Question

1. If $x:y=3:5$,find the ratio $3x+4y:8x+5y$.
2. If $x:y=8:9$ find the ratio $(7x-4y):3x+2y$
3. If two numbers are in the ratio $6:13$ and their l.c.m. is $312$, find the numbers.
4. What should be added to each term of the ratio $7:13$ so that the ratio becomes $2:3$

Answer 1

1). Let $x=3\lambda$

$y=5\lambda$

So, $\frac{3x+4y}{3x+5y}=\frac{3\left(3\lambda \right)+4\left(5\lambda \right)}{8\left(3\lambda \right)+5\left(5\lambda \right)}=\frac{9\lambda +20\lambda }{24\lambda +25\lambda }$

$=\frac{29\lambda }{49\lambda }=29:49$

2). Let $x=8\lambda$

$y=9\lambda$

So, $\frac{7x-4y}{3x+2y}=\frac{7\left(8\lambda \right)-4\left(9\lambda \right)}{3\left(8\lambda \right)+2\left(9\lambda \right)}=\frac{56\lambda -36\lambda }{24\lambda +18\lambda }=\frac{20\lambda }{42\lambda }=\frac{10}{21}=10:21$

3). Let the numbers be $6x$ and $13x$

Now, Since ${}^{\prime }{x}^{\prime }$ is a common multiple

$\therefore$ LCM of $6x$ and $13x=13\times 6\times x$

$=78x$

Now, $78x=312\Rightarrow x=\frac{312}{78}=4$

Hence, Required numbers are $\Rightarrow 6x=6\times 4=24$

$13x=13\times 4=52$

4). Let the number to be added by $y$

Such that, $\frac{7+y}{13+y}=\frac{21}{3}\Rightarrow 21+3y=26+2y\Rightarrow y=5$