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Brain teaser – A distribution manager’s dilemma

John the distribution manager for a leading mobile phone company has a dilemma. The warehouse has 10 unlabelled rows of pallets, each row contains thousands of phones destined for different countries. Each 100g mobile phone is exactly the same except for those in the row destined for Japan, which have a “special” 2g chip encased within the phone to make sure they work on the Japanese networks.

All the trucks are waiting outside ready to go their separate ways, how can John make sure the right phones go to Japan in the quickest time possible? All John has at his disposal is a digital balance.

Posted by on Mar 22nd, 2010 and filed under Discussions. You can follow any responses to this entry through the RSS 2.0. You can skip to the end and leave a response. Pinging is currently not allowed.

6 Responses for “Brain teaser – A distribution manager’s dilemma”

  1. Santhosh says:

    Weigh R1 n R2..if both are of equal weight, use either R1 or R2 against the other rows to find the odd one out..
    if R1 n R2 are of unequal weight, use R1 against R3,
    if the balance is equal then R2 is the odd one out…
    if the balance is again unequal then R1 is the odd one out..

  2. magicdrshoon says:

    Take 1 phone from row 1, 2 phones from row 2, 3 phones from row 3 etc.
    The total weight if all the phones were 100g each would be 10!, or 550g.

    X = Total weight
    Y = Row number destined for Japan

    Y = (X – 10!)/2

  3. Olie says:

    Assuming a balance scale can only John tell John which of two weights is heavier (or lighter), he can identify the heavier (102g) Japanese row from the others (100g) in 2-3 measurements using one phone from each row (R1-R10).
    1. Determine which group is heavier between R1-R5 and R6-R10. Keep the heavier group (e.g., R1-R5).
    2. Determine which pair is heavier between R1-R2 and R3-R4. Keep the heavier pair (e.g., R1-R2).
    3. Determine which phone is heavier between R1 and R2. The heavier phone is Japanese. 3 measurements.
    4. If the two pairs are equal weight in step 2, assume R5 is the heavier Japanese phone. 2 measurements.

  4. John should take a different number of phones from each 9 rows and put them all on the digital balance. Since there are 10 rows, he should get a weight of 4500 grams (by doing [9+8+7+6+5+4+3+2+1] multiplied by weight of 100g) plus the extra weight of 2g per phone from the Japanese row. So if, for example, he gets a total weight of 4516 grams, then he knows the row that he took 8 phones from is the row destined for Japan (since 8*2 equals the extra 16g in the weight.) If the weight comes out to be exactly 4500 grams, then John will know that the row which he did not take out any phone from is the one destined to Japan, since no extra weight of 2g was added to the scale.
    Let me know if this makes sense ;)

  5. Walker says:

    I like the brain teasers!

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