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

`y=(56)/(3)`

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`

`(4)/(6)=(k\cdot 6)/(6)`

`(2)/(3)=k`

`y=(2)/(3)\cdot x\cdot z`

Now, we will solve for y by substituting
`k=(2)/(3)`,
`x=7` and
`z=4`

`y=(2)/(3)\cdot 7\cdot 4`

`y=(56)/(3)`

Therefore,
`y=(56)/(3)`.

Answer 2

The value of y is
`(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* 6* 1`

`\Rightarrow 4=6k`

`\Rightarrow k=(4)/(6)`

`\Rightarrow k=(2)/(3)`

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

`\Rightarrow y=(2)/(3)* 7* 4`

`\Rightarrow y=(56)/(3)`

Therefore, the value of y is
`(56)/(3)` when x=7 and z=4.