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# 3 hard problems

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Intern
Joined: 13 Feb 2009
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13 Feb 2009, 03:04
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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
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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.
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13 Feb 2009, 05:09
CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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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.
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13 Feb 2009, 08:19
1
KUDOS
Expert's post
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=z-2:

x^2+(z-2)^2=z^2 --> x^2=2*2*(z-1) ----> so z-1 is perfect square and is even. Let's take z-1=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
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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=z-2:

x^2+(z-2)^2=z^2 --> x^2=2*2*(z-1) ----> so z-1 is perfect square and is even. Let's take z-1=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.

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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.. .....

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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.
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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?
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14 Feb 2009, 12:08
Just in case, if x^2 + y^2=z^2 and x, y, z are integers, then x=m^2-n^2, y=2mn, z=m^2+ n^2 or
x=2mn, y=m^2-n^2, z=m^2 +n^2
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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

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.

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14 Feb 2009, 14:27
1
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Expert's post
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
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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.
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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=L-L/3-L/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.
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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!
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14 Feb 2009, 15:07
Don't worry, I don't think they are 2-min GMAT problems. Anyway they are very tricky but useful on the way to Q50+.
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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

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-----------------------
tusharvk

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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.
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tusharvk

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14 Feb 2009, 22:46
Walker!
But sum of 4 the least from 5 in ours case always < 5 and ........?
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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

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# 3 hard problems

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