Author 
Message 
Intern
Joined: 13 Feb 2009
Posts: 21

3 hard problems [#permalink]
Show Tags
13 Feb 2009, 03:04
2
This post was BOOKMARKED
This topic is locked. If you want to discuss this question please repost it in the respective forum.
In the triangle ABC AB=x, BC=y, CA=z ; x, y, z are positive integers and 0<x<y<z. Is triangle ABC a right? 1. x is odd. 2. y is odd. _______________________________________________________________________________ A car traveled from town A to town B by traveling for third of time of the trip at speed of x miles per hour, then the car traveled third of the distance at speed of y miles per hour and at speed of z miles per hour for the remaining half of the trip. What is the average speed of the trip? 1. y=70 2. x=50
_____________________________________________________________________ What is the sum of the five integers m,n,p,k,f? 1. m+n+p+k+f =m*n*p*k*f 2. m>0 ,n>0 ,p>0 ,k>0 ,f >0



Manager
Joined: 27 May 2008
Posts: 200

Re: 3 hard problems [#permalink]
Show Tags
13 Feb 2009, 03:44
matematikconsultant wrote: In the triangle ABC AB=x, BC=y, CA=z ; x, y, z are positive integers and 0<x<y<z. Is triangle ABC a right? 1. x is odd. 2. y is odd. _______________________________________________________________________________ A car traveled from town A to town B by traveling for third of time of the trip at speed of x miles per hour, then the car traveled third of the distance at speed of y miles per hour and at speed of z miles per hour for the remaining half of the trip. What is the average speed of the trip? 1. y=70 2. x=50
_____________________________________________________________________ What is the sum of the five integers m,n,p,k,f? 1. m+n+p+k+f =m*n*p*k*f 2. m>0 ,n>0 ,p>0 ,k>0 ,f >0 1)E for right angle, z^2 = x^2 + y^2 x and y is odd doesn't give any clue tht the triangle is right angle. 3) A sum is 0.



Intern
Joined: 13 Feb 2009
Posts: 21

Re: 3 hard problems [#permalink]
Show Tags
13 Feb 2009, 05:09
Answers are incorrect.



CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: 3 hard problems [#permalink]
Show Tags
13 Feb 2009, 05:23
3. sum also can be 8 (or 8): 1+1+2+2+2=1*1*2*2*2=8. Therefore, A is incorrect.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: 3 hard problems [#permalink]
Show Tags
13 Feb 2009, 08:19
1) C. Nice problem, but I don't think it is true GMAT one that we can solve under 2 min. We can build a triangle knowing all three sides. So, can we construct a right triangle that satisfy our condition(s)? 1. x is odd > 3:4:5 is a good example of a right triangle. Therefore, we can construct a right triangle with odd x and the condition is insufficient. 2. y is odd. It is harder as I don't remember any right triangle with odd y. So, let's consider a right triangle for which x is even, y is odd and z is odd. Moreover, y=z2: x^2+(z2)^2=z^2 > x^2=2*2*(z1) > so z1 is perfect square and is even. Let's take z1=16 > 8^2+15^2=17^2. Yeah, we've constructed a right triangle. Insufficient. 1&2. Let's x=2a+1 and y=2b+1 > x^2+y^2=4a^2+4a+1+4b^2+4+1 > 4*(a^2+a+b^2+b)+2 > is divisible by 2 but not 4. Therefore, it couldn't be a perfect square and there is no a right triangle with odd x and odd y. Sufficient
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



