3
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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 7

5
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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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6
  • 1
    $\begingroup$ Let others at least think....:P too fast :) $\endgroup$ May 6, 2015 at 13:39
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    $\begingroup$ @user2408578 Ahah fast and furious! $\endgroup$
    – leoll2
    May 6, 2015 at 13:40
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    $\begingroup$ Perhaps a more (or at least another) intuitive way of phrasing it would be the simple substitution of A + (B - A) - B? $\endgroup$
    – Bailey M
    May 6, 2015 at 13:41
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    $\begingroup$ 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. $\endgroup$
    – Nejc
    May 6, 2015 at 13:48
  • $\begingroup$ @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! $\endgroup$
    – leoll2
    May 6, 2015 at 15:37
4
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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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2
  • $\begingroup$ Well, but this doesn't explain why this solution is true. $\endgroup$
    – leoll2
    May 6, 2015 at 16:40
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    $\begingroup$ Is this a Troll? $\endgroup$
    – Daedric
    May 6, 2015 at 17:31
1
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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
  • 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 M
    May 6, 2015 at 15:30
1
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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 $-C+C=0$

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1
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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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0
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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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3
  • $\begingroup$ 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! $\endgroup$
    – F1Krazy
    Jul 21, 2017 at 8:23
  • $\begingroup$ Welcome to Puzzling! (Take the Tour!) How does your answer add to the identical ones already given? You should always look at existing answers before providing one of your own, to ensure you are not just adding a duplicate. $\endgroup$
    – Rubio
    Jul 23, 2017 at 11:01
  • $\begingroup$ Is there any problem with my answer? Or is it duplicate? $\endgroup$ Jul 24, 2017 at 14:38
0
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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

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