Search found 77 matches
IMO C (28) You have 8 teams (A,B,C,D,E,F,G,H) that play against one another. A vs. B A vs. C A vs. D A vs. E A vs. F A vs. G A vs. H Since no team can play against itself, A plays 7 games. B plays also 7 games, but now we have counted the game A vs. B and B vs. A which is the same. Therefore we must...
- by Nermal
Sun Sep 13, 2009 1:59 am- Forum: Problem Solving
- Topic: OG12: There are 8 teams in a certain league
- Replies: 5
- Views: 9008
I tried all the possible outcomes for this equation. There are 8 different combinations of positive and negative values of |x|-|y|=|x+y| You only get two that are not zeros. x+y=-x-y => xy=-1 -x-y=x+y => xy=-1 Therefore: xy<0 (E) But this way takes too long, maybe someone can show a quicker way of s...
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Sun Sep 13, 2009 1:52 am- Forum: Problem Solving
- Topic: Number properties
- Replies: 6
- Views: 1376
IMO D (50) You get four different equations: 1. y=-x+5 2. y=-x-5 3. y=x-5 4. y=x+5 These form a square with the y-axis as its diagonal of length 10. This diagonal divides the square into two isoceles triangles of the form 45-45-90. The ratio of the sides in this kind of triangle is x : x : x*sqroot2...
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Sat Sep 12, 2009 12:14 pm- Forum: Problem Solving
- Topic: area of that region
- Replies: 3
- Views: 1313
Kaplan: multiples and ratios
The “connection� between any two positive integers a and b is the ratio of the smallest common multiple of a and b to the product of a and b. For instance, the smallest common multiple of 8 and 12 is 24, and the product of 8 and 12 is 96, so the connection between 8 and 12 is 24/96 = 1/4 The pos...
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Sat Sep 12, 2009 8:51 am- Forum: Problem Solving
- Topic: Kaplan: multiples and ratios
- Replies: 5
- Views: 1415
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Sat Sep 12, 2009 7:07 am- Forum: Problem Solving
- Topic: semicircle
- Replies: 2
- Views: 1797
OA is B.rohan_vus wrote:IMO B .. Let me know if its correct ? I will give expln later
Good job, rohan_vus!
- by Nermal
Sat Sep 12, 2009 6:57 am- Forum: Problem Solving
- Topic: MGMAT: exponents
- Replies: 4
- Views: 1185
IMO C According to statement 1, x can either be 3 or 1/3; so it is not sufficient. Statement 2 only says that |x-3| not equal to 0, it doesn't give any information on x except that it cannot be 3. Not sufficient therefore. Taking both together we see that x can only be 1/3. This is information enoug...
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Sat Sep 12, 2009 2:50 am- Forum: Data Sufficiency
- Topic: absolute value sufficiency
- Replies: 4
- Views: 1259
I don't understand the logic of those statements.
statement 1 says they both earned the same amount of interest in a given year, statement 2 contradicts statement 1 by given different amounts for A and B in the same given year.
Am I missing something?
- by Nermal
Sat Sep 12, 2009 2:39 am- Forum: Data Sufficiency
- Topic: pl solve ..exam in few days
- Replies: 4
- Views: 1347
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Sat Sep 12, 2009 2:28 am- Forum: Data Sufficiency
- Topic: Daniel loses big time
- Replies: 2
- Views: 1755
MGMAT: multiples
If a and b are both single-digit positive integers, is a + b a multiple of 3?
(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.
(2) a - 2b is a multiple of 3.
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Sat Sep 12, 2009 1:53 am- Forum: Data Sufficiency
- Topic: MGMAT: multiples
- Replies: 5
- Views: 2818
MGMAT: exponents
If n is a non-negative integer such that 12^n is a divisor of 3,176,793, what is the value of n^12 - 12^n ?
A) - 11
B) - 1
C) 0
D) 1
E) 11
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Sat Sep 12, 2009 1:43 am- Forum: Problem Solving
- Topic: MGMAT: exponents
- Replies: 4
- Views: 1185
MGMAT: area of a triangle
In the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?
(1) DA = 4
(2) Angle ABD = 30 degrees
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Sat Sep 12, 2009 1:35 am- Forum: Data Sufficiency
- Topic: MGMAT: area of a triangle
- Replies: 7
- Views: 2679
I think it becomes clear when thinking of sqroot 2
as 2^(1/2)
then you see that sqroot2*sqroot2=2^(1/2)*2^(1/2)
Now you have exponents with the same base so in order to multiply them you have to keep the base and add the exponents:
2^((1/2)+(1/2))=2^1=2
- by Nermal
Sat Sep 12, 2009 1:13 am- Forum: Problem Solving
- Topic: OG 12 Diagnostic Square Root Question D17
- Replies: 5
- Views: 3767
The first two terms definitely each yield an integer. 12!/6! ---> 6! is part of 12!, so in the numerator you acutally only have the integers from 12 to 7 inclusive. If you multiply any two integers you will always get an integer as the result. Same reasoning for 12!/8!. 12!/(7!5!): a little trickier...
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Fri Sep 11, 2009 12:57 pm- Forum: Problem Solving
- Topic: PS - integer factorial
- Replies: 5
- Views: 4076
IMO A (16)
median is 7
arithmetic mean is median/(6/7)=7/(6/7)=7*(7/6)=49/6
(4+5+6+8+10+x)/6=49/6
(33+x)/6=49/6
33+x=49
x=16
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Fri Sep 11, 2009 12:46 pm- Forum: Problem Solving
- Topic: PS - median
- Replies: 4
- Views: 1459