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Post by Kimmy on Feb 22, 2010 11:17:36 GMT
After all that running about kicking balls I am ready for a fag.  Can you find a 9-letter English word where it is possible to sequentially remove a single letter to form a 8-letter word, then a 7-letter word .... right down to a single-letter word? All the words thus formed must be correct English - and actually none of then are particularly obscure. The letter removed at each stage, can come from anywhere within the word. Try starting with a word containing some of the most commonly used letters and then seeing if you can add letters to it and take them away to keep forming new words. Sounds obvious, I know, and you might suddenly find the solution is staring you in the face. There might be more than one answer to this.  |
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Post by BC on Feb 23, 2010 0:18:45 GMT
After all that running about kicking balls I am ready for a fag. Can you find a 9-letter English word where it is possible to sequentially remove a single letter to form a 8-letter word, then a 7-letter word .... right down to a single-letter word? All the words thus formed must be correct English - and actually none of then are particularly obscure. The letter removed at each stage, can come from anywhere within the word. Try starting with a word containing some of the most commonly used letters and then seeing if you can add letters to it and take them away to keep forming new words. Sounds obvious, I know, and you might suddenly find the solution is staring you in the face. There might be more than one answer to this. I found the answer to be quite startling. Startling Starting Staring String Sting Sing Sin In I Nice clue.  BC  |
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Post by Kimmy on Feb 23, 2010 9:59:50 GMT
After all that running about kicking balls I am ready for a fag. Can you find a 9-letter English word where it is possible to sequentially remove a single letter to form a 8-letter word, then a 7-letter word .... right down to a single-letter word? All the words thus formed must be correct English - and actually none of then are particularly obscure. The letter removed at each stage, can come from anywhere within the word. Try starting with a word containing some of the most commonly used letters and then seeing if you can add letters to it and take them away to keep forming new words. Sounds obvious, I know, and you might suddenly find the solution is staring you in the face. There might be more than one answer to this. I found the answer to be quite startling. Startling Starting Staring String Sting Sing Sin In I Nice clue. BC  |
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Post by Kimmy on Feb 23, 2010 10:06:50 GMT
That was a startling answer to that last question.  Did you get it?  I know you didn't or you would have told us.  Right. Time to go  Puzzle 91 A man drove from 'A' to 'B'. On the first day he travelled 1/3 of the distance. On day two he travelled 1/2 of the remaining distance. On day three he travelled 2/3 of the remaining distance. On day four, after covering 3/4 of the remaining distance, he was still 5 miles away from 'B'. How many miles had he covered so far? |
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Post by BC on Feb 23, 2010 23:22:54 GMT
Puzzle 91 A man drove from 'A' to 'B'. On the first day he travelled 1/3 of the distance. On day two he travelled 1/2 of the remaining distance. On day three he travelled 2/3 of the remaining distance. On day four, after covering 3/4 of the remaining distance, he was still 5 miles away from 'B'. How many miles had he covered so far? 180 miles to travel Day 1 covers 1/3 leaves 120 Day 2 covers 1/2 leaves 60 Day 3 covers 2/3 leaves 20 Day 4 covers 3/4 leaves 5 So the total distance is 180 miles. But the question asks how far had he travelled up to when there was 5 miles left. So the answer is... 175 miles. BC |
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Post by liz on Feb 24, 2010 8:36:45 GMT
Beat me to it.
