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Difficult Math Problem #114 - Combinations
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420
B. 2520
C. 168
D. 90
E. 105
from diff math doc, ans coming when some ppl respond with explanations
- by 800guy
Wed Apr 04, 2007 2:50 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #114 - Combinations
- Replies: 12
- Views: 3270
oa
here's the oa: The sum of digits of a multiple of 3 should be div by 3. for a 5 digit number to be div by 3, the sum of digits (given the digits here) can be only 12 or 15. For a sum of 12, the digits that can be used : 0,1,2,4,5 for a sum of 15: 1,2,3,4,5 Number of numbers from the first set = 4.4!...
- by 800guy
Wed Apr 04, 2007 2:49 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #113 - Permutations
- Replies: 4
- Views: 2203
Difficult Math Problem #113 - Permutations
How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
(A) 15
(B) 96
(C) 216
(D) 120
(E) 180
from diff math doc, ans coming when peopple respond with explanations
- by 800guy
Mon Apr 02, 2007 12:04 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #113 - Permutations
- Replies: 4
- Views: 2203
oa
oa:
C is correct.
The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5
- by 800guy
Mon Apr 02, 2007 12:03 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #112 - Number Theory
- Replies: 7
- Views: 2440
Difficult Math Problem #112 - Number Theory
How many 5-digit positive integers exist where no two consecutive digits are the same?
A.) 9*9*8*7*6
B.) 9*9*8*8*8
C.) 9^5
D.) 9*8^4
E.) 10*9^4
oa coming when a few people respond with explanations. from difficult math doc.
- by 800guy
Thu Mar 29, 2007 10:37 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #112 - Number Theory
- Replies: 7
- Views: 2440
oa
oa:
let’s assume Ben has already been chosen. Then I have to choose 3 more people from the remaining, excluding Kelly, that is, three from six people, that’s 6c3.
so the total is (1c1.6c3)/8c4 which is 2/7
- by 800guy
Thu Mar 29, 2007 10:34 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #111 - Probability
- Replies: 7
- Views: 3216
Difficult Math Problem #111 - Probability
There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?
from diff math doc, oa coming when people respond with explantions
- by 800guy
Mon Mar 19, 2007 9:44 am- Forum: Problem Solving
- Topic: Difficult Math Problem #111 - Probability
- Replies: 7
- Views: 3216
OA
OA: Let the divisor be a. x = a*n + 11 ---- (1) y = a*m + 21 ----- (2) also given, (x+y) = a*p + 4 ------ (3) adding the first 2 equations. (x+y) = a*(n+m) + 32 ----- (4) equate 3 and 4. a*p + 4 = a*(n+m) + 32 or a*p + 4 = [a*(n+m) + 28] + 4 cancel 4 on both sides. u will end up with. a*p = a*(n+m) ...
- by 800guy
Mon Mar 19, 2007 9:43 am- Forum: Problem Solving
- Topic: Difficult Math Problem #110 - Arithmetic
- Replies: 7
- Views: 2438
Difficult Math Problem #110 - Arithmetic
Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor? A. 36 B. 28 C. 12 D. 9 E. None from diff math doc. answer coming when a some people respond with explan...
- by 800guy
Fri Mar 16, 2007 9:00 am- Forum: Problem Solving
- Topic: Difficult Math Problem #110 - Arithmetic
- Replies: 7
- Views: 2438
oa
here's the oa: 1000 - multiples of 2 and/or 5 multiples of 2 = 500 (all even #) multiples of 5 = (995 -5)/10 + 1 [ Using AP formula] = 100 Answer = 1000 - (500 + 100) = 400 You cannot calculate for all multiples of 5 because you have already removed all even integers (including 10, 20, and 30). The ...
- by 800guy
Fri Mar 16, 2007 8:59 am- Forum: Problem Solving
- Topic: Difficult Math Problem #109 - Number Theory
- Replies: 2
- Views: 2278
Re: Expert comments
Any expert comments if the clause is restrictive or non-restrictive ..pl? It's my understanding that clauses beginning with "that" are restrictive and should not be separated by a comma: http://www.ucalgary.ca/UofC/eduweb/grammar/course/punctuation/3_4c.htm thats how i understand it too
- by 800guy
Wed Mar 14, 2007 10:00 am- Forum: Sentence Correction
- Topic: When is it safe to use the word "that?"
- Replies: 11
- Views: 2930
Difficult Math Problem #109 - Number Theory
How many integers less than 1000 have no factors (other than 1) in common with 1000?
(1) 400
(2) 410
(3) 411
(4) 412
(5) None of the above
from diff math doc. oa coming after some people respond with explanations
- by 800guy
Wed Mar 14, 2007 8:42 am- Forum: Problem Solving
- Topic: Difficult Math Problem #109 - Number Theory
- Replies: 2
- Views: 2278
- by 800guy
Wed Mar 14, 2007 8:41 am- Forum: Problem Solving
- Topic: Difficult Math Problem #108 - Probability
- Replies: 3
- Views: 3559
oa
here's the oa from the doc: look at the conditions; it says that the first person who tosses a head wins. Let’s say A tosses first. what is the probability that he wins H + TTH + TTTTH + TTTTTTH + TTTTTTTTH i.e. either the first toss is head, or the first time A tosses the coin he gets a tail and B ...
- by 800guy
Wed Mar 14, 2007 8:41 am- Forum: Problem Solving
- Topic: Difficult Math Problem #108 - Probability
- Replies: 3
- Views: 3559
i think your right, ash. thanks!!ash_maverick wrote:The Oa ur r talking abt is not right. That gives the summation of all the numbers in that series.
But here it is not asking asking about the summation of all the number in tht series. Just 2*98+14 should give the answer.
- by 800guy
Tue Mar 13, 2007 11:25 am- Forum: Problem Solving
- Topic: Difficult Math Problem #107 - Algebra
- Replies: 7
- Views: 2569