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If $3 \leq X \leq 5$ and $8 \leq Y \leq 11$ then which of the following options is TRUE?

1. $\left(\dfrac{3}{5} \leq \dfrac{X}{Y} \leq \dfrac{8}{5}\right)$
2. $\left(\dfrac{3}{11} \leq \dfrac{X}{Y} \leq \dfrac{5}{8}\right)$
3. $\left(\dfrac{3}{11} \leq \dfrac{X}{Y} \leq \dfrac{8}{5}\right)$
4. $\left(\dfrac{3}{5} \leq \dfrac{X}{Y} \leq \dfrac{8}{11}\right)$

$\frac{X}{Y}$ is minimum for the given range when $X$ is minimum possible and $Y$ is maximum possible i.e., $3/11$

$\frac{X}{Y}$ is maximum for the given range when $X$ is maximum possible and $Y$ is minimum possible i.e., $5/8$

we can also say option A ,C,D 8/5,8/5,8/11 which is in the form of Y/X and we want X/Y so these three are wrong so only option is B . by the way ur answer is correct
but in that case c also holds the condition please answer me

This can be solved by taking the particular cases.

case 1: take X=4 and Y=8 --------------> by solving you can eliminate option a) and option (d)

Now calculate the ratio of left options then you will be find option (b) is answer .

You can also take others as example or actually for option (c)   .27 <=  X/Y   <=   1..6  -----------> this is never possible because ratio exceeds 1 but Y values is always greater than X values.So (c ) is wrong ,

Hence option (B )  is right

by

$3\leq X\leq 5 ----- eqn. 1$

Now using multiplicative inverse property on Y

When a and b are both positive or both negative: (Direction of inequality changes)

If a < b then 1/a > 1/b

If a > b then 1/a < 1/b

But when either a or b is negative (not both) the direction stays the same:

If a < b then 1/a < 1/b

If a > b then 1/a > 1/b

$\therefore Y becomes \frac{1}{11}\leq \frac{1}{Y}\leq \frac{1}{8} ------ eqn. 2$

Now multiplying eqn.1 and eqn. 2 we get

$\frac{3}{11}\leq \frac{X}{Y}\leq \frac{5}{8}$

Hence option B is correct.

Here we saw X does not change but Y is inverse so it value will always inverse.

if   P=(2:3) then its inverse is 1/P=(3:2)

so 8≤Y≤11 then 11≤1/Y≤8.

now X/Y become  (3/11≤X/Y≤5/8)

so option B is right.