This blog is the online extension of our intermediate classrooms. Our goal is to enhance and document our learning experience throughout the school year, and share this journey with teachers, parents and students. We welcome your constructive feedback, and we look forward to learning with you!
Grade 8 POTW: - Each bolt has a mass of 7.0g. - Can be heavier or lighter by as much as 2.14% - Mass of bolts = exactly 1kg or 1000g - Maximum and minimum number of bolts?
Dealing with sevenths is actually the saddest thing ever since the decimal expansion is pretty annoying to deal with. This is why I’m going to assume that the mass of the bolts can be smaller than 1000g, but never larger.
Maximum number (must have the lowest possible weight):
7 - (7*0.0214) = 6.8502g 1000/6.8502 = 145.981139
I DID NOT ROUND BECAUSE I WANT MY ANSWER TO BE AS EXACT AS POSSIBLE AT THE END.
Since we are in the Integers unit and because we cannot have 0.981139 of a bolt, we must round down. This is because 146 bolts would exceed the maximum 1 kg.
Minimum number (must have the highest possible weight):
7 + (7 x 0.0214) = 7.1498 1000/7.1498 = 139.864052
But you notice that the decimal following the 139 is very close to 140. And if we double check with 129 bolts, we would get a mass of 993 point something. This is pretty off from 1 kg. Instead, we could try to upgrade the 139 to 140. With the same mass of 7.1498, 140 bolts would be over by around 0.972g. Dividing this by 140, we can subtract this from the overall mass of the original number of bolts. Now its closer to 1kg.
The maximum number of bolts is 145 and the minimum number of bolts is 139 if you would accept a 6 point something difference from 1000g. If not the answer is 140 bolts.
Since the clock gains 12 min every hour, we can find the number of hours between the time Dante needs to leave and the time he set his clock to the right time. 12:30 - 5:30 - 7 hours 12 minutes x 7 hours = 84 minute change 1:30 - 84 minutes = 12:06 pm 12:30 - 12:06 = 24 minutes
Dante has 24 minutes left until he needs to return to work. But I suggest he leave now so he can go to Walmart and get a new clock.
I think you made a mistake when finding the difference between the moment he left and the moment he looked at his clock during lunch. It should've been 1:30 not 12:30.
Dante B. Late is not going to be late, even though his name implies it.
Method 1 of solving: Tediously make a table You could go with this method and plot the change every hour. 5:30 -> 6:30 = 6:42 (12 minutes ahead) 6:42 -> 7:42 = 7:54 (24 mins) 7:54 -> 8:54 = 9:06 (36 mins) ... 11:18 = 11:30.
Obviously this could go on until you figured out the actual exact time, but this would be tedious and time-consuming. Instead...
Method 2: Get to know your x. If we let x = hour and y = clock time, This is represented by the equation y + (5+1/2) = (1+1/6)x + (5+1/2), which we can then solve when y=13.5. 13.5 = (1+1/6)x Solving this with long division, or just with a calculator, we obtain 11:57:1... in real time. That means he still has 0:33:96... left to eat.
Correction: The actual equation should have been 8 = (5/6)x, but I did a flop and put [+ (5+1/2)] on both sides, so the real answer is 6:40 after 5:30, which is 12:10 if I did the math right.
Grade 8 POTW Okay, so each bolt weighs somewhere between: 7 x 0.0214 - 7 and 7 x 0.0214 + 7 or 6.8502g and 7.1498g The box’s total weight is 1 kg or 1000g. To find out the minimum number of bolts, we just need to divide the box’s total weight (1000g) by the highest possible bolt weight (7.1498g). To find out the maximum number of bolts, we just need to divide the box’s total weight (1000g) by the lowest possible bolt weight (6.8502g). Minimum bolts: 1000/7.1498 = 139.864052 Maximum bolts: 1000/6.8502 = 145.981139 We will review both situations. For the minimum number of bolts, we got 139.864052. If we round down, we’ll be off by a little bit (139 x 7.1498 = 993.8222, 1000 - 993.8222 = 6.1778). Instead we can round up to 140 bolts. To make it exact, we can just replace each 7.1498 bolt by ones with a smaller value until it reaches 1kg. For the maximum number of bolts, we got 145.981139. This is a similar situation to before. Rounding up with make it too large, so we just round down to 145. Therefore, the minimum number of bolts is 140 and maximum is 145.
Grade 7 POTW This question is pretty simple. We know he leaves his house at 5:30 a.m. Okay, so his clock gains 12 minutes per hour. The rate would be at 12/60 = 1/5 of an hour. This means that in any situation after the clock has been set correctly, the new time would be h + (1/5)h = (6/5)h. From when Dante left at 5:30 to when he came back at “1:30”, 8 hours has passed. This would mean that (6/5)h = 8 and we just need to solve: (6/5)h = 8 (6/5)h/(6/5) = 8/(6/5) h = 6.666… = 6h + (2/3)h = 6 hours and 40 minutes If we add the 6 hours and 40 minutes to the original 5:30 a.m. time, we get: 5:30 + 6 hours = 11:30 11:30 + 40 minutes = 12:10 The correct time is 12:10. Between 12:10 and 12:30 is 20 minutes. Therefore, Dante still has 20 minutes before he needs to leave his home to go back to work.
