Question

If y varies directly as x and z and y=4 when x=6 and z=1 find y when x=7 and z=4

Answer 1

Answer-

The value of y is $\dfrac{56}{3}$ when x=7 and z=4.

Solution-

As given in the question, y is directly proportional to x, so

$y\propto x$   -----------------1

And also y is directly proportional to z, so

$y\propto z$   -----------------2

Combining equation 1 and 2,

$\Rightarrow y\propto x.z$

$\Rightarrow y=k.x.z$

Where,

k = proportionality constant

When x=6 and z=1, y=4. Putting theses values,

$\Rightarrow 4=k\times 6\times 1$

$\Rightarrow 4=6k$

$\Rightarrow k=\dfrac{4}{6}$

$\Rightarrow k=\dfrac{2}{3}$

Now, we have to find the value of y, when x=7 and z=4

$\Rightarrow y=\dfrac{2}{3}\times 7\times 4$

$\Rightarrow y=\dfrac{56}{3}$

Therefore, the value of y is $\dfrac{56}{3}$ when x=7 and z=4.

Answer 2

Answer:

$y=\frac{56}{3}$

Step-by-step explanation:

We have been given that y varies directly as x and z.

We know that equation of two directly proportional quantities is in form $y=kx$, where, k is constant of proportionality.

Since y varies directly as x and z, so our equation would be $y=k\cdot x\cdot z$

We are also told that $y=4$, $x=6$ and $z=1$. Let us find constant of proportionality as:

$4=k\cdot 6\cdot 1$

$4=k\cdot 6$

$\frac{4}{6}=\frac{k\cdot 6}{6}$

$\frac{2}{3}=k$

$y=\frac{2}{3}\cdot x\cdot z$

Now, we will solve for y by substituting $k=\frac{2}{3}$, $x=7$ and $z=4$

$y=\frac{2}{3}\cdot 7\cdot 4$

$y=\frac{56}{3}$

Therefore, $y=\frac{56}{3}$.