Hi,
I was doing Test Prep 1 with the official software and I'm pretty stumpped with one question.
If the operation @ is defined for all integers a and b by a@b = a+b-ab, which of the following statements must be true for all integers a, b, and c?
I. a@b = b@a
II. a@0 = a
III. (a@b)@c = a@(b@c)
Answers:
I only
II only
I and II only
I and III only
I, II, and III (CORRECT ANSWER)
Thanks for any help you can give me!
V.
GMAT Prep Test - Strange Operator Question
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Hey V_schame,
This is a function question, like f(x) = 2x, only they made up a funny symbol to make it more confusing. All it means is that whatever two numbers go on either side of the @ symbol, you need to run them through that little program (add them together, and subtract the product).
I. a@b = a + b - ab
b@a = b + a - ba
Does: a + b - ab = b + a - ba? Indeed it does!
II. a@0= a + 0 - a*0 = a Hey, that one is true, too!
III. This is a complicated one. Let's start with just the left side, and save ourselves time by remembering we already know what a@b is:
(a@b)@c = (a + b - ab)@c
Now, remember that the (a + b - ab) will represent just the "a", in terms of the original function, while the c represents the "b" in the original function:
(a + b - ab)@c = (a + b - ab) + c - (a + b - ab) * c
Phew! That's the left side done! The right will be the same, only with the letters scrambled around:
(b@c) = b + c - bc
So now notice that (b + c - bc) represents the "b" part of the original equation, while the "a" is just the same old a:
a@(b + c - bc) = a + (b + c - bc) - a * (b + c - bc)
Okay, now we have both sides, so let's see if they're equal:
(a + b - ab) + c - (a + b - ab) * c = a + (b + c - bc) - a * (b + c - bc)
a + b - ab + c - ac - bc + abc = a + b + c - bc - ab - ac + abc
If you start doing some adding and subtracting, you'll see that the exact same terms are on both sides. They're equal!
That was a toughie!
-t
This is a function question, like f(x) = 2x, only they made up a funny symbol to make it more confusing. All it means is that whatever two numbers go on either side of the @ symbol, you need to run them through that little program (add them together, and subtract the product).
I. a@b = a + b - ab
b@a = b + a - ba
Does: a + b - ab = b + a - ba? Indeed it does!
II. a@0= a + 0 - a*0 = a Hey, that one is true, too!
III. This is a complicated one. Let's start with just the left side, and save ourselves time by remembering we already know what a@b is:
(a@b)@c = (a + b - ab)@c
Now, remember that the (a + b - ab) will represent just the "a", in terms of the original function, while the c represents the "b" in the original function:
(a + b - ab)@c = (a + b - ab) + c - (a + b - ab) * c
Phew! That's the left side done! The right will be the same, only with the letters scrambled around:
(b@c) = b + c - bc
So now notice that (b + c - bc) represents the "b" part of the original equation, while the "a" is just the same old a:
a@(b + c - bc) = a + (b + c - bc) - a * (b + c - bc)
Okay, now we have both sides, so let's see if they're equal:
(a + b - ab) + c - (a + b - ab) * c = a + (b + c - bc) - a * (b + c - bc)
a + b - ab + c - ac - bc + abc = a + b + c - bc - ab - ac + abc
If you start doing some adding and subtracting, you'll see that the exact same terms are on both sides. They're equal!
That was a toughie!
-t
Tommy Wallach, Company Expert
ManhattanGMAT
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a@b = a + b - ab.v_schame wrote:Hi,
I was doing Test Prep 1 with the official software and I'm pretty stumpped with one question.
If the operation @ is defined for all integers a and b by a@b = a+b-ab, which of the following statements must be true for all integers a, b, and c?
I. a@b = b@a
II. a@0 = a
III. (a@b) @ c = a @ (b@c)
Answers:
I only
II only
I and II only
I and III only
I, II, and III
In other words, a@b = SUM - PRODUCT.
Statement I is included in four of the five answer choices.
Thus, it is almost certain that statement I must be true.
Otherwise, a test-taker will be able to eliminate four answer choices simply by evaluating statement I.
To save time, start with statement II.
To make the process easier, plug in values.
Let a=2, b=3, and c=10.
Statement II: a@0 = a
2@0 = 2
2+0 - (2*0) = 2
2 = 2.
On the left side, we simply added and subtracted 0 from the value of a=2.
From this example, we can see that statement II will be true for any integer value of a.
Eliminate A and D, which do not include statement II.
Statement III: (a@b) @ c = a @ (b@c)
First calculate the values INSIDE THE PARENTHESES.
(2@3) @ 10 = 2 @ (3@10)
(2+3 - 2*3) @ 10 = 2 @ (3+10 - 3*10)
-1 @ 10 = 2 @ -17
-1 + 10 - (-1*10) = 2 + (-17) - (2)(-17)
19 = 19.
When a=2, b=3 and c=10, statement III is true.
While not a definitive proof, it seems VERY unlikely that these 3 randomly selected values would prove to be an exceptional case.
Eliminate B and C, which do not include statement III.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Hello Mitch,
Ran into this questions just yesterday on a practice test and got it wrong. I also went for the plug in approach but then I got a little nervous because I thought I might need to test other values like negative integers in order to prove it.
I notice you went for numbers other than 0 or 1 that tend to skew answers when plugging in on certain cases.
Is there any other information that made you think that plugging negative integers was not necessary?
Thanks!
Ran into this questions just yesterday on a practice test and got it wrong. I also went for the plug in approach but then I got a little nervous because I thought I might need to test other values like negative integers in order to prove it.
I notice you went for numbers other than 0 or 1 that tend to skew answers when plugging in on certain cases.
Is there any other information that made you think that plugging negative integers was not necessary?
Thanks!
GMATGuruNY wrote:a@b = a + b - ab.v_schame wrote:Hi,
I was doing Test Prep 1 with the official software and I'm pretty stumpped with one question.
If the operation @ is defined for all integers a and b by a@b = a+b-ab, which of the following statements must be true for all integers a, b, and c?
I. a@b = b@a
II. a@0 = a
III. (a@b) @ c = a @ (b@c)
Answers:
I only
II only
I and II only
I and III only
I, II, and III
In other words, a@b = SUM - PRODUCT.
Statement I is included in four of the five answer choices.
Thus, it is almost certain that statement I must be true.
Otherwise, a test-taker will be able to eliminate four answer choices simply by evaluating statement I.
To save time, start with statement II.
To make the process easier, plug in values.
Let a=2, b=3, and c=10.
Statement II: a@0 = a
2@0 = 2
2+0 - (2*0) = 2
2 = 2.
On the left side, we simply added and subtracted 0 from the value of a=2.
From this example, we can see that statement II will be true for any integer value of a.
Eliminate A and D, which do not include statement II.
Statement III: (a@b) @ c = a @ (b@c)
First calculate the values INSIDE THE PARENTHESES.
(2@3) @ 10 = 2 @ (3@10)
(2+3 - 2*3) @ 10 = 2 @ (3+10 - 3*10)
-1 @ 10 = 2 @ -17
-1 + 10 - (-1*10) = 2 + (-17) - (2)(-17)
19 = 19.
When a=2, b=3 and c=10, statement III is true.
While not a definitive proof, it seems VERY unlikely that these 3 randomly selected values would prove to be an exceptional case.
Eliminate B and C, which do not include statement III.
The correct answer is E.