Difference between revisions of "2002 AMC 10A Problems/Problem 16"

(New page: == Problem == Let <math>\text{a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5}</math>. What is <math>\text{a + b + c + d}</math>? <math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \...)
 
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== Problem ==
 
== Problem ==
  
Let <math>\text{a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5}</math>. What is <math>\text{a + b + c + d}</math>?
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Let <math>a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. What is <math>a + b + c + d</math>?
  
 
<math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5</math>
 
<math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5</math>
  
 
==Solution==
 
==Solution==
Let <math>\text{x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5}</math>. Since one of the sums involves a,b,c, and d, it makes sense to consider 4x. <math>\text{4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20}</math>. Rearranging, we have <math>\text{3(a+b+c+d)=-10}</math>, so <math>\text{a+b+c+d}=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>.
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Let <math>x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have <math>4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20</math>. Rearranging, we have <math>3(a+b+c+d)=-10</math>, so <math>a+b+c+d=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 23:22, 26 December 2008

Problem

Let $a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$. What is $a + b + c + d$?

$\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5$

Solution

Let $x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have $4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20$. Rearranging, we have $3(a+b+c+d)=-10$, so $a+b+c+d=\frac{-10}{3}$. Thus, our answer is $\boxed{\text{(B)}\ -10/3}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions