Difference between revisions of "2010 AMC 10B Problems/Problem 9"
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<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math> | <math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math> | ||
− | + | ==Solution 1== | |
− | |||
Simplify the expression <math> a-(b-(c-(d+e))) </math>. I recommend to start with the innermost parenthesis and work your way out. | Simplify the expression <math> a-(b-(c-(d+e))) </math>. I recommend to start with the innermost parenthesis and work your way out. | ||
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Our answer is <math>3 \Rightarrow \boxed{\textbf{(D)}}</math> | Our answer is <math>3 \Rightarrow \boxed{\textbf{(D)}}</math> | ||
− | + | ==Solution 2== | |
Lucky Larry had not been aware of the parenthesis and would have done the following operations: | Lucky Larry had not been aware of the parenthesis and would have done the following operations: | ||
<math>1-2-3-4+e=e-8</math> | <math>1-2-3-4+e=e-8</math> |
Revision as of 22:18, 9 February 2014
Contents
Problem
Lucky Larry's teacher asked him to substitute numbers for ,
,
,
, and
in the expression
and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry sustitued for
,
,
, and
were
,
,
, and
, respectively. What number did Larry substitude for
?
Solution 1
Simplify the expression . I recommend to start with the innermost parenthesis and work your way out.
So you get:
Henry substituted with
respectively.
We have to find the value of , such that
(the same expression without parenthesis).
Substituting and simplifying we get:
So Henry must have used the value for
.
Our answer is
Solution 2
Lucky Larry had not been aware of the parenthesis and would have done the following operations:
The correct way he should have done the operations is:
Therefore we have the equation
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.