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If the relations shown hold for the operation ⊕ and the numbers m, n,

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Re: If the relations shown hold for the operation ⊕ and the numbers m, n, [#permalink]
gmatt1476 wrote:
$$m ⊕ p =n$$

$$n ⊕ r =m$$

$$n ⊕ q =q$$

$$p ⊕ q =p$$

$$q ⊕ p =r$$

If the relations shown hold for the operation ⊕ and the numbers m, n, p, q, and r, then $$[(m⊕p)⊕ q]⊕ p =$$

(A) m
(B) n
(C) p
(D) q
(E) r

PS45430.02

Just plug in the values from the given relations above
[(m⊕p)⊕q]⊕p
=>[(n)⊕q]⊕p
=> [q]⊕p
=> r
Option E
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If the relations shown hold for the operation ⊕ and the numbers m, n, [#permalink]
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SOLUTION :

A NEW OG 2021 QUESTION

(m ⊕p)⊕q]⊕p

=(n⊕ q]⊕p (Using, m ⊕ p=n)

=q ⊕p (Using, n ⊕q=q)

=r (Given, q ⊕p=r)

(OPTION E)

Hope this helps
Devmitra Sen (Math)
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Re: If the relations shown hold for the operation ⊕ and the numbers m, n, [#permalink]
gmatt1476 wrote:
$$m ⊕ p =n$$

$$n ⊕ r =m$$

$$n ⊕ q =q$$

$$p ⊕ q =p$$

$$q ⊕ p =r$$

If the relations shown hold for the operation ⊕ and the numbers m, n, p, q, and r, then $$[(m⊕p)⊕ q]⊕ p =$$

(A) m
(B) n
(C) p
(D) q
(E) r

PS45430.02

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Re: If the relations shown hold for the operation and the numbers m, n, [#permalink]
Top Contributor
Given that $$m ⊕ p =n$$ , $$n ⊕ r =m$$, $$n ⊕ q =q$$, $$p ⊕ q =p$$, $$q ⊕ p =r$$ and we need to find the value of $$[(m⊕p)⊕ q]⊕ p =$$

$$[(m⊕p)⊕ q]⊕ p$$ = $$[n⊕ q]⊕ p$$ (As $$m ⊕ p =n$$)

= $$q ⊕ p$$ (As $$n ⊕ q =q$$)

= r (As $$q ⊕ p =r$$)

Hope it helps!

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Re: If the relations shown hold for the operation and the numbers m, n, [#permalink]
Hi,
My answer was "p", because I got as far as m+p=n so
[(m+p) + q] + p = [n+q] + p.

Then I tried to see if I can express "n" or "q" with "p," so I looked over at the relations. I noticed n+q=q. Instead realizing that I should be using this equation to make (q + p), I thought, that this means n=0. Then I noticed that p+q=p. So q=0, too.

Therefore, (n+q) + p = 0 + p.

in addition, if q=0 then q+p=r means that r=p. I pretty sure I'm wrong but I don't see how and it gives me a headache. Can someone please explain why is this wrong?
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Re: If the relations shown hold for the operation and the numbers m, n, [#permalink]
Annapanna wrote:
$$m ⊕ p =n$$

$$n ⊕ r =m$$

$$n ⊕ q =q$$

$$p ⊕ q =p$$

$$q ⊕ p =r$$

If the relations shown hold for the operation ⊕ and the numbers m, n, p, q, and r, then $$[(m⊕p)⊕ q]⊕ p =$$

(A) m
(B) n
(C) p
(D) q
(E) r

Hi,
My answer was "p", because I got as far as m+p=n so
[(m+p) + q] + p = [n+q] + p.

Then I tried to see if I can express "n" or "q" with "p," so I looked over at the relations. I noticed n+q=q. Instead realizing that I should be using this equation to make (q + p), I thought, that this means n=0. Then I noticed that p+q=p. So q=0, too.

Therefore, (n+q) + p = 0 + p.

in addition, if q=0 then q+p=r means that r=p. I pretty sure I'm wrong but I don't see how and it gives me a headache. Can someone please explain why is this wrong?

The problem with your solution is that you mistake the ⊕ sign for addition. It's not; it's a function. Now, taking this into consideration, reread the question and review the solutions above. Hope it helps.
Re: If the relations shown hold for the operation and the numbers m, n, [#permalink]
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