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If (a-b)(b-c)/a-c
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08 Apr 2023, 10:17
08 Apr 2023, 10:17
Expert Reply
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Question Stats:
56% (01:37) correct
43% (02:10) wrong based on 44 sessions
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If \(\frac{(a-b)(b-c)}{(a-c)}= 1\) and b=1, then which of the following cannot be the value of a?
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If (a-b)(b-c)/a-c
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27 May 2023, 03:29
27 May 2023, 03:29
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Expert Reply
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OE
Given \(\frac{(a-b)(b-c)}{(a-c)}= 1\) (equation 1)
Then \((a-b)(b-c)= (a-c)\)
c(2-a)=1 (equation 2)
Since, the answer choices represent the values of a so let us put them in the eqn.2
and check which of them cannot be the value of a.
A. 0
If a = 0, the eqn.2 reduces to
c(2 − 0) = 1
i.e. c =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If a = 1, the eqn.2 reduces to
c (2 − 1) = 1
i.e. c = 1
If we put a = 1 and c = 1 in eqn.1 then we are getting denominator 0 which is possible.’
Hence, a ≠ 1
Since this question has only one answer, so we do not need to check rest of
the answer choices.
Ans. (B)
a cant also be 2
Given \(\frac{(a-b)(b-c)}{(a-c)}= 1\) (equation 1)
Then \((a-b)(b-c)= (a-c)\)
c(2-a)=1 (equation 2)
Since, the answer choices represent the values of a so let us put them in the eqn.2
and check which of them cannot be the value of a.
A. 0
If a = 0, the eqn.2 reduces to
c(2 − 0) = 1
i.e. c =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If a = 1, the eqn.2 reduces to
c (2 − 1) = 1
i.e. c = 1
If we put a = 1 and c = 1 in eqn.1 then we are getting denominator 0 which is possible.’
Hence, a ≠ 1
Since this question has only one answer, so we do not need to check rest of
the answer choices.
Ans. (B)
a cant also be 2
Re: If (a-b)(b-c)/a-c
[#permalink]
21 Oct 2023, 17:35
21 Oct 2023, 17:35
1
This is a ridiculous question since a also cannot be 2. Even by your own logic c(2-a)=1 can never be true if a=2.
Carcass wrote:
OE
Given \(\frac{(a-b)(b-c)}{(a-c)}= 1\) (equation 1)
Then \((a-b)(b-c)= (a-c)\)
c(2-a)=1 (equation 2)
Since, the answer choices represent the values of a so let us put them in the eqn.2
and check which of them cannot be the value of a.
A. 0
If 𝑎 = 0, the eqn.2 reduces to
𝑐(2 − 0) = 1
i.e. 𝑐 =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If 𝑎 = 1, the eqn.2 reduces to
𝑐 (2 − 1) = 1
i.e. 𝑐 = 1
If we put 𝑎 = 1 and 𝑐 = 1 in eqn.1 then we are getting denominator 0 which is possible.’
Hence, 𝑎 ≠ 1
Since this question has only one answer, so we do not need to check rest of
the answer choices.
Ans. (B)
Given \(\frac{(a-b)(b-c)}{(a-c)}= 1\) (equation 1)
Then \((a-b)(b-c)= (a-c)\)
c(2-a)=1 (equation 2)
Since, the answer choices represent the values of a so let us put them in the eqn.2
and check which of them cannot be the value of a.
A. 0
If 𝑎 = 0, the eqn.2 reduces to
𝑐(2 − 0) = 1
i.e. 𝑐 =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If 𝑎 = 1, the eqn.2 reduces to
𝑐 (2 − 1) = 1
i.e. 𝑐 = 1
If we put 𝑎 = 1 and 𝑐 = 1 in eqn.1 then we are getting denominator 0 which is possible.’
Hence, 𝑎 ≠ 1
Since this question has only one answer, so we do not need to check rest of
the answer choices.
Ans. (B)
