Korean Job Discussion Forums

silkhighway · Joined: 24 Oct 2010 Location: Canada

EDIT: Corrections to proper suit in bold below:

Well I failed on that one, so I better make this one a good one. Johnny and Jimmy are going to play a game, so they sit down and clearly spell out the rules.

1) Jimmy will have three hidden cards. Two of the cards are spades, and the other is a heart. For convenience sake, these cards are numbered 1, 2, and 3. Jimmy will take a quick peak so he knows what cards are what. Johnny doesn't know.
2) Johnny will pick a card, but not look at it yet.
3) Jimmy will then show a spade from the *other* 2 cards. If there are two spades, Jimmy can choose to show whatever one he wants. Jimmy then will offer Johnny to switch cards.
4) Johnny can stay with the original card he's chosen or he can switch to the remaining card.
5) After Johnny stays or goes, they look at the card. If it's a heart, Jimmyy gives Johnny a dollar. If it's a spade, he gives him a punch in the face.

So they start to play..Jimmyy shuffles up the three cards and then offers Johnny a choice. "Card number 2!" Johnny says. "Ok then..", Jimmy says, "in that case, card number 3 is a spade." Jimmy flips over card number 3. "So do you want to switch to card number 1 or stay with card number 2?"

So what should Johnny do? Should he keep his original card, switch, or does it matter?

PS: This is a variation of a popular problem. If you know it just put the initials or some hint that you know it.

crisdean · Joined: 04 Feb 2010 Location: Seoul Special City

silkhighway · Joined: 24 Oct 2010 Location: Canada

Ahhhh sorry, I messed up again. There are two spades and one heart in the three cards. I'm going to correct the assumptions.

crisdean · Joined: 04 Feb 2010 Location: Seoul Special City

dirtmcgirt · Joined: 08 Dec 2005

The answer is 2.

What google calculator probably doesn't take into account is the distributive property and when the theorem applies. When a number is tied to two other numbers inside a bracket you must do that first. a(b+c)=ab+ac

48/2(9+3) = 48/ 2(9+3)=(2*9+2*3)=(18+6)=24

At this point you then go left to right. 48/24=2

I think the key here is the distributive property and the use of it. At the end of the day it is a poorly set up question. Also someone said that the 2(9+3) is in the numerator but that's not true it's in the denominator. I have no idea how it could be in the numerator. My two bits.

Dirt

Chris.Quigley · Joined: 20 Apr 2009 Location: Belfast. N Ireland

The answer is 288 based on the way it is written. (In my opinion)

48/2(9+3) = 48/2*(9+3) = 288

Brackets, Exponents, Division, Multiplication, Addition, Subtraction.

Enter this into excel:

"=SUM(48/2*(9+3))"

Division comes before multiplication.

Strangely my financial calculator gets 2, while my graphing calculator gets 288. Obviously my calculators disagree. But when I change the equation to 48/2*(3+9) both calculators get 288. Confusing.

interestedinhanguk · Joined: 23 Aug 2010

mcb1180 · Joined: 16 Feb 2009 Location: North Carolina

I too want to beat the dead horse.

48
2(9+3)

and

48�2(9+3)=

are two different equations.

One is written as a fraction with a numerator and a denominator while the other is a simple equation.

I was always taught that if an equation used a / rather than a � sign it was to be considered a fraction, with everything after the / being in the denominator. This of course mainly occurred in the time when there were no programs to type equations in the same manner that we would right them.

Such as math problems where exponents were written like such

9X2-5X2+7=3X2

Where the 2s are exponents.