Question

solve this problem please with its method​

Answer 1

Question 27. If $\sf{x:y=3:5}$, find $\sf{(2x+3y):(5x+7y)}$.

Question 28. If $\sf{(3a+2b):(2a+5b)=12:19}$, find $\sf{a:b}$.

How to solve ratio questions?

The ratio can be simplified if we multiply the extremes and the means.

The extremes are placed outside.

The means are placed inside.

Question 28.

We get $\sf{3y=5x}$ from the given ratio. The condition between two variables is $\sf{y=\dfrac{5}{3} x}$.

So we substitute it into the ratio.

$\sf{(2x+3y):(5x+7y)}$

$\sf{=(2x+3\times\dfrac{5}{3} x):(5x+7\times\dfrac{5}{3} x)}$

$\sf{=(2x+5x):(5x+\dfrac{35}{3} x)}$

$\sf{=(7x):(\dfrac{50}{3} x)}$

$\sf{=(3\times7):(3\times\dfrac{50}{3} )}$

$\sf{=21:50}$

Question 29.

The means product of means and extremes products are equal.

$\sf{19(3a+2b)=12(2a+5b)}$

$\implies\sf{57a+28b=24a+60b}$

$\implies\sf{33a=32b}$

Let's divide both sides by b.

$\implies\sf{33\times\dfrac{a}{b}=32}$

$\implies\sf{\dfrac{a}{b} =\dfrac{32}{33} }$

$\implies\sf{a:b=32:33}$

More information:

Assume we have a ratio $\sf{a:b=c:d}$.

This is equivalent to $\sf{\dfrac{a}{b} =\dfrac{c}{d} }$.

Let's multiply on both sides.

$\implies\sf{\dfrac{a}{b} \times bd=\dfrac{c}{d} \times bd}$

$\implies\sf{ad=bc}$

Each $ad$ and $bc$ is the product of extremes and means. So, the products are equal.

Answer 2

Answer:

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