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Re: a couple from the gmat prep I took today [#permalink]
Picking numbers method could reduce the time taken to answer Ques 2.. took me about 80 secs.
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Re: a couple from the gmat prep I took today [#permalink]
abhi758 wrote:
Picking numbers method could reduce the time taken to answer Ques 2.. took me about 80 secs.


well, took me less than 25 seconds...
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Re: a couple from the gmat prep I took today [#permalink]
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).
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Re: a couple from the gmat prep I took today [#permalink]
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
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Re: a couple from the gmat prep I took today [#permalink]
The semicircle problem;

If we draw a perpendicular line through the point P to x axis at point A (-sqrt[3], 0), we have POA as 30-60-90 Triangle, where PA=2; PA =1; AO=sqrt[3]

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[3]; OQ=2 (Since POQ is right angle triangle)

We have “s” =OB= 1
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Re: a couple from the gmat prep I took today [#permalink]
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)
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Re: a couple from the gmat prep I took today [#permalink]
Solution for 2:


Let P = 100

1.1 * S/100 = S1/90

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

Answer - B
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Re: a couple from the gmat prep I took today [#permalink]
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
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Re: a couple from the gmat prep I took today [#permalink]
Attachment:
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""



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