Ozymandias Posted September 8, 2020 #351 Share Posted September 8, 2020 6 minutes ago, ant0n said: Oh no, I was just having the solution. You revealed it too early. Sorry, Anton. It has been up there since last night and I thought you had enough of it. I gave that hint last night at 10.53pm in message #349! 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 8, 2020 #352 Share Posted September 8, 2020 (edited) 6 minutes ago, ant0n said: Why give a hint if you reveal it all 11 minutes later? I was about to typewrite my solution when I realized you revealed it all 5 minutes earlier. I gave the same hint in message #349 last night. You have been posting here a number of times since then without coming back to this problem so I assumed you had lost interest. Sorry again. Edited September 8, 2020 by Ozymandias 1 Link to comment Share on other sites More sharing options...
ant0n Posted September 8, 2020 #353 Share Posted September 8, 2020 1 minute ago, Ozymandias said: Sorry, Anton. It has been up there since last night and I thought you had enough of it. I gave that hint last night at 10.53pm in message #349! Well, that's wasted. I liked that teaser. Anyway... You're way too impatient! ^^ I'm going to post a new mathematical brainer teaser, oK? 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 8, 2020 #354 Share Posted September 8, 2020 Just now, ant0n said: Well, that's wasted. I liked that teaser. Anyway... You're way too impatient! ^^ I'm going to post a new mathematical brainer teaser, oK? I am impatient. All my friends tell me that!! By all means post another problem. Thanks. 1 Link to comment Share on other sites More sharing options...
ant0n Posted September 8, 2020 #355 Share Posted September 8, 2020 (edited) 5 minutes ago, Ozymandias said: I am impatient. All my friends tell me that!! By all means post another problem. Thanks. Well, that's not a big deal, ok? I loved your teaser, really, and I enjoyed solving it. Well, I was impatient too this evening (it's 12:13 am here right now). Since that solving that problem was very time consuming for me, I didn't want to spend time on it today until this evening/night. I think you just add to privately ask me whether I was about to answer or not. As simple as that. A new teaser is coming... Edited September 8, 2020 by ant0n Link to comment Share on other sites More sharing options...
Ozymandias Posted September 8, 2020 #356 Share Posted September 8, 2020 1 minute ago, ant0n said: Well, that's not a big deal, ok? I loved your teaser, really, and I enjoyed solving it. Well, I was impatient too this evening (it's 12:13 am here right now). Since that solving that problem was very time consuming for me, I didn't want to spend time on it today until this evening/night. I think you just add to privately ask me whether I was about to answer or not. As simple as that. A new teaser is coming... Point taken, Anton. And since it is so late for you and I am off to bed now myself, maybe you should leave posting a new problem until tomorrow. Good night for now. 1 Link to comment Share on other sites More sharing options...
ant0n Posted September 8, 2020 #357 Share Posted September 8, 2020 (edited) 19 minutes ago, Ozymandias said: I am impatient. All my friends tell me that!! By all means post another problem. Thanks. Patience is sacred. (sorry, I got confused with the dates the hints got published) "Earthlings usually have 10 fingers. That explains why they preferably use the decimal system to count. An astrobiologist recently found an inhabited planet which inhabitants usually have only 6 fingers. He discovered they count in base 6 (0, 1, 2, 3, 4, 5, 10, 11, ...). How do they write 15,812 (base 10) in base 6?" Please EXPLAIN the reasoning that leads you to the result. Edited September 8, 2020 by ant0n Link to comment Share on other sites More sharing options...
Ozymandias Posted September 9, 2020 #358 Share Posted September 9, 2020 18 hours ago, ant0n said: An astrobiologist recently found an inhabited planet which inhabitants usually have only 6 fingers. He discovered they count in base 6 (0, 1, 2, 3, 4, 5, 10, 11, ...). How do they write 15,812 (base 10) in base 6?" Please EXPLAIN the reasoning that leads you to the result. The expression of any number like the base ten 15,812 depends on us all agreeing and knowing that the position (or place) occupied by each digit in that number is significant; e.g. the 8 is in the 100s place, the two is in the unit place, etc. In terms of the base (10 in this case) the 8 = 8 x 100 = 8 x 10^2 = 800; the 2 = 2 x 1 = 2 x 10^0 = 2; the 5 = 5 x 1000 = 5 x 10^3 = 5000, etc. Moving right to left the digits in the number occupy place-holders in ascending powers of the base (starting with zero). To convert from base ten (denary) to base 6, just divide 15,812 repeatedly by six. 15812 / 6 = 2635 remainder 2 gives the number in the 6^0 place (the units in base 6) >> 2 x 1 = 2 2635 / 6 = 439 remainder 1 gives the number in the 6^1 place (the 6s in base 6) >> 1 x 6 = 6 439 / 6 = 73 remainder 1 gives the number in the 6^2 place (the 36s in base 6) >> 1 x 36 = 36 73 / 6 = 12 remainder 1 gives the number in the 6^3 place (the 216s in base 6) >> 1 x 216 = 216 12 / 6 = 2 remainder 0 gives the number in the 6^4 place (the 1296s in base 6) >> 0 x 1296 = 0 2 / 6 = 0 remainder 2 gives the number in the 6^5 place (the 7776s in base 6) >> 2 x 7776 = 15552 The new number in the new base is these remainders expressed in the reverse order to the way they were generated. Therefore, 15,812 (base 10) is equal to 201,112 (base 6). Try applying this technique to 'convert' 15,812 to base 10. The remainders will be 2, 1, 8, 5 and 1 !! 1 Link to comment Share on other sites More sharing options...
