Let them keep their job

paiz:
Let them keep their job

Lol.... Epic

paiz:
Let them keep their job

Lolz

MasterKim:
1. There are 25 horses. What is the minimum number of race needed so you can find the fastest 3 horses?
You can race upto 5 horses at a time but you don't have to watch

2. Here is the math puzzle:
— Peter, how old are your children?
— Well, Thomas, there are 3 of them, and the product of their ages is 36.
— That is not enough...
— The sum of their ages is exactly the number of beers we have drunk today.
— That is still not enough.
— OK, the last thing is that my oldest child wears ared hat.

How old were each of Peter’s children?

1. 6 races

For question 2
Possible combinations for the age of the children are:
(4,3,3), (9,2,2), (9,4,1), (6,6,1), (12,3,1), (18,2,1), (36,1,1), (6,3,2)
Now, the hint is sum of the ages is being told and correct ages can't be guessed. So, the possible choices left are:
(9,2,2), (6,6,1).
Now, the hint is oldest is wearing a Red hat. So, the only answer left is:
(9,2,2)
9 years, 2 years and 2 years. The man had twins
Note: The eldest was wearing a red hat to differentiate himself from the twins.

Where's Lalasticlala?

MasterKim:

How did u get ur answer?

Check my answer below

MasterKim:
1. There are 25 horses. What is the minimum number of race needed so you can find the fastest 3 horses?
You can race upto 5 horses at a time but you don't have to watch

limitations
No have a stopwatch so we can’t time the horses (meaning that we can’t compare the race times of each horse).
We can only race 5 horses in each race.

Let's name and arrange the horses 25 horses A - Y in 5 groups in table
A B C D E
F G H I J
K L M N O
P Q R S T
U V W X Y

Race 1 - 5
Let the horses in each row compete with each other, meaning we'll be having 5 races.

For convenience sake, we assume fastest horses are listed in ascending order, from the slowest (extreme left) to the fastest (extreme right), that is, the fastest in each rows are E, J, O, T and Y.

Let's eliminate the 2 slowest horses in each group and redraw the table

C D E
H I J
M N O
R S T
W X Y

Race 6
We know the 5 fastest horses from each group are not necessarily the 5 fastest horses. We race the fastest in group to determine the fastest of all the horses, that is, horses E, J, O, T and Y. Let's assume E is the fastest, followed by J, then O, in that order. Therefore, the overall fastest has been determine to be E. Automatically we can eliminate T and Y since those 2 are definitely not in the top 3. Then, we can automatically eliminate all the horses that T and Y competed against in the preliminary races which are R&S and W&X respectively.

Now, we know that E is the fastest horse in the group since it was the fastest horse out of the 5 group leaders. So, we don’t need to race it anymore.

Further elimination;
If J and O are the 2nd and 3rd fastest in the group leaders, then we should be able to eliminate H since J raced against it and he was in 3rd place in that race.
Also, I could only possibly be the 3rd fastest, and since H is slower than I, we can safely eliminate H. We can also eliminate M and N since O was the 3rd fastest in the group leaders, and N and M were slower than O.

Race 7
In order to determine the second and third fastest horses, the remaining horses are:

C D
I J
O

The horses can then go on the seventh race and the first and second fastest, assuming they're D and J will be determined.


So, in addition to the overall first position we've determined to be E in race 1, the second and third are D and J determine from race 7.


Therefore the minimum of races needed to determined the 3 fastest horses is 7

2. Here is the math puzzle:
— Peter, how old are your children?
— Well, Thomas, there are 3 of them, and the product of their ages is 36.
— That is not enough...
— The sum of their ages is exactly the number of beers we have drunk today.
— That is still not enough.
— OK, the last thing is that my oldest child wears ared hat.

How old were each of Peter’s children?

