by Guest » Tue Nov 26, 2019 11:14 am
Wow! I doubt it would have been possible to make this harder to read! It is turned sideways and too large to fit on a screen. Surely it would have been simpler for you to just type it in?
I think it says:
Seventy-eight sets of siblings, twins, triplets, and quads, took part in a scientific study, making a total of 199 subject. If there were six times as many sets of twins as there were quads, how many sets of triplets were there involved in the study?
Surely just typing that is easier than taking a snapshot, uploading it, the expecting people to go through gymnastics in order to help you!
Let x= number of sets of twins, y= number of sets of triplets, z= number of sets of quads.
There were 78 sets so x+ y+ z= 78. There were a total of 199 subjects so, since each set of twins means two people, each set of triplets three, and each set of quads four, we have 2x+ 3y+ 4z= 199. That gives two equations. We need one more equation to solve for the three unknowns.
Unfortunately, I find the next statement, "there were six times as many sets of twins as there were quads", ambiguous. It specifically refers to "sets of twins" but doesn't say "sets of quads". Was that supposed to be assumed, so that x= 6z, or is a "quad". as opposed to "set of quad" an individual person, so that x= 6(4z)= 24z?
Try each one. Solve x+ y+ z= 78, 2x+ 3y+ 4z= 199, and x= 6z. Of course, z must be a positive integer.
Or solve x+ y+ z= 78, 2x+ 3y+ 4z= 199, x= 24z. Is z an integer here?
And don't forget that the question is asking for the value of z. the number of sets of quads. It is fairly easy to eliminate y from the first two equations giving an equation in x and z, then eliminate x from that and the last equation, allowing you to solve for z without needing to find x and y.