3
$\begingroup$

What will come at the "?"

Follow the pattern.

A + B = E

B - A = C

A + D = Q

D - B = L

then

A + C - B = ?

$\endgroup$
5
$\begingroup$

The answer is

0

Because

we know that
$B-A=C$
so
$A-B=(-C)$
$A+C-B=A-B+C=(-C)+C=0$

$\endgroup$
  • 1
    $\begingroup$ Let others at least think....:P too fast :) $\endgroup$ – user2408578 May 6 '15 at 13:39
  • 1
    $\begingroup$ @user2408578 Ahah fast and furious! $\endgroup$ – leoll2 May 6 '15 at 13:40
  • 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 M May 6 '15 at 13:41
  • 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. $\endgroup$ – Nejc May 6 '15 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 '15 at 15:37
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Well, but this doesn't explain why this solution is true. $\endgroup$ – leoll2 May 6 '15 at 16:40
  • 2
    $\begingroup$ Is this a Troll? $\endgroup$ – Daedric May 6 '15 at 17:31
1
$\begingroup$

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 = ?

$\endgroup$
  • 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 '15 at 15:30
1
$\begingroup$

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 $-C+C=0$

$\endgroup$
1
$\begingroup$

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)

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '17 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 '17 at 11:01
  • $\begingroup$ Is there any problem with my answer? Or is it duplicate? $\endgroup$ – Rohit-Pandey Jul 24 '17 at 14:38
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.