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Official Guide 12th Data Sufficiency Section, OG 12 DS

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Official Guide 12th Data Sufficiency Section, OG 12 DS

by MrCleantek » Mon Oct 10, 2011 8:51 pm
Hi All,
I have seen a section for OG 12 PS problems in PS section of this forum. But there is none for DS.
So I started this because I was havign problems with Q109 in DS. 12th OG.
Here on we can use this to put OG12 DS questions.

Here is the question. And I did not like the expl in OG12 on this.

109. In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS?
(1) The area of triangular region ABX is 32
(2) The length of one of the altitudes of triangle ABC is 8.
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by MrCleantek » Mon Oct 10, 2011 8:53 pm
And the answer is A
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by GmatMathPro » Mon Oct 10, 2011 9:16 pm
Image

I just made up some numbers to indicate the various midpoints in this figure. Click https://gmatmathpro.com/wp-content/uploads/DS109.jpg to view the full figure.

When you connect the midpoints of two sides of a triangle, you create a 2nd, smaller triangle that is similar to the larger one. Specifically, the sides are 1/2 as big as the corresponding sides of the big triangle. This is done twice in this problem, so we have triangle RCS which has sides that are 1/4 the size of the corresponding sides of the big triangles. There is also a rule that says if two triangles are similar with ratios of corresponding sides a/b, then the ratio of their areas will be a^2/b^2. In this case, the ratio of the sides is 1/4, so the ratio of the areas will be 1/16. That is, triangle RCS has an area that is 1/16 the area of triangle ABC.

Statement 1: Triangular region ABX constitutes half of the area of the big triangle, so with this info we know triangle ABC has area of 64. RCS is 1/16 that, so we know the area is 4. SUFFICIENT.

Statement 2: This tells us nothing useful by itself. We need to be able to calculate an area, but we only have one altitude and we don't even know which base it goes with. INSUFFICIENT.
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by MrCleantek » Mon Oct 10, 2011 9:54 pm
Pete,
I think assuming the values is wrong here. What if AX and BY are not equal to 2? Then the lines XY and RS would not be parallel. Then the correspondence wil not hold. Also how ABX is half of ABC?
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by GmatMathPro » Mon Oct 10, 2011 10:13 pm
I'm not assuming any values exactly, I just used those numbers to indicate that those points are the midpoints as specified in the problem. Forget about the exact values. Just remember that X and Y are midpoints of AC and BC respectively, and R and S are midpoints of XC and YC respectively. From this follows the fact that XY and RS and AB are all parallel to each other, and AB is half of XY which is half of RS. The theorem I am referencing is illustrated here: https://planetmath.org/encyclopedia/Tria ... eorem.html

As for your other question, if you wanted to calculate the area of ABC, you could drop an altitude from point B(let's call it h) to base AC(let's call it b). Then the area would be (1/2)bh. FInding the area of triangle ABX can be done in the same way, except because X is the midpoint of AC, AX is half the length of AC. So comparing the two triangles, the heights are the same, but the base of ABX is one-half the base of ABC, so triangle ABX has an area that is half the area of ABC
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by MrCleantek » Mon Oct 10, 2011 10:59 pm
My 1st question is answered through that theorem. :)
On my 2nd question. I thought about it for so long and read some material on the web to conclude that heights are same for both!!! Dumb me. Height can be outside the triangle but should be no the plane of the base. I was confused that it should be on the base.
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by MrCleantek » Wed Oct 12, 2011 9:43 am
I have another question.
I am not understanding the answer expl from og12.
Q61: At a certain picnic, each guest was served either a single or double scoop of ice cream. How many of the guestes were served a double scoop of ice cream?
(1) 60% of the guests were served double scoop.
(2) Total of 120 scoops of ice cream were served to all the guests at the picnic.
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by GmatMathPro » Wed Oct 12, 2011 10:06 am
Statement 1: If 60% of the group was served a double scoop, that means that 40% of the group was served a single scoop. We don't know the size of the group, so we can't tell how many people got the double scoop. However, we do know that D/S=60/40 or D/S=3/2, where D represents the number of people who got a double scoop and S is the number of people who got a single scoop. INSUFFICIENT

Statement 2: If the total number of scoops is 120, then 2D+S=120. This would have any different possible solutions. For example, S=2 D=59 or S=4 D=58.

Statements 1&2: Now we have two equations and two variables so we can set up a system of equations and solve. D/S=3/2 implies D=3S/2. Making a substitution to the other equation, 2(3S/2)+S=120---->4S=120--->S=30 so D=45. SUFFICIENT.
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by MrCleantek » Wed Oct 12, 2011 10:37 am
Hey Pete,
I got it.
But in the 2nd statement expl, I think you should put equation as D+S=120 and not as 2D+S=120. Since D is double scoop already!
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by MrCleantek » Wed Oct 12, 2011 10:44 am
But then S=48 and D=96. D+S is not equal to 120
My word problem skills suck! :(
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by GmatMathPro » Wed Oct 12, 2011 12:11 pm
Yeah you just have to think very precisely when you set these things up. D is the number of people who got double scoops. It doesn't represent anything about scoops itself.

If you're not sure, write out a few terms to make sure your expression makes sense.

1 person who gets double scoop = 2 scoops
2 people.......=4 scoops
3 people.......=6 scoops
.
.
D people.......=2D scoops
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by MrCleantek » Wed Oct 12, 2011 7:42 pm
10-4. Over and out!
Thanks.
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