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Difficult Math Problem #121 - Interest
A person put 1000 dollars in a bank at a compound interest 6 years ago. What percentage of the initial sum is the interest if after the first three years the accrued interest amounted to 19% of the initial sum? A) 38% B) 42% C) 19% D) 40% E) 50% from diff math problem doc, oa coming when ppl respon...
- by 800guy
Fri Apr 20, 2007 8:44 am- Forum: Problem Solving
- Topic: Difficult Math Problem #121 - Interest
- Replies: 5
- Views: 3295
OA
OA:
a2 = k-1 ; b2 = k+1
a3= (k+1)-(k-1) = 2 ; b3 = (k+1)+(k-1) = 2k
a4 = 2k - 2 = 2(k-1) = 2(b1-a1)
- by 800guy
Fri Apr 20, 2007 8:43 am- Forum: Problem Solving
- Topic: Difficult Math Problem #120 - Sequences
- Replies: 4
- Views: 4351
Difficult Math Problem #120 - Sequences
Sequence A and B. a1=1, b1=k. an=b(n-1)-a(n-1) bn=b(n-1)+a(n-1). What is a4=?
from difficult math problems doc, oa coming after people answer with questions
- by 800guy
Wed Apr 18, 2007 8:18 am- Forum: Problem Solving
- Topic: Difficult Math Problem #120 - Sequences
- Replies: 4
- Views: 4351
OA
OA:
(x+y)30/100 = x*40/100 + y*25/100
30x + 30y = 40x + 25y
y = 2x or y/x = 2/1 or y:x = 2:1 hence x = 33 1/3%
- by 800guy
Wed Apr 18, 2007 8:15 am- Forum: Problem Solving
- Topic: Difficult Math Problem #119 - Arithmetic
- Replies: 7
- Views: 3791
- by 800guy
Mon Apr 16, 2007 8:06 am- Forum: GMAT Verbal & Essays
- Topic: Verbal Strategy book
- Replies: 22
- Views: 35219
Difficult Math Problem #119 - Arithmetic
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ? (A) 10% (B) 33 1/3% (C) 40% (D) 50% (E) 66 2/3% from diff ma...
- by 800guy
Mon Apr 16, 2007 7:56 am- Forum: Problem Solving
- Topic: Difficult Math Problem #119 - Arithmetic
- Replies: 7
- Views: 3791
oa
OA: Number of integers that divide 3: the range is 100-150 Relevant to this case, we take 102 - 150 (since 102 is the first to div 3) 102 = 34*3 150= 50*3, so we have 50-34+1 = 17 multiples of 3 For multiples of 5, 100=5*20 150=5*30 30-20+1 =11 Now we have a total of 27 integers, but we double count...
- by 800guy
Mon Apr 16, 2007 7:55 am- Forum: Problem Solving
- Topic: Difficult Math Problem #118 - Number Theory
- Replies: 3
- Views: 20111
Difficult Math Problem #118 - Number Theory
How many integers between 100 and 150, inclusive can be evenly divided by neither 3 nor 5?
from diff math doc, ans coming after ppl respond with explanations
- by 800guy
Fri Apr 13, 2007 9:29 am- Forum: Problem Solving
- Topic: Difficult Math Problem #118 - Number Theory
- Replies: 3
- Views: 20111
oa
oa:
Chance of drawing a blue on the first draw = 2/8, so chance of not drawing a blue on the first draw is 6/8
similarly chance of not drawing blue on second draw = 5/7
Therefore probability of not drawing blue in 2 draws = 6/8*5/7 = 15/28
- by 800guy
Fri Apr 13, 2007 9:28 am- Forum: Problem Solving
- Topic: Difficult Math Problem #117 - Probability
- Replies: 8
- Views: 3027
Difficult Math Problem #117 - Probability
A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue? A. 15/28 B. 1/4 C. 9/16 D. 1/32 E. 1/16 oa coming after some people respond, from diff math doc
- by 800guy
Wed Apr 11, 2007 12:42 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #117 - Probability
- Replies: 8
- Views: 3027
OA:
OA:
right triangle with sides x<y<z and area of 1 => z = hypotenuse and xy/2 = 1
i.e xy = 2
If x were equal to y, we would have had xy = y^2 = 2. And y = root2
But, x<y and so y>root2.
- by 800guy
Wed Apr 11, 2007 12:41 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #116 - Triangles
- Replies: 4
- Views: 2126
Difficult Math Problem #116 - Triangles
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? A. y > ROOT2 B. ROOT3/2 < y < ROOT2 C. ROOT2/3 < y < ROOT3/2 D. ROOT3/4 < y < ROOT2/3 E. y < ROOT3/4 from diff ma...
- by 800guy
Mon Apr 09, 2007 7:42 am- Forum: Problem Solving
- Topic: Difficult Math Problem #116 - Triangles
- Replies: 4
- Views: 2126
oa
oa: there are actually 20 ways to interchange the letters, namely, the first letter could be one of 5, and the other letter could be one of 4 left. So total pairs by product rule = 20. Now, there are two cases when it wouldn’t change the name. First, keeping them all the same. Second, interchanging ...
- by 800guy
Mon Apr 09, 2007 7:41 am- Forum: Problem Solving
- Topic: Difficult Math Problem #115 - Probability
- Replies: 5
- Views: 3930
Difficult Math Problem #115 - Probability
My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
a) 5%
b) 10%
c) 20%
d) 25%
e) 30%
oa coming after some people respond w/explanations. from diff math doc
- by 800guy
Fri Apr 06, 2007 8:51 am- Forum: Problem Solving
- Topic: Difficult Math Problem #115 - Probability
- Replies: 5
- Views: 3930
oa
oa:
out of 8 people one team can be formed in 8c2 ways.
8c2*6c2*4c2*2c2= 2520.
The answer is 105. Divide 2520 by 4! to remove the multiples ( for example: (A,B) is same as ( B,A) )
- by 800guy
Fri Apr 06, 2007 8:50 am- Forum: Problem Solving
- Topic: Difficult Math Problem #114 - Combinations
- Replies: 12
- Views: 3270