Gmatprep PS Ques

Question

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

x2suresh · Answer

1)

Equation of line passing through P y = -1/sqrt(3) x 
please observe ratio of y:x = 1:sqrt(3) 
Hypotenuse = sqrt(3+1) =2 =radius
 
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

x2suresh · Answer

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

arjtryarjtry · Answer

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)

greatchap · Answer

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.

bigfernhead · Answer

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.

greatchap · Answer

Hi guys,

Thanks for your help.

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

scthakur · Answer

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,   *   = -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.

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