Math

The ratio of boys to girls at a party was 4:3 respectively. When 60 boys and 150 girls came to join in, the ratio of boys to girls became 2:3. How many boys were there at first?

b = boys

g = girls

\(\small{ \begin{array}{lrcll} (1) & \frac{b}{g} &=& \frac43 \\ & 3b &=& 4g \\ & 3b - 4g &=& 0 \\\\ (2) & \frac{b+60}{g+150} &=& \frac23 \\ & 3\cdot (b+60) &=& 2\cdot (g+150)\\ & 3b+180 &=& 2g + 300 \\ & 3b - 2g &=& 300 - 180 \\ & 3b - 2g &=& 120 \\ \\ \hline \\ (1) & 3b - 4g &=& 0 \\ (2) & 3b - 2g &=& 120 \\\\ (2)-(1) & 3b - 2g - (3b - 4g) &=& 120 -0 \\ &3b - 2g - 3b + 4g &=& 120 \\ & 2g &=& 120 \\ & \mathbf{g} &\mathbf{=}& \mathbf{60} \\\\ & 3b &=& 4g \\ & 3b &=& 4\cdot 60 \\ & 3b &=& 240 \\ & \mathbf{b} &\mathbf{=}& \mathbf{80} \end{array} }\)

At start

x is boys

y is girls

ratio=4/3

4/3=x/y 

===4y=3x

after:

x+60/y+150=2/3

by solving this,

y=60

x=80

Let x be the original number of boys and y be the original number of girls.....so

x / y  =  4/3   implies that   3x = 4y   implies that y  = (3/4)x

So......after 60 boys and  150 boys are added, we have

[x + 60] / [ (3/4)x + 150 ]  = 2 / 3   cross-multiply

3[x + 60] = 2 [ (3/4)x + 150]   simplify

3x + 180  = (3/2)x + 300  

(3/2)x = 120

x = 80   =  number of boys  originally

(3/4)(80)  = 60  = number of girls originally