Explanation for Question 31 From the Math (Calc) Section on the Official Sat Practice Test 8
So the question 31 tells us that there are two atoms of hydrogen and 2 one atom of oxygen and one molecule of water. 3 And then we're asked how many atoms of hydrogen are there in 51 molecules 4 of water. And so this is a proportions question. 5 And the reason that we know this is because we're given a rate in 6 the first part of the question, and the second part of the questions were 7 asked to solve for a component of that rate or of that ratio. 8 You can think of it as right. And so we're told that we're given 9 51 molecules of water and for every one molecule of water, 10 we have two hydrogens was that, 11 which I'll just write is H and if we have 51 waters 12 where you're solving for X, right, how many hydrogens there are. 13 So here's what I'm solving for it. And here's the proportion that I've read 14 an important note to make is that I have both of my hydrogens and 15 the denominator are both of my waters in the numerator. 16 Now you could flip that and have your hydrogens in the numerator and your 17 waters and the denominator that doesn't actually matter. 18 All that matters is that your hydrogens are next to each other and your 19 waters are next to each other. Now I can solve this first by cross 20 multiplying. So I'll do X times one, 21 which is X, and I'll do two times 51, 22 which is 102. And so now I've solved for it, 23 right? If I have 51 water molecules, I have 100 into hydrogens.