Lesson: Strange Operators

Comment on Strange Operators

If xθy=(x+1)/(y−1) for all y≠1, and aθb=1, then a=

A. b−2
B. b−1
C. b
D. b+1
E. b+2

I did (a+1)/(b-1) = 1
I don't understand what should I do to take out a
Thanks in advance
gmat-admin's picture

Let's take it from (a+1)/(b-1) = 1
Multiply both sides of the equation by (b-1) to get: (a+1) = 1(b-1)
Expand right side: a + 1 = b - 1
Subtract 1 from both sides: a = b - 2

Answer: A

Hi Brent, for the below question I am getting the different answer zero, here's why:
(√5 + √5)² - (√5 - √5)² = using the property a² - b² = (a+b)(a-b) I got, (√5 + √5) (√5 - √5) = (2√5)(0) = 0


If x ¤ y = (x + y)² - (x - y)². Then √5 ¤ √5 =

A. 0
B. 5
C. 10
D. 15
E. 20
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-x-y-x-y-x-y-then-218476.html

You made a small mistake in your solution.

You're using the property that says a² - b² = (a + b)(a - b)

So, in the expression, (√5 + √5)² - (√5 - √5)², what is a, and what is b?
Well, a = (√5 + √5) and b = (√5 - √5)

This means (a + b)(a - b) = [(√5 + √5) + (√5 - √5)][(√5 + √5) - (√5 - √5)]
= [(√5 + √5) + 0][(√5 + √5) - 0]
= (√5 + √5)(√5 + √5)
= (2√5)(2√5)
= 4√25
= 20

I have two different solutions posted here: https://gmatclub.com/forum/if-x-y-x-y-x-y-then-218476.html#p1727422

Does that help?

Cheers,
Brent

Hi Brent,

I was referring to your solution : https://gmatclub.com/forum/for-all-positive-integers-m-3m-when-m-is-odd-125937-20.html#p2048884

Am I right to infer that 81 = [27] or 81 = [162] and since [162] is not an option that is provided we may mark [27] as correct?

Thanks & Regards,
Abhirup
gmat-admin's picture

Solution link: https://gmatclub.com/forum/for-all-positive-integers-m-3m-when-m-is-odd-...

Great point!
[162] would also be correct.

Cheers,
Brent

Thanks Brent!!

Hi Brent,

Could you please explain this question a bit more (https://gmatclub.com/forum/for-any-real-number-x-the-operator-is-defined-as-x-x-1-x-213328.html). I'm not sure how the connection between the & equation with x and & equation with p were made. Thank you.
gmat-admin's picture

Hi Swatato,

Here's the solution you're referring to: https://gmatclub.com/forum/for-any-real-number-x-the-operator-is-defined...

I spent the first part of the solution, showing that &(p + 1) = (p + 1)(-p) [upon simplification]
ASIDE: This step is crucial, because once we know that &(p + 1) and (p + 1)(-p) are EQUAL, we can later replace &(p + 1) with (p + 1)(-p).

Then I dealt with the given information that says: p + 1 = &(p + 1)

So, I took the part on the right side of the equation, &(p + 1), and replaced it with its equivalent value of (p + 1)(-p)

This results in the equation p + 1 = (p + 1)(-p), which we can now solve for p.

Does that help?

Cheers,
Brent

Hi Brent,

For question https://gmatclub.com/forum/an-operation-is-defined-by-the-equation-x-y-x-246192.html
Can we not factor the equation: x²/4 - xy - y² as (x/2 - y)²? as it will lead to the same solution when you solve the parenthesis.
Why did you first common out (1/4) and then solved the equation?
Thanks in advance!
gmat-admin's picture

Question link: https://gmatclub.com/forum/an-operation-is-defined-by-the-equation-x-y-x...

You're correct to say that x²/4 - xy - y² = (x/2 - y)².
However, I think it's safe to say that most students will not automatically see that factorization.
So, as an intermediate step, I factored out the 1/4 so that the resulting quadratic is easier to recognize as a special product.

Cheers,
Brent

Hi Brent,

question https://gmatclub.com/forum/if-denotes-the-least-integer-greater-than-or-equal-to-x-132223.html

In your solution you have
"REPHRASED target question: Is -1 ≤ x < 0". Shouldn't it be -1 < x < 0. My understanding is that if x = -1 than [x]=-1. And if we allow x to be greater or equal to -1 and less than 0, then possible answers for [x] could be -1 or 0, hence we have to restrict it to only greater than -1 and less than 0. Am I missing something here?

Thanks.
gmat-admin's picture

Good catch!

You're partially right about how I should have rephrased the target question.

Notice that:
[2.1] = 3, since 3 is the smallest integer that's greater than or equal to 2.1.
[0] = 0, since 0 is the smallest integer that's greater than OR EQUAL TO 0
[1] = 1, since 1 is the smallest integer that's greater than or equal to 1.

The target question asks whether [x] = 0
This will occur if -1 < x ≤ 0

So, the REPHRASED target question should be "Is -1 < x ≤ 0"

I've edited my answer here: https://gmatclub.com/forum/if-denotes-the-least-integer-greater-than-or-...

Thanks again for the heads up!

Cheers,
Brent

Brent, hi.
Could you please explain one more time.
For any real number x, the operator & is defined as:

&(x) = x(1 − x)
If p + 1 = &(p + 1), then p = ?
How did we come to: &(p+1) = (p+1)[1 − (p+1)] = (p + 1)(-p) ?
gmat-admin's picture

GIVEN: &(x) = x(1 − x)

So, the "&" symbol next to a number tells us to take that number and multiply it by that number subtracted from 1.

In other words, &number = number(1 - number)
So, &7 = 7(1 - 7)
And &88.3 = 88.3(1 - 88.3)
And &w = w(1 - w)
And &2k = 2k(1 - 2k)
And &j² = j²(1 - j²)
And &(p+1) = (p+1)[1 − (p+1)]

In all cases, we're taking the number (be it 7 or 88.3 or w or 2k or j²) and multiplying it by that number subtracted from 1.

Does that help?

Cheers.
Brent

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