Thread: [SOLVED] SAT math page 737 #16

1. [SOLVED] SAT math page 737 #16

If $\displaystyle y=\frac{5x^3}{z}$ what happens to the value of y when both x and z are doubled?

a) y is not changed
b) y is halved
c) y is doubled
d) y is tripled
e) y is multiplied by 4

when both the denominator and numerator are both doubled doesn't it both cancel out and therefore y remains the same?

2. Originally Posted by fabxx
If $\displaystyle y=\frac{5x^3}{z}$ what happens to the value of y when both x and z are doubled?

a) y is not changed
b) y is halved
c) y is doubled
d) y is tripled
e) y is multiplied by 4

when both the denominator and numerator are both doubled doesn't it both cancel out and therefore y remains the same?

replace x and z with 2x and 2z respectively. what happens to the y-value?

3. Originally Posted by Jhevon
replace x and z with 2x and 2z respectively. what happens to the y-value?
I got $\displaystyle \frac{10x^3}{2z}$
which is the same as $\displaystyle \frac{5x^3}{z}$ isn't it? But how come the correct answer is e) y is multiplied by 4?

4. Originally Posted by fabxx
I got $\displaystyle \frac{10x^3}{2z}$
which is the same as $\displaystyle \frac{5x^3}{z}$ isn't it? But how come the correct answer is e) y is multiplied by 4?
that is because $\displaystyle \frac {10x^3}{2z}$ is not what you're supposed to get. i said replace x with 2x and z with 2z. what you did was multiply by 2/2, which is 1, of course you got the same thing back

5. grr. I still don't get it. Doesn't the question say to double BOTH x and z?

6. Originally Posted by fabxx
grr. I still don't get it. Doesn't the question say to double BOTH x and z?
yes. that means you change x to 2x and z to 2z as i have been saying

thus $\displaystyle x^3 \to (2x)^3 = 8x^3$ and $\displaystyle z \to 2z$

when it is all said and done, we have $\displaystyle y_{\text{new}} = \frac {5 (8)x^3}{2z} = 4 \cdot \frac {5x^3}z = 4y$

thus y is multiplied by 4