# Solve this simple direct and inverse variation

**Admin**Administrator- Posts : 180

Solve this simple maths. Just direct and inverse variation.

P varies directly as Q^2 and inversely as root R. When Q=8 and R=25,P=16. Find P when Q=5 and R=9.

P varies directly as Q^2 and inversely as root R. When Q=8 and R=25,P=16. Find P when Q=5 and R=9.

MathsBoard Forum Admin

You can avoid reality, but you cannot avoid the consequences of avoiding reality.

**Lillian**Group Moderator- Posts : 48

P ¤

√R

When we apply K in the above would have

P=

√R

16=

√25

16=

5

when you cross multiply we would have

16 x 5 = k x 64

80=64k

64 = 64

k=1.25

In order to find P, we will still use the formula above in step one.

P=

√9

P=

√9

P=

3

__Q__^{2}√R

**STEP 1: Applying the constant K in the variation formula above**When we apply K in the above would have

P=

__K x Q__^{2}√R

**STEP 2: Substituting with the given values**16=

__K x 8__^{2}√25

16=

__K x 64__5

**STEP 3: Cross multiplying the value**when you cross multiply we would have

16 x 5 = k x 64

80=64k

**STEP 4: Dividing both sides by 64**__80__=__64k__64 = 64

k=1.25

**STEP 5: Finding P in the given question**In order to find P, we will still use the formula above in step one.

P=

__k x 5__^{2}√9

P=

__1.25 x 5__^{2}√9

P=

__1.25 x 25__3

**P=10.417 answer****Admin**Administrator- Posts : 180

@Lillian you really did wel well

MathsBoard Forum Admin

You can avoid reality, but you cannot avoid the consequences of avoiding reality.

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