Saturday, 27 July 2013

The One "(Work) In Progress"

While supervising the construction of Metro in Technolasia, Pavan came up with quite a weird pattern of placing the barricades near the “Work in progress” board. There are 2 pillars, Left and Right, separated by 8m long stretch. He chose seven equidistant positions in this stretch to place the barricades. But the length of the rope required to connect these depends on their order of placement. Once a barricade is placed, it is connected to its immediate left and right barricades, if there is none, then to the pillar, with a stretched rope.
Assuming Pavan bought a 57m of rope.

Question 1:
What’s the minimum amount of wire he could have used?
1)28     2) 20   3) 26   4) 24

Question 2:
If he has no bar on using the fund. What’s the length of wire he could have used (not wasted)?
(not wasted)?
1) 27    2) 28   3) 35    4) 36

Question 3:
If he applies to another such stretch, 9m long, and with 9 barricades. Then what
would be his minimum wastage? (in case of 57 m long rope)
1) 3      2) 2      3) 0     4) 1



Posted by at
Labels: , ,

8 comments:

  1. Vinaya Patil28 July 2013 at 00:37

    I hv few doubts regarding this Q..
    1)Pavan chose 7 EQUIDISTANT positions IN the trench to place the barricades rt?.. moreover all the barricades are identical.. then how does their order of placement matter??
    2)since he is placing the barricades in the trench which is 8m long.. shouldn't 8m be the min length of rope required (neglecting the part wound around the barricades to connect them) ??
    3)if 8m is the min length of rope wouldn't anything more than 8m be called wasted??
    4) what does min wastage mean??? if you say 'x' is the min wastage.. you are saying that there exists a way which does not have any wastage..
    so min wastage should be 0 rt??

    ReplyDelete
  2. rishabh chaudhary28 July 2013 at 16:21

    ans is :
    the baricades are not placed all together at single time
    for minimum wastage we put a barricade in center
    than 4+4m rope is used
    than we place one baricode to right it will take 3+1m than on its right it will take 2+1m then finally 2m is used:
    this is to be done on right also
    hence the ans :
    2(4+3+2)+4*2=26

    ReplyDelete
  3. Anonymous29 July 2013 at 10:36

    Q.1:24
    | | | | | | | | |
    Pillar 1 2 3 4 5 6 7 Pillar
    First he places the middle(4th) barricade, using 8m wire.
    Then ,on the left of this barricade,he first places the 2nd barricade( between pillar and middle barricade), using 2+2=4m wire.
    Then ,he places 1st and 3rd barricades,each using 2m(1+1=2) wire.
    Similarly,placing the barricades on the right,
    he uses:
    8+2*(4+2+2) = 24m wire

    ReplyDelete
    Replies
    1. Anonymous29 July 2013 at 20:37

      thumbs up man you found a way which is out of the box
      this seems as the correct ans

      Delete
  4. Anonymous29 July 2013 at 10:49

    Q.2 : 35
    If he has enough funds,he doesn't use his brain much and simply arranges the barricades one by one.
    For 1st barricade he uses 8m,for 2nd,he uses 7m,for 3rd,he uses 6m, and so on.
    Hence he uses :
    8+7+6+5+4+3+2=35m rope.

    ReplyDelete
  5. Anonymous29 July 2013 at 11:18

    Q.3 : No answer coming
    Minimum wastage implies maximum usage of the 57m long rope.
    Maximum usage is when he places the barricades one by one starting from the left and ending on the right.
    Thus ,he uses:
    9+8.1+7.2+6.3+....+1.8=48.6m
    Hence , the minimum wastage is
    57-48.6=8.4m ,which is not an option.

    If the stretch is 10m long with 9 barricades and we apply similar argument ,we get minimum wastage as 3m.
    So,in the question,it should have been a 10m long stretch.

    ReplyDelete
  6. Unknown31 July 2013 at 01:09

    ....answers to the question sholud be
    1. 24 m
    2. 35 m
    3. 1 m....

    ReplyDelete

Subscribe to: Post Comments (Atom)