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

(Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
==Problem 11==
+
==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>?
Line 16: Line 16:
  
 
== See also ==
 
== See also ==
{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
+
{{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]]
 +
{{MAA Notice}}

Latest revision as of 00:29, 5 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 (ProblemsAnswer KeyResources)
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