eqn 2 into eqn 1:

f(x) + 2f(2002/x) = 3x
f(x) + 2*(6006/x - 2f(x)) = 3x
f(x) + (12012/x) - 4f(x) = 3x
(12012/x) - 3x = 3f(x)
f(x) = (4004/x) - x

Therefore f(0.5) = (4004/0.5) - (0.5) = 8007.5

Choose C.

-BM-

maihuna wrote:are suresh bala p***a gaya hai kya, kis lihaj se ye 780+ gmat q hai...kaha padha raha hai gmat faculty ke roop me...kya score hai?

If you have a problem with the questions, dont solve them. But do not use such language on this board again. Also English is the only language you can use to communicate on this board.

If you have any particular concern about a thread or a poster contact one of the site admins or moderators, but do not at any cost disturb a thread with crude messages like above.

Regards

OK Lets look at it from another way. We have a quadilateral ABCO whose opp angles are 90degrees and 2 adjescent sides AO and OC are equal. That makes it a square, hence BO diagonal and hance angle ABO 45deg.
This is a " fig not drawn to scale types"

OK Lets look at it from another way. We have a quadilateral ABCO whose opp angles are 90degrees and 2 adjescent sides AO and OC are equal. That makes it a square, hence BO diagonal and hance angle ABO 45deg.
This is a " fig not drawn to scale types"

Let f(x) be a real-valued function such that f(x) + 2f(2002/x) = 3x for all x>0. Find the value of f(1/2).

A. 8005.5
B. 2002.5
C. 8007.5
D. 4006.5
E. 4004.5

if x=1/2 the equation turns out to be
f(1/2) + 2f(2002/(1/2)) = 3(1/2)
f(1/2) + 2f(4004)= 3/2 let us call this as equation 1

to eliminate f(4004) lets us calculate f(4004)
f(4004)+2f(2002/4004)= 3(4004)
f(4004)+ 2f(1/2) = 12012 let us call this as equation 2

sloving for f(1/2) from two equations
2(equation 2) - (equation1)3f(1/2) = 24024-3/2
f(1/2)= 8008-1/2= 8007.5
hence the answer

Folks, both Naresh and Bluementor gave the correct explanation to this question and I am sure everyone understood them.

I suggest you to look into the solution provided by bluementor because he generalized the value of f(x) which will help us to answer this question even if it is asked in terms of DS. Also the solution provided by Naresh is quick enough.

Anyway, I have simplified a bit of calculation...look at the below image