SVP
Joined: 29 Aug 2007
Posts: 2467

Re: 3 hard problems [#permalink]
Show Tags
13 Feb 2009, 22:01
walker wrote: 1) C. Nice problem, but I don't think it is true GMAT one that we can solve under 2 min.
We can build a triangle knowing all three sides. So, can we construct a right triangle that satisfy our condition(s)?
1. x is odd > 3:4:5 is a good example of a right triangle. Therefore, we can construct a right triangle with odd x and the condition is insufficient.
2. y is odd. It is harder as I don't remember any right triangle with odd y. So, let's consider a right triangle for which x is even, y is odd and z is odd. Moreover, y=z2:
x^2+(z2)^2=z^2 > x^2=2*2*(z1) > so z1 is perfect square and is even. Let's take z1=16 > 8^2+15^2=17^2. Yeah, we've constructed a right triangle. Insufficient.
1&2. Let's x=2a+1 and y=2b+1 > x^2+y^2=4a^2+4a+1+4b^2+4+1 > 4*(a^2+a+b^2+b)+2 > is divisible by 2 but not 4. Therefore, it couldn't be a perfect square and there is no a right triangle with odd x and odd y. Sufficient For me, Q1 looks flawed. From statement 1 and 2, x and y are odds. Sum of squares of odds is even. I do not know any even that is a sum of the squares of two odds is a perfect square. If so, z cannot be an integer where as the question stem says that x, y and z are +ve integers.. ..... matematikconsultant wrote: In the triangle ABC AB=x, BC=y, CA=z ; x, y, z are positive integers and 0<x<y<z. Is triangle ABC a right? 1. x is odd. 2. y is odd.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: 3 hard problems [#permalink]
Show Tags
13 Feb 2009, 23:22
Hi, Tiger! The problem isn't flawed. " In the triangle ABC AB=x, BC=y, CA=z ; x, y, z are positive integers and 0<x<y<z." Can we construct triangle with x,y odd and z even? Yes. But " Is triangle ABC a right?" No, the triangle cannot be right as you say there is no right triangle with x,y odd and z even. So, the answer is NO and two statements together are sufficient. GMAT TIGER wrote: For me, Q1 looks flawed. From statement 1 and 2, x and y are odds. Sum of squares of odds is even. I do not know any even that is a sum of the squares of two odds is a perfect square. If so, z cannot be an integer where as the question stem says that x, y and z are +ve integers.. .....
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



SVP
Joined: 29 Aug 2007
Posts: 2467

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 07:20
walker wrote: Hi, Tiger! The problem isn't flawed. " In the triangle ABC AB=x, BC=y, CA=z ; x, y, z are positive integers and 0<x<y<z." Can we construct triangle with x,y odd and z even? Yes. But " Is triangle ABC a right?" No, the triangle cannot be right as you say there is no right triangle with x,y odd and z even. So, the answer is NO and two statements together are sufficient. GMAT TIGER wrote: For me, Q1 looks flawed. From statement 1 and 2, x and y are odds. Sum of squares of odds is even. I do not know any even that is a sum of the squares of two odds is a perfect square. If so, z cannot be an integer where as the question stem says that x, y and z are +ve integers.. ..... Oh Yeah. Missed that one.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Intern
Joined: 13 Feb 2009
Posts: 21

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 11:48
GT, you are right. Yours solution whith x=2a+1, y=2b+1 is very good! And others 2 problems?



Intern
Joined: 13 Feb 2009
Posts: 21

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 12:08
Just in case, if x^2 + y^2=z^2 and x, y, z are integers, then x=m^2n^2, y=2mn, z=m^2+ n^2 or x=2mn, y=m^2n^2, z=m^2 +n^2



SVP
Joined: 29 Aug 2007
Posts: 2467

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 14:22
matematikconsultant wrote: ____________________________________________________________________ 3. What is the sum of the five integers m, n, p, k, f? 1. m+n+p+k+f =m*n*p*k*f 2. m>0, n>0, p>0, k>0, f >0 This is already answered by walker. should be C. walker wrote: 3. sum also can be 8 (or 8): 1+1+2+2+2=1*1*2*2*2=8. Therefore, A is incorrect.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 14:27
3) Let's consider the case when both conditions are true: S=m+n+p+k+f =m*n*p*k*f. Looking a bit at this equation we can see that all numbers cannot be larger as sum increases more slowly than product, excluding case when one of the numbers is 0). Let's begin with the minimum possible sum: S=5: all numbers are 1 and sum (5) is not equal product (1); S=6: (2,1,1,1,1) and sum (6) is not equal product (2); S=7: 7 is a prime number, so our product must contain it but it is impossible as sum will be always larger than product for any prime number. S=8: 8=2*2*2  (2,2,2,1,1). Bingo! it is what we were looking for. S=9: 9=3*3  (3,3,1,1,1). Hm?! it also satisfies both conditions. So, it is E
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



