Question

If the ratio of boys to girls in a class is B and the ratio of the girls to boys is G, then 3(B+G) is
[a] Equal to 3
[b] Less than 3
[c] More than 3
[d] Less than $\dfrac{1}{3}$

Answer 1

Hint: Assume that the number of girls in the class is g and the number of boys in the class is b. Hence find the ratios B and G. Observe that $BG=1$. Hence prove that $B+G=B+\dfrac{1}{B}$. Add and subtract 2 on both sides of the equation and use the fact that ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ to prove that $B+G={{\left( \sqrt{B}-\dfrac{1}{\sqrt{B}} \right)}^{2}}+2$. Hence find the range of $3\left( B+G \right)$ and hence find which of the options is correct.

Complete step-by-step answer:
Let the number of boys in the class is b, and the number of girls in the class is g.
Since B is the ratio of the boys to the girls, we have
$B=\dfrac{b}{g}$
Since G is the ratio of the number of girls to the number of boys, we have
$G=\dfrac{g}{b}$
Hence, we have
$BG=\dfrac{b}{g}\times \dfrac{g}{b}=1$
Dividing both sides by B, we get
$G=\dfrac{1}{B}$
Hence, we have
$B+G=B+\dfrac{1}{B}$
Adding and subtracting 2 on RHS, we get
$B+G=B+\dfrac{1}{B}-2+2$
Hence, we have
$B+G={{\left( \sqrt{B} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{B}} \right)}^{2}}-2\times \sqrt{B}\times \dfrac{1}{\sqrt{B}}+2$
We know that ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$
Hence, we have
$B+G={{\left( \sqrt{B}-\dfrac{1}{\sqrt{B}} \right)}^{2}}+2$
We know that $\forall x\in \mathbb{R},{{x}^{2}}\ge 0$
Hence, we have
${{\left( \sqrt{B}-\dfrac{1}{\sqrt{B}} \right)}^{2}}\ge 0$
Adding 2 on both sides, we get
${{\left( \sqrt{B}-\dfrac{1}{\sqrt{B}} \right)}^{2}}+2\ge 2$
Hence, we have
$B+G\ge 2$
Multiplying both sides by 3, we get
$3\left( B+G \right)\ge 6$
Hence, we conclude that option [c] is correct.

Note: Alternative solution:
We know that $AM\ge GM$, where AM is the arithmetic mean, and GM is the geometric mean.
Hence, we have
$\dfrac{B+G}{2}\ge \sqrt{BG}$
Since BG = 1, we have
$\dfrac{B+G}{2}\ge 1$
Multiplying both sides by 2, we get
$B+G\ge 2$
Multiplying both sides by 3, we get
$3\left( B+G \right)\ge 6$, which is the same as obtained above.
Hence option [c] is correct.