Possible combinations
a*b*c = 36

1*1*36 (sum is 38)
1*2*18 (sum is 21)
1*3*12 (sum is 16)
1*4*9 (sum is 14)
1*6*6 (sum is 13)
2*2*9 (sum is 13)
2*3*6 (sum is 11)
3*3*4 (sum is 10)

Since The sum of their ages is exactly the number of beers they have drunk today, and is still not enough, that means the answer is either 1*6*6 (sum is 13)
2*2*9 (sum is 13).

Since the last thing is that the oldest child wears a red hat, therefore the combination that allows maximum age (oldest) is 2,2&9

For #2

For product of 36, the possible combinations are
6-3-2 (sum=11)
9-2-2 (sum=13)
4-3-3 (sum=10)
9-4-1 (sum=14)
12-3-1 (sum=16)
18-2-1 (sum=21)
6-6-1 (sum=13)


Now the second tip would give him a clue,the no of beers he drank would give him his answer but if the number of beers he drank is 13, we have two options, 9-2-2 and 6-6-1

That's where the eldest son comes in, since there's an elder one, it can't be 6-6-2, we have 9-2-2.

For #1

We divide the horses into groups of 5.. We have five groups,.
Let's label each group A,B,C,D,E
Next step is to race each group and select the winners of each group label them as[b]A1,B1,...[/b]

We've had 5 have races.
The next race is between the winners. A1,B1,C1,D1 and E1 (race 6)

Let's assume they finish in the order a,b,c,d,e
Since A1 is the winner, its the fastest of all 25. But we can't say the same thing for B and C yet because other racers from A can actually be faster than B and C

Next race(7) is between groupA (excluding A1 because its the fastest,A4,A5) and B1 i.e (A2,A3,B1 and C1)

We didn't put A4 and A5 because we need only 3 fastest, if A2 and A3 beat B1, the get the position, no need to include them

PS: A4 simply means he is in group A, and finished forth the race.. Same goes to others

Cc:masterkim
Lalasticlala

MasterKim:

2. Here is the math puzzle: — Peter, how old are your children? — Well, Thomas, there are 3 of them, and the product of their ages is 36. — That is not enough... — The sum of their ages is exactly the number of beers we have drunk today. — That is still not enough. — OK, the last thing is that my oldest child wears ared hat.
How old were each of Peter’s children?

3

Drniyi4u:



7 races

How did u get ur answer?

ikbnice:
For question 2
Possible combinations for the age of the children are:
(4,3,3), (9,2,2), (9,4,1), (6,6,1), (12,3,1), (18,2,1), (36,1,1), (6,3,2)
Now, the hint is sum of the ages is being told and correct ages can't be guessed. So, the possible choices left are:
(9,2,2), (6,6,1).
Now, the hint is oldest is wearing a Red hat. So, the only answer left is:
(9,2,2)
9 years, 2 years and 2 years. The man had twins
Note: The eldest was wearing a red hat to differentiate himself from the twins.

The eldest must be one of the twins and is wearing the red hat to differentiate them. You will not need to differentiate a 9yo from 2yo's, would you? It follows therefore that the twins are older than the last child, hence 6,6,1 is more appropriate.

Nicolars:
For #2

For product of 36, the possible combinations are
6-3-2 (sum=11)
9-2-2 (sum=13)
4-3-3 (sum=10)
9-4-1 (sum=14)
12-3-1 (sum=16)
18-2-1 (sum=21)
6-6-1 (sum=13)


Now the second tip would give him a clue,the no of beers he drank would give him his answer but if the number of beers he drank is 13, we have two options, 9-2-2 and 6-6-1

That's where the eldest son comes in, since there's an elder one, it can't be 6-6-2, we have 9-2-2.

Why can't it be 662? Abi there is no older twin ni?

The eldest must be one of the twins and is wearing the red hat to differentiate them. You will not need to differentiate a 9yo from 2yo's, would you? It follows
therefore that the twins are older than the last child, hence 6,6,1 is more appropriate.

paiz:
Let them keep their job

Abeg abeg......For this buhari era wey person nor dey chop belle full na him dem dey bring this kind questions wey dey make brain hot?? We are ok with the wheelbarrow we dey push rite now

lobell:


Why can't it be 662? Abi there is no older twin ni?

