# SAT problem of the week, August 25, 2014 Solution

Hint for process of elimination: Since \$latex y &s=1\$ and \$latex xy &s=1\$ are both positive, \$latex x &s=1\$ is also positive. Therefore \$latex y-x<y=15 &s=1\$. This eliminates choice E, leaving you with four choices. Some awareness about how big or small the answer should be, for example noting that \$latex x=8 &s=1\$, may help you eliminate choice D for being too big, and choice A for being too small.

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Answer: B

Solution: Since \$latex xy = 120 &s=1\$ and \$latex y = 15 &s=1\$, it follows that \$latex 15x = 120 &s=1\$. Solving this equation gives \$latex x = 8 &s=1\$. Therefore \$latex y-x = 15 – 8 = 7 &s=1\$

Pitfalls:

A. \$latex 5 = x – 3 &s=1\$, not \$latex y – x &s=1\$

C. \$latex 8 = x &s=1\$, not \$latex y-x &s=1\$. A great deal of students solve for \$latex x &s=1\$ and do not bother to check whether the problem was asking for \$latex x &s=1\$ or simply something involving \$latex x &s=1\$. Don’t fall into this trap. Re-read the last sentence of every math problem before choosing your answer.

D. \$latex 12 = y-3 &s=1\$, not \$latex y-x &s=1\$.

E. \$latex 23 = y+x &s=1\$. Make sure you do the correct operation. Don’t let caught thinking every operation is addition.