1. ## word problem

jolibee is younger than Mcdo who is older than Burger king. Burger king is younger than Mr. mappy who is older than Jolibee. Burger king is older than Jolibee.Mcdo is younger than Mr. Mappy.

Who is the youngest?

2. Originally Posted by ranthrave
Scoooby is having a great party. He made a technique to get lots of people to attend.
He invited his 5 best friends and he said that they could each invite 4 people.

Each of those could invite 3.
Each of those could invite 2.
Each of those could invite 1.

Overall, how many people did Scooby invite to his party?

Each one of those invites 4 guests (20 in all)
Each of those invites 3 guests (60 in all)
Each of those invites 2 guests (120 in all)
Each of those invites 1 guest (120 in all)
Total attendees = 325

Unless this is more compicated than I think!

3. Hello,
Originally Posted by ranthrave
jolibee is younger than Mcdo who is older than Burger king. Burger king is younger than Mr. mappy who is older than Jolibee. Burger king is older than Jolibee.Mcdo is younger than Mr. Mappy.

Who is the youngest?
The ages will be represented by the letters of their names.

«Jolibee (J) is younger than McDo (Mc)»
$\displaystyle (1) ~:~ \text{J} < \text{Mc}$

«Mcdo (Mc) who is older than Burger king (BK)»
$\displaystyle (2) ~:~ \text{BK} < \text{Mc}$

«Burger king (BK) is younger than Mr. mappy (Ma)»
$\displaystyle (3) ~:~ \text{BK} < \text{Ma}$

«Mr. mappy who is older than Jolibee»
$\displaystyle (4) ~:~ \text{J} < \text{Ma}$

«Burger king is older than Jolibee»
$\displaystyle (5) ~:~ \text{J} < \text{BK}$

«Mcdo is younger than Mr. Mappy»
$\displaystyle (6) ~:~ \text{Mc} < \text{Ma}$

---------------------------------------------------
Lazy way :
Which of Mc, Ma, J and BK is not on the right of an inequation ? (this would mean it is older than one of the others)

---------------------------------------------------
Long way :
We know from (6) that Mc < Ma.

So getting back to (2) and (3), where Mc and Ma are older than a common person, we can say that (3) is not useful, since from (2) and (6), we get BK < Mc < Ma. This includes (3)

Same reasoning for (1) and (4), (4) is not useful.

So we're left with (1), (2), (5), (6)
From (5), combined with the others, we get :

(1) : J < Mc
(2)+(5) : J < BK < Mc. This includes (1), so (1) is not useful anymore.
adding the information from (6) :
J < BK < Mc < Ma

Done.

4. Originally Posted by ranthrave
jolibee is younger than Mcdo who is older than Burger king. Burger king is younger than Mr. mappy who is older than Jolibee. Burger king is older than Jolibee.Mcdo is younger than Mr. Mappy.

Who is the youngest?

Start another thread when you ask another question. Don't edit over your last question with a new one.

5. Mcdo is younger than Mr. Mappy.

Without thinking about it too much, it seems prudent to split up the confusing statements.

jolibee is younger than Mcdo who is older than Burger king.

Jol < McD
McD > Bur

Burger king is younger than Mr. mappy who is older than Jolibee.

Bur < MrM
MrM > Jol

Burger king is older than Jolibee.

Bur > Jol

Mcdo is younger than Mr. Mappy.

McD < MrM

Collect all the statements

Jol < McD
McD > Bur
Bur < MrM
MrM > Jol
Bur > Jol
McD < MrM

Make all the inequalities face the same direction.

Jol < McD
Bur < McD
Bur < MrM
Jol < MrM
Jol < Bur
McD < MrM

That was too easy. Not a single inference is required. Who NEVER appears on the right side of an inequality?

6. Hello, ranthrave!

JoliBee is younger than McDo who is older than Burger King.
Burger King is younger than Mr. Mappy who is older than JoliBee.
Burger King is older than JoliBee.
McDo is younger than Mr. Mappy.

Who is the youngest?

$\displaystyle \begin{array}{cccc}\text{JoliBee is younger than McDo: } & JB < MD & (1) \\ \text{McDo is older than Burger King:} & BK < MD & (2)\\ \text{Burger King is younger than Mr. Mappy:} & BK < MM & (3) \\ \text{Mr. Mappy is older than JoliBee:} & JB < MM & (4) \\ \text{Burger King is older than JoliBee:} & JB < BK & (5) \\ \text{McDo is younger than Mr. Mappy:} & MD < MM & (6)\end{array}$

And we have: .$\displaystyle \boxed{\begin{array}{c|ccccccc} (5) & JB & < & BK \\ \hline (2) & & & BK & < & MD \\ \hline (6) & & & & & MD & < & MM \end{array}}$

$\displaystyle \text{Therefore: }\; \underbrace{JB}_{\text{youngest}} \:< \:BK \:< \:MD \:<\: MM$

Edit: Too fast for me TKH !
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