When the value of a quantity is proportional to two or more other quantities, we say that a joint variation has occurred.

The only quantity is thereby said to vary jointly as the two other quantities, for instance, the value of a cylinder V varies jointly as the height h and also directly as the square of the radius r of the base of the cylinder. This is written as

V x r^{2}h

Therefore, V = ñr^{2}h

Where ñ is a constant of proportionality

In solving problems which involve joint variations, the same process of finding the constant is followed as under direct and inverse variations

With this foundation we have laid down, let solve some problems

Contents

**Example 1:**

If p varies directly as the square of q and inversely as r. if p = 36, q=3 and r = 4, find p when q =5

And r = 2.

**Solution**

Let interpret the statement

P varies directly as the square of q and inversely as r

P varies directly as the square of a P x 9^{2}

P = kq^{2}

P varies inversely as r

P x 1/r when x is the sign of proportionality

P = K/r

So the statement becomes

P = kq^{2}/r

Which is the required equation

N/B: inverse = 1/a

E.g inverse of a = 1/a

If p = 36, q = 3 and r =4

Use the data given to solve for B (other constant)

Substitute the values into the equation

36 = R x 3^{2} /4

36×4 = 9x

144/4 – 9k/7

K = 16

Find p when q = 5 and r =2

Again, substitute the values into the formula

P2 = -16×5/2

P = 200

**Example 2:**

P varies directly as q^{2} and inversely as Tr. When q = 8 and r = 25, p = 16. Find p when q=5 and r=9

**Solution**

P varies directly as q^{2}

P x q^{2}

P varies inversely as Tr

P x 1/Tr

The relation becomes

P x q^{2}x 1/tr

Remove the sign of proportionality and introduce a constant K

P = Kq^{2} /Tr

When q =8, r = 25, p =16

Substitute into the formula

16 = Kx8^{2}/

16 = K x 64/5

Cross multiply

16 x 5 = 64K

80/64 = 64/64

K = 80/64 = 10/8 = 5/4

Therefore K = 5

When 9 =5, r = 9 find p

P = 5/4 x 3^{2} / = 5 x 25/3

P = 1.25 x 25/3 = 3+25/3 = 10

**Example 3:**

If P Q and R are related in such a way that p ∞ Q^{2}/R when p = 36, Q = 3

And R = 4, calculate Q when P = 200 and r =2

**Solution**

The equation relation given me is

P x Q^{2}/R

Remote the sign of proportional ∞ and introduce a constant K

P = KQ^{2} /R

When p = 36, 4=3 and r 4

36= K x 3^{2} /R

36 = 9K/1 4

Cross multiply and make K the subject

K = 36 x 4/9 = 16

Q when p = 2—and q=2

200 = 16xQ^{2}/2

Cross multiply

200 x 2 = 16Q^{2}

400/16 = 16Q^{2}/16

2= Q^{2}

Take square root of both sides

= Q^{2×12}

∞ = 5

**Read Also:**

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**Example 4:**

X, Y, Z are related quantities, such that X varies directly as Y and inversely as the square root of Z. when X = 300, Y = 75 and Z = 25, calculate the values of X when Y=468 and Z= 144

**Solution**

Let’s form the equation first X varies directly as Y

X x Y

X varies inversely as the square root of Z

X x 1/

The relation becomes

X x Y x 1/

X = Ky/ where K is the sign of proportionality when X = 300, Y = 75 and Z = 25 let substitute into the formula

300 = K x 75/

300 = 75K/5

Cross multiply

300 x 5 = 75K

1500/75 = 75K/75

K = 20

Therefore, K = 20

Find X when Y = 468 and X = 144

X = 20 x 468/

X = 9360/12 = 780

X = 780

**Example 5:**

M varies directly as a and inversely as the square of P if M = 3 when Y = 2 and P = 1, find M in time if n and P

Solution

M varies directly as n

M x N

M varies inversely as square of P

M x 1/P^{2}

M = Kn/P^{2} (where K is the sign of proportionality of M = 3, n = 2 and p =1

3/3 = 2K/2

Therefore, K = 3/2

Substitute K = 3/2 into the equation

M = 3/2 n

M = 3n/2p^{2}

**Conclusion**

As the name implies, joint variation is the combination of direct and inverse variations so when you encounter problems on this topic of variation.

Firstly establish the direct variation and then the inverse variation and then join them together with the introduction of a constant if you are able to do that, no problems will be an obstacle in joint variations.