Practice Test 1 #10.



P and Q are lie on the circle, O is the center. What is the value of S?

The answer is 1. I can get this answer but it's the long way using distance formula, equation of a circle, and the listed answer choices. Is there a faster way to derive the answer?


Practice Test 1 #24

At a dinner party 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group?

A) 5 b) 10 C) 24 D) 32 E) 120

Answer is C 24. I couldn't think of way to solve this using the anagrams or combinatrics.

Practice Test 1 #19

In the xy-plane, does the line with equation y = 3x + 2 contain the point (r,s)

A) (3r + 2 - s )(4r + 9 - s ) = 0

B) (4r - 6 - s)(3r + 2 - s ) = 0

I noticed that if you plugged in s for y and r for x you get 3r + 2 - s

So I figured D. But the answer is C.

The permuation problem is actually very easy. You can use the circular permuation formula of (n-1)!

The Data Sufficiency problem regarding y = 3x + 2 is because if you combine both equations then you know that 3r + 2 - s = 0 in order for both equations to be true. 3r + 2 - s = 0 is thus s = 3r + 2 which is y = 3x + 2.

I'm still wondering how to solve the circle problem quickly.

a7lee,

Here is how I solved it: -

I used the two properties of geometry to calculate 's'. The properties are:-

1) Line segment PO & QO are perpendicular, So the product of the slopes of the line segments should be -1.

Hence, [(1-0)/(-sqrt(3)-1)]x(t/s) = -1

Solving above, t = sqrt(3)xs or s = t/sqrt(3)


2) Triangle POQ is a right triangle so applied pythagoras theorem.

In triangle POQ,

PQ^2 = PO^2 + OQ^2

Now we know PO = OQ = Radius of the circle = 2 (Distance of between the vertices (-sqrt(3),-1) and (0,0))

Similarly PQ^2 = (s + sqrt(3))^2 + (t+1)^2 (Distance between vertices (-sqrt(3),-1) and (s,t))

So putting all the values in pythagoras theorem, also substitute s = t/sqrt(3) from property (1)

(s+ sqrt(3))^2 + (t-1)^2 = 4+4

[t/sqrt(3) + sqrt(3)]^2 + (t-1)^2 = 8

(t^2)/3 + 3 +2t + t^2 + 1 -2t = 8, Solving more: -

t^2 = 3, from this equation t can have two values sqrt(3) or -(sqrt(3)) but we know t is in first quadrant so t = sqrt(3)

Now solving for s from property 1

s = t/sqrt(3) = sqrt(3)/sqrt(3) = 1

The above calculation looks more cumbersome then they are, downpoint of typing the solution from the keyboard.

Hope it helps

GMAT 2007

P = (-sqrt3, 1)
Q = (s, t)
Equation of a circle is x^2 + y^2 = r^2 where (x,y) is any point on the circle.

3 + 1 = s^2 + t^2
s^2 + t^2 = 4 ---------------------(1)

Now, product of slopes is -1 when lines are perpendicular to each other.
[(1-0)/(-sqrt3 - 0)] * [(t-0)/(s-0)] = -1
t = s*sqrt3

Substituting this in (1),
s^2 + 3s^2 = 4
s^2 = 1
s = 1 since P is in first quadrant.

Thanks, knowing the slope of the line makes the problem a lot faster to solve.