If the ratio of boys to girls in a class is B and the ratio of girls to boys is G the B G is

Both rates and ratios are a comparison of two numbers.

A rate is simply a specific type of ratio.

The difference is that a rate is a comparison of two numbers with different units, whereas a ratio compares two numbers with the same unit.

For example, in a room full of students, there are 10 boys and 5 girls. This means the ratio of boys to girls is 10:5.

If we simplify the ratio, we see that the ratio of boys to girls is 2:1, since #10 -: 5 = 2# and #5 -: 5 = 1#. Thus, there are 2 boys in the room for every 1 girl.

Let's say we'd like to buy a soda for each student in the classroom. The local pizza place offers a group discount on soda purchases: $10 for 20 sodas.

Since we only need 15 sodas, we're wondering how much each soda costs at the discounted rate. To find out, we'll set up two rates, because we have 2 different units, dollars and number of sodas:

If #(20 " sodas")/($10)#, then #(1 " soda")/(x)#

We cross multiply for #x#:

#20x = $10#

#x= 10-:20#

#x = 1/2#

Since our unit is $, we'll convert #1/2# to dollars. One half of a dollar is $0.50.

Thus, the discounted rate for the sodas is $.50/1 soda.

Here is a helpful video on simplifying rates and ratios:

Correct option is (B) greater than 1

Let there are x number of girls and y number of boys in the class.

The ratio of boys to girls in the class is B.

\(\therefore\) \(\frac xy\) = B         _________(1)

The ratio of girls to boys in the class is G.

\(\therefore\) \(\frac yx\) = G        _________(2)

Now, B+G \(=\frac xy+\frac yx\)

\(=\frac{x^2+y^2}{xy}\)

\(\because(x-y)^2\geq0\)

\(\Rightarrow x^2+y^2-2xy\geq0\)

\(\Rightarrow x^2+y^2\geq2xy\)

\(\Rightarrow\frac{x^2+y^2}{xy}\geq2>1\)

\(\Rightarrow\frac{x^2+y^2}{xy}>1\)

Hence, B+G > 1

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

Verified

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.

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