How To Solve Direct Variation | See Shortcut

Variation is used to describe the relations which exist between two quantities.

When two quantities H and R are related, to the extent that

H₁/R₁ = H₂/R₂

Given that R₁ and R₂ are any two value of R and H₁ also H₁and H₂ are the corresponding value of H₁ therefore, H is said to vary directly as R. the symbol “X” is used to represent proportionality therefore in the above example, we can write that H ∞ R i.e H varies directly as R

If A ∞ B, then for A to be equal to B, there must be a quantity to multiply B so as to be equal to A.

Having known want direct variation is all about, let solve some problems

Example 1:

If y varies directly as the square of x, and y – 98 when x = 7, calculate y when x = 5

Solution

y ∞ x²

y = Kx² where K is constant

when y = 98, x =7

we substitute into the formula to obtain K

y = Kx²

98 = Kx(7)²

98 = K x 49

98/49 = 49K/49

K = 2

find y when x = 5

y = 2 x(5)²

y = 2 x 25

y = 100

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Example 2:

Given that P varies directly as square root of  q and p = 12 when q =3, find the value of P when q = 5

Solution

P ∞ q²

P – Kq²where K is constant

P = 12 when q = , substitute into the equation and solve for K first

12 = K x (3)²

12 = K x 9

12/9 = 9k/9

K = 12/9 = 4/3

Find P when q = 5

P = kq²

P = 4/3 x (5)²

P = 4/3 x 25

P = 100/3

Example 3:

If A varies directly as the cube root of B and if A = 3 when B = 64, find the formula connecting the variables hence fine B when A = 15/4

Solution

N/B: Cube root of B =

A ∞

When A = 3, B = 64

Substitute A = K  where K is constant

3 = K x

Cube root of 64 is a number that multiply itself 4 times to give 64 the number is 4

3 = K x 4

3/4 = 4K/4

K = ¾

Find B when A = 15/4

Let put it into the equation

A = 15/4 = ¾  x

15/4 = 3 /4

We cross multiply

3  x 4 = 15 x 4

12 /12 = 60/12

 = 5

Cube both sides

(B)^{1/3 x 3} = 5³

B = 5³ = 5x5x5

B = 125

Example 4:

If x -3 is directly proportional to the square of y and x =5 when y =2. Find x when y =6

Solution

Let interpret the statement first

x – 3 x y²

x – 3  = Ky²

Where K is constant

When x= 5, y = 2

Substitute into the formula to obtain K

5 – 3 = K (2)²

2 = K x 4

2/4 = 4K/4

K = 2/4 = ½

Therefore, K = 1/2

Find x when y = 6

x – 3 = ½ x (6)²

x – 3 = 6 x 6/2

x – 3 = 36/2

x -3 = 18

x = 18 + 3

x = 21

Example 5:

The time of a pendulum varies as the square roof its length. If the length of a pendulum which beats 15 seconds is 9cm. find

• The length that beats 80 seconds
• The time of a pendulum with length 36cm

Solution

Let time = T

Length = L

That means

T ∞

When T = 15 seconds, L = 9cm

T = K  where K is constant

Let substitute x into the formula

15 = K

15 = K x 3

15/3 = 3K/3

K = 15/3 = 5

• When T = 80 seconds, find L

80 = 5 x

80/5 = 5 /5

 = 16

Square both sides of the equation

(L)^1/2+2 = 16²

L = 256cm

• Find T when L = 36cm

T = 5 x

T = 5 x 6

T = 30 seconds

Therefore, the time is 30 seconds

As the name implies, direct variation shows a direct relationship between two quantities. This makes problems under it to be simple.

If you are able to relate quantities with respect to the sign of proportionality, you will be able to solve any problems.