The eldest must be one of the twins and is wearing the red hat to differentiate them. You will not need to differentiate a 9yo from 2yo's, would you? It follows
therefore that the twins are older than the last child, hence 6,6,1 is more appropriate.

No.. It is not more appropriate

Yes, it is simple, just add that and that, subtract that from that, multiple this one with this and finally divide, you see simple something.

MasterKim:
1. There are 25 horses. What is the minimum number of race needed so you can find the fastest 3 horses?
You can race upto 5 horses at a time but you don't have to watch

2. Here is the math puzzle:
— Peter, how old are your children?
— Well, Thomas, there are 3 of them, and the product of their ages is 36.
— That is not enough...
— The sum of their ages is exactly the number of beers we have drunk today.
— That is still not enough.
— OK, the last thing is that my oldest child wears ared hat.

How old were each of Peter’s children?

1. Make group of 5 horses and run 5 races. Suppose five groups are a,b,c,d,e and next alphabet is its individual rank in tis group(of 5 horses).for eg. d3 means horse in group d and has rank 3rd in his group. [ 5 RACES DONE ]
a1 b1 c1 d1 e1
a2 b2 c2 d2 e2
a3 b3 c3 d3 e3
a4 b4 c4 d4 e4
a5 b5 c5 d5 e5

Now make a race of (a1,b1,c1,d1,e1).[RACE 6 DONE] suppose result is a1>b1>c1>d1>e1
which implies a1 must be FIRST.
b1 and c1 MAY BE(but not must be) 2nd and 3rd.
FOR II position, horse will be either b1 or a2
(we have to find top 3 horse therefore we choose horses b1,b2,a2,a3,c1 do racing among them [RACE 7 DONE].
The only possibilities are :
c1 may be third
b1 may be second or third
b2 may be third
a2 may be second or third
a3 may be third
The final result will give ANSWER. suppose result is a2>a3>b1>c1>b2
then answer is a1,a2,a3,b1,c1.
HENCE ANSWER is 7 RACES

2. The first clue is that the product is 36. Note that 36=3*3*2*2, so all posible combinations are:

6-3-2 (sum=11)
9-2-2 (sum=13)
4-3-3 (sum=10)
9-4-1 (sum=14)
12-3-1 (sum=16)
18-2-1 (sum=21)
6-6-1 (sum=13)

Note that only 9-2-2 and 6-6-1 give the same amount when added.
Obviously, the second man must know how many beers thay have drunk, so he knows the sum.
If he cannot decide yet it's only because the sum is 13 and there are two combinations that add up to 13.
And as we learned further that the oldest son wears a hat, it is clear that the correct combination of ages is 2-2-9, where there is exactly one of them the oldest one.
So the Age of Peter's Children is 2,2,9

ikbnice:
For question 2
Possible combinations for the age of the children are:
(4,3,3), (9,2,2), (9,4,1), (6,6,1), (12,3,1), (18,2,1), (36,1,1), (6,3,2)
Now, the hint is sum of the ages is being told and correct ages can't be guessed. So, the possible choices left are:
(9,2,2), (6,6,1).
Now, the hint is oldest is wearing a Red hat. So, the only answer left is:
(9,2,2)
9 years, 2 years and 2 years. The man had twins
Note: The eldest was wearing a red hat to differentiate himself from the twins.

Since a 9 year old foes not need a hat to differentiate himself from 2 year olds, don't you think the more practical one would be (4,3,3) as there won't be much difference in their height and then the hat can truly serve a purpose? Plus, 10 bottles of beer between two friends is practical

I guess one race is enough

u guys should never visit the nairaland jobs section abeg.. too much brain cells here my battery don drain

GoodDieYoung:



Since a 9 year old foes not need a hat to differentiate himself from 2 year olds, don't you think the more practical one would be (4,3,3) as there won't be much difference in their height and then the hat can truly serve a purpose? Plus, 10 bottles of beer between two friends is practical

Considering the order of how the questions were asked, 4,3,3 will give a unique sum of ages that equals 10.

