What will come at the "?"
Follow the pattern.
A + B = E
B - A = C
A + D = Q
D - B = L
then
A + C - B = ?
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7 Answers
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6
The answer is
0
Because
we know that
$B-A=C$
so
$A-B=(-C)$
$A+C-B=A-B+C=(-C)+C=0$
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1$\begingroup$ Let others at least think....:P too fast :) $\endgroup$ Commented May 6, 2015 at 13:39
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1$\begingroup$ @user2408578 Ahah fast and furious! $\endgroup$– leoll2Commented May 6, 2015 at 13:40
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1$\begingroup$ Perhaps a more (or at least another) intuitive way of phrasing it would be the simple substitution of A + (B - A) - B? $\endgroup$– Bailey MCommented May 6, 2015 at 13:41
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3$\begingroup$ This is true, if you assume A, B, C are numbers, and +,- are addition and subtraction of numbers. You are assuming commutative and distributive property of objects. $\endgroup$– NejcCommented May 6, 2015 at 13:48
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$\begingroup$ @Nejc The author didn't mention about operators not being used the "standard" way, so I assume that plus means addition. Did he also mention that each letter corresponds to one number? No! $\endgroup$– leoll2Commented May 6, 2015 at 15:37
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2
Similar to Leoll2's answer. It can also be:
$\int\limits_{-\infty}^{+\infty}\frac{(\sqrt{-0} \times cosh(f(x)))^5}{π^e}dx$
because of:
using [B - A = C], the answer becomes [A + (B - A) - B] which all cancels out to nothing
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$\begingroup$ Well, but this doesn't explain why this solution is true. $\endgroup$– leoll2Commented May 6, 2015 at 16:40
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2
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1
I came up with
E - C
Flip the equations so you get A, B, and C by themselves. Of course these are other possibilities, but these are the ones that lead to the shortest answer.
A = E - B
B = C + A
C = B - A
Expand original question using newly found values and eliminate variables that cancel each other.
(E - B) + (B - A) - (C + A) = ?
E-B+B-A-C+A = ?
E-C = ?
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1$\begingroup$ You need to apply the minus to both C and A in your third term, which would make your second term 'E-B+B-A-C-A' and your third term 'E-A-C-A'. A much different answer. $\endgroup$– Bailey MCommented May 6, 2015 at 15:30
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The answer is:
0
Because
We know that
$B-A=C$
so
$B-C=A$
Substituting $A$ in $A + C-B = ?$
We have
$B-C+C-B=0$
Because $B-B = 0$ and $-C+C=0$
answered May 6, 2015 at 17:00
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Ok I know this isn't the right answer, but I got something different that still works with a pattern. My answer is
F, or 6
The "rules" for the calculation is
Turn the letters into their corresponding numbers (A=1, B=2), and square them if they're on the left side of the equation.
That is, the first equations becomes
1(A) + 4(B, or 2^2) = 5
Where
5 corresponds to the 5th letter, E.
Do this for the remaining equations and the pattern holds up:
4(B) - 1(A) = 3(C)
1(A) + 16(D) = 17(Q)
16(D) - 4(B) = 12(L)
Thus, the final equation becomes
1(A) + 9(C) - 4(B) = 6(F)
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3
The answer is
0.
Because
Answer 1
1. A+C-B Asked.
2. C=B-A Given.
3. A+B-A-B=0 After putting Value in 1.
Answer 2
1. A+C-B Asked.
2. C=B-A Given.
3. B=C+A From 2nd.
4. A+C-C-A=0 After putting Value in 1.
Answer 3
1. A+C-B Asked.
2. A=Q-D From 3rd.
3. B=D-L From 4th.
4. C=B-A From 2nd.
5. Q-D+C-(D-L).
6. Q-D+B-A-D+L.
7. Q-D+D-L-(Q-D)-D+L.
8. Q-D+D-L-Q+D-D+L=0.
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$\begingroup$ Hi! Welcome to PSE! Would you mind putting your answer in spoiler tags, to avoid ruining the solution for anyone who wants to try the puzzle for themselves? Thanks! $\endgroup$– F1KrazyCommented Jul 21, 2017 at 8:23
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$\begingroup$ Is there any problem with my answer? Or is it duplicate? $\endgroup$ Commented Jul 24, 2017 at 14:38
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The answer is
0
because
B - A = C
and
-A = C - B
and
0 = C - B + A
and
0 = A + C - A
how do i multiline spoiler, plz help. i had to use "and" to separate
answered Jul 24, 2017 at 17:51