
Suppose \[y\] varies jointly as \[x\] and \[z\]. How do you find \[y\] when \[x = 6\] and \[z = 8\], if \[y = 6\] when \[x\] is \[4\] and \[z\] is \[2\]?
Answer
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Hint: In the question, it is given that the variable \[y\] varies jointly with variables \[x\] and \[z\]. This means that \[y\] is proportional to both the variables \[x\] and \[z\]. On combining these proportionalities, and replacing the sign of proportionality with the equality sign, we will obtain a relation between \[y\], \[x\], and \[z\] in the form of a constant. Substituting the values in the obtained relation, we will get the required answer.
Complete step-by-step solution:
According to the question, the variable \[y\] varies jointly with variables \[x\] and \[z\]. This means that the variable \[y\] is proportional to both the variables \[x\] and \[z\]. Writing this statement mathematically, we have
\[y \propto x\] and \[y \propto z\]
On combining the above two proportionalities, we have
\[y \propto xz\]
Now, for replacing the sign of proportionality with the sign of equality, we introduce a constant \[k\] as in the below equation
\[y = kxz\]…………………………\[\left( 1 \right)\]
According to the question, we have \[y = 6\] when \[x\] is \[4\] and \[z\] is \[2\]. Therefore substituting \[y = 6\], \[x = 4\] and \[z = 2\] in the above equation, we get
\[ \Rightarrow 6 = k\left( 4 \right)\left( 2 \right)\]
\[ \Rightarrow 6 = 8k\]
On dividing both the sides by \[8\], we get
\[ \Rightarrow k = \dfrac{6}{8}\]
\[ \Rightarrow k = \dfrac{3}{4}\]
Substituting this value of constant in the equation \[\left( 1 \right)\], we get
\[y = \dfrac{3}{4}xz\]
Now, according to the question, we need to find out the value of \[y\] when \[x = 6\] and \[z = 8\]. Therefore on substituting \[x = 6\] and \[z = 8\] in the above equation, we get
\[ \Rightarrow y = \dfrac{3}{4}\left( 6 \right)\left( 8 \right)\]
Multiplying the terms, we get
\[ \Rightarrow y = 36\]
Hence, the value of \[y\] when \[x = 6\] and \[z = 8\] is equal to \[36\].
Note:
The joint variation need not necessarily mean the direct proportion. The joint variation can occur when one quantity varies directly or inversely with the other quantity. But since in this question we were not given the information about the direct or indirect proportionality, we assumed the joint variation to be direct proportionality.
Complete step-by-step solution:
According to the question, the variable \[y\] varies jointly with variables \[x\] and \[z\]. This means that the variable \[y\] is proportional to both the variables \[x\] and \[z\]. Writing this statement mathematically, we have
\[y \propto x\] and \[y \propto z\]
On combining the above two proportionalities, we have
\[y \propto xz\]
Now, for replacing the sign of proportionality with the sign of equality, we introduce a constant \[k\] as in the below equation
\[y = kxz\]…………………………\[\left( 1 \right)\]
According to the question, we have \[y = 6\] when \[x\] is \[4\] and \[z\] is \[2\]. Therefore substituting \[y = 6\], \[x = 4\] and \[z = 2\] in the above equation, we get
\[ \Rightarrow 6 = k\left( 4 \right)\left( 2 \right)\]
\[ \Rightarrow 6 = 8k\]
On dividing both the sides by \[8\], we get
\[ \Rightarrow k = \dfrac{6}{8}\]
\[ \Rightarrow k = \dfrac{3}{4}\]
Substituting this value of constant in the equation \[\left( 1 \right)\], we get
\[y = \dfrac{3}{4}xz\]
Now, according to the question, we need to find out the value of \[y\] when \[x = 6\] and \[z = 8\]. Therefore on substituting \[x = 6\] and \[z = 8\] in the above equation, we get
\[ \Rightarrow y = \dfrac{3}{4}\left( 6 \right)\left( 8 \right)\]
Multiplying the terms, we get
\[ \Rightarrow y = 36\]
Hence, the value of \[y\] when \[x = 6\] and \[z = 8\] is equal to \[36\].
Note:
The joint variation need not necessarily mean the direct proportion. The joint variation can occur when one quantity varies directly or inversely with the other quantity. But since in this question we were not given the information about the direct or indirect proportionality, we assumed the joint variation to be direct proportionality.
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