### Confusing Math Puzzle problem! 90% will fail to attempt this logical brain teaser math puzzle. “The Emoji Puzzle” Most interesting and confusing logic puzzle! Find the correct answer

Whatsup guys! 🙂

Who love Emojis? What a stupid question 😈 of course we all do. Here’s Emoji day special challenge for you.

Welcome back to Genius Puzzle Series, Puzzle #14 & #15. Another interesting brain-teaser logical puzzle for you. Ready? Play with facebook emoji and find out the correct answer.

The Puzzle divided into two parts. The easy one, and the hard level.

Let’s go to the puzzle without wasting your time.

### The Emoji Puzzle: Genius Puzzle Series #14: Level Easy

### The Emoji Puzzle: Genius Puzzle Series #15: Level Hard

Post your answer in the comment section below. The answer will be updated soon, till now share this awesome & interesting Emoji puzzle with your Whatsapp and facebook group friends, let see how many of your friends can solve this confusing logical puzzle.

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**Answer Updates**

### The Emoji Puzzle Answer #14

**Correct answer is 17**

#### Solution for #14:

Let’s assume the puzzle icon or equations as follow:

1eq. ⇒ A + B × B = C

2eq. ⇒ D + D + B = 10

3eq. ⇒ E + E + A = 12

4eq. ⇒ C + A = 20 = 4(E)

5eq. ⇒ E + (D+D) × A = ??

#### In 4eq. =>

4E = 20, E = 20/4 = 5, **E = 5**

#### In 3eq. =>

E + E + A = 12, (put E value from 4eq.), We get **A = 2**

#### In 4eq. =>

C + A = 20, (put A value from 3eq.), We get **C = 18**

#### In 1eq. =>

A + B × B = C, (put A & C value from 3eq. and 4eq.), We get **B = 4**

#### In 2eq. =>

D + D + B = 10, (put B value from 1eq.), We get **D = 3**

#### Last OR 5eq. =>

E + (D+D) × A = ??

5 + (3+3) × 2 = 5 + 6 × 2 = **17 Ans.**

### The Emoji Puzzle Answer #15

**13 Answer**

Solution: Let’s assume the puzzle icon or equations as follow:

1eq. ⇒ A + B + C = 4D

2eq. ⇒ A + C = B = E

3eq. ⇒ A + E = 8 + C

4eq. ⇒ C + C + C = B

5eq. ⇒ A + C × E – D = ??

#### In 4eq. =>

⇒ 3C = B = E (B=E from 2eq.)

#### In 3eq. =>

⇒ A + E = 8 + C (We know E = 3C from 4th eq.)

⇒ A + 3C = 8 + C

⇒ A + 3C – C = 8

⇒ A + 2C = 8

#### In 2eq. =>

⇒ A + C = B (We know B = 3C from 4th eq.)

⇒ A + C = 3C

⇒ A – 2C = 0

Now Compare latest 2nd & 3rd eq. with each other, you will get the value of A and C.

⇒ A + 2C = 8 (from 3rd eq.) Vs

⇒ A – 2C = 0 (from 2rd eq.)

**A = 4 , C = 2**

#### In 2eq. =>

⇒ A + C = B = E

⇒ 4 + 2 = 6

**B = E = 6**

#### In 1eq. =>

⇒ A + B + C = 4D (put their values)

⇒ 4 + 6 + 2 = 4D

⇒ 12 = 4D

⇒ D = 12/4 = 3

⇒ **D = 3**

#### Last OR 5th Eq. =>

⇒ A + C × E – D = ?? (We know **A = 4 , C = 2, B = E = 6, D = 3**)

⇒ 4 + 2 × 6 – 3

⇒ 4 + 12 – 3

⇒ **13 Ans.**

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17 and 13

How 13