Challenging Maths Problem

I was wondering if anyone could figure out how to solve this problem: it is a question off BMAT:

During the Beijing olympics, Australia won 46 medals in total. In the official ranking system, Gold medals are weighted to 3 points, silver to 2 and bronze to 1. If Australia was ranked with 89 points, what is the minimum amount of silver medals that they won?

This has proved to be the most frustrating problem I have ever encountered, and I have absolutely no clue as to how to solve it since I have never learnt this before. Please someone relieve me of my insanity!!

Reply 1

0x2a

2

Simulatenous equations?

Let no. of gold medals = x, silver = y, bronze = z.

x + y + z = 46

3x + 2y + z = 89

(edited 10 years ago)

Reply 2

james22

16

Trail and error works best, since they got 89 points they must have at most 44 silvers, however if they had 44 silvers then even with the otehr 2 being bronze they would go over their points value. This tells you that they must have gotten 43 or lower. Keep working down to see when you can find a possible combination.

Original post by 0x2a

Simulatenous equations?

Let no. of gold medals = x, silver = y, bronze = z.

x + y + z = 46

3x + 2y + z = 89

Since the solutions must be integers this becomes an integer programing problem which is not easy to solve systimatically, and impossible with A level knowledge.

(edited 10 years ago)

Reply 3

0x2a

2

Original post by james22

Since the solutions must be integers this becomes an integer programing problem which is not easy to solve systimatically, and impossible with A level knowledge.

Yup tried to solve it with no avail. Going down from 43 to a possible combination will take too long though.

Obviously the number the gold medals must be less than 29. I've reduce the number of gold medals down the only 3,2,1 (I haven't checked yet).

Reply 4

james22

16

Original post by 0x2a

Yup tried to solve it with no avail. Going down from 43 to a possible combination will take too long though.

Obviously the number the gold medals must be less than 29. I've reduce the number of gold medals down the only 3,2,1 (I haven't checked yet).

It really doesn't take long, try it. You will hit the answer in less than a minute if you don't make a mistake.

Reply 5

0x2a

2

Original post by james22

It really doesn't take long, try it. You will hit the answer in less than a minute if you don't make a mistake.

Is it obvious that the answer is the minimum? (I'm just curious, since I don't want to try it atm ><)

Reply 6

Exon

15

There's a very simple way of doing it involving trial and error and simultaneous equations.

Start with 0 silvers, let x = number of bronzes and y = number of golds.
x + 3y = 89
x + y = 46
2y = 43
y = 21.5 (not an integer so Australia could not have won 0 silvers)

1 silver:
x + 3y = 87
x + y = 45
2y = 42
y = 21

x + 21 = 45
x = 24

Therefore the minimum amount of silvers won by Australia is 1.

EDIT: Made a small error in my working. Fixed.

(edited 10 years ago)

Reply 7

james22

16

Original post by 0x2a

Is it obvious that the answer is the minimum? (I'm just curious, since I don't want to try it atm ><)

I missread, I though it said maximum not minimum. The logic still works though (just start at 0 and work up).

Reply 8

Hamza357

0?

(edited 10 years ago)

Reply 9

Exon

15

Original post by james22

I missread, I though it said maximum not minimum. The logic still works though (just start at 0 and work up).

Yeah I noticed. No big deal it's easy to misread a question.

Reply 10

Exon

15

Original post by Hamza357

Minimum they can possible have is 0 because

89/3 = 29, remainder 2 which could be both bronze, or all could be bronze

No...

It clearly says in the question that Australia won 46 medals overall. You have to account for that as well.

Reply 11

Hamza357

I just noticed, the answer would still be the same. The minimum they could win is 0 because they all could be bronze.

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Reply 12

Exon

15

Original post by Hamza357

I just noticed, the answer would still be the same. The minimum they could win is 0 because they all could be bronze.

If they were all bronze they would've won 89 medals not 46 :confused:

Reply 13

Prepotency

7

Don't use trial and error.

Use simultaneous equations and the answer is straight forward.

Answer:

Spoiler

Reply 14

Dalek1099

21

x+y+z=46
3x+2y+z=89
2x+y=43 2x has to be an even number so y has to be at least 1 to make the odd number 43.

Reply 15

Exon

15

Original post by Dalek1099

2x+y=43 2x has to be an even number so y has to be at least 1 to make the odd number 43.

Forgot you could do that. This is what happens after not touching maths for ages. :redface:

Reply 16

123formyabc

12

Original post by Hamza357

I just noticed, the answer would still be the same. The minimum they could win is 0 because they all could be bronze.

Posted from TSR Mobile

U wot mate??

Reply 17

CardiffCpl

Original post by louislallouette

I was wondering if anyone could figure out how to solve this problem: it is a question off BMAT:

During the Beijing olympics, Australia won 46 medals in total. In the official ranking system, Gold medals are weighted to 3 points, silver to 2 and bronze to 1. If Australia was ranked with 89 points, what is the minimum amount of silver medals that they won?

This has proved to be the most frustrating problem I have ever encountered, and I have absolutely no clue as to how to solve it since I have never learnt this before. Please someone relieve me of my insanity!!

min silver 17.x

Reply 18

CardiffCpl

Min 17 silver the equation is between the gold and bronze.x

Reply 19

Hamza357

Original post by louislallouette

I was wondering if anyone could figure out how to solve this problem: it is a question off BMAT:

During the Beijing olympics, Australia won 46 medals in total. In the official ranking system, Gold medals are weighted to 3 points, silver to 2 and bronze to 1. If Australia was ranked with 89 points, what is the minimum amount of silver medals that they won?

This has proved to be the most frustrating problem I have ever encountered, and I have absolutely no clue as to how to solve it since I have never learnt this before. Please someone relieve me of my insanity!!

89/3= 29. R2
89/2= 44. R1
89/1= 89.

x/3 + b/2 +c/1 =46

xbc=89
So...
51/3 + 0/2 + 38/1= 55 invalid
54/3 + 0/2 + 35/1= 53 invalid
60/3 + 0/2 + 29/1= 49 invalid
63/3 + 2/2 + 24/1= 46 Correct
66/3 + 0/2 + 23/1= 45 invalid
69/3 + 0/2 + 20/1= 43 invalid
69/3 + 2/2 + 18/1= 43 invalid

Minimum of 1 silver

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Challenging Maths Problem

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