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.
If (a-b)(b-c)/a-c
[#permalink]
08 Apr 2023, 10:17
08 Apr 2023, 10:17
Expert Reply
3
Bookmarks
00:00
Question Stats:
56% (01:34) correct
43% (01:50) wrong based on 23 sessions
Hide Show timer Statistics
If \(\frac{(a-b)(b-c)}{(a-c)}= 1\) and b=1, then which of the following cannot be the value of a?
0
1
2
3
4
0
1
2
3
4
Re: If (a-b)(b-c)/a-c
[#permalink]
27 May 2023, 03:29
27 May 2023, 03:29
Expert Reply
1
Bookmarks
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 ๐ = 0, the eqn.2 reduces to
๐(2 โ 0) = 1
i.e. ๐ =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If ๐ = 1, the eqn.2 reduces to
๐ (2 โ 1) = 1
i.e. ๐ = 1
If we put ๐ = 1 and ๐ = 1 in eqn.1 then we are getting denominator 0 which is possible.โ
Hence, ๐ โ 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 ๐ = 0, the eqn.2 reduces to
๐(2 โ 0) = 1
i.e. ๐ =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If ๐ = 1, the eqn.2 reduces to
๐ (2 โ 1) = 1
i.e. ๐ = 1
If we put ๐ = 1 and ๐ = 1 in eqn.1 then we are getting denominator 0 which is possible.โ
Hence, ๐ โ 1
Since this question has only one answer, so we do not need to check rest of
the answer choices.
Ans. (B)
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 ๐ = 0, the eqn.2 reduces to
๐(2 โ 0) = 1
i.e. ๐ =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If ๐ = 1, the eqn.2 reduces to
๐ (2 โ 1) = 1
i.e. ๐ = 1
If we put ๐ = 1 and ๐ = 1 in eqn.1 then we are getting denominator 0 which is possible.โ
Hence, ๐ โ 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 ๐ = 0, the eqn.2 reduces to
๐(2 โ 0) = 1
i.e. ๐ =1/2
which is satisfying eqn.1
Hence, this can be the value of a.
B. 1
If ๐ = 1, the eqn.2 reduces to
๐ (2 โ 1) = 1
i.e. ๐ = 1
If we put ๐ = 1 and ๐ = 1 in eqn.1 then we are getting denominator 0 which is possible.โ
Hence, ๐ โ 1
Since this question has only one answer, so we do not need to check rest of
the answer choices.
Ans. (B)
