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A few GMAT Prep Problems

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A few GMAT Prep Problems [#permalink]  28 Aug 2006, 11:33
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Intern
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Question 1

Ans D
for statement 1 (QPR=30)
QPR+PQR=PRS(EXTERIOR ANGLE THEOREM)
30+PQR=PRS necessary equation obtained

For statement2
PRQ+PQR=150
IMPLIES QPR=30 ........thus again follow step 1

Thus both sufficient
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Ans C
letme know if its correct before my explanation
Manager
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The answer for the 1 question is D

We require angle PQR - angle PRS

S1: angle qpr = 30.
angle PQR + angle PRQ = 150
angle PQR + angle PRQ = 150

S2: angle PQR + angle PRQ = 150
angle PRQ + angle PRS = 180

Susbstituting in the 1 equation we get
angle PQR +180- angle PRS = 150

angle PQR -angle PRS = -30

S2: angle PQR + angle PRQ = 150
angle PRQ + angle PRS = 180

Susbstituting in the 1 equation we get
angle PQR +180- angle PRS = 150

angle PQR -angle PRS = -30

so D.
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Q1. D (either is sufficient).

The first clue is that both of them are saying the same thing. If <QPR = 30 degrees thats the same as saying the sum of hte other two is 150. So now we are already down to either D or E.

Since both PQS and PRS are right triangles, the sum of hte other two angles = 90.
i.e. RPS + SRP = 90
and RPS + 30 + PQR = 90

This can tell you by how much PQR is smaller (30 degrees).

Q2. C

y = (x+a) (x+b)
When it intersects the x-axis, y = 0.

Therefore, x = -a or -b, and the points are (0, -a) and (0, -b).
2. Gives us (0, -6) as one answer but we still dont know the value of the other (a or b).
1. tells us the relation between a and b. Now we have all hte information needed to solve.

MG
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Q3 [#permalink]  28 Aug 2006, 13:12
I dont know how to make a sq root! any tips?

I think the answer is (B) 1

Anyway, so we have two radii of the circle forming a right angled triangle. From sqrt(3) and 1, we get the value of the radius = 2.

So the hypotenuse of the right triangle is 2(sqrt)2.

We can get the value of S by subtracting sqrt(3) from this.

Which is (approx) 2*1.414 - 1.732
is approximately 1.09.

MG[/u][/code]
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for coordinate2...

y=(x+a)(x+b)

Expanding y=x^2 + ax + bx + ab

S1: a+b =-1

Not sufficient we don't know a and b.

S2: Putting x=0 and y = -6 we get ab= -6
but we don't know a and b.
So insufficient.

Combining S1 and S2 we get.
y=x^2 - x -6

Solving we get y = 3 and -2
So C.
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coordinate.jpeg

OP = 2
OQ= s^2+t^2=4................................1

Also OPQ is an right angled triangle
so
QP^2=OQ^2+OP^2
(s+3^1/2)^2+(t-1)^2 = 2^2+2^2
Solving we get

s^2+ 3+ 2s3^1/2 +t^2+1 -2t=8
Putting the value of s^2 + t^2 and solving we get
s3^1/2 =t
Putting in equation 1 we get

s=+-1

It cannot be -1

so 1.

B it is.....
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Re: Q3 [#permalink]  28 Aug 2006, 15:14
mikki0000 wrote:
I dont know how to make a sq root! any tips?

I think the answer is (B) 1

Anyway, so we have two radii of the circle forming a right angled triangle. From sqrt(3) and 1, we get the value of the radius = 2.

So the hypotenuse of the right triangle is 2(sqrt)2.

We can get the value of S by subtracting sqrt(3) from this.

Which is (approx) 2*1.414 - 1.732
is approximately 1.09.

MG[/u][/code]

I took the same route. calculating the radii of the circle.

If the Y -axis were given to bisect the angle then we could use the easier route, reflection about an axis.
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Re: Q3 [#permalink]  28 Aug 2006, 19:38
mikki0000 wrote:
I dont know how to make a sq root! any tips?

I think the answer is (B) 1

Anyway, so we have two radii of the circle forming a right angled triangle. From sqrt(3) and 1, we get the value of the radius = 2.

So the hypotenuse of the right triangle is 2(sqrt)2.

We can get the value of S by subtracting sqrt(3) from this.

Which is (approx) 2*1.414 - 1.732
is approximately 1.09.

MG[/u][/code]

I took the same route. calculating the radii of the circle.

If the Y -axis were given to bisect the angle then we could use the easier route, reflection about an axis.
Manager
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Problem#3

Hypotenuse on negative side of x axis OP = 2

using trignometry sin(30deg)= opp/hyp = 1/2

Hence angle of OP with -x axis = 30deg
Since OP and OQ has angle of 90deg between them

The angle between OQ and + xaxis must be 60 deg

hence adjacent side = 1 = s

Hence B

Heman
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Problem#1

Let PRS=x and PQS=y
then we need to find x-y

(1) QPR=30
In the big triangle
180= 90+ y + QPS
90-y = QPS
90-y = QPR + RPS
90-y =30 + RPS
60-y = RPS---(A)

In smaller triangle
180 = 90 + x + RPS
90 = x + RPS
sub for RPS from(A)
90 = x + 60-y
30 = x-y Suff

(2)PQR + PRQ = 150
y +180-x = 150
30=x-y
Suff

Hence D

Heman
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1. D
2. C
3. B

Although in the semicircle question, i thought we could use the distance between two points in a coordinate plane formula to get to the answer.

Distance between 2 points (P and Q) =

sqrt[(s-(sqrt3))^2 + (t-1)^2) = 2sqrt(2)
(s-(sqrt3))^2 + (t-1)^2 = 8
s^2+t^2+2sqrt(3)s-2t = 4 ------------- (1)

Distance between 2 points (O and Q) =

sqrt[(s-(0))^2 + (t-0)^2) = 2
s^2 + t^2 = 4 ----------------- (2)

Using eq 1 and eq 2, we get

2sqrt(3)s-2t = 0
2sqrt(3)s = 2t
sqrt(3)s = t
s/t = 1/sqrt(3)

Therefore, s = 1.

Please correct me if i am wrong.

SHA
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heman wrote:
Problem#3

Hypotenuse on negative side of x axis OP = 2

using trignometry sin(30deg)= opp/hyp = 1/2

Hence angle of OP with -x axis = 30deg
Since OP and OQ has angle of 90deg between them

The angle between OQ and + xaxis must be 60 deg

hence adjacent side = 1 = s

Hence B

Heman

Heman, is there a rule that says that "The angle between OQ and + xaxis must be 60 deg". Because that i believe is an assumption.

SHA
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SuperHumanAmit wrote:

Heman, is there a rule that says that "The angle between OQ and + xaxis must be 60 deg". Because that i believe is an assumption.

SHA

A straight line would always have angles adding up to 180 deg
In this case I found the ange between OP and -ve x axis = 60 deg
I did not assume that the angle was 60 deg

The angle between OP and OQ = 90 deg (given in Qstem)
Hence the remaining angle ie between OQ and +ve x axis must be= 180-60-90=30 deg

Heman
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