### Brain Teasers

# The Four-Stone Ring

While Christmas shopping, I chose a gold ring set with stones for my mother.

A ring set with 2 rubies and a diamond would have cost £3,000. One set with 4 amethysts and a diamond would come to £2,000. And 3 emeralds, 1 amethyst and a diamond would be £1,400.

Being a thoughtful son, I chose a ring that would have sentimental value. As my mother called her children Ellen, David, Richard and Andy, I have chosen a ring with one of each stone, to represent her children's names.

How much will my ring containing 1 emerald, 1 diamond, 1 ruby and 1 amethyst cost me?

A ring set with 2 rubies and a diamond would have cost £3,000. One set with 4 amethysts and a diamond would come to £2,000. And 3 emeralds, 1 amethyst and a diamond would be £1,400.

Being a thoughtful son, I chose a ring that would have sentimental value. As my mother called her children Ellen, David, Richard and Andy, I have chosen a ring with one of each stone, to represent her children's names.

How much will my ring containing 1 emerald, 1 diamond, 1 ruby and 1 amethyst cost me?

### Hint

Finding the cost of each gem is not possible. It is still possible to find the value of the ring.### Answer

The three rings can be expressed as:0E + 0A + 2R + 1D = £3,000

0E + 4A + 0R + 1D = £2,000

3E + 1A + 0R + 1D = £1,400

If we multiply the first line by 1.5 and divide the second line by 2, then total all three rings we get:

0E + 0A + 3R + 1.5D = £4,500

0E + 2A + 0R + .5D = £1,000

3E + 1A + 0R + 1D = £1,400

_________________________

3E + 3A + 3R + 3D = £6,900

£6,900 / 3 = £2,300

From this, we can see that a ring with one of each stone will cost me £2,300.

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## Comments

Excellent math question! There is a typo of an extra zero in the total on the last line of the answer though but the answer is still clearly understood. I liked the way you solved this one, although I am a bit confused as to why you kept changing the variables (E’s become G’s, A’s become B’s, R’s stay R and D’s become W’s). I like to keep variables constant throughout a solution like this, but maybe that is just me. Anyway, I happened to get the right answer in a less graceful way. I just played with different values for each stone for a few minutes until I came across R=1000, D=1000, A=250, and E=50. Plug them in and each line works. By adding them up I got the same answer as you. I call it the luck method of problem solving and I do not recommend it for everyone. 8^)

You may notice if you use excel that you have an infinite number of workable solutions. This is because you've got 3 equations and 4 unknown variables. e-150 r-1200 a-350 d-600 or e-160 r-1220 a-360 d-560 also work. if you add an extra ring it should be possible to solve with only one correct answer.

Since everyone else's answers differ to mine, I came up with:

Diamond: 800

Rubies: 1100

Amethysts: 300

Emeralds: 100

Diamond: 800

Rubies: 1100

Amethysts: 300

Emeralds: 100

Good one, but you can get it just by random guessing. I DID!!

It seems people are missing the point of the hint and solution - these are not "random guesses" or "luck" they are trial and error solutions to the system of 4 equations given in the solution (the 3 equations given in the teaser and the sum of the 3 equations in the answer). So, the unknown - the solution - 4 numbers that sum to 2300 - was what the "random" and "lucky" solutions all had in common. BTW there are NOT an infinite number of solutions only one - the puzzler asked for the sum of the 4 gems - 2300 not the value of each. Thank you for a nice math puzzle and well presented solution.

I just went through and did it the easy way, (se dever said you couldn't have half pounds),

r:1000$

d:1000$

a:250$

e:37.5$

(sorry... american... don't have pound symbol on keyboard)

r:1000$

d:1000$

a:250$

e:37.5$

(sorry... american... don't have pound symbol on keyboard)

This teaser didn't make much sense to me. Why did you multiply the first line and divide the second. Don't you have to preform the same mathmatical equation to each line when you have unknown variables?

