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>
  
===Alternate Solution===
+
==Alternate Solution==
 
We have <math>3a+7b=1977</math> and <math>5a+b=2007</math>. Notice the symmetry we have if we add the two equations together: <math>8a+8b=3984</math>. Dividing by 8, we have <math>a+b=498</math>. <math>\boxed{\mathrm{B}}</math>
 
We have <math>3a+7b=1977</math> and <math>5a+b=2007</math>. Notice the symmetry we have if we add the two equations together: <math>8a+8b=3984</math>. Dividing by 8, we have <math>a+b=498</math>. <math>\boxed{\mathrm{B}}</math>
  

Latest revision as of 20:05, 4 February 2023

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}}$

Alternate Solution

We have $3a+7b=1977$ and $5a+b=2007$. Notice the symmetry we have if we add the two equations together: $8a+8b=3984$. Dividing by 8, we have $a+b=498$. $\boxed{\mathrm{B}}$

See Also

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