ant0n Posted September 9, 2020 #359 Share Posted September 9, 2020 7 minutes ago, Ozymandias said: The expression of any number like the base ten 15,812 depends on us all agreeing and knowing that the position (or place) occupied by each digit in that number is significant; e.g. the 8 is in the 100s place, the two is in the unit place, etc. In terms of the base (10 in this case) the 8 = 8 x 100 = 8 x 10^2 = 800; the 2 = 2 x 1 = 2 x 10^0 = 2; the 5 = 5 x 1000 = 5 x 10^3 = 5000, etc. Moving right to left the digits in the number occupy place-holders in ascending powers of the base (starting with zero). To convert from base ten (denary) to base 6, just divide 15,812 repeatedly by six. 15812 / 6 = 2635 remainder 2 gives the number in the 6^0 place (the units in base 6) >> 2 x 1 = 2 2635 / 6 = 439 remainder 1 gives the number in the 6^1 place (the 6s in base 6) >> 1 x 6 = 6 439 / 6 = 73 remainder 1 gives the number in the 6^2 place (the 36s in base 6) >> 1 x 36 = 36 73 / 6 = 12 remainder 1 gives the number in the 6^3 place (the 216s in base 6) >> 1 x 216 = 216 12 / 6 = 2 remainder 0 gives the number in the 6^4 place (the 1296s in base 6) >> 0 x 1296 = 0 2 / 6 = 0 remainder 2 gives the number in the 6^5 place (the 7776s in base 6) >> 2 x 7776 = 15552 The new number in the new base is these remainders expressed in the reverse order to the way they were generated. Therefore, 15,812 (base 10) is equal to 201,112 (base 6). Try applying this technique to 'convert' 15,812 to base 10. The remainders will be 2, 1, 8, 5 and 1 !! Great! Well done! Your turn. Link to comment Share on other sites More sharing options...
Ozymandias Posted September 9, 2020 #360 Share Posted September 9, 2020 It would be nice if others joined the thread but experience tells me otherwise. Anyway, here's my new problem: Problem: a mining company know that 3 identical water pumps working simultaneously together can draw 41,250 litres of water in 1.5 hours. They have a disused quarry containing an estimated 500,000 litres of water and want to empty it in 8 hours. How many of these pumps do they require? Link to comment Share on other sites More sharing options...
danydandan Posted September 9, 2020 Author #361 Share Posted September 9, 2020 31 minutes ago, Ozymandias said: It would be nice if others joined the thread but experience tells me otherwise. Anyway, here's my new problem: Problem: a mining company know that 3 identical water pumps working simultaneously together can draw 41,250 litres of water in 1.5 hours. They have a disused quarry containing an estimated 500,000 litres of water and want to empty it in 8 hours. How many of these pumps do they require? Hey sorry for my lateness. Every time I get an email the question is already answered. Seven pumps would be the answer I think? 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 9, 2020 #362 Share Posted September 9, 2020 (edited) 43 minutes ago, danydandan said: Hey sorry for my lateness. Every time I get an email the question is already answered. Seven pumps would be the answer I think? Yes, Dan, that is correct (6.8, in fact - but you can't operate 0.8 of a pump!!). I think you should also give the solution method for the benefit of others (although the 'others' are just Anton and I, and he probably knows it anyway). Welcome back and it's .... Your turn. Edited September 9, 2020 by Ozymandias 1 Link to comment Share on other sites More sharing options...
