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Difference between revisions of "2007 AMC 10B Problems/Problem 5"

(Created page with "==Problem== In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these fa...")
 
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\textbf{(C) } \text{All Arogs are Crups and are Dramps.}\\
 
\textbf{(C) } \text{All Arogs are Crups and are Dramps.}\\
 
\textbf{(D) } \text{All Crups are Arogs and are Brafs.}\\
 
\textbf{(D) } \text{All Crups are Arogs and are Brafs.}\\
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\textbf{(E) } \text{All Arogs are Dramps and some Arogs may not be Crumps.}</math>
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\textbf{(E) } \text{All Arogs are Dramps and some Arogs may not be Crups.}</math>
  
 
==Solution==
 
==Solution==
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</asy></center>
 
</asy></center>
  
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It may be easier to visualize this by drawing some sort of diagram. From the first statement, you can draw an Arog circle inside of the Braf circle, since all Arogs are Brafs, but no all Brafs are Arogs. Ignore the second statement for now, and draw a Dramp circle in the Arog circle and a Crup circle in the Dramp circle. You can see the second statement is already true because all Crups are Arogs. As you can see, the only statement that is true is <math>\boxed{\mathrm{(D)}}</math>
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It may be easier to visualize this by drawing some sort of diagram. From the first statement, you can draw an Arog circle inside of the Braf circle, since all Arogs are Brafs, but not all Brafs are Arogs. Ignore the second statement for now, and draw a Dramp circle in the Arog circle and a Crup circle in the Dramp circle. You can see the second statement is already true because all Crups are Arogs. As you can see, the only statement that is true is <math>\boxed{\mathrm{(D)}}</math>
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 +
==Solution 2==
 +
Let's replace the terms with some more familiar terms. Let "Arogs" be math problems, "Brafs" be problems, "Crups" be 2007 AMC 10 problems, and "Dramps" be AMC 10 problems. All the statements then hold. Let us now analyze the claims individually.
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<math>\textbf{(A)}</math> All AMC 10 problems are problems and 2007 AMC 10 problems. This statement is false.
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<math>\textbf{(B)}</math> All problems are 2007 AMC 10 problems and AMC 10 problems. This statement is false.
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<math>\textbf{(C)}</math> All math problems are 2007 AMC 10 problems and AMC 10 problems. This statement is false.
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<math>\textbf{(D)}</math> All 2007 AMC 10 problems are math problems and problems. This statement is true.
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<math>\textbf{(E)}</math> All math problems are AMC 10 problems and some math problems may not be 2007 AMC 10 problems. This statement is false.
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Our answer is <math>\boxed{\textbf{(D)}~\text{All Crups are Arogs and are Brafs.}}</math>
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~ cxsmi
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2007|ab=B|num-b=4|num-a=6}}
 
{{AMC10 box|year=2007|ab=B|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 22:59, 18 May 2024

Problem

In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these facts?

$\textbf{(A) } \text{All Dramps are Brafs and are Crups.}\\ \textbf{(B) } \text{All Brafs are Crups and are Dramps.}\\ \textbf{(C) } \text{All Arogs are Crups and are Dramps.}\\ \textbf{(D) } \text{All Crups are Arogs and are Brafs.}\\ \textbf{(E) } \text{All Arogs are Dramps and some Arogs may not be Crups.}$

Solution

[asy] unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(9pt)); dotfactor=4; real r1=1, r2=2, r3=3, r4=4; pair O1=(0,0), O2=(0,-0.5), O3=(0,-1), O4=(0,-1.5); path circleA=Circle(O1,r1); draw(circleA); path circleB=Circle(O2,r2); draw(circleB); path circleC=Circle(O3,r3); draw(circleC); path circleD=Circle(O4,r4); draw(circleD); label("$Crups$",(0,-.5)); label("$Dramps$",(0,-2)); label("$Arogs$",(0,-3.5)); label("$Brafs$",(0,-5)); [/asy]

It may be easier to visualize this by drawing some sort of diagram. From the first statement, you can draw an Arog circle inside of the Braf circle, since all Arogs are Brafs, but not all Brafs are Arogs. Ignore the second statement for now, and draw a Dramp circle in the Arog circle and a Crup circle in the Dramp circle. You can see the second statement is already true because all Crups are Arogs. As you can see, the only statement that is true is $\boxed{\mathrm{(D)}}$

Solution 2

Let's replace the terms with some more familiar terms. Let "Arogs" be math problems, "Brafs" be problems, "Crups" be 2007 AMC 10 problems, and "Dramps" be AMC 10 problems. All the statements then hold. Let us now analyze the claims individually.

$\textbf{(A)}$ All AMC 10 problems are problems and 2007 AMC 10 problems. This statement is false.

$\textbf{(B)}$ All problems are 2007 AMC 10 problems and AMC 10 problems. This statement is false.

$\textbf{(C)}$ All math problems are 2007 AMC 10 problems and AMC 10 problems. This statement is false.

$\textbf{(D)}$ All 2007 AMC 10 problems are math problems and problems. This statement is true.

$\textbf{(E)}$ All math problems are AMC 10 problems and some math problems may not be 2007 AMC 10 problems. This statement is false.

Our answer is $\boxed{\textbf{(D)}~\text{All Crups are Arogs and are Brafs.}}$

~ cxsmi

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
All AMC 10 Problems and Solutions

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