If [tex]\( r \)[/tex] varies jointly as [tex]\( s \)[/tex] and [tex]\( t \)[/tex], and [tex]\( r = 8 \)[/tex] when [tex]\( s = 6 \)[/tex] and [tex]\( t = 4 \)[/tex], find [tex]\( s \)[/tex] when [tex]\( t = 16 \)[/tex] and [tex]\( r = 24 \)[/tex].



Answer :

To solve the problem, we need to follow the concept of joint variation, which means [tex]\( r \)[/tex] varies jointly as [tex]\( s \)[/tex] and [tex]\( t \)[/tex]. This relationship can be expressed with the equation:

[tex]\[ r = k \cdot s \cdot t \][/tex]

where [tex]\( k \)[/tex] is the constant of variation.

1. Find the constant of variation [tex]\( k \)[/tex]

We're given that [tex]\( r = 8 \)[/tex] when [tex]\( s = 6 \)[/tex] and [tex]\( t = 4 \)[/tex]. Substituting these values into the equation:

[tex]\[ 8 = k \cdot 6 \cdot 4 \][/tex]

Simplify:

[tex]\[ 8 = k \cdot 24 \][/tex]

Solve for [tex]\( k \)[/tex]:

[tex]\[ k = \frac{8}{24} \][/tex]

[tex]\[ k = \frac{1}{3} \][/tex]

2. Use the constant [tex]\( k \)[/tex] to find the new value of [tex]\( s \)[/tex] when [tex]\( r = 24 \)[/tex] and [tex]\( t = 16 \)[/tex]

Substituting the known values and constant [tex]\( k \)[/tex] into the joint variation equation:

[tex]\[ 24 = \frac{1}{3} \cdot s \cdot 16 \][/tex]

Simplify the right-hand side:

[tex]\[ 24 = \frac{16}{3} \cdot s \][/tex]

Multiply both sides by 3 to clear the fraction:

[tex]\[ 72 = 16 \cdot s \][/tex]

Solve for [tex]\( s \)[/tex]:

[tex]\[ s = \frac{72}{16} \][/tex]

[tex]\[ s = 4.5 \][/tex]

Therefore, when [tex]\( r = 24 \)[/tex] and [tex]\( t = 16 \)[/tex], the value of [tex]\( s \)[/tex] is [tex]\( 4.5 \)[/tex].