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Sharp Brains, can you answer?

Classik

Classik

Member
Sharp brains!!! Can you answer?

I went to buy a cellphone currently sold at £50. I didn't have it. I went to my mom and borrowed £25 and also £25 from my dad. Now I have £50.Luckily for me, when I went to buy the cellphone I bought it at £45. I have £5 left. I went back to my parents and gave £1 back to my mom and £1 to my dad. I am now left with £3. I still owe my parents £24 each. Mathematically, £24 + £24 = £48k + £3 (the remainder after I had given a pound back to each)= £51

I.e., £24 + £24 + £3 = £48 + £3 = £51

Now, where did the extra £1 come from since I only had £50

Prove your brain!!!!
 
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Classik said:
Sharp brains!!! Can you answer?

I went to buy a cellphone currently sold at £50. I didn't have it. I went to my mom and borrowed £25 and also £25 from my dad. Now I have £50.Luckily for me, when I went to buy the cellphone I bought it at £45. I have £5 left. I went back to my parents and gave £1 back to my mom and £1 to my dad. I am now left with £3. I still owe my parents £24 each. Mathematically, £24 + £24 = £48k + £3 (the remainder after I had given a pound back to each)= £51

I.e., £24 + £24 + £3 = £48 + £3 = £51

Now, where did the extra £1 come from since I only had £50

Prove your brain!!!!
Click to expand...

(I'm using $ because I have no clue how to show pounds sterling)

You're looking at the situation incorrectly.

$50 was involved, originally in cash.
$45 of that was exchanged for the phone and now exists as such; $5 remains.
$2 of the $5 was repaid, and the final $3 remains in your pocket. No extra money exists.
 
Classik said:
Sharp brains!!! Can you answer?

I went to buy a cellphone currently sold at £50. I didn't have it. I went to my mom and borrowed £25 and also £25 from my dad. Now I have £50.Luckily for me, when I went to buy the cellphone I bought it at £45. I have £5 left. I went back to my parents and gave £1 back to my mom and £1 to my dad. I am now left with £3. I still owe my parents £24 each. Mathematically, £24 + £24 = £48k + £3 (the remainder after I had given a pound back to each)= £51

I.e., £24 + £24 + £3 = £48 + £3 = £51

Now, where did the extra £1 come from since I only had £50

Prove your brain!!!!
Click to expand...

Classik: You can't add the £3 because that's still their money.

It would be more correct to say £45 (cell phone) + £2 (you paid) + £3 (you did not pay) = £50 the total money in the system.

So now you owe £45 (cell phone) +£3 (you did not pay) = £48.

In other words, only £48 remains in the actual debt yet, so why you would add £3 to that debt (which is still part of the debt) to get the whole total of the system is beyond me.

This brain-teaser is similar to the grocery store trick I pulled, er, I mean thought of. :lol

You pay for a $7 item with a twenty bill and get $13 back as 2 fives and 3 ones. Then you say to the cashier, hey, I have exact bills right here so give me the $20 back and I'll give you $7 the exact amount so that you don't have to give me change! :lol That might work with someone still foggy from a hang-over.

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<style type="text/css"> <!-- @page { margin: 0.79in } P { margin-bottom: 0.08in } --> </style>
 
tim-from-pa said:
Classik: You can't add the £3 because that's still their money. It would be more correct to say £45 (cell phone) + £2 (you paid) + £3 (you did not pay) = £50 the total money in the system. So now you owe £45 (cell phone) +£3 (you did not pay) = £48. In other words, only £48 remains in the actual debt yet, so why you would add £3 to that debt (which is still part of the debt) to get the whole total of the system is beyond me. This brain-teaser is similar to the grocery store trick I pulled, er, I mean thought of. You pay for a $7 item with a twenty bill and get $13 back as 2 fives and 3 ones. Then you say to the cashier, hey, I have exact bills right here so give me the $20 back and I'll give you $7 the exact amount so that you don't have to give me change! That might work with someone still foggy from a hang-over. <!-- @page { margin: 0.79in } P { margin-bottom: 0.08in } --> <!-- @page { margin: 0.79in } P { margin-bottom: 0.08in } -->
Click to expand...
:lol But the condition is obvious. :thumbsup (....though by logic it's still 51 pounds)
 
This is similar to a case in Ohms law: 2 resistors are connected in series (R1 and R2)to a voltage source of 5V. Series current is 1A. What are the voltage drops across R1 and R2 if R1 and R2 are 10ohms and 20ohms respectively. Now compare Vs to Vdrops. Lol.. :lol How possible is that practically?
 
