Difference between revisions of "2007 iTest Problems/Problem 2"

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Thus, <math>b=2007-5a</math>, and substituting, <math>3a+14049-35a=1977\Rightarrow</math><math>-32a=-12072\Rightarrow a=377.25</math>. Thus, <math>b=2007-1886.25\Rightarrow b=120.75</math>. Thus, <math>a+b=377.25+120.75=498</math><math>\Rightarrow \boxed{\mathrm{B}}</math>
 
Thus, <math>b=2007-5a</math>, and substituting, <math>3a+14049-35a=1977\Rightarrow</math><math>-32a=-12072\Rightarrow a=377.25</math>. Thus, <math>b=2007-1886.25\Rightarrow b=120.75</math>. Thus, <math>a+b=377.25+120.75=498</math><math>\Rightarrow \boxed{\mathrm{B}}</math>
  
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==See Also==
 
{{iTest box|year=2007|num-b=1|num-a=3}}
 
{{iTest box|year=2007|num-b=1|num-a=3}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 14:23, 11 December 2007

Problem

Find $a + b$ if $a$ and $b$ satisfy $3a + 7b = 1977$ and $5a + b = 2007$.

$\mathrm{(A)}\, 488\quad\mathrm{(B)}\, 498$

Solution

$3a + 7b = 1977$ and $5a + b = 2007$.

Thus, $b=2007-5a$, and substituting, $3a+14049-35a=1977\Rightarrow$$-32a=-12072\Rightarrow a=377.25$. Thus, $b=2007-1886.25\Rightarrow b=120.75$. Thus, $a+b=377.25+120.75=498$$\Rightarrow \boxed{\mathrm{B}}$

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 1
Followed by:
Problem 3
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