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.
a,b and c are distinct numbers such that abc≠0
[#permalink]
Updated on: 12 Jul 2020, 11:05
Updated on: 12 Jul 2020, 11:05
00:00
Question Stats:
47% (01:16) correct
52% (00:44) wrong based on 38 sessions
Hide Show timer Statistics
a,b and c are distinct numbers such that abc≠0
My channel URL: https://www.youtube.com/channel/UCvdY0k ... EOsT5PMFRQ
Quantity A |
Quantity B |
a^2 + b^2 + c^2 - ab -bc - ca |
0 |
My channel URL: https://www.youtube.com/channel/UCvdY0k ... EOsT5PMFRQ
Re: abc≠0
[#permalink]
12 Jul 2020, 10:53
12 Jul 2020, 10:53
1
I might be wrong, but it seems like the answer imbedded in the question is wrong based on the specified assumptions
a^2 + b^2 + c^2 - ab -bc - ca
First we can combine a^2 and b^2 together using the square of difference (a-b)^2 = a^2 -2ab +b^2
Thus a^2 + b^2 - ab part of the equation can be expressed as (a-b)^2 + ab (the plus ab term is needed to make -2ab equal to -ab)
Then we are left with: (a-b)^2 + ab -bc - ca + c^2
This can be rewritten as: (a-b)^2 + b(a -c) + c(c-a)
Rewrite c(c-a) as -c(a-c) ; yielding (a-b)^2 + b(a -c) - c(a-c)
Since the second and third term of the equation contain a common element (a-c) we can further rewrite as:
(a-b)^2 + (b-c)(a -c)
Now, the (a-b)^2 is always positive regardless of values of a or b (because squared)
(b-c) and (a-c) terms can also be positive when a and b are larger than c
However, nowhere in the problem is it indicated that a, b, and c are distinct numbers, or that they cannot equal to each other.
Thus one of the solutions to the equation is the scenario where a=b=c
Under these conditions the entire polynomial will equal to zero
Thus without additional specifications, the answer should be D (not enough info provided)
a^2 + b^2 + c^2 - ab -bc - ca
First we can combine a^2 and b^2 together using the square of difference (a-b)^2 = a^2 -2ab +b^2
Thus a^2 + b^2 - ab part of the equation can be expressed as (a-b)^2 + ab (the plus ab term is needed to make -2ab equal to -ab)
Then we are left with: (a-b)^2 + ab -bc - ca + c^2
This can be rewritten as: (a-b)^2 + b(a -c) + c(c-a)
Rewrite c(c-a) as -c(a-c) ; yielding (a-b)^2 + b(a -c) - c(a-c)
Since the second and third term of the equation contain a common element (a-c) we can further rewrite as:
(a-b)^2 + (b-c)(a -c)
Now, the (a-b)^2 is always positive regardless of values of a or b (because squared)
(b-c) and (a-c) terms can also be positive when a and b are larger than c
However, nowhere in the problem is it indicated that a, b, and c are distinct numbers, or that they cannot equal to each other.
Thus one of the solutions to the equation is the scenario where a=b=c
Under these conditions the entire polynomial will equal to zero
Thus without additional specifications, the answer should be D (not enough info provided)
Re: a,b and c are distinct numbers such that abc≠0
[#permalink]
12 Jul 2020, 16:32
12 Jul 2020, 16:32
1
The answer is D.
a^2 + b^2 + c^2 - ab -bc - ca = (a+b+c)^2- 3(ab+bc+ca)
Put, a= 0.1, b= 0.2, c=0.3, then this becomes negative. So, Quantity B is greater
Put, a= 1, b= 5, c= 6, then this becomes positive. So, Quantity A is greater
Finally, answer is D
[Do not advertise yourself with wrong questions and answers please]
a^2 + b^2 + c^2 - ab -bc - ca = (a+b+c)^2- 3(ab+bc+ca)
Put, a= 0.1, b= 0.2, c=0.3, then this becomes negative. So, Quantity B is greater
Put, a= 1, b= 5, c= 6, then this becomes positive. So, Quantity A is greater
Finally, answer is D
[Do not advertise yourself with wrong questions and answers please]