Classik said:
This is similar to a case in Ohms law: 2 resistors are connected in series (R1 and R2)to a voltage source of 5V. Series current is 1A. What are the voltage drops across R1 and R2 if R1 and R2 are 10ohms and 20ohms respectively. Now compare Vs to Vdrops. Lol.. :lol How possible is that practically?
Click to expand...
case in ohm law.lol.

after r2 its zero volts as you are on the ground side.
 
This dialectics was what got so many Christians stumped about the Trinity. How can three be one - fully three and fully one?
 
Classik said:
This is similar to a case in Ohms law: 2 resistors are connected in series (R1 and R2)to a voltage source of 5V. Series current is 1A. What are the voltage drops across R1 and R2 if R1 and R2 are 10ohms and 20ohms respectively. Now compare Vs to Vdrops. Lol.. :lol How possible is that practically?
Click to expand...

That's making no sense. Current is determined by the electromotive force and the system resistance (or impedance). The total series resistance is 30 ohms, so therefore according to ohms law is 5/30 = 1/6 amp. The 10 ohms has 10 X 1/6 = 5/3v and the other is twice that or 20 x 1/6 = 10/3v. The total voltage is 5/3 + 10/3 = 15/3 or 5v.
 
:toofunny :toofunny :toofunny:toofunny:toofunny:toofunny:toofunny:toofunny. I suceeded in boring through your imagination too. I can't end laughing
Jason and Tim, both tricked. (Well, at the same time both of u have tried to read into the OP). I will kind of give a + to quest. she was close to the OP true answer.
tim-from-pa said:
It would be more correct to say £45 (cell phone) + £2 (you paid) + £3 (you did not pay) = £50 the total money in the system.
Click to expand...
And Tim, :lol, take a look at your comment and compare it with:
Now compare Vs to Vdrops. Lol..:lol How possible is that practically?
Click to expand...
...and also look at Post#9 now in parallel (same 'I' source).

So, Tim and Jason look at post#9 again - (not verifying it tho)...try both conditions. When you do and get my trick you can get the OP right. This is my biggest brain-teaser of the millenium. Once again, looooooooooool:toofunny.


gooood luckkkkk
jason would have been more appropriate if I had said 'paralled' first. Figure out the OP and post#9.
 
I really don't follow you at all Classik. The puzzles are using reasoning that makes no sense whatsoever. If not following incoherence is considered tricked to you, then I guess so be it.

This must be one of those types of jokes "If you have 20 sick sheep and one dies, how many do you have left?" (A person maybe hears 26 and says 25)
 
tim-from-pa said:
I really don't follow you at all Classik. The puzzles are using reasoning that makes no sense whatsoever. If not following incoherence is considered tricked to you, then I guess so be it. This must be one of those types of jokes "If you have 20 sick sheep and one dies, how many do you have left?" (A person maybe hears 26 and says 25)
Click to expand...
:lol
 
tim-from-pa said:
That's making no sense. Current is determined by the electromotive force and the system resistance (or impedance). The total series resistance is 30 ohms, so therefore according to ohms law is 5/30 = 1/6 amp. The 10 ohms has 10 X 1/6 = 5/3v and the other is twice that or 20 x 1/6 = 10/3v. The total voltage is 5/3 + 10/3 = 15/3 or 5v.
Click to expand...

Actually Jasoncran is correct, the voltage drop across R1 is 1.66v or 3.33v tap out, and the rest is dropped across R2 as R2 must go to ground 0v.
 
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if one uses a test light and hits the ground side after a load it must be zero as one is one the ground side. the loads on a circuit must consume all the voltage. if they dont then after them theres another resistance.ie take a test light and put it on the ground side of your car's head light and it will dim.
 
rrowell said:
Actually Jasoncran is correct, the voltage drop across R1 is 1.66v or 3.33v tap out, and the rest is dropped across R2 as R2 must go to ground 0v.
Click to expand...

Hi. Welcome:wave. I doubt if we had had a conversation before. Great attempt:)
what do you think about the OP? Can you attempt it? It got some funny reactions from people when it was posted on fb.:D
 
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