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more logical problems

The 12-coin problem has appeared many many times in different puzzle publications and probably a lot of students or tutors know the answer.
I will post here some new logical problems, and if the tutors and students like to answer them, I will post more of them in future.
Here is the first two problems:

1- We ordered 9 boxes 100-ohm resistors and one box of 120- ohm resistors. When we received them, they looked same size and same appearance resistors. Each box contained 25 similar resistors.
We called our electrical engineer to identify the 120-ohm resistor box. How many resistance measurements he shoud make to identify that box. ?

2- In a family, there is 1 grandmother, 1 grandfather, 2 fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters, 2 sons, 2 daughters, 1 father-in-law, 1 mother-in-law, 1 daughter-in-law. Wat is the smallest possible number of people in this family?

822 day(s) ago

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Answers (12)

matherapist

well done!! It's much better to work out the solution with a bit of help than have the answer given to you on a silver platter, I assume you can generalize the approach to other situations.

M

Posted 818 dy ago

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math_fanatic

Okay, I think I figured it out the resistor problem:

1) Put one box aside
2) Take 1 resistor from the first box, 2 resistors from the second box, 3 resistors from the third box, 4 resistors from the fourth box, 5 resistors from the fifth box, 6 resistors from the sixth box, 7 resistors from the seventh box, 8 resistors from the eight box, and 9 resistors from the ninth box. Put them in a series.

3) Take one measurement of the series.

If the 120 Ohm resistors are NOT in the series the total resistance of the series will add up to 4500 ohms. If you get this result, the120's are in the box you put aside.

If your 120's are in the series you will be able to tell which box they're in based on the total resistance given by your one and only measurement:

Let TR = total resistance of the series

If the TR is 4520, the 120's are in the 1st box.
If the TR is 4540, the 120's are in the 2nd box.
If the TR is 4560, the 120's are in the 3rd box.
If the TR is 4580, the 120's are in the 4th box.
If the TR is 4600, the 120's are in the 5th box.
If the TR is 4620, the 120's are in the 6th box.
If the TR is 4640, the 120's are in the 7th box.
If the TR is 4660, the 120's are in the 8th box.
If the TR is 4680, the 120's are in the 9th box.

An easy way to figure out which box it is without having to know the precise measurement you would need in order to determine whether the 120's are from which box is to subtract 4500 from your TR and divide by 20. You'll get an answer of 1, 2, 3, 4, 5, 6, 7, 8, or 9. These numbers correspond to the box your 120's are in.

Hope that's right.

Posted 818 dy ago

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math_fanatic

I don't get the second balance problem posted by raj but I want to try to answer the resistor problem again using maththerapist's hints and knowing now that I can be sure after one measurement of which box the resistors are in. It's not as important to me to figure out the fewest number of measurements as to understand WHY I only have to make one measurement. Please give me a couple of days before posting the logic behind the answer.

Posted 818 dy ago

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matherapist

I need to know what you mean by a "balance " scale. Is it one in which known weights are put on one side of the scale and the unknown on the other, or is it used as a relative measure of weight (one side is heavier or lighter).? The first use is normally referred to as a triple beam balance which has one platform the unknown and can measure to the 1/10's of grams.

For the case of the triple beam or putting known weights, The answer is the same logic as your #1 problem and involves taking a different number of coins from each bag. ie 1 weighing.

For the relative measue, much like the 12 coin problem, make use of the fact that one bag is lighter.
Comapare 3 bags against 3 other bags on a pan balance. If they are equal (balance), the lighter bag is in the other 3. put one of these 3 reamaining on each side (using 2 of the 3) and if one side goes up.that's the one-if they balance , the remaining one is it. I will leave it to the student to figure out how to do it if the first measurement of 3bags vs 3 bags does not balance.

For the record, I have not seen this one either but the principles are the same.

Posted 819 dy ago

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I will post here a different version of problem # 1, may be matherapist will recognize it.