Grade 8 POTW:
ReplyDelete- Each bolt has a mass of 7.0g.
- Can be heavier or lighter by as much as 2.14%
- Mass of bolts = exactly 1kg or 1000g
- Maximum and minimum number of bolts?
Dealing with sevenths is actually the saddest thing ever since the decimal expansion is pretty annoying to deal with. This is why I’m going to assume that the mass of the bolts can be smaller than 1000g, but never larger.
Maximum number (must have the lowest possible weight):
7 - (7*0.0214) = 6.8502g
1000/6.8502 = 145.981139
I DID NOT ROUND BECAUSE I WANT MY ANSWER TO BE AS EXACT AS POSSIBLE AT THE END.
Since we are in the Integers unit and because we cannot have 0.981139 of a bolt, we must round down. This is because 146 bolts would exceed the maximum 1 kg.
Minimum number (must have the highest possible weight):
7 + (7 x 0.0214) = 7.1498
1000/7.1498 = 139.864052
But you notice that the decimal following the 139 is very close to 140. And if we double check with 129 bolts, we would get a mass of 993 point something. This is pretty off from 1 kg. Instead, we could try to upgrade the 139 to 140. With the same mass of 7.1498, 140 bolts would be over by around 0.972g. Dividing this by 140, we can subtract this from the overall mass of the original number of bolts. Now its closer to 1kg.
The maximum number of bolts is 145 and the minimum number of bolts is 139 if you would accept a 6 point something difference from 1000g. If not the answer is 140 bolts.
Grade 7 POTW:
ReplyDeleteSince the clock gains 12 min every hour, we can find the number of hours between the time Dante needs to leave and the time he set his clock to the right time.
12:30 - 5:30 - 7 hours
12 minutes x 7 hours = 84 minute change
1:30 - 84 minutes = 12:06 pm
12:30 - 12:06 = 24 minutes
Dante has 24 minutes left until he needs to return to work. But I suggest he leave now so he can go to Walmart and get a new clock.
I think you made a mistake when finding the difference between the moment he left and the moment he looked at his clock during lunch. It should've been 1:30 not 12:30.
Deleteoh no. ive made an oopsie
DeleteDante B. Late is not going to be late, even though his name implies it.
ReplyDeleteMethod 1 of solving: Tediously make a table
You could go with this method and plot the change every hour.
5:30 -> 6:30 = 6:42 (12 minutes ahead)
6:42 -> 7:42 = 7:54 (24 mins)
7:54 -> 8:54 = 9:06 (36 mins)
...
11:18 = 11:30.
Obviously this could go on until you figured out the actual exact time, but this would be tedious and time-consuming. Instead...
Method 2: Get to know your x.
If we let x = hour and y = clock time,
This is represented by the equation y + (5+1/2) = (1+1/6)x + (5+1/2), which we can then solve when y=13.5.
13.5 = (1+1/6)x
Solving this with long division, or just with a calculator, we obtain
11:57:1... in real time.
That means he still has 0:33:96... left to eat.
Correction: The actual equation should have been 8 = (5/6)x, but I did a flop and put [+ (5+1/2)] on both sides, so the real answer is 6:40 after 5:30, which is 12:10 if I did the math right.
ReplyDeleteGrade 8 POTW
ReplyDeleteOkay, so each bolt weighs somewhere between:
7 x 0.0214 - 7 and 7 x 0.0214 + 7 or 6.8502g and 7.1498g
The box’s total weight is 1 kg or 1000g. To find out the minimum number of bolts, we just need to divide the box’s total weight (1000g) by the highest possible bolt weight (7.1498g). To find out the maximum number of bolts, we just need to divide the box’s total weight (1000g) by the lowest possible bolt weight (6.8502g).
Minimum bolts: 1000/7.1498 = 139.864052
Maximum bolts: 1000/6.8502 = 145.981139
We will review both situations. For the minimum number of bolts, we got 139.864052. If we round down, we’ll be off by a little bit (139 x 7.1498 = 993.8222, 1000 - 993.8222 = 6.1778). Instead we can round up to 140 bolts. To make it exact, we can just replace each 7.1498 bolt by ones with a smaller value until it reaches 1kg.
For the maximum number of bolts, we got 145.981139. This is a similar situation to before. Rounding up with make it too large, so we just round down to 145.
Therefore, the minimum number of bolts is 140 and maximum is 145.
Grade 7 POTW
ReplyDeleteThis question is pretty simple. We know he leaves his house at 5:30 a.m. Okay, so his clock gains 12 minutes per hour. The rate would be at 12/60 = 1/5 of an hour. This means that in any situation after the clock has been set correctly, the new time would be h + (1/5)h = (6/5)h. From when Dante left at 5:30 to when he came back at “1:30”, 8 hours has passed. This would mean that (6/5)h = 8 and we just need to solve:
(6/5)h = 8
(6/5)h/(6/5) = 8/(6/5)
h = 6.666… = 6h + (2/3)h = 6 hours and 40 minutes
If we add the 6 hours and 40 minutes to the original 5:30 a.m. time, we get:
5:30 + 6 hours = 11:30
11:30 + 40 minutes = 12:10
The correct time is 12:10. Between 12:10 and 12:30 is 20 minutes.
Therefore, Dante still has 20 minutes before he needs to leave his home to go back to work.