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

This topic has expert replies
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

[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

User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pm
Thanked: 59 times
Followed by:33 members

by fskilnik@GMATH » Fri Jan 11, 2019 6:11 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Max@Math Revolution wrote:[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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

by Max@Math Revolution » Sun Jan 13, 2019 4:34 pm

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

=>

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.


Image

Therefore, the answer is C.
Answer: C