SVP
Joined: 29 Aug 2007
Posts: 2467

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 14:34
matematikconsultant wrote: 2: A car traveled from town A to town B by traveling for third of time of the trip at speed of x miles per hour, then the car traveled third of the distance at speed of y miles per hour and at speed of z miles per hour for the remaining half of the trip. What is the average speed of the trip?
1. y=70 2. x=50 _____________________________________________________________________ I am opt for E as z, the speed for half of the distance, is not given. Need to find average speed, d/t. 1: It says the speed for t/3 hours was 70mph. what about rest 2t/3 hours? do not know. 2: It says the speed for d/3 distance 50mph. what about 2d/3 distance?do not know. togather: do not know. so E should be it.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 14:58
2) B I hope I was accurate with symbols.... Let's L is distance between A and B L1, L2, L3  all distances along path. t is time of the trip. t1, t2, t3  all times along path. Average=L/t For each part of the trip we can write out: L1=1/3t*x; L/3=t2*y; L/2=t3*z Now, let's make a pause and look at tree formulas above. Do you see? We can easily find L1=LL/3L/2=L/6. And our first formula is L/6=t/3 * x > L/t=2x. It is average speed. so A is sufficient. What about A) y=70? After playing with formulas I haven't found possibility to solve the problem. So, It is B.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



SVP
Joined: 29 Aug 2007
Posts: 2467

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 15:00
walker wrote: 3) Let's consider the case when both conditions are true: S=m+n+p+k+f =m*n*p*k*f. Looking a bit at this equation we can see that all numbers cannot be larger as sum increases more slowly than product, excluding case when one of the numbers is 0). Let's begin with the minimum possible sum:
S=5: all numbers are 1 and sum (5) is not equal product (1); S=6: (2,1,1,1,1) and sum (6) is not equal product (2); S=7: 7 is a prime number, so our product must contain it but it is impossible as sum will be always larger than product for any prime number. S=8: 8=2*2*2  (2,2,2,1,1). Bingo! it is what we were looking for. S=9: 9=3*3  (3,3,1,1,1). Hm?! it also satisfies both conditions. So, it is E Thats like playing a chess!
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 15:07
Don't worry, I don't think they are 2min GMAT problems. Anyway they are very tricky but useful on the way to Q50+.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



Manager
Joined: 04 Jan 2009
Posts: 237

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 19:11
Nice problems. For the first one, I recalled that in pythagorean triplets, atleast one of x and y is even. So, both together eliminate the possibility that ABC is right. Hence, C. matematikconsultant wrote: In the triangle ABC AB=x, BC=y, CA=z ; x, y, z are positive integers and 0<x<y<z. Is triangle ABC a right? 1. x is odd. 2. y is odd. _______________________________________________________________________________ A car traveled from town A to town B by traveling for third of time of the trip at speed of x miles per hour, then the car traveled third of the distance at speed of y miles per hour and at speed of z miles per hour for the remaining half of the trip. What is the average speed of the trip? 1. y=70 2. x=50
_____________________________________________________________________ What is the sum of the five integers m,n,p,k,f? 1. m+n+p+k+f =m*n*p*k*f 2. m>0 ,n>0 ,p>0 ,k>0 ,f >0
_________________
 tusharvk



Manager
Joined: 04 Jan 2009
Posts: 237

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 19:16
walker wrote: 3) Let's consider the case when both conditions are true: S=m+n+p+k+f =m*n*p*k*f. Looking a bit at this equation we can see that all numbers cannot be larger as sum increases more slowly than product, excluding case when one of the numbers is 0). Let's begin with the minimum possible sum:
S=5: all numbers are 1 and sum (5) is not equal product (1); S=6: (2,1,1,1,1) and sum (6) is not equal product (2); S=7: 7 is a prime number, so our product must contain it but it is impossible as sum will be always larger than product for any prime number. S=8: 8=2*2*2  (2,2,2,1,1). Bingo! it is what we were looking for. S=9: 9=3*3  (3,3,1,1,1). Hm?! it also satisfies both conditions. So, it is E walker Nice explanation. I am not sure I understood the problem though. If they are asking us to find all the five integers so that the sum can be inferred, we certainly need more equations (5 equns to solve 5 unknowns). If we are just to find what the sum is equal to then just 1 is sufficient. How do I distinguish this? If faced with this question (without the subject line saying hard problems), I would definitely have answered E because we need 5 equations.
_________________
 tusharvk



Intern
Joined: 13 Feb 2009
Posts: 21

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 22:46
Walker! But sum of 4 the least from 5 in ours case always < 5 and ........?



Intern
Joined: 13 Feb 2009
Posts: 21

Re: 3 hard problems [#permalink]
Show Tags
14 Feb 2009, 22:55
Walker and GT! It is the second impotant condition in the speed problem




Re: 3 hard problems
[#permalink]
14 Feb 2009, 22:55



Go to page
1 2
Next
[ 35 posts ]



