Peekay
-
Posts
72 -
Joined
-
Last visited
Content Type
Profiles
Forums
Events
Gallery
Blogs
Posts posted by Peekay
-
-
-
Not only did I miss something, I missed a lot - combinations of the 2 step options! The actual ways would be much more.Unless I missed something...
All 1 steps - 1 way
One 2 step - 8 ways
Two 2 steps - 6 ways
Three 2 steps - 4 ways
Four 2 steps - 2 ways
Five 2 steps - 1 way
Total = 22 ways
-
Unless I missed something...
All 1 steps - 1 way
One 2 step - 8 ways
Two 2 steps - 6 ways
Three 2 steps - 4 ways
Four 2 steps - 2 ways
Five 2 steps - 1 way
Total = 22 ways
-
1. 3:36pm
2. 2.5cents
3. 4100
4. 3 cuts
5. 3 additional people
-
Start both hourglasses at the same time.
When 4 is done, turn the 4 - elapsed time 4mins
When 7 is done, turn both (1min left in 4) - elapsed time 7mins
When 4 is done, turn 7 (1min in 7) - elapsed time 8 mins
When 7 is done, your egg is cooked to perfection!
-
Following the conventional method:
Dad=Howard (H) or George (G)
Mom=Virgina (V) or Dorothy (D)
Option 1: Dad=H,Mom=V leads to statements 1,2 & 4 being T and 3 could be either T or F. Does not work.
Option 2: Dad=G, Mom=V leads to statements 1,4 being T and 2,3 being F. We can stop here since we have the solution.
But just to be exhaustive -
Option 3: Dad=H, Mom=D leads to 3 T statements
Option 4: Dad=G, Mom=D leads to potentially 4 F statements unless "incestuous" relationships are considered, in which case there could be 2 T and 2 F statements. We can dismiss this as "ambiguous".
-
From the first part of the problem it is clear that the only time you can have the same percentage of wine in both vessels is if the percentage is 50%. The only way this can be done is to transfer the entire wine into the water and then transfer back half of the mixture (same volume as step 1).
-
1/3 There are 2 black and 2 red cards on the table. The first coin can be put on any card (p=1). Once that is done, there are 3 cards left out of which one is the same color as the one with the first coin. So probability = 1/3.
-
If I did it correctly, temp will reach 32 deg at 12:50
-
The problem states that "At 12:00pm the pool is half covered by the shade of a tree" and again later states that "It can be assumed that the shade covers half the pool at 12:30". Is the pool HALF covered by the shade at 12:00 or 12:30? Please clarify.
-
The best way to do this is to take the Negative of the stated conditions:
Not Men = 100-59 = 41
Not FBCC = 100-72 = 28
Not CG = 100-81 = 19
Not RH = 100-89 = 11
So, a total of 99 people might NOT be have one or more of the above characteristics.
Which leaves us with a MINIMUM number of 1 person who will definitely have the above characteristics.
-
You are wrong on the 5th step where you casually mention canceling the powers on both sides!
(-1/2)^2 = (+1/2)^2 does not yield -1/2 = +1/2
But, definitely a good one!
-
I agree. We have to differentiate between the weight (non-contact force due to gravitational attraction) and the feeling of weightlessness (contact force from normal reaction). So, while the person will have a weight he will feel weightless similar to when you have a free fall in an amusement park ride.
-
N
Just a variation of OTTFFSSENT!
-
Vanessa and Eve have to be of opposite kinds.
If V is True -> E is Lie -> A is True -> M the gf and a liar -> V is sister which leads to contradiction since V & A have to be innocent
So, V is Liar -> E is True -> A is Liar -> M is True -> V is not sister. Since E & M are innocent, that leaves Amanda as sister.
-
Start both hourglasses
When the 4min runs out, turn it over - 4mins elapsed
When the 7min runs out, turn both over - 7mins elapsed (4mins has 1min)
When the 1min in the 4min runs out, turn over the 7min - 8mins elapsed (7min has 1min)
Let the 1min in the 7min run out - 9mins elapsed
-
The password is "different"
-
Prancer, Cupid, Rudolph, Dasher, Blitzen, Vixen, Comet, Donder, Dancer.
-
envelope
-
Consider the 20 people in a ring pulled tight at two ends. So, you have 2 rows of 10 people each (1 facing 20, 2 facing 19, etc.)
Next, you rotate the ring once (say clockwise). So, 20 moves onto the top row and faces 19, 1 faces 18, etc.
Continue this operation 19 times. Every person meets everyone else once and you have the schedule.
For an odd number of people, you will have to add the "dummy" one to pair them off.
-
Since the mass of both bodies are the same, the normal force would be the same and so would the force of friction. They would start to slide at the same angle.
Note: The frictional force is independent of surface area of contact.
-
8 oranges divided into 3 parts = 8/3 each (I am discounting the soda)
John contributed 5 oranges but consumed 8/3 of it, so his net contribution is (5 - 8/3 = 7/3)
Peter contributed 3 oranges but consumed 8/3 of it, so his net contribution is (3 - 8/3 = 1/3)
Paul consumed the net contribution from John and Peter : 7/3 + 1/3 = 8/3
So, the money paid by Paul should be divided in the ratio of 7/3:1/3 or 7:1. John gets $7, Peter gets $1.
-
Without even going into equations, it is apparent from the problem statement that 20 beetles were just keeping up with the growth while the rest were devouring the original height (or length) of the plant. So, assuming that the growth rate of the plant is independent of the height of the plant, 20 beetles can happily feed on the plant for ever.
Why can't we get a source of food like this growing in our backyard?
-
88
Solve the equation:
((x - 1) / 3) + 87 = 4 * (((x + 8) / 4) + 5)
The 44th Riddle
in New Logic/Math Puzzles
Posted
Assuming that the golden ingots cannot be cut/ sub-divided and that everyone wants to get fair share while maximizing his/her own, the best outcome is 2 people getting 3 ingots while 3 people get 2 each.
In this case, it is best for the first elder to suggest an allocation of 2, 3, 3, 2, 2. He will then win the votes of the next 2 elders and the allocation will be ratified.