Difference between revisions of "1995 AHSME Problems/Problem 3"

(Undo revision 41558 by Talkinaway (talk))
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</math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }  <math>
 
</math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }  <math>
  
== Solution 1==
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== Solution ==
</math>99.99 - (29.98*3 + 9.98) = 99.99 - 99.92 = .07 \Rightarrow \mathrm{(B)}<math>
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</math>99.99 - (29.98*3 + 9.98) = 99.99 - 99.92 = .07 \Rightarrow \mathrm{(B)}$
 
 
==Solution 2==
 
 
 
Alternately, the in-store price is </math>1<math> cent below </math>100.00<math> dollars.
 
 
 
The television price is </math>3\cdot 2 + 2 = 8<math> cents below </math>30.00 \cddot 3 + 10.00 = 100.00<math> dollars.
 
 
 
Therefore, the difference is </math>8 - 1 = 7<math> cents, and the answer is </math>\mathrm{(B)}$
 
  
 
== See also ==
 
== See also ==

Revision as of 00:59, 18 August 2011

Problem

The total in-store price for an appliance is $&#036;$99.99$. A television commercial advertises the same product for three easy payments of$$$29.98$ and a one-time shipping and handling charge of $&#036;$9.98$. How many cents are saved by buying the appliance from the television advertiser?$ \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 } $== Solution ==$99.99 - (29.98*3 + 9.98) = 99.99 - 99.92 = .07 \Rightarrow \mathrm{(B)}$

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 26 27 28 29 30
All AHSME Problems and Solutions