In the xy-plane, a circle has center (0,0) and radius 5. Is

[Max@Math Revolution]

[Math Revolution GMAT math practice question]

In the xy-plane, a circle has center (0,0) and radius 5. Is the point (r,s) inside or on the circle?

1) -3 < r < 3
2) -4 < s < 4

[fskilnik@GMATH]

[Math Revolution GMAT math practice question]

In the xy-plane, a circle has center (0,0) and radius 5. Is the point (r,s) inside or on the circle?

1) -3 < r < 3
2) -4 < s < 4

$$\left( {r,s} \right)\,\,\,\mathop \in \limits^? \,\,\,\left\{ {\,\left( {x,y} \right)\,\,:\,\,{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2} \leqslant {5^2}\,} \right\}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,\,{r^2} + {s^2}\,\,\,\mathop \leqslant \limits^? \,\,\,25\,}\,\,$$
$$\left( 1 \right)\,\,\,\,\left| r \right| < 3\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,6} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,\,\left| s \right| < 4\,\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {6,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,\,\left\{ \matrix{
\,{r^2} = {\left| r \right|^2} < {3^2} \hfill \cr
\,{s^2} = {\left| s \right|^2} < {4^2} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{r^2} + {s^2} < 25\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

[Max@Math Revolution]

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The inequality satisfied by points inside or on the circle is r^2+s^2 ≤ 5^2=25.

Since we have 2 variables (r and s) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first.

Conditions 1) & 2):
Since -3 < r < 3 and -4 < s < 4, we have 0 ≤ r^2 < 3^2=9 and 0 ≤ s^2 < 4^2=16. Thus, 0 ≤ r^2+s^2 < 9+16=25 and both conditions together are sufficient.




Therefore, the answer is C.
Answer: C