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# Gmatprep PS Ques

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Manager
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04 Sep 2008, 06:07
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Hello Everyone,

I came across 2 GMATPrep PS Questions which I couldn't solve. I have attached an image which shows the two questions.

Q-1 is regarding co-ordinate geometry. Can we solve it using distance or slope.

Q-2 is operations on rational numbers.. how do we take out values mention like cube root of 4 or 4 root of 4 (dunno how to mention that) (Though I got this ques correct but it was a fluke)

Help will be appreciated.

Thanks a lot,

Cheers,
GR
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SVP
Joined: 07 Nov 2007
Posts: 1806
Location: New York
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Kudos [?]: 934 [0], given: 5

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04 Sep 2008, 06:28
greatchap wrote:
Hello Everyone,

I came across 2 GMATPrep PS Questions which I couldn't solve. I have attached an image which shows the two questions.

Q-1 is regarding co-ordinate geometry. Can we solve it using distance or slope.

Q-2 is operations on rational numbers.. how do we take out values mention like cube root of 4 or 4 root of 4 (dunno how to mention that) (Though I got this ques correct but it was a fluke)

Help will be appreciated.

Thanks a lot,

Cheers,
GR

1)

Equation of line passing through P y = -1/sqrt(3) x
please observe ratio of y:x = 1:sqrt(3)

equation of line passing through Q (Perpendicular to P) is y=sqrt(3)x
y:x ratio sqrt(3):1 = sqrt(3)a:a
2= sqrt(sqrt(3)a ^2+a^q) = 2a
--> a=1

point Q (1,sqrt(3))

s=1
Ans B
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SVP
Joined: 07 Nov 2007
Posts: 1806
Location: New York
Followers: 38

Kudos [?]: 934 [0], given: 5

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04 Sep 2008, 06:37
greatchap wrote:
Hello Everyone,

I came across 2 GMATPrep PS Questions which I couldn't solve. I have attached an image which shows the two questions.

Q-1 is regarding co-ordinate geometry. Can we solve it using distance or slope.

Q-2 is operations on rational numbers.. how do we take out values mention like cube root of 4 or 4 root of 4 (dunno how to mention that) (Though I got this ques correct but it was a fluke)

Help will be appreciated.

Thanks a lot,

Cheers,
GR

2) $$M= {4}^{1/2} + {4}^{1/3} + 4^{1/4}$$
$$= 2+ 2^{2/3}+2^{2/4}$$
= 2+ greater than 1 + greater than 1
= greather than 4

please note that 2^0 it self 1. 2 or 4 power any positive number is greater than 1.
No calculations required to solve this problem.

I hope it is clear
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Current Student
Joined: 11 May 2008
Posts: 556
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Kudos [?]: 190 [0], given: 0

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04 Sep 2008, 06:40
M = 4^1/2+4^1/3+4^1/4
=2 + 4^1/3 + 2^2^1/4
= 2+ 4^1/3+ square root(2)
= 2+4^1/3+1.414(if u remember )
= 3.414+some value whihc lies betn 2 and 1.414.
so max can be 5.414 and min is 4.414, both of which are >4
ans is E

it is good if u can rem roots of nos. upto 10. makes ur work that much easier
here is the list
root of
1 = 1
2=1.414
3=1.732
4=2
5=2.24
6=2.44
7=2.64
see the diff is 0.20 bet'n each.
8=2.82
9=3
10=3.16(remember pi?.. its the same)
Manager
Joined: 14 May 2007
Posts: 182
Location: India
Followers: 2

Kudos [?]: 74 [0], given: 11

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04 Sep 2008, 10:32
x2suresh wrote:

2) $$M= {4}^{1/2} + {4}^{1/3} + 4^{1/4}$$
$$= 2+ 2^{2/3}+2^{2/4}$$
= 2+ greater than 1 + greater than 1
= greather than 4

please note that 2^0 it self 1. 2 or 4 power any positive number is greater than 1.
No calculations required to solve this problem.

I hope it is clear

Hmm, I am a bit confused. Isnt 2^2/4 same as 2^1/2. And how do i get to know value of 2^2/3.

I am sorry, I didnt understand question 1 as well.
Retired Moderator
Joined: 18 Jul 2008
Posts: 975
Followers: 10

Kudos [?]: 219 [0], given: 5

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04 Sep 2008, 11:12
Is solving for the hypotenuse necessary?

I found the slope perpendicular to P via O (slope = -1/sqrt(3)), and took the negative reciprical to get the slope of Q (slope = sqrt(3)).

To satisfy the slope of Q, the only coordinates for s,t would be (1, sqrt(3))

Thus, s = 1.

x2suresh wrote:
greatchap wrote:
Hello Everyone,

I came across 2 GMATPrep PS Questions which I couldn't solve. I have attached an image which shows the two questions.

Q-1 is regarding co-ordinate geometry. Can we solve it using distance or slope.

Q-2 is operations on rational numbers.. how do we take out values mention like cube root of 4 or 4 root of 4 (dunno how to mention that) (Though I got this ques correct but it was a fluke)

Help will be appreciated.

Thanks a lot,

Cheers,
GR

1)

Equation of line passing through P y = -1/sqrt(3) x
please observe ratio of y:x = 1:sqrt(3)

equation of line passing through Q (Perpendicular to P) is y=sqrt(3)x
y:x ratio sqrt(3):1 = sqrt(3)a:a
2= sqrt(sqrt(3)a ^2+a^q) = 2a
--> a=1

point Q (1,sqrt(3))

s=1
Ans B
Manager
Joined: 14 May 2007
Posts: 182
Location: India
Followers: 2

Kudos [?]: 74 [0], given: 11

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12 Sep 2008, 03:03
Hi guys,

I understood the Q2 which is quadratic question.

I still didnt get a hang of the coordinate geometry ques. how did this equation come -
Equation of line passing through P y = -1/sqrt(3) x

Is there an easier method.

Cheers,
GR
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Joined: 17 Jun 2008
Posts: 1553
Followers: 11

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12 Sep 2008, 04:30
greatchap wrote:
Hi guys,

I understood the Q2 which is quadratic question.

I still didnt get a hang of the coordinate geometry ques. how did this equation come -
Equation of line passing through P y = -1/sqrt(3) x

Is there an easier method.

Cheers,
GR

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

Hence, s^2 + t^2 = 4 ......................(1)

Now, OP and OQ are perpendicular to each other and hence multiplication of their slope will be -1.

Hence, [(t-0)/(s-0)] * [(1-0)/(sqrt3-0)] = -1 or t = (sqrt3)s

Substitute this value of t in equation (1) above.....s will be 1 or -1....but since s is in the first quadrant, it will be 1.

Hope, this helps.
Re: Gmatprep PS Ques   [#permalink] 12 Sep 2008, 04:30
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