Simplify $$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$

[BTGmoderatorLU]

Simplify,

$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$

$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$

$$(B)12^{y+1}$$

$$(C)16^y+9^y$$

$$(D)12^y$$

$$(E)4^y*12^y$$

The OA is B.

Can any explain this PS question Please? I don't understand it. Thanks.

[DavidG@VeritasPrep]

Simplify,

$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$

$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$

$$(B)12^{y+1}$$

$$(C)16^y+9^y$$

$$(D)12^y$$

$$(E)4^y*12^y$$

The OA is B.

Can any explain this PS question Please? I don't understand it. Thanks.

4^y + 4^y + 4^y + 4^y = 4 * 4^y = 4^(y+1)
3^y + 3^y + 3^y = 3 * 3^y = 3^(y + 1)
4^(y+1) * 3^(y+1) = 12^(y+1)
The answer is B

[GMATGuruNY]

Simplify,

$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$

$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$

$$(B)12^{y+1}$$

$$(C)16^y+9^y$$

$$(D)12^y$$

$$(E)4^y*12^y$$

Let y=-1.
Plugging y=-1 into the given expression, we get:
(4¯¹ + 4¯¹ + 4¯¹ + 4¯¹)(3¯¹ + 3¯¹ + 3¯¹) = (1/4 + 1/4 + 1/4 + 1/4)(1/3 + 1/3 + 1/3) = (1)(1) = 1.

The target value is 1.
Now plug y=-1 into the answers to see which yields the target value of 1.
A: 4¯� + 3¯³ = 1/(4�) + 1/(3³) ≠ 1.
B: 12� = 1.
C: 16¯¹ + 9¯¹ = (1/16) + (1/9) ≠ 1.
D: 12¯¹ = 1/12 ≠ 1.
E: 4¯¹ * 12¯¹ = (1/4)(1/12) ≠ 1.

Only B works.

The correct answer is B.

[EconomistGMATTutor]

Simplify,

$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$

$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$

$$(B)12^{y+1}$$

$$(C)16^y+9^y$$

$$(D)12^y$$

$$(E)4^y*12^y$$

Hi LUANDATO,
Let's take a look at your question.

$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$

Factor out 4^y and 3^y
$$=\ 4^y\left(1+1+1+1\right).3^y\left(1+1+1\right)$$
$$=\ 4^y\left(4\right).3^y\left(3\right)$$
$$=\ 4^{y+1}.3^{y+1}$$
$$=\ \left(4\times3\right)^{y+1}$$
$$=\ 12^{y+1}$$

Therefore, Option B is correct.
Hope this helps.
I am available if you'd like any follow up.

[Scott@TargetTestPrep]

Simplify,

$$(4^y+4^y+4^y+4^y)\cdot(3^y+3^y+3^y)$$

$$\left(A\right)\ 4^{4y}\cdot3^{3y}$$

$$(B)12^{y+1}$$

$$(C)16^y+9^y$$

$$(D)12^y$$

$$(E)4^y*12^y$$

Note that 4^y is common to the first factor, and 3^y is common to the second factor. Thus, we have:

(4^y + 4^y + 4^y + 4^y) x (3^y + 3^y + 3^y)

4^y(1 + 1 + 1 + 1) x 3^y(1 + 1 + 1)

4^y(4) x 3^y(3)

4^(y+1) x 3^(y+1)

12^(y+1)

Answer: B