If \(x\) is the number of rulers bought and \(y\) is the number of pens bought, and the total number of things bought (which includes only rulers and pens) is 200, then the sum of the number of rules bought and the numbers of pens bought must be 200. Can you state this as an equation?I have tried I just cant do it.I have done the ones before and after .
No, that's not an equation, and it doesn't relate to the total number of items. What we want is:50x + 20y + 200 is this right but still stuck. This is the 1st time I have covered these questions.
80 what? (We always report units, when answering story problems.)
No, this is not at all correct. What you've written there is only an expression. Suppose Barry bought five rulers (i.e x = 5) and three pens (i.e. y= 3). Then we'd have 50(5) + 20(3) + 200 = 510... but what are we doing with this number? What does it mean? How does having the number 510 help us solve the problem?50x + 20y + 200 is this right but still stuck. This is the 1st time I have covered these questions. thank you.
Thank you I will reread this tomorrow and have a go, because I need to master it. what level GCSE maths would this be.No, this is not at all correct. What you've written there is only an expression. Suppose Barry bought five rulers (i.e x = 5) and three pens (i.e. y= 3). Then we'd have 50(5) + 20(3) + 200 = 510... but what are we doing with this number? What does it mean? How does having the number 510 help us solve the problem?
Rather, let's return to the very basics. In my example where Barry bought five rulers and three pens, we know that he bought eight total pieces of stationary. But how did we know this? What operation did we use? Well, we used addition to deduce that Barry bought 5 + 3 = 8 total pieces of stationary. However, you're probably noticing a huge problem right about now. The problem specifically states that Barry bought 200 total pieces of stationary, so the solution of (5 rulers, 3 pens) is obviously not a valid one. But (100 rulers, 100 pens) would be, as would (25 rulers, 175 pens), (50 rulers, 150 pens), and so on...
Now, using the exact same line of reasoning and the exact same operation (addition), can you see that if Barry bought three pens and x rulers, he must have bought x + 3 total pieces of stationary? Similarly, if he bought five rulers and y pens, he must have bought 5 + y total pieces of stationary. Combining these two and letting both the number of rulers and number of pens be unknown variables, we see that Barry must have bought x + y pieces of stationary. Since the problem specifies that the total pieces of stationary is 200, what equation can you create?
At this point, we have one equation, but there's nothing more we can do with just this - in order to solve for two variables, we need two equations. So, what other information do we know, that might help us create another equation? We know how much each ruler costs and how much each pen costs. Can we do something with that? Suppose that Barry bought 125 rulers and 75 pens. Since each ruler costs 50p, we know he must have spent 6250p = £62.50 on rulers. Likewise, he must have spent 1500p = £15.00 on pens. But how did we know these values? What operation did we use? Well, we used multiplication to deduce that Barry spent 125 * 50p = 6250p on rulers and 75 * 20p = 1500p on pens. From there, we can use addition to find that the total amount spent is £62.50 + £15.00 = £77.50.
What would happen if we let the number of rulers be an unknown? If Barry bought x rulers, it only makes sense, using the exact same logic and exact same operation as before (multiplication), that he must have spent x * 50p on rulers. Likewise, if Barry bought y pens, he must have spent y * 20p on pens. How much, then, did he spend in total on stationary? Given that the problem specifies that Barry spent £76 = 7600p in total, what second equation can you create?
Given these two equations, it should be clear enough how to solve the resulting system and find out how many rulers and how many pens Barry bought. So give it a go and see what you come up with.
Hi. No problem. (But you could have mentioned the method you're supposed to use, in your first post.)We have only been taught the elimination method …