Question

13. There is an 80 gram mixture of material X, Y, and Z. The ratio X :Y:Z is 4/7 / 5 . If material Z is removed, what is the new weight, in gram, of the mixture?A. 11B.64C. 25D.45

Answer 1

The ratio given in the problem implies that for every 4 gr of X in the mixture, there are 7 gr of Y and 5 gr of Z. Notice that:

$\begin{gathered} 4+7+5=16 \\ \frac{80}{16}=5 \end{gathered}$

Then, we need to multiply all the previous quantities by 5. Then, in the 80gr mixture, there are 20gr of X, 35gr of Y, and 25gr of Z.

If we remove Z from the mixture, we will end up with a total mass of:

$80-25=55$

55 gr is the answer to the question, despite not being among the options

Prove that the options given are not possible:

Notice that options A and C imply that most of the mixture consists of Z substance, which is not possible according to the relation.

The only possibility would be C, in that case:

$\begin{gathered} 4+7=11 \\ \frac{64}{11}=5.8181\ldots \\ \Rightarrow Z\approx29.0909\ldots \\ 64+29.0909\approx93.0909 \end{gathered}$

Which is not equal to 100gr