GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha
[#permalink]
17 Mar 2018, 07:26
17 Mar 2018, 07:26
Expert Reply
00:00
Question Stats:
36% (02:03) correct
63% (02:04) wrong based on 19 sessions
Hide Show timer Statistics
−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most accurately describes the range of \(a^2\) ?
A. −16 < \(a^2\) < 11
B. −4 < \(a^2\) < 11
C. 0 < \(a^2\) < 16
D. 0 < \(a^2\) < 121
E. 16 < \(a^2\) < 121
A. −16 < \(a^2\) < 11
B. −4 < \(a^2\) < 11
C. 0 < \(a^2\) < 16
D. 0 < \(a^2\) < 121
E. 16 < \(a^2\) < 121
Drill 2
Question: 5
Page: 497
Question: 5
Page: 497
Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha
[#permalink]
18 Mar 2018, 07:00
18 Mar 2018, 07:00
Expert Reply
Explanation
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha
[#permalink]
22 Mar 2018, 00:52
22 Mar 2018, 00:52
I don't understand how the answer can be D.
If b = -3 then we have -4 < a < 7.
If b = 1 then we have 0 < a < 11.
So - 4 < a < 11.
Squaring the inequality yields, 16 < a^{2} < 121.
a^{2} > 16 > 0.
If b = -3 then we have -4 < a < 7.
If b = 1 then we have 0 < a < 11.
So - 4 < a < 11.
Squaring the inequality yields, 16 < a^{2} < 121.
a^{2} > 16 > 0.
sandy wrote:
Explanation
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
