A few GMAT Prep Problems : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 09:28

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A few GMAT Prep Problems

Author Message
Intern
Joined: 24 Jun 2006
Posts: 48
Followers: 0

Kudos [?]: 0 [0], given: 0

A few GMAT Prep Problems [#permalink]

### Show Tags

28 Aug 2006, 11:33
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Attachments

triangle.jpeg [ 22.11 KiB | Viewed 844 times ]

coordinate2.jpeg [ 37.27 KiB | Viewed 842 times ]

coordinate.jpeg [ 17.28 KiB | Viewed 842 times ]

Intern
Joined: 28 Aug 2006
Posts: 3
Location: Kuwait
Schools: imd
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

28 Aug 2006, 12:29
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
Intern
Joined: 28 Aug 2006
Posts: 3
Location: Kuwait
Schools: imd
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

28 Aug 2006, 12:33
Ans C
letme know if its correct before my explanation
Manager
Joined: 30 Jan 2006
Posts: 62
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

28 Aug 2006, 12:55
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.
Manager
Joined: 09 Aug 2005
Posts: 72
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

28 Aug 2006, 13:04
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
Manager
Joined: 09 Aug 2005
Posts: 72
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

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]
Manager
Joined: 30 Jan 2006
Posts: 62
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

28 Aug 2006, 14:44
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.
Manager
Joined: 30 Jan 2006
Posts: 62
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

28 Aug 2006, 14:56
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.....
Intern
Joined: 13 May 2006
Posts: 21
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

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.
Intern
Joined: 13 May 2006
Posts: 21
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

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
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

28 Aug 2006, 21:02
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
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

28 Aug 2006, 21:11
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
Intern
Joined: 24 Jun 2006
Posts: 48
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

28 Aug 2006, 23:16

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
Intern
Joined: 24 Jun 2006
Posts: 48
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

28 Aug 2006, 23:18
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
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

29 Aug 2006, 09:58
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
29 Aug 2006, 09:58
Display posts from previous: Sort by