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8 votes

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

- $\left(\dfrac{3}{5} \leq \dfrac{X}{Y} \leq \dfrac{8}{5}\right)$
- $\left(\dfrac{3}{11} \leq \dfrac{X}{Y} \leq \dfrac{5}{8}\right)$
- $\left(\dfrac{3}{11} \leq \dfrac{X}{Y} \leq \dfrac{8}{5}\right)$
- $\left(\dfrac{3}{5} \leq \dfrac{X}{Y} \leq \dfrac{8}{11}\right)$

22 votes

Best answer

1 vote

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**

1 vote

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

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.