An Arab Sultan orders 10 bags of coins, each bag supposed to contan 100 coins of 24 carat pure gold, and each coin weighing 50 grams exactly. However, the jeweller had only 9 bags of real gold, so he adds1 bag of gold plated copper coins looking exactly similar to the real gold bags, but each coin weighing 49.5 grams instead of 50. The Arab Sultan, willing to check the trustworthiness the jeweller, wants to use an honest balance scale to see if the coins are all 50 gram each. If he knows that all 100 coins in each bag have similar weights, how many weighings he has to do to find the gold-plated coin bag?

This version of problem # 1 posted before is solved using exactly the same logical procedure.

Posted 820 dy ago

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matherapist

I will answer question 1 of Professor rajbaba78's problems. Suffice it to say I agree with him that the answer is one, and I do not recall seeing a problem of this type in the past (it may have been too easy for us 1950ish Cornell engineers). The answer occured to me while watching the Olympic pairs dancing.

I will take you through the thought process and give you a hint or two that will enable you, the student, to solve this.

First of all I concluded that any solution that involved taking only 1 item from each box was the obvious thing to do, but would not work. Using series and parallel resistor theory, I could do it in 3 or less, but that's when I thought there were 10 boxes, not 9.

Thus I concluded that if I took different numbers from each box, and used the sum of an arithmetic series, I could do the problem and the number of boxes was immaterial as long as the number of items in each box was greater than the number of boxes. Using this hint, perhaps students can solve this. It doesn't matter if the boxed items are resistors, coins, or marbles

Matherapist.

Problem 2 has been answered, and once i saw that gandpa and grandma could be inlaws, 7 followed fairly quickly. It sort of reminded me of an old song "I'm my own gandpa" which you can find on youtube.

Posted 820 dy ago

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math_fanatic has answered the second question correctly which is 7 people in the family. The first problem's answer is just one measurement. It is not just choosing the correct box by luck, it is a sound logical procedure.
I will wait a couple more days, and if I don't see the correct anawer, then I will post it myself. matherapist should know the answer, because different versions of this problem were available that were posted with 12 coin-problem many years ago!.

Posted 820 dy ago

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I saw matherapist's note that he saw the 12-coin's problem in1955 as a college student. It seems that matherapist is 10 years younger than I am, because I saw the 12-coin problem in 1945 in college.
These days, you can see its solution online at several WEB addresses, just go to Google page and type
12 coin or 12 ball problem and read more than a dozen solutions indicating that 3 weighing would be sufficient.

Posted 821 dy ago

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math_fanatic

MFB, Obviously if you just pick a random box and test the resistance of a resistor and you just happen to choose correctly, you can do it in one measurement. The point is to find a systematic approach to minimize the number of measurements you have to take. With my method I've ensured that the maximum it will require is 4. If there is someone out there who can come up with a lower maximum of measurements without using trial and error, please show us how.

Matherapist and Raj, I have not heard many of these problems. I heard about a coin problem similar to the one you posted a few weeks ago but it did not require me to determine whether the counterfeit was lighter or heavier and I solved it on my own without looking it up. I had to come up with a new solution to the problem to deal with determining whether the weight was heavier or lighter but I understood the concept because I just solved this problem a few weeks ago.

The resistance problem is not that much different than the coin problem except that it requires you to have some knowledge about E&M. I don't happen to know much about it but I was fortunate to find the Wikipedia page about measuring the resistance of resistors in a series. I still do not know if I did the problem correctly.

I would not even bother to answer these questions if there wasn't some challenge in it for me. So please feel free to post whatever brain teasers you like even if they're old. Not all of us have heard them.

Posted 821 dy ago

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math_fanatic

1) To answer the first question, I need to know something about how resistance is measured. I looked it up on wikipedia and it looks like you can put resistors in a series and then test the total resistance of a group of resistors. I'm going to go on that assumption since it seems the most likely scenario to solve this type of problem.

First: Divide the boxes into two groups. Take a resistor from each box in group 1 and put them in a series. If the resistance is 520 (the sum of the resistances of all of the resistors in the series), you know the 120 ohm box is in that group. If the total resistance is 500, the 120's are in the other group of 5 boxes.

