A variable y is partly constant and partly varies directly as \sqrt{x}. Given that y = 26 … — JAMB UTME

A variable $y$ is partly constant and partly varies directly as $\sqrt{x}$ . Given that $y = 26$ when $x = 16$ , and $y = 38$ when $x = 49$ , what is the value of $y$ when $x = 64$ ?

Answer: 42

Answer

The value of

y

when

x = 64

42

The relationship is given by the partial variation equation

y = a + b\sqrt{x}

. Substituting the pairs

(16, 26)

and

(49, 38)

yields the linear system

a + 4b = 26

and

a + 7b = 38

. Solving this system gives the constants

a = 10

and

b = 4

. Substituting

x = 64

into

y = 10 + 4\sqrt{64}

results in

y = 10 + 4(8) = 42

Step-by-Step Solution

Formulate the partial variation equation.

y = a + b\sqrt{x}

, where

a

and

b

are constants of variation.

Partial variation implies

y

is the sum of a constant term

a

and a term directly proportional to

\sqrt{x}

Set up simultaneous equations using the given pairs of

(x, y)

Equation 1:

a + 4b = 26

Equation 2:

a + 7b = 38

Evaluating

\sqrt{16} = 4

and

\sqrt{49} = 7

simplifies the relationship into two linear equations in two unknowns.

Solve for constants

a

and

b

b = 4

and

a = 10

Subtracting Equation 1 from Equation 2 yields

3b = 12 \implies b = 4

, and substituting

b = 4

back into Equation 1 gives

a = 10

Calculate

y

when

x = 64

y = 10 + 4(8) = 42

Using the specific formula

y = 10 + 4\sqrt{x}

for

x = 64

gives

y = 10 + 32 = 42

Key Concept

Partial Variation with Simultaneous Equations

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