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Updated on: 05 Nov 2023, 11:36
Originally posted by ChandlerBong on 18 Apr 2023, 08:19.
Last edited by ChandlerBong on 05 Nov 2023, 11:36, edited 2 times in total.
Last edited by ChandlerBong on 05 Nov 2023, 11:36, edited 2 times in total.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (02:01) correct
69%
(01:35)
wrong
based on 42
sessions
History
Date
Time
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Not Attempted Yet
If b is a positive number, is \(a^2\) − 2a + b > 0?
(1) a > 2
(2) b > 1
(1) a > 2
(2) b > 1
18 Apr 2023, 12:35
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If b is a positive number, is a^2 − 2a + b > 0?
(1) a > 2
(2) b > 1
Given: b>0, since this is Yes No type of data sufficiency question with variable in the question stem, we can plug in values for the variables and check whether statements are giving single answer.
St 1: a>2 Let a = 3 then a^2 = 9, 2a = 2 x 3 = 6 => a^2 − 2a = 9 - 6 = 3 >0
b is also >0, so a^2 − 2a + b > 0--- Yes
In fact, since a>2, a2 - 2a will always be >0.
So, St 1 is sufficient
St 2 : b > 1 but we have no info on a
a can be +ve, -ve or fraction.
If a=0, then a^2 − 2a + b > 0
If a<0, then also a^2 − 2a + b > 0 ( a= -5 then 25+10+b >0)
If a>0, then also a^2 − 2a + b > 0
If a is + fraction say 1/2, then 1/4 - 1 + b = -3/4 + b >0 as b>1, say a= -1/5 then 1/25 - 2/5 + b = -9/25 + b which is again > 0.
So, if b>1, for all values of a
a^2 − 2a + b > 0
St 2 sufficient
Ans : D ( both statements are independently sufficient)
(1) a > 2
(2) b > 1
Given: b>0, since this is Yes No type of data sufficiency question with variable in the question stem, we can plug in values for the variables and check whether statements are giving single answer.
St 1: a>2 Let a = 3 then a^2 = 9, 2a = 2 x 3 = 6 => a^2 − 2a = 9 - 6 = 3 >0
b is also >0, so a^2 − 2a + b > 0--- Yes
In fact, since a>2, a2 - 2a will always be >0.
So, St 1 is sufficient
St 2 : b > 1 but we have no info on a
a can be +ve, -ve or fraction.
If a=0, then a^2 − 2a + b > 0
If a<0, then also a^2 − 2a + b > 0 ( a= -5 then 25+10+b >0)
If a>0, then also a^2 − 2a + b > 0
If a is + fraction say 1/2, then 1/4 - 1 + b = -3/4 + b >0 as b>1, say a= -1/5 then 1/25 - 2/5 + b = -9/25 + b which is again > 0.
So, if b>1, for all values of a
a^2 − 2a + b > 0
St 2 sufficient
Ans : D ( both statements are independently sufficient)
18 Apr 2023, 14:33
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ChandlerBong
A couple of observations along the way can help us solve this question without having to test numbers.
Statement 1
(1) a > 2
Two inferences that we can make here -
- a is positive
- \(a^2 - 2a\) is always positive in thi case.
Reasoning: Both \(a^2\) and 2a have 'a' in common. The value of \(a^2\) is obtained by multiplying 'a' with 'a', while the value of 2a is obtained by multiplying 2 with 'a'. While this may seem trivial at first, we are given that a > 2, hence when we multiply 'a' with 'a', we are multiplying a value greater than 2 with 'a', on the other hand, in obtaining '2a' we are multiplying a value equal to 2 with a. Hence \(a^2\) will always be greater than 2a and the difference between \(a^2\) and 2a, i.e. \(a^2 - 2a\) will be always positive.
Thus, \(a^2 − 2a + b\) = positive + b = positive
The statement is sufficient, and we can eliminate B, C, and E.
Statement 2
(2) b > 1
The expression \(a^2 − 2a\) will have its minimum value = -1, when a = 1. Hence if b > 1, the sum of \((a^2 − 2a)\) and b will always be greater than 0.
In other words, \(a^2 − 2a + b > 0\)
This statement is also sufficient to answer the question.
Option D
