A man is buying a gold ring set with stones for his wife on her birthday. A ring set with 4 amethysts and 1 diamond comes to US$2,000. One with 3 emeralds, 1 amethyst and 1 diamond would be US$1,400. And one set with 2 rubies and 1 diamond would cost US$3,000. Being a thoughtful husband, he choses a ring with 4 stones, each representing one of their 4 children.

As their children are named Andy, David, Ellen, and Richard, how much in US$ will the ring containing 1 amethyst, 1 diamond, 1 emerald, and 1 ruby cost him?

## Thursday, November 19, 2009

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I get US$2,300:

ReplyDelete1)4a + 1d = 2000

2)3e + 1a + 1d = 1400

3)2r + 1d = 3000

______________________

1a + 1d + 1e + 1r = x

(*6)

--->

6a + 6d + 6e + 6r = 6x

(- 3*eq. 3)

--->

6a + 3d + 6e = 6x - 9000

(- 2* eq. 2)

--->

4a + d = 6x - 11800

(- eq. 1)

--->

0 = 6x - 13800

--->

13800 = 6x

--->

x = 2,300

I got the same answer, but with way less math and mostly lucky guessing.

ReplyDeleteI took the last equation: 2 rubies + 1 diamond = 3,000 and guessed, "What if the rubies are $1K each, and the diamonds are, as well?"

Took it from there to conclude that if the rubies and diamonds are each $1K, then the first equation 4a + d = 2000 meant that amethysts cost $250, and finally plugging that all into the second equation to figure emeralds cost $50.

250 + 1,000 + 1,000 + 50 = 2,300

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ReplyDeleteAnswer is 2300

ReplyDeleteTo solve this by simultaneous equations , we need 4 equations since we have 4 variables,

Oudeis created the 4 equation, since it was not readily given in the question.

Tabitha got it the easy way, guessing, I did the same thing, but started with very low values.

www.guessthelogo.blogspot.comInteresting note though, guessthelogo:

ReplyDeleteWe actually have 5 unknowns and 4 variables (note the "x" in the equation I start with), so none of these gems have a definite value. What I mean by that is: There are multiple solutions for the values of the gems themselves; only their sum is definite.

I'll illustrate by solving the equation set for "a", then providing two distinct solutions:

1)4a + 1d = 2000

2)3e + 1a + 1d = 1400

3)2r + 1d = 3000

Solve in terms of a:

d = 2000 - 4a

r = 1500 - d/2

= 1500 - 1000 + 2a

= 500 + 2a

e = (1400 - a - d)/3

= 1400/3 - a/3 - (2000 - 4a)/3

= -600/3 + a

= -200 + a

Solution # 1:

let a = 250

d = 1000

r = 1000

e = 50

(Verify:

1: 4a + d = 1000 + 1000 = 2000

2: 3e + a + d = 150 + 250 + 1000 = 1400

3: 2r + d = 2000 + 1000 = 3000)

Solution # 2:

let a = 300

d = 800

r = 1100

e = 100

(Verify:

1: 4a + d = 1200 + 800 = 2000

2: 3e + a + d = 300 + 300 + 800 = 1400

3: 2r + d = 2200 + 800 = 3000)

It would seem then, that the gold band is free.

ReplyDeleteI will take 20 without any stones please.