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?
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.com
Interesting 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.