# Question #0d7d0

Feb 23, 2017

$D .$

#### Explanation:

In each Option, the L.H.S. is $\frac{6}{8} = \frac{\left(3\right) \left(2\right)}{\left(4\right) \left(2\right)} = \frac{3}{4.}$

Now, in $A , \text{ the R.H.S.} = \frac{51}{68} = \frac{\left(17\right) \left(3\right)}{\left(17\right) \left(4\right)} = \frac{3}{4.}$

In $B , \text{the R.H.S.=} \frac{39}{52} = \frac{\left(13\right) \left(3\right)}{\left(13\right) \left(4\right)} = \frac{3}{4.}$

In $C , \text{ the R.H.S.=} \frac{21}{28} = \frac{\left(7\right) \left(3\right)}{\left(7\right) \left(4\right)} = \frac{3}{4.}$

In $D , \text{ the R.H.S.=} \frac{15}{24} = \frac{\left(3\right) \left(5\right)}{\left(3\right) \left(8\right)} = \frac{3}{8.}$

So, $D$ is not a true proportion.