Second: After you've decided which group of 5 boxes your 120 ohm resistors are in, take a resistor from each box then put one aside. Place the remaining 4 in 2 series of 2 each. Test the single resistor first. If it's 120, you're done in 2 measurements. If not, the 120's are in one of the series of 2 and you must take 2 additional measurements.

Third: Test one of the series of two resistors. If you get 220, the 120's are in that group of two. If you get 200, they're in the other group of two.

Fourth: After you identify the series of 2 that your 120's are in, test 1 of the resistors in that series. If the resistance is 120, that one belongs to the box of 120's. If not, the one you didn't measure is from the box of 120's.

This solution requires either 2 or 4 measurements. If there is something I'm missing about measuring resistance, please explain. I'm not a physicist or an electrical engineer.

2) I don't know what you mean by a "family". Do you mean these are the only people in the family (which is not very likely) or do you just mean that these are people that are in a family with many people not listed here? I'm going to try to answer both questions, starting with the latter.

SOLUTION 1:
A female can be a grandmother, a mother, a child, a grandchild, a sister, a daughter, a mother-in-law, and a daughter -in-law.

A male can be a grandfather, a father, a child, a grandchild, a brother, a son, and a father - in -law.

For the females this leaves a mother, 2 children, 1 grandchild, 1 sister, and 1 daughter.
A female can be a mother, a child, a grandchild, a sister, and a daughter.

For the males this leaves a father, 1 child, and a son. A male can be all of these things.

So the minimum number of people in an extended family with these titles is 4.

SOLUTION 2:

Under the condition that these are the ONLY people in the family and they must all be related to each other by the titles you've listed, the answer is 7. Here is the breakdown.

GM = GRANDMOTHER, GF = GRANDFATHER, F1 = FATHER 1, F2 = FATHER 2
M1 = MOTHER 1, M2 = MOTHER 2, C1 = CHILD 1, C2 = CHILD 2, C3 = CHILD 3, C4 = CHILD 4
G1 = GRANDCHILD 1, G2 = GRANDCHILD 2, G3 = GRANDCHILD 3, B = BROTHER
SS1 = SISTER 1, SS2 = SISTER 2, S1 = SON 1, S2 = SON 2, D1 = DAUGHTER 1, D2 = DAUGHTER 2
FIL = FATHER IN LAW, MIL = MOTHER IN LAW, DIL = DAUGHTER IN LAW

Person 1 is: GM, M1, MIL
Person 2 is: GF, F1, FIL
Person 3 is: F2, C1, S1
Person 4 is: M2, DIL
Person 5 is: C2, G1, SS1, D1
Person 6 is: C3, G2, SS2, D2
Person 7 is: C4, G3, B, S2

Family tree:
P1(Person 1) is the grandmother of P5, P6, P7. P1 is the mother of P3, and mother in law of P4.
P2 is the grandfather of P5, P6, P7. P2 is the father of P3, and father in law of P4.
P3 is the father of P5, P6, P7. P3 is the child and son of P1 and P2.
P4 is the mother of P5, P6, P7. P4 is the daughter in law of P1 and P2.
P5 is the child and daughter of P3 and P4. P5 is the grandchild of P1 and P2. P5 is the sister of P6,P7.
P6 is the child and daughter of P3 and P4. P6 is the grandchild of P1 and P2. P6 is the sister of P5,P7.
P7 is the child and son of P3 and P4. P7 is the grandchild of P1 and P2. P7 is the brother of P5,P6.

Posted 821 dy ago

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MFB13

For your first problem, if your lucky, only one measurement would be needed. But if you are not lucky at all, the maximum number of measurements would be eight. So you would need from 1 to 8 measurements to be sure.

For the second problem, I think that the minimum number of people in this family would be 7. I am assuming that of a conservative type family setup.

Posted 821 dy ago

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matherapist

I first saw the 12 coin problem in 1955 at college. None of these problems are original, and It's mainly for those who havent seen them. Most can probably be looked up.

Posted 821 dy ago

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