ant0n Posted September 9, 2020 #363 Share Posted September 9, 2020 11 minutes ago, Ozymandias said: Yes, Dan, that is correct (6.8, in fact - but you can't operate 0.8 of a pump!!). I think you should also give the solution method for the benefit of others (although the 'others' are just Anton and I, and he probably knows it anyway). Welcome back and it's .... Your turn. Yes, I got 6.8 -> 7 pumps too ^^ (I was about one hour too late this time ^^) I do appreciate when the solver explicitly explains how he/she got to the solution. I look forward to Danydandan's new mathematics brain teaser Link to comment Share on other sites More sharing options...
danydandan Posted September 9, 2020 Author #364 Share Posted September 9, 2020 (edited) 20 minutes ago, Ozymandias said: Yes, Dan, that is correct (6.8, in fact - but you can't operate 0.8 of a pump!!). I think you should also give the solution method for the benefit of others (although the 'others' are just Anton and I, and he probably knows it anyway). Welcome back and it's .... Your turn. 41250litre with three pumps over 1.5 hours = one pump doing 13750 per 1.5 hours = 153l/m from one pump. 480 minutes in 8 hours. 500000/153=3268. 3268/480= 6.8. Thus 7 seven pumps, i had to round up as I love doing these in my head. ( And hence why I am always late.) My question is based on work. Not too sure it fits the brief but I will post another one if you guys do not think it fits the brief. A differential wheel and axle, with a wheel radius of 24cm and a radii of the axle are 4cm and 6cm. It requires an effort of 144kg N to lift 2240kg. What is the efficiency of the machine for this load? Edited September 9, 2020 by danydandan Link to comment Share on other sites More sharing options...
Ozymandias Posted September 9, 2020 #365 Share Posted September 9, 2020 2 minutes ago, danydandan said: 41250litre with three pumps over 1.5 hours = one pump doing 13750 per 1.5 hours = 153l/m from one pump. 480 minutes in 8 hours. 500000/153=3268. 3268/480= 6.8. Thus 7 seven pumps, i had to round up as I love doing these in my head. My question is based on work. A differential wheel and axle, with a wheel radius of 24cm and a radii of the axle are 4cm and 6cm. It requires an effort of 144kg N to lift 2240kg. What is the efficiency of the machine for this load? I'm an engineer and I don't clearly understand what you are asking! Normally, a differential wheel and axle mechanism uses the wheel to input the effort and the axle to lift the load. Why are their two axle radii? Also, re 144kgN , what is a kgN. Effort is measured in newtons (N), mass is measured in kg. A little more info and clarity would be helpful. Also, this is a mechanical engineering problem more than a maths one, Dan. 1 Link to comment Share on other sites More sharing options...
danydandan Posted September 9, 2020 Author #366 Share Posted September 9, 2020 5 minutes ago, Ozymandias said: I'm an engineer and I don't clearly understand what you are asking! Normally, a differential wheel and axle mechanism uses the wheel to input the effort and the axle to lift the load. Why are their two axle radii? Also, re 144kgN , what is a kgN. Effort is measured in newtons (N), mass is measured in kg. A little more info and clarity would be helpful. Also, this is a mechanical engineering problem more than a maths one, Dan. What I had in my head was a differential wheel and an axle that consists of two portions of different radii. The weight is lifted by a pulley which is carried in a loop of rope which wound around the two portions of the axle in different direction. The 144kg N is a typo...... meant to be 144g to N (1.4 N give or take) as a conversion. I will go back to basics. We can leave this one be, the answer is 12.5%. Statistics: A light rod has particles of unit mass fixed at a distance of 3, 7 ,8, 12 & 15 cm from the end of the rod. Find the distance of the centre of gravity of the system from the end of the rod. Link to comment Share on other sites More sharing options...
