For the DS question:
Each statement alone is clearly insufficient.
Taking the combined statement:
Case 1: p=10 and n= 6--->(10+6)(10-6) 16:5 R=1 and 4:3 R=1 and (16*4):15 R=4
Case 2: p=5 and n=1--->(5+1)(5-1) 6:5 R=1 and 4:3 R=1 and (6*4):15 R=9
Since we can get different answers we should choose E.
GMAT Prep Test Math doubts
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macattack
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Now for the geometry problem here is how it goes:
Step 1: calculate the radius of the circle--->R=sqrt((sqrt(3)^2)+1^2)=2
Step 2: Calculate the legth of the hypotenuse of the isosceles right triangle= R*sqrt(2)=2sqrt(2)
Now set up the equations:
Equation 1: s^2+t^2=4 (pythagorean theorem)
Equation 2: sqrt((s+sqrt(3))^2+(t-1)^2)=2sqrt(2) (distance between P and Q = legth of hypotenuse)
Step 3: square Equation 2:
(s+sqrt(3))^2+(t-1)^2=8
s^2+t^2+3+1+2sqrt(3)s-2t=8
Step 4: substitute equation one into equation 2 (s^2+t^2=4)
--->2sqrt(3)s-2t=0
Step 5: Rearrange the terms: t=sqrt(3)*s
Step 6: Plug the possible answer choice and check which one verifies s^2+t^2=4
The only possible answer is 1!
Hope that helped
Step 1: calculate the radius of the circle--->R=sqrt((sqrt(3)^2)+1^2)=2
Step 2: Calculate the legth of the hypotenuse of the isosceles right triangle= R*sqrt(2)=2sqrt(2)
Now set up the equations:
Equation 1: s^2+t^2=4 (pythagorean theorem)
Equation 2: sqrt((s+sqrt(3))^2+(t-1)^2)=2sqrt(2) (distance between P and Q = legth of hypotenuse)
Step 3: square Equation 2:
(s+sqrt(3))^2+(t-1)^2=8
s^2+t^2+3+1+2sqrt(3)s-2t=8
Step 4: substitute equation one into equation 2 (s^2+t^2=4)
--->2sqrt(3)s-2t=0
Step 5: Rearrange the terms: t=sqrt(3)*s
Step 6: Plug the possible answer choice and check which one verifies s^2+t^2=4
The only possible answer is 1!
Hope that helped
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Brent@GMATPrepNow
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Here's my solution to the geometry question.
So, as you can see s = 1, so the answer is B
Cheers,
Brent
Here's my solution to the geometry question.
So, as you can see s = 1, so the answer is B
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