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Post by Kimmy on Feb 24, 2010 13:05:44 GMT
175 miles: the total trip was 180 miles. Working backwards: Day 4: travelled 3/4 leaving 5 miles, which means that 5 miles is 1/4, so 20 miles were left at the start of Day 4. Day 3: travelled 2/3 leaving 20 miles, which means that 20 miles is 1/3, so 60 miles were left at the start of Day 3. Day 2: travelled 1/2 leaving 60 miles, which means that 60 miles is 1/2, so 120 miles were left at the start of Day 2. Day 1: travelled 1/3 leaving 120 miles, which means that 120 miles is 2/3, so 180 miles were left at the start of Day 1. Of the 180 miles overall, 175 were covered in the 4 days. |
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Post by Kimmy on Feb 24, 2010 13:07:50 GMT
Right back home again lets see whats inside this cube.  Puzzle 92 You are given a 10x10x10 cube composed of 1x1x1 "mini-cubes" glued together. If the outer most layer falls off, how many "mini-cubes" would have fallen off? |
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Post by BC on Feb 24, 2010 23:23:32 GMT
Right back home again lets see whats inside this cube. Puzzle 92 You are given a 10x10x10 cube composed of 1x1x1 "mini-cubes" glued together. If the outer most layer falls off, how many "mini-cubes" would have fallen off? This is quite hard to "see". There are six sides. A and B face each other, and are each 10x10 (100). They drop off = 200 C and D face each other, and are now only 8x10 (80) each. They drop off = 160 E and F face each other, and are now only 8x8 (64) each. They drop off = 128 200 + 160 + 128 = 488That's what I recon anyway. BC |
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Post by Kimmy on Feb 25, 2010 12:26:58 GMT
488. |
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Post by Kimmy on Feb 25, 2010 12:28:04 GMT
Where did I put that bucket?  Puzzle 93 At dawn on Monday a snail fell into a bucket that was 12 inches deep. During the day it climbed up 3 inches, however, during the night it fell back 2 inches. On what day did the snail finally manage to climb out of the bucket? |
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Post by BC on Feb 25, 2010 22:51:06 GMT
Where did I put that bucket? Puzzle 93 At dawn on Monday a snail fell into a bucket that was 12 inches deep. During the day it climbed up 3 inches, however, during the night it fell back 2 inches. On what day did the snail finally manage to climb out of the bucket? One to print out and do in bed with a cuppa. Will post my answer on the morro. Night night.  |
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Post by BC on Feb 26, 2010 10:10:39 GMT
Where did I put that bucket? Puzzle 93 At dawn on Monday a snail fell into a bucket that was 12 inches deep. During the day it climbed up 3 inches, however, during the night it fell back 2 inches. On what day did the snail finally manage to climb out of the bucket? I recon the answer is either Wednesday. Or Thursday.  Monday 0 + 3 = 3 Midnight 3 - 2 = 1 Tuesday 1 + 3 = 4 Midnight 4 - 2 = 2 Wednesday 2 + 3 = 5 Midnight 5 - 2 = 3 Thursday 3 ...and so on... Friday 4 ...and so on... Saturday 5 ...and so on... Sunday 6 ...and so on... Monday 7 ...and so on... Tuesday 8 ...and so on... Wednesday 9 + 3 = 12 Midnight 12 - 2 = 10 Thursday 10 + 3 = 13 Up to midnight on the Wednesday, the snail gets 12 inches up the bucket. But is he out? I think he is at the top of the bucket, but not over the rim. So although I think that will probably be the answer you're looking for, I am going for the Thursday, when he is definitely out. BC The world keeps on spinning... |
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Post by Kimmy on Feb 26, 2010 13:13:05 GMT
The following Wednesday, nine days later. |
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Post by Kimmy on Feb 26, 2010 13:16:47 GMT
No snails pace today.  Come on all you lurkers. Pull into the pit lane and post your answer or else I will do this to your computer.  Puzzle 94 At the local F1 race track, Eddie was testing the latest car. During the eight test laps, he managed to average the first two laps at 100mph, the next two laps at 102mph, the next two laps at 140mph and the last two laps at 150mph. What was his average speed for the entire 8 laps? |
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Post by liz on Feb 27, 2010 8:28:54 GMT
94. 123mph
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Post by BC on Feb 27, 2010 16:17:12 GMT
94 Using Kimmy's formula, (see page 3 of this thread), I think the answer may be 119mphTaking a lap as 10 miles, so 8 laps is 80 miles... 80/(2*((10/100)+(10/102)+(10/140)+(10/150))) = 119mph BC |
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Post by Kimmy on Feb 27, 2010 16:49:31 GMT
119mph: set the track length to be a random number, e.g. 5 miles. Therefore the first two laps took 10 miles / 100 mph = 10 / 100 hours. Similarly, the second two laps took 10 / 102 hours, etc. The total time taken for 40 miles is therefore, 10/100 + 10/102 + 10/140 + 10/150 hours. Therefore the average speed for the entire 8 laps is 40 / (10/100 + 10/102 + 10/140 + 10/150) = 119 mph. |