The last clue about the oldest putting on red hat wouldn't have been necessary since the puzzle's focus was on the ages of his kids.

The correct answer remain 9,2,2. Mind you, age and physical appearance does not always tally

LaEvilIMiss:
u guys should never visit the nairaland jobs section abeg.. too much brain cells here my battery don drain

Come and use my charger

LaEvilIMiss:
u guys should never visit the nairaland jobs section abeg.. too much brain cells here my battery don drain

hahaha

Start to dey charge am sister

maths!! na wetin ah dey fear pass women

Drniyi4u:


limitations
No have a stopwatch so we can’t time the horses (meaning that we can’t compare the race times of each horse).
We can only race 5 horses in each race.

Let's name and arrange the horses 25 horses A - Y in 5 groups in table
A B C D E
F G H I J
K L M N O
P Q R S T
U V W X Y

Race 1 - 5
Let the horses in each row compete with each other, meaning we'll be having 5 races.

For convenience sake, we assume fastest horses are listed in ascending order, from the slowest (extreme left) to the fastest (extreme right), that is, the fastest in each rows are E, J, O, T and Y.

Let's eliminate the 2 slowest horses in each group and redraw the table

C D E
H I J
M N O
R S T
W X Y

Race 6
We know the 5 fastest horses from each group are not necessarily the 5 fastest horses. We race the fastest in group to determine the fastest of all the horses, that is, horses E, J, O, T and Y. Let's assume E is the fastest, followed by J, then O, in that order. Therefore, the overall fastest has been determine to be E. Automatically we can eliminate T and Y since those 2 are definitely not in the top 3. Then, we can automatically eliminate all the horses that T and Y competed against in the preliminary races which are R&S and W&X respectively.

Now, we know that E is the fastest horse in the group since it was the fastest horse out of the 5 group leaders. So, we don’t need to race it anymore.

Further elimination;
If J and O are the 2nd and 3rd fastest in the group leaders, then we should be able to eliminate H since J raced against it and he was in 3rd place in that race.
Also, I could only possibly be the 3rd fastest, and since H is slower than I, we can safely eliminate H. We can also eliminate M and N since O was the 3rd fastest in the group leaders, and N and M were slower than O.

Race 7
In order to determine the second and third fastest horses, the remaining horses are:

C D
I J
O

The horses can then go on the seventh race and the first and second fastest, assuming they're D and J will be determined.


So, in addition to the overall first position we've determined to be E in race 1, the second and third are D and J determine from race 7.


Therefore the minimum of races needed to determined the 3 fastest horses is 7



Possible combinations
a*b*c = 36

1*1*36 (sum is 38)
1*2*18 (sum is 21)
1*3*12 (sum is 16)
1*4*9 (sum is 14)
1*6*6 (sum is 13)
2*2*9 (sum is 13)
2*3*6 (sum is 11)
3*3*4 (sum is 10)

Since The sum of their ages is exactly the number of beers they have drunk today, and is still not enough, that means the answer is either 1*6*6 (sum is 13)
2*2*9 (sum is 13).

Since the last thing is that the oldest child wears a red hat, therefore the combination that allows maximum age (oldest) is 2,2&9

Chisos

Yahooplusplus:
Yes, it is simple, just add that and that, subtract that from that, multiple this one with this and finally divide, you see simple something.

Na so

ikbnice:
For question 2
Possible combinations for the age of the children are:
(4,3,3), (9,2,2), (9,4,1), (6,6,1), (12,3,1), (18,2,1), (36,1,1), (6,3,2)
Now, the hint is sum of the ages is being told and correct ages can't be guessed. So, the possible choices left are:
(9,2,2), (6,6,1).
Now, the hint is oldest is wearing a Red hat. So, the only answer left is:
(9,2,2)
9 years, 2 years and 2 years. The man had twins
Note: The eldest was wearing a red hat to differentiate himself from the twins.

if he wasn't a twin would he need to differentiate himself? I think not... it is safe to say that the first two are twins then using your logic