Ozymandias Posted September 10, 2020 #367 Share Posted September 10, 2020 (edited) 15 hours ago, danydandan said: What I had in my head was a differential wheel and an axle that consists of two portions of different radii. The weight is lifted by a pulley which is carried in a loop of rope which wound around the two portions of the axle in different direction. The 144kg N is a typo...... meant to be 144g to N (1.4 N give or take) as a conversion. I will go back to basics. We can leave this one be, the answer is 12.5%. Statistics: A light rod has particles of unit mass fixed at a distance of 3, 7 ,8, 12 & 15 cm from the end of the rod. Find the distance of the centre of gravity of the system from the end of the rod. That's another mechanical engineering question rather than a pure maths one but I'll answer it. We have to assume from the lack of information that the 'light' rod is massless and is to be ignored in the analysis. This is a simple application of the Principle of Moments, so make one end your chosen fulcrum and we will call the Centre of Gravity (CofG) of the whole system from that fulcrum D (in centimetres). Moments are measured in newtonmetres (Nm) but I am going to take a shortcut here and leave everything in kilogrammes (kg) and centimetres (cm) because the units will cancel each other out, although that is not good practice when teaching. Normally kg x 9.81m/s^2 = N and cm/100 = m. So,.......... Sum of the Anticlockwise Moments = Sum of the Clockwise Moments (Total Mass of System) x (Distance of CofG to Fulcrum) = (Sum of Individual Moments of each of the Masses) (1kg + 1kg + 1kg + 1kg + 1kg) x C = (1kg x 3cm) + ( 1kg x 7cm) + (1kg x 8cm) + (1kg x 12cm) + ( 1kg x 15cm) 5kg x Dcm = 3kgcm + 7kgcm + 8kgcm + 12kgcm + 15kgcm 5Dkgcm = 45kgcm Dcm = 45kgcm/5kg Dcm = 9cm [the kg unit cancelled out] Answer: The Centre of Gravity of the system lies 9cm along the rod from the chosen fulcrum Edit: That was not a maths problem, Dan. Edited September 10, 2020 by Ozymandias 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 10, 2020 #368 Share Posted September 10, 2020 (edited) Problem: A chain used to lay out a right-angle triangle on a building site has 192 identical links in it. The first and last links in the chain are hooked together over a single spike or peg and driven into the ground. Through which other links in the chain must the other two spikes be driven to create the right-angled triangle. Edited September 10, 2020 by Ozymandias Link to comment Share on other sites More sharing options...
danydandan Posted September 10, 2020 Author #369 Share Posted September 10, 2020 7 hours ago, Ozymandias said: Problem: A chain used to lay out a right-angle triangle on a building site has 192 identical links in it. The first and last links in the chain are hooked together over a single spike or peg and driven into the ground. Through which other links in the chain must the other two spikes be driven to create the right-angled triangle. I am struggling with this, Is the correct method k = 192/sin45 + sin45 + sin90 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 10, 2020 #370 Share Posted September 10, 2020 (edited) 58 minutes ago, danydandan said: I am struggling with this, Is the correct method k = 192/sin45 + sin45 + sin90 It has to be a right-angled triangle so the spikes go through those links in the chain that divide it into three sides of a triangle having ratios to each other of some Pythagorean triple. Edited September 10, 2020 by Ozymandias Link to comment Share on other sites More sharing options...
danydandan Posted September 11, 2020 Author #371 Share Posted September 11, 2020 8 hours ago, Ozymandias said: It has to be a right-angled triangle so the spikes go through those links in the chain that divide it into three sides of a triangle having ratios to each other of some Pythagorean triple. So using a 3:4:5 ratio I'm getting 56 for second pin, 112 for next pin. Third pin has link 1 and 192. Link to comment Share on other sites More sharing options...
Ozymandias Posted September 11, 2020 #372 Share Posted September 11, 2020 3 hours ago, danydandan said: So using a 3:4:5 ratio I'm getting 56 for second pin, 112 for next pin. Third pin has link 1 and 192. Certainly on the right track, Dan. The 112 pin is correct. Will you show us your solution method. Link to comment Share on other sites More sharing options...
danydandan Posted September 11, 2020 Author #373 Share Posted September 11, 2020 7 hours ago, Ozymandias said: Certainly on the right track, Dan. The 112 pin is correct. Will you show us your solution method. Hi Ozy, I done this in my head last night and I am pretty sure 112 is a fluke because I cannot remember how I got it. Let me try below, typing out loud so to speak. Assuming a perimeter of 192. 3x+4x+5x=192 12x=192 192/12=16=x So. 3x=3.16=48. 4x=4.16=64. 5x=5.16=80. 48+64+80=192. So pin one goes into 1 and 192, pin two goes into 48 (not sure where I got 56 from and subsequently getting 112 to be honest.) and third pin goes into 112. Link to comment Share on other sites More sharing options...
Ozymandias Posted September 11, 2020 #374 Share Posted September 11, 2020 1 hour ago, danydandan said: Hi Ozy, I done this in my head last night and I am pretty sure 112 is a fluke because I cannot remember how I got it. Let me try below, typing out loud so to speak. Assuming a perimeter of 192. 3x+4x+5x=192 12x=192 192/12=16=x So. 3x=3.16=48. 4x=4.16=64. 5x=5.16=80. 48+64+80=192. So pin one goes into 1 and 192, pin two goes into 48 (not sure where I got 56 from and subsequently getting 112 to be honest.) and third pin goes into 112. That's it, Dan. Your go! Link to comment Share on other sites More sharing options...
Ozymandias Posted September 12, 2020 #375 Share Posted September 12, 2020 !!!!!!!!!!!!!!!!!!!!!!!!! Link to comment Share on other sites More sharing options...
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