I thought it was kinda tough!

Dec 14, 2003

I loved this one - thanks.

I am no mathematician so I had to resort to trial and error. I think my answer works (but I am open to correction). It doesn't say that all 4 stones had to be different values so my answer is

Amethyst=400, Diamond=400, Emerald=200, Ruby=1300

I am no mathematician so I had to resort to trial and error. I think my answer works (but I am open to correction). It doesn't say that all 4 stones had to be different values so my answer is

Amethyst=400, Diamond=400, Emerald=200, Ruby=1300

i don't get it

i think a lot of people got the question wrong. its not asking for the price of each stone. bcoz in that case, there's an unlimited number of answers (4 variables using only 3 equations).

but if your looking for the price of the ring containing 1 emerald, 1 diamond, 1 ruby and 1 amethyst, then there's only 1 answer. and that is Â£2300!

i think this is a very clever teaser. my initial impulse was to solve for the price of each stone, but saw that its impossible. but it didn't fool me. still got the correct answer pretty easily.

but if your looking for the price of the ring containing 1 emerald, 1 diamond, 1 ruby and 1 amethyst, then there's only 1 answer. and that is Â£2300!

i think this is a very clever teaser. my initial impulse was to solve for the price of each stone, but saw that its impossible. but it didn't fool me. still got the correct answer pretty easily.

Brilliant teaser! So many teasers are trivial or simple recombinations of the same algebra problem.

Very nice solution and explanation also.

I solved it similarly, although somewhat differently. I solved for R and E in terms of A:

2R + D - 3000 = 4A + D - 2000

R = 2A + 500

3E + A + D - 1400 = 4A + D - 2000

E = A - 200

Then I replaced these in the second equation:

4A = (R - 500) + (E + 200) + A

4A = R + E + A - 300

R + E + A + D - 300 = 2000

R + E + A + D = 2300

The relationships between A and R and between A and E were easy to deduce on sight, so the whole thing solved rather easily in my head.

(R - 500) + (

Very nice solution and explanation also.

I solved it similarly, although somewhat differently. I solved for R and E in terms of A:

2R + D - 3000 = 4A + D - 2000

R = 2A + 500

3E + A + D - 1400 = 4A + D - 2000

E = A - 200

Then I replaced these in the second equation:

4A = (R - 500) + (E + 200) + A

4A = R + E + A - 300

R + E + A + D - 300 = 2000

R + E + A + D = 2300

The relationships between A and R and between A and E were easy to deduce on sight, so the whole thing solved rather easily in my head.

(R - 500) + (

Ignore the extra text before the smiley in my last comment.

For a start everyone seems to have ignored the cost of the gold ring. Surely that should figure?

So I had G (Gold) R (Ruby) A (Amethyst) E (Emerald) and D(Diamond).

G+2R+D=3000 [1]

G+4a+D=2000 [2]

Subtracting 2R - 4A = 100 or 2A=R-500

Then G+3E+A+D=1400 [3]

Adding [1] [2] and [3]

3G+3E+2R+5A+3d=6400

Now plug in for 2A of the 5A

3G+3E+3R-500+3A+3D=6400

3(G+E+R+A+D)=6900

G+R+E+A+D=2300

It is interesting that the answer is constant whether you consider the value of the gold setting or not.

So I had G (Gold) R (Ruby) A (Amethyst) E (Emerald) and D(Diamond).

G+2R+D=3000 [1]

G+4a+D=2000 [2]

Subtracting 2R - 4A = 100 or 2A=R-500

Then G+3E+A+D=1400 [3]

Adding [1] [2] and [3]

3G+3E+2R+5A+3d=6400

Now plug in for 2A of the 5A

3G+3E+3R-500+3A+3D=6400

3(G+E+R+A+D)=6900

G+R+E+A+D=2300

It is interesting that the answer is constant whether you consider the value of the gold setting or not.

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