[Math Revolution GMAT math practice question]
Is xy<1?
1) x^2+y^2 < 1
2) x + y < 1
Is xy<1?
This topic has expert replies
- Max@Math Revolution
- 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
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
\[\,xy\,\,\mathop < \limits^? \,\,\,1\]Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is xy<1?
1) x^2+y^2 < 1
2) x + y < 1
(1) Points (x,y) that satisfy statement (1) are precisely the points inside the circle with center at the origin and radius 1, therefore we have
\[0 \leqslant \,\,\left| x \right|\, < \,\,1\] AND \[0 \leqslant \,\,\left| y \right|\, < \,\,1\]
Conclusion:
\[xy\,\, \leqslant \,\,\left| {xy} \right|\,\, = \,\,\left| x \right| \cdot \left| y \right|\,\, < \,\,1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]
(2) Insufficient:
> Take (x,y) = (0,0) to answer in the affirmative
> Take (x,y) = (-1,-1) to answer in the negative
The above follows the notations and rationale taught in the GMATH method.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
- Max@Math Revolution
- 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
=>
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.
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2):
x^2 + y^2 < 1
=> x^2< 1 - y^2 ≤ 1 since y^2 ≥ 0
=> x^2< 1
=> -1 < x < 1
and
x^2 + y^2 < 1
=> y^2< 1 - x^2 ≤ 1 since x^2 ≥ 0
=> y^2< 1
=> -1 < y < 1
Combining these two inequalities yields -1 < xy < 1, so xy < 1. Both conditions are sufficient, when taken together.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
The argument showing that conditions 1 and 2 are sufficient, when taken together, only used condition 1. Therefore, condition 1 is sufficient.
Using the same argument as above,
x^2 + y^2 < 1
=> x^2< 1 - y^2 ≤ 1 since y^2 ≥ 0
=> x^2< 1
=> -1 < x < 1
and
x^2 + y^2 < 1
=> y^2< 1 - x^2 ≤ 1 since x^2 ≥ 0
=> y^2< 1
=> -1 < y < 1.
So, -1 < xy < 1, and condition 1) is sufficient.
Condition 2)
If x = 1/3 and y = 1/3, then xy = 1/9 < 1 and the answer is 'yes'
If x = -2 and y = -2, then xy = 4 > 1 and the answer is 'no'
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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.
Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2):
x^2 + y^2 < 1
=> x^2< 1 - y^2 ≤ 1 since y^2 ≥ 0
=> x^2< 1
=> -1 < x < 1
and
x^2 + y^2 < 1
=> y^2< 1 - x^2 ≤ 1 since x^2 ≥ 0
=> y^2< 1
=> -1 < y < 1
Combining these two inequalities yields -1 < xy < 1, so xy < 1. Both conditions are sufficient, when taken together.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
The argument showing that conditions 1 and 2 are sufficient, when taken together, only used condition 1. Therefore, condition 1 is sufficient.
Using the same argument as above,
x^2 + y^2 < 1
=> x^2< 1 - y^2 ≤ 1 since y^2 ≥ 0
=> x^2< 1
=> -1 < x < 1
and
x^2 + y^2 < 1
=> y^2< 1 - x^2 ≤ 1 since x^2 ≥ 0
=> y^2< 1
=> -1 < y < 1.
So, -1 < xy < 1, and condition 1) is sufficient.
Condition 2)
If x = 1/3 and y = 1/3, then xy = 1/9 < 1 and the answer is 'yes'
If x = -2 and y = -2, then xy = 4 > 1 and the answer is 'no'
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]