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Post by Kimmy on Feb 27, 2010 16:57:15 GMT
Just got this box out of the car, now not sure if its big enough for what I want.  Puzzle 96 I have a box. The top's area is 240 square inches, the front's area is 300 square inches and the end's area is 180 square inches. What are the dimensions of the box? |
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Post by BC on Feb 28, 2010 13:08:40 GMT
Just got this box out of the car, now not sure if its big enough for what I want. Puzzle 96 I have a box. The top's area is 240 square inches, the front's area is 300 square inches and the end's area is 180 square inches. What are the dimensions of the box? Top A x B = 240 = 12 x 20 Front B x C = 300 = 20 x 15 End A x C = 180 = 12 x 15 So... A = 12 B = 20 C = 15 BC |
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Post by Kimmy on Feb 28, 2010 13:20:54 GMT
Length=20, depth=12, height=15: This is easily solved with a little simple algebra. If we call the three sides A, B, and C we have: A x B = 240 A x C = 300 B x C = 180 Multiplying these together we get: A2 x B2 x C2 = 12,960,000 Which means that: A x B x C = 3600 Substituting A x B = 240 back in gives C = 15 and the rest follow. |
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Post by Kimmy on Feb 28, 2010 13:22:30 GMT
I feel like a party. Care to join me?  Puzzle 97 How many people must be at a party before you are likely to have two having the same birthday (but not necessarily the same year)? |
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Post by BC on Feb 28, 2010 23:18:21 GMT
I think I've heard something like this when I've been at a large dinner party. I think it is something like 30 odd. I don't know what the maths are to work that out though. I did a quick google search and wasn't far off. Accoring to one contributor "once you have 23 people in a room the odds of two people having the same birthday is 50%. Once you have 57 people the odds jump to over 99%". BC |
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Post by Kimmy on Mar 1, 2010 12:47:02 GMT
23. By likely, we mean greater than 50% chance. With one person there is a 0 percent chance that you'll have two people with the same birthday. With two people the probability that they won't share a birthday is 364/365. The probability that they will share a birthday is therefore 1 - (364/365). With three people the probability that they won't share a birthday is the same as for two people, times 363/365. So the probability that three people will share a birthday is 1 - (364/365) * (363/365). Notice that with each additional person added, the probability that they shares a birthday with one of the previous persons goes up, because there are fewer "free" days remaining. We keep adding people until the %age is greater than 50%. When we have 23 people the %age is 50729%. |
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Post by Kimmy on Mar 1, 2010 12:49:20 GMT
Now let me see. Have I time to post another question?  Puzzle 102 I really must get a new watch. The one I have loses exactly 20 minutes every hour. It now shows 4:00am and I know that it was correct at midnight, when I last set it. I happen to know that the watch stopped 4 hours ago, so what is the correct time now? |
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Post by BC on Mar 1, 2010 22:58:06 GMT
I really must get a new watch. The one I have loses exactly 20 minutes every hour. It now shows 4:00am and I know that it was correct at midnight, when I last set it. I happen to know that the watch stopped 4 hours ago, so what is the correct time now? Good. A nice easy one as I have to work late again tonight. | Real time | Watch time | | midnight | midnight | | 0100 | 1240 | | 0200 | 0120 | | 0300 | 0200 | | 0400 | 0240 | | 0500 | 0320 | | 0600 | 0400 |
So when the watch stopped at 0400, it was really 0600. As it is now 4 hours later, the time now is 10 a.m.BC |
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Post by Kimmy on Mar 2, 2010 10:54:07 GMT
10.00am: since the watch is losing 20 minutes every hour, for every real hour that has passed, the watch will only show 40 minutes. Since the watch shows 4.00am, we know that 240 watch minutes have passed. This therefore equals 360 real minutes and hence 6 hrs. The watch stopped 240 minutes ago, therefore the time must now be 10.00am. |
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Post by Kimmy on Mar 2, 2010 10:56:34 GMT
Better day today so I think I will take a bus ride.  Puzzle 103 A local country bus service travels from Biddulph to Leek. Along the way the bus stops at 10 different places. So, in total, there are 12 stops including the start and end stops. How many different tickets can be given out by the bus driver. For example, Biddulph to Leek, Rushton to Leek, Biddulph to Rushton, etc. |
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Post by liz on Mar 2, 2010 13:41:05 GMT
103. 66 tickets
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Post by BC on Mar 3, 2010 1:20:48 GMT
I agree. How is it I needed to modify a post that contained just 6 letters? Bloomin' keyboard.  |
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