a couple from the gmat prep I took today

Question

semicircle.png 99.png I'm really interested in an explanation to the semicircle one, I still can't figure out

(I'm terrible at geometry).

hgp2k · Answer

Triangle POQ is a right angle triangle. So  PQ^2 = PO^2 + OQ^2 
 (s+\sqrt{3})^2 + (t-1)^2 = ((-\sqrt{3}-0)^2 + (1-0)^2) + ((s-0)^2 + (t-0)^2) 
 s^2+2s\sqrt{3}+3+t^2+1-2t = 4+s^2+t^2 
 t=s\sqrt{3}

We also know that  PO = OQ 
so  s^2+t^2 = 4 
 s^2+3s^2 = 4 
 s=1

hgp2k · Answer

Old number of employees =  x 
Old avg salary =  s 
So Old total salary =  s*x

New number of employees =  0.9*x 
New avg salary =  1.1*s 
So New total salary =  1.1*s*0.9*x = 0.99*s*x

% = (0.99*s*x/s*x)*100 = 99%

abhi758 · Answer

Picking numbers method could reduce the time taken to answer Ques 2.. took me about 80 secs.

hgp2k · Answer

well, took me less than 25 seconds...

nitronori41 · Answer

for the second question, the only trick is in making sense of the convoluted language in the question stem. 
took me a minute to make sense of all the decrease... 10%, before and after july 1..... jumble.

After that, the actual calculation takes very little time regardless of the method used (algebra / number substitution).

gmattokyo · Answer

The semicircle one was really dodgy. hgp2k, nice solution and with abhi758, amazed @ ur speeds

great going! saruba, if you don't mind my asking, how much was the quant score for this prep test? wondering if it is a 750 level Q

jade3 · Answer

The semicircle problem;

If we draw a perpendicular line through the point P to x axis at point A (-sqrt , 0), we have POA as 30-60-90 Triangle, where PA=2; PA =1; AO=sqrt

Similarly we draw a perpendicular line through the point Q to x axis at point B; we have QOB as 60-30-90 Triangle , where OB =1; QB=sqrt ; OQ=2 (Since POQ is right angle triangle)

We have “s” =OB= 1

mrsmarthi · Answer

Here is another solution for Question 1.

OP and OQ are perpendicular. Two lines are perpendicular, then the product of the slopes = -1.

Slope of OP = 1/sqrt(3) (Diff in Y coordinates / diff in X coordinates) 
Slope of OQ = t/-s

Product of the slopes = t/(-s*sqrt(3)) = -1 ==> t = s*sqrt(3)

Since it is given that the fig is a semi circle, OP = OQ. Applying diff between two points forumula, we have

4 = s^2 + t^2.

Substitute t = s*sqrt(3) and we get 4s^2 = 4 ==> s^2 = 1 ==> s = +/- 1. Among the option that we have, s = 1. (Ans B)

subhashghosh · Answer

Solution for 2:

Let P = 100

1.1 * S/100 = S1/90

S1/S * 100 = 1.1 * 90/100 * 100 = 99

Answer - B

valencia · Answer

Solving for the semi-circle problem. 
I used trignometry to solve the problem. We know that x^2+y^2= 3+1=4, therefore the radii OP is 2.

This radii OP is the hypotenuse for the right triangle formed by height 1 and sqrt3. solving for the exterior angle i.e. POX. 
Using sine theta formula: 
Sine theta = opp/hyp = 1/2. Therefore the angle is 30. once you know this angle is 30, then the angle QOX on the other side is 60.

Now the radii OQ is 2. 
Cos theta = Adj side/hyp 
Cos 60 = s/2
s = 1, since cos 60 is 1/2

fluke · Answer

99.png Employees before July 1st: 100 Total salary before july 1st: T_b Average before July 1st: \frac{T_b}{100} Employees after July 1st: 90 Total salary after july 1st: T_a Average after July 1st: \frac{T_a}{90} \frac{T_a}{90} = 1.1*\frac{T_b}{100} \frac{T_a}{T_b} = 1.1*\frac{90}{100} = \frac{9.9}{10} = \frac{99}{100} = 0.99 = 99% Ans: "B""

a couple from the gmat prep I took today

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615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

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