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Difference between revisions of "1998 AJHSME Problems/Problem 11"

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==Problem 11==
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==Problem==
  
 
Harry has 3 sisters and 5 brothers.  His sister Harriet has <math>\text{S}</math> sisters and <math>\text{B}</math> brothers.  What is the product of <math>\text{S}</math> and <math>\text{B}</math>?
 
Harry has 3 sisters and 5 brothers.  His sister Harriet has <math>\text{S}</math> sisters and <math>\text{B}</math> brothers.  What is the product of <math>\text{S}</math> and <math>\text{B}</math>?
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Harry has 3 sisters and 5 brothers. His sister, being a girl, would have 1 less sister and 1 more brother.
 
Harry has 3 sisters and 5 brothers. His sister, being a girl, would have 1 less sister and 1 more brother.
  
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<math>3-1=2</math>
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<math>S = 3-1=2 </math>
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<math>5+1=6</math>
 
  
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<math>2\times6=12=\boxed{C}</math>
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<math>B = 5+1=6</math>
  
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<math>S\cdot B = 2\times6=12=\boxed{C}</math>
  
 
== See also ==
 
== See also ==
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{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
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{{AJHSME box|year=1998|num-b=10|num-a=12}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 23:29, 4 July 2013

Problem

Harry has 3 sisters and 5 brothers. His sister Harriet has $\text{S}$ sisters and $\text{B}$ brothers. What is the product of $\text{S}$ and $\text{B}$?

$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$

Solution

Harry has 3 sisters and 5 brothers. His sister, being a girl, would have 1 less sister and 1 more brother.

$S = 3-1=2$

$B = 5+1=6$

$S\cdot B = 2\times6=12=\boxed{C}$

See also

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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