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I was looking thru a book when i came across these problems. I managed to work most of them out. I'll put some here to see if anyone has lots of free time and do them to keep them occupied.
1. You are given an unlimited supply of 1 x 1, 2 x 2, 3 x 3, 4 x 4, 5 x 5 and 6 x 6 squares.
Two sets of squares are considered the same if the squares can be paired, one from each set, so that the squares in each pair are the same size. For example {1 x 1, 2 x 2, 4 x 4} is the same as {2 x 2, 1 x 1, 4 x 4} but not the same as
{1 x 1, 1 x 1, 2 x 2, 4 x 4}
a) Make a list of all the different sets of 10 squares whose areas add to 48; justify that your list contains all the different sets.
B) Make a list of all the different sets of 10 squares that can be fitted together to cover (without overlaps) a 6 x 8 rectangle; justify that your list contains all the different sets.
2. When Indiana walks into a cave, the door shuts behind him. He looks around and sees chests full of diamonds, emeralds, rubies and sapphires. There is a set of scales near the chests. The following statements are written on the door:
· Each jewel of the same type weighs the same.
· No two different types of jewel weigh the same.
· One sapphire weighs the same as one diamond and one emerald together.
· Three diamonds weigh the same as one sapphire and one ruby.
· One sapphire weighs more than one emerald and three rubies.
· Fourteen rubies weigh more than one sapphire.
· The door will open and all the jewels will be yours if at your first attempt you place one diamond on one tray and other types of jewel on the other tray to make the scales balance.
Of course, Indiana was able to get out of the cave with all the jewels.
a) List the four types of jewel in order from heaviest to lightest. Explain why there is only one answer.
B) If a diamond was placed on a tray scale, what did Indiana place on the other to open the door? Show that this is the only way he could have escapes.
3. Fourteen pebbles is a game played with fourteen pebbles in a jar. To play the game you first roll two dice and add the numbers showing. You remove that number of pebbles. You then choose to roll either one or two dice. If you roll a total equal to the number of pebbles remaining in the jar, you win. Your score is the total you rolled with the second roll multiplied by the number of dice you rolled. If you lose, you score zero.
For example, Bam Bam rolls a 9 with his first roll of two dice. He removes nine pebbles. If he then chooses to roll one die and rolls a 5, he scores 5 x 1 = 5 points.
If he rolls two dice and gets a total of 5, he scores 5 x 2 = 10 points. If he does not roll a total of 5, he loses and scores 0.
a) What non-zero scores are possible in a game of Fourteen Pebbles?
B) Fred and Wilma are playing a game of Fourteen Pebbles and discuss what to do if the first roll is an 11. Fred says “I think you should always choose to roll one die if you have removed eleven pebbles.” Wilma says “No, I think it is better to roll two dice.”
i) If you want to win, that is, to successfully remove the remaining pebbles, as often as possible, who is correct? Explain your answer.
ii) If you want to gain as many points as possible over a long period of time, who is correct? Explain your answer.
c) To obtain the maximum possible score over a long period of time, for what number(s) of pebbles remaining is it better to choose one die for the second roll? For what number(s) of pebbles remaining is it better to choose two dice for the second roll?
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That's a smart way of getting your math done ;)
Heh, I suck at it so I cant help you :(
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It's been a long time since I did any math, but this one looks fun.
"2. When Indiana walks into a cave, the door shuts behind him. He looks around and sees chests full of diamonds, emeralds, rubies and sapphires. There is a set of scales near the chests. The following statements are written on the door:"
d e r s
"· Each jewel of the same type weighs the same." Good.
"· No two different types of jewel weigh the same." Okay.
"· One sapphire weighs the same as one diamond and one emerald together."
s=d+e
"· Three diamonds weigh the same as one sapphire and one ruby."
3*d=s+r or just 3d=s+r or it could be d+d+d=s+r
"· One sapphire weighs more than one emerald and three rubies."
s>(e+3r) I don't understand this one. But I'll get back to it.
"· Fourteen rubies weigh more than one sapphire."
14r>s
"· The door will open and all the jewels will be yours if at your first attempt you place one diamond on one tray and other types of jewel on the other tray to make the scales balance."
d=? Basically we need to know what combo weighs the same as a diamond.
We know the following:
s=d+e
3d=s+r
s>(e+3r)
14r>s
I'll mess with them in order, trying to solve for d.
s=d+e
s-s=d+e-s Which equals 0=d+e-s
0-d=d+e-s-d Which equals -d=e-s
-1*-d=-1(e-s) Which equals d=s-e
So we know that one sapphire weighs more than one diamond Back to the first problem. I will be solving for e.
s=d+e
s-d=d+e-d which equals s-d=e
Now I will put s-e in place of d.
s-s-e=e
s-s-e+e=e+e which equals 0=e which doesn't make any freaking sense. :) But we do know this:
e=s-d
&
d=s-e
So I will try replacing e with s-d in the problem d=s-e, but e is negative, so...
d=s-1(s-d) which becomes d=s-s+d which becomes d=d which tells us nothing.
I should've solved for r. I'll do that now.
3d=s+r
3d-s=s+r-s which becomes r=3d-s
So now we know:
r=3d-s
s=d+e
e=s-d
d=s-e
Why can't he just put a diamond on each scale?
Or use a bag of sand. I'm tired of thinking, maybe someone can use this. I forgot how to solve those ‘more than' ‘less than' problems. It's been a long time since math
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I will not help you to do your homework.
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2.a) Sapphire, Emerald, Diamond, Ruby
2.B) He put 4 rubies on the other tray
why he can't put a diamond on the other tray? read the question better:
The door will open and all the jewels will be yours if at your first attempt you place one diamond on one tray and other types of jewel on the other tray to make the scales balance.
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i can just say... amazing...
