Search found 98 matches
Thanks Stuart and Ian for providing the solution and pointing out my mistake. I tend to double count sometimes making the problem more difficult than it actually is. Like Harsh mentioned this is not the first time I am double counting. If there are ways to determine or realize that I have double cou...
- by ace_gre
Thu Feb 04, 2010 7:59 pm- Forum: Problem Solving
- Topic: Flag with stripes
- Replies: 9
- Views: 11152
IMO E . Here is my approach.. Since two adjacent stripes cannot be of same color, minimum number of colors needed for a flag is 2 and max is 4. No. of ways of choosing 2 out of 4 = 4C2 = 6. with two colors, first stripe can be any of the two colors and rest all can be of only one color. 2 *1 * 1* 1*...
- by ace_gre
Thu Feb 04, 2010 2:18 pm- Forum: Problem Solving
- Topic: Flag with stripes
- Replies: 9
- Views: 11152
Let x be the total number of matches. 2/3 *x = 20( i.e 17 +3) x = 30. So 10 more matches remain. 3/4 * x = 3/4 * 30 ~ 22.5. Rounding this 23 matches out of 30 need to be won. of the remaining 10 matches 23-17 = 6 need to be won. So 4 additional matches can still be lost. (total losses can be 7) IMO A
- by ace_gre
Thu Feb 04, 2010 1:52 pm- Forum: Problem Solving
- Topic: P & C
- Replies: 3
- Views: 1351
Thanks for replying. It could have been a typo. Looking at your approaches and given that all your answers coincide, I think 42 is the answer.
- by ace_gre
Thu Feb 04, 2010 10:33 am- Forum: Problem Solving
- Topic: interesting work problem
- Replies: 5
- Views: 1393
another work problem
Deep and Ana decided to work together & complete the construction of a road.However, as Deep went out of town, Ana started work alone and finished 1/4 th of the road and took 'x' days more than what they would have taken if they worked together. After this, Ana fell ill and Deep working alone, c...
- by ace_gre
Thu Feb 04, 2010 1:15 am- Forum: Problem Solving
- Topic: another work problem
- Replies: 2
- Views: 1049
interesting work problem
Hi, I found this question in another forum and was not convinced with the solution. Can you please solve? Thanks! I do not have the OA. 13 men and 7 boys can finish a job in 7 days, while 6 boys and 13 women can finish the same job in 6 days. In how many days can 1 man, 1 boy and 1 woman working tog...
- by ace_gre
Thu Feb 04, 2010 12:49 am- Forum: Problem Solving
- Topic: interesting work problem
- Replies: 5
- Views: 1393
- by ace_gre
Wed Feb 03, 2010 3:23 pm- Forum: Problem Solving
- Topic: Saving Money
- Replies: 2
- Views: 2485
Hey ajith, 0.3 = 3/10 and not 1/3..and probability of not raining is 7/10 not 2/3 ? I guess you are tired
- by ace_gre
Wed Feb 03, 2010 3:20 pm- Forum: Problem Solving
- Topic: probability question
- Replies: 6
- Views: 4113
Hi, Here is my approach. Lets create a table with the days of the week. Consider a week from Sun - Sat. Days that rain are represented by R and no rain be X There are 5 ways to rain on three consecutive days in a week. Su M T W Th F Sa R R R X X X X X R R R X X X X X R R R X X X X X R R R X X X X X ...
- by ace_gre
Wed Feb 03, 2010 12:50 pm- Forum: Problem Solving
- Topic: probability question
- Replies: 6
- Views: 4113
Proceeding the same way as metalhead,
288 = 2^5 * 3^2
x = 5
y = 2
2^x-1 = 2^4 = 16
3^y-2 = 3^0 = 1
IMO A
- by ace_gre
Fri Jan 29, 2010 5:14 pm- Forum: Problem Solving
- Topic: If (2^x)(3^y)=288, where x and y are positive integers then
- Replies: 6
- Views: 6536
- by ace_gre
Fri Jan 29, 2010 5:08 pm- Forum: Problem Solving
- Topic: Seating - Probability Question
- Replies: 13
- Views: 14126
Let the cost of a mango be x. Reduced price is 4/5 x. (i.e. 0.8 x)
(40/0.8 x) - (40/x) = 25
10/x = 25
x = 10/25 = 2/5
Cost of 200 mangoes at reduced price of 4/5x = 4/5 * 2/5 * 200 = 64.
IMO D
- by ace_gre
Fri Jan 29, 2010 5:06 pm- Forum: Problem Solving
- Topic: need clarification
- Replies: 2
- Views: 2643
Thanks Sanju your approach is very good.. Basic idea is the same as Sanju's but used combinations for finding "no pair" draws. B -> Blue right glove (3) B' ->Blue left glove (3) B*-> Blue gloves can be left or right R -> Red right glove (2) R' -> Red left glove (2) R*->Red gloves can be le...
- by ace_gre
Fri Jan 29, 2010 12:54 pm- Forum: Problem Solving
- Topic: probability- need help
- Replies: 5
- Views: 4520
Since girls cannot be seated together, lets seat the boys first. 6 boys can be seated in 6! ways.
Seating 6 boys creates 7 spaces between them for seating 4 girls.
_B1_B2_B3_B4_B5_B6_
So 4 girls can be seated in any of the 7 spaces==> 7P4 = 7!/ 3!
Total # of ways = 6! * 7! / 3!
- by ace_gre
Thu Jan 28, 2010 3:26 pm- Forum: Problem Solving
- Topic: Seating - Probability Question
- Replies: 13
- Views: 14126
Total number of people in the contest = A, B, C and 6 others. Here order does not matter only people matter. Total number of ways of selecting 3 out of 9 = 9C3 = 84 No. of ways of choosing 2 out of 3 triplets = 3C2 No. of ways of choosing 1 out of other 6 = 6C1 Ways of choosing 2 of 3 and one other ...
- by ace_gre
Thu Jan 28, 2010 3:17 pm- Forum: Problem Solving
- Topic: probability- confusing ques
- Replies: 9
- Views: 5323