Difference between revisions of "2005 CEMC Gauss (Grade 7) Problems/Problem 23"
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Therefore, since <math>4\square</math> balance <math>2\bigcirc</math>, which weigh <math>4 kg</math> combined, then each <math>\square</math> weighs <math>1 kg</math>. | Therefore, since <math>4\square</math> balance <math>2\bigcirc</math>, which weigh <math>4 kg</math> combined, then each <math>\square</math> weighs <math>1 kg</math>. | ||
We then look at each of the remaining combinations. | We then look at each of the remaining combinations. | ||
+ | |||
<math>1\triangle</math>, <math>1\bigcirc</math>, and <math>1\square</math> weigh <math>3 + 2 + 1 = 6 kg</math>. | <math>1\triangle</math>, <math>1\bigcirc</math>, and <math>1\square</math> weigh <math>3 + 2 + 1 = 6 kg</math>. | ||
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<math>3\square</math> and <math>1\triangle</math> weigh <math>3 + 3 = 6 kg</math>. | <math>3\square</math> and <math>1\triangle</math> weigh <math>3 + 3 = 6 kg</math>. | ||
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<math>2\square</math> and <math>2\bigcirc</math> weigh <math>2 + 2\times 2 = 6 kg</math>. | <math>2\square</math> and <math>2\bigcirc</math> weigh <math>2 + 2\times 2 = 6 kg</math>. | ||
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<math>2\triangle</math> and <math>1\square</math> weigh <math>2\times 3 + 1 = 7 kg</math>. | <math>2\triangle</math> and <math>1\square</math> weigh <math>2\times 3 + 1 = 7 kg</math>. | ||
+ | |||
<math>1\bigcirc</math> and <math>4\square</math> weigh <math>2 + 4 = 6 kg</math>. | <math>1\bigcirc</math> and <math>4\square</math> weigh <math>2 + 4 = 6 kg</math>. | ||
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Therefore, it is the combination of <math>2\triangle</math> and <math>1\square</math> which will not balance the other combinations. The answer is <math>D</math>. | Therefore, it is the combination of <math>2\triangle</math> and <math>1\square</math> which will not balance the other combinations. The answer is <math>D</math>. | ||
== See Also == | == See Also == | ||
− | {{CEMC box|year=2005|competition=Gauss (Grade 7)|num-b= | + | {{CEMC box|year=2005|competition=Gauss (Grade 7)|num-b=22|num-a=24}} |
Latest revision as of 02:41, 24 October 2014
Problem
Using an equal-armed balance, if balances
and
balances
, which of the following would not balance
?
Solution 1
If balance
, then
would balance the equivalent of
.
Similarly,
would balance the equivalent of
.
If we take each of the answers and convert them to an equivalent number of
, we would have:
Therefore, and
do not balance the required. The answer is
.
Solution 2
Since balance
, then
would balance
.
Therefore,
would balance
, so since
balance
, then
would balance
, or
would balance
.
We can now express every combination in terms of only.
,
, and
equals
.
and
equals
.
and
equals
.
and
equals
.
and
equals
.
Therefore, since ,
, and
equals
, then it is
and
which will not balance with this combination. Thus, the answer is
.
Solution 3
We try assigning weights to the different shapes.
Since balance
, assume that each
weighs
and each
weighs
.
Therefore, since
balance
, which weigh
combined, then each
weighs
.
We then look at each of the remaining combinations.
,
, and
weigh
.
and
weigh
.
and
weigh
.
and
weigh
.
and
weigh
.
Therefore, it is the combination of and
which will not balance the other combinations. The answer is
.
See Also
2005 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
CEMC Gauss (Grade 7) |