Number Properties Riddle

Question

Here is timetrader again, making up another non-GMAT question for those who want to give a break to GMAT thinking and tickle their brain with something different. This one is easy!

:arrow:

Story:
In a factory that produces auto parts one out of ten machines started malfunctioning. Shift manager found out that some of the parts in a batch were defected. He also realized that the quickest way to see the difference between the defected part and the normal one is to check the weight. Normal part weights 10 pounds while the broken one weights 9 pounds. Luckily the manager had the electronic scales handy.
:?:

Question: How many times the manager will have to use the scales to find out which machine started malfunctioning?

Answer choices:

A)1
B)3
C)6
D)9
E)10

P.S. As always - Please explain your answer!

: Question: How many times the manager will have to use the scales to find out which machine started malfunctioning? Answer choices: A)1 B)3 C)6 D)9 E)10 P.S. As always - Please explain your answer!

bhanushalinikhil · Answer

Good question for a change. But I wonder, aren't there a few assumptions that we need to consider like the maximum capacity of the weight or something

Anyways, I would go with 10. Assuming that the manager would test each part from the machines, the number of which is a multiple of 10. Though, I am pretty sure that I am wrong. Eagerly waiting for the a good detailed explaination.

rohansherry · Answer

not sure but just a try.......

the ans is one (1)

the only mulitiple of 9 and 10 is 90 or 180 .......

while figure can only disguise when defected items are multiple of 10......

supose if we have 15 items 10 defective and 6 correct...the weight will come 150 for 16 items....this is only possible with the taken pair...same all others can be done in a single time...

investasi · Answer

i'm guessing 6 times (C).

- Divide the 10 machines into 4 batches. 
- Weigh each batches (4 tests).
- If none of the machines are broken the scale should read:
30 30 30 10
- If, for example, the broken machine is on the 2nd batch the scale should read:
30 29 30 10
then we need 2 additional tests to find out which machine is broken: 
10 9 10
- 4 tests + 2 tests = 6 tests.

bhushan252 · Answer

max no of times one needs to chk is 6 times "investasi" is right 
I agree with C...+1 to you

Ibodullo · Answer

i would go with three steps.

Let's divide 10 parts into two 1) 6 and 2) 4.
Step1. Weigh 1), let's imagine that defected part in this batch and electornic scale showed 59 pounds.
Step2. Divide those 6 parts into 3 and 3. and weigh the first three for example. let's imagine we have got the defected part in the weighed batch. (even if we don't have the defect in the batch, that wouldn't change number of steps, cause we have two equal number of parts in both batches)
Step3. Out of 3 weigh 2, and if the defected part is one of two parts just weighed, remove one from electronic scale and you will be able to figure out which one is defected.

But if in Step 1 above, the wighed batch will be ok, then we know that the defected part is among 4 others.
Then, Step2. Divide those 4 parts into 2 and 2. and weigh the first pair for example. let's imagine we have got the defected part in the weighed batch. (even if we don't have the defected part in the batch, that wouldn't change number of steps, cause we have two equal number of parts in both batches)
Step3. Weigh one of the parts, and if it weighs 10 pounds, than the other is defeted.

I hope its clear

Ibodullo · Answer

I think the problem is asking min number of checks. Am i right? Max could be 10 also

prabu · Answer

Lets assume 10 machine

1)Divide half 5+5, so any one of 5 batch will have less weight. - here we need to use 1 times 
2)Out of 5, separate 2 and 3 - here we need to use 1 times
3)if 2 machine batch has malfunction then - here we need to use 1 time 
4)otherwise if 3 machine batch -here we need to use 2 times

Worst case we need to use 4 times.
best case we need to use only 3 times

/Prabu

timetrader · Answer

Great to see some thoughts!!!

To clarify. Yes it is the minimum number of times the manager will have to use the scales. (Note: NOT the least POSSIBLE number of times but the minimum number of times the scales have to be used in order to figure out, full proof, which is a broken machine)

@rohansherry You are the closest to the truth. Though I think my question wasn't clear to you. The manager needs to find the broken machine, not the broken parts. )) (he will deal with the broken parts later :lol:

)

P.S the answer is not 10. Investasi - good try. But it is possible to do less, the question is how?

) P.S the answer is not 10. Investasi - good try. But it is possible to do less, the question is how?

rohansherry · Answer

Man i cant tell you how many doors i have knocked to ask this ques from people and guess what i have got all the answers from different sources....some said 2 also which isnt in the option..... i just hope you post the answer soon....

timetrader · Answer

You won't believe how simple it seems after you do find the solution :-D

investasi · Answer

Another try: Answer is A (1 test)
Let the 1st machine produce 10^10 tools
Let the 2nd machine produce 10^9 tools
Let the 3rd machine produce 10^8 tools
Let the 4th machine produce 10^7 tools
Let the 5th machine produce 10^6 tools
Let the 6th machine produce 10^5 tools
Let the 7th machine produce 10^4 tools
Let the 8th machine produce 10^3 tools
Let the 9th machine produce 10^2 tools
Let the 10th machine produce 10^1 tools
Weigh all the tools
If none of the machines are broken the scale should read 111111111100
If for example the 10th machine is broken the scale should read 111111111090
If for example the 9th machine is broken the scale should read 111111111000
etc

rohansherry · Answer

this is the 6th time i am coming to this thread.... checking again and again..i am over curious to know the answer....I have my D day after 3 days dont do this to me :evil:

.... pls PM me the answer..... :lol:

....just kidding.... the answer should be 3 .... but how..

timetrader · Answer

Great!  investasi  - it is definitely one of the ways to do it! Kudos...
Very close to my explanation... however it can be done in a more simple way. Yes, the solution is to take different number of parts from the machines, but you don't need that many parts. But definitely that way works. The OA is A.

timetrader · Answer

To put it simply,the manager would need only:

1 part from the first machine
2 parts from the second machine
3 parts from the third machine
 
10 parts from the tenth machine

.....put all the parts on a scale.....

if the weight finishes with 9 - its the first machine
if the weight finishes with 8 - its the second machine
if the weight finishes with 7 - its the third machine

I think you got it by now :-D

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