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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 :thumbsup:
Devmitra Sen (Math)
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Re: If the relations shown hold for the operation ⊕ and the numbers m, n, [#permalink]
Expert Reply
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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Answer: Option E

Video solution by GMATinsight



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Re: If the relations shown hold for the operation and the numbers m, n, [#permalink]
Expert Reply
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 \(q ⊕ p =r \))

So, Answer will be E
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]
Expert Reply
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.
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Re: If the relations shown hold for the operation and the numbers m, n, [#permalink]
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