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If A varies directly as the square root of B and inversely as the cube of C, and if A = 3 when B = 256 and C = 2, find B when A = 24 and C = $\dfrac{1}{2}$.

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Last updated date: 27th Mar 2024
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MVSAT 2024
Answer
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Hint: In this question we have to find value of B when A = 24 and C = $\dfrac{1}{2}$ where A varies directly as square root of B and inversely as the cube root of C. So, by combining the given first two conditions we will introduce proportionality constant K by removing proportionality sign and then we will get the relation of A, B, C and K. We get the value of K by using the remaining condition and then substitute the value of A and C in the relation to find the value of B.
Complete step- by-step answer:
According to the question,
A varies directly as the square root of B,
$\therefore {\rm{A\;}} \propto {\rm{\;}}\sqrt {\rm{B}} $
A varies inversely as the cube of C,
$\therefore {\rm{A\;}} \propto \dfrac{1}{{{{\rm{C}}^3}}}$
According to question, A varies directly as the square root of B and inversely as the cube of C,
Therefore, ${\rm{A\;}} \propto \dfrac{{\sqrt {\rm{B}} }}{{{{\rm{C}}^3}}}$
As we know if we remove proportionality sign then we have to multiple by a constant term,
$ \Rightarrow {\rm{A}} = {\rm{K}} \times \dfrac{{\sqrt {\rm{B}} }}{{{{\rm{C}}^3}}}$ …. (1)
Where K is a constant (It is also called proportionality constant)
Given: A = 3 when B = 256 and C = 2
Put the value of A, B and C in equation 1, we get
$ \Rightarrow 3 = {\rm{K}} \times \dfrac{{\sqrt {256} }}{{{2^3}}}$
Using cross multiplication, we get
$ \Rightarrow 3 \times 8 = {\rm{K}} \times 16$
$ \Rightarrow {\rm{K\;}} = \dfrac{{3{\rm{\;}} \times {\rm{\;}}8}}{{16}} = \dfrac{3}{2}$
Put the value of K in equation 1, we get
$ \Rightarrow {\rm{A}} = \dfrac{3}{2} \times \dfrac{{\sqrt {\rm{B}} }}{{{{\rm{C}}^3}}}$
Now we have to find the value of B when A = 24 and C = $\dfrac{1}{2}$
$ \Rightarrow 24 = \dfrac{3}{2} \times \dfrac{{\sqrt {\rm{B}} }}{{{{\left( {\dfrac{1}{2}} \right)}^3}}}$
$ \Rightarrow 24 \times \dfrac{2}{3} = \dfrac{{\sqrt {\rm{B}} }}{{\left( {\dfrac{1}{8}} \right)}}$
Using cross multiplication, we get
$ \Rightarrow 24 \times \dfrac{2}{3} \times \dfrac{1}{8} = \sqrt {\rm{B}} $
$ \Rightarrow 24 \times \dfrac{2}{{24}} = \sqrt {\rm{B}} $
$ \Rightarrow \sqrt {\rm{B}} = 2$
Squaring both sides, we get
$ \Rightarrow {\left( {\sqrt {\rm{B}} } \right)^2} = {2^2}$
⇒ B = 4
When A = 24 and C = $\dfrac{1}{2}$ then B = 4.
Note: Whenever we face such types of questions, the best approach is to combine the given conditions and remove the proportionality sign in order to get an equation which contains all the unknown variables. Students make mistakes by not combining both proportionality conditions.