A long thin strip of width 7cm is kept on a flat surface.
Another identical strip is kept on it in an overlapping
manner such that the combined width of the two is 11 cm.
What is the width of 55 such strips kept like this?

• 11-7=4 cm
  
  4*54=216
  
  216+7=223 cm
  
  ans= 223
• 13 years agoHelpfull: Yes(19) No(3)
• I believe..
  first strip width is 7cm
  And after keeping another strip the width become 11cm
  so now both are different strip.
  now 55 such strip are kept.means 27*11+7=309(as 54/2=27 and the width of last one is 7)
• 13 years agoHelpfull: Yes(16) No(9)
• first strip width is 7cm,
  second strip width is 4cm
  according to this, the combined 54 strips width is 594cm,55th strip is 7cm
  so that 55 strips width is 601cm
• 13 years agoHelpfull: Yes(2) No(5)
• Their is 55 strips..width of single strip is 7cm and pair having 11 cm width.So
  Given 55 strip and their 27 pair and one single strip.
  so that 27*11+11=304
• 13 years agoHelpfull: Yes(2) No(1)
• 11*55=605cm
• 13 years agoHelpfull: Yes(1) No(6)
• 304..
  two strips=>11cms
  27*2+1=54+1=55
  so 27*11+7=304cms
• 13 years agoHelpfull: Yes(1) No(1)
• 11-7=4
  4*54=216
  216+7=223
• 13 years agoHelpfull: Yes(1) No(1)
• (11/2)*55 i have done this
  
• 13 years agoHelpfull: Yes(0) No(4)
• after putting 2nd strip extra width is 11-7 = 4cm
  we are putting 54 such strip over by over on 55th strip and so on.....
  then total width will be = 7cm + 54* 4 = 223 cm
  so the ANS is 223 cm.
• 13 years agoHelpfull: Yes(0) No(0)
• Width of two strips(overlapping)=11cm
  width of 54 such strips(taking 2 at a time i.e.54/2=27)=27*11=297cm
  adding one more strip(width of each is 7cm)=297+7=304cm
  ANS=304cm
• 13 years agoHelpfull: Yes(0) No(1)
• 605...as we multiply 55*11=605
• 12 years agoHelpfull: Yes(0) No(1)
• if their are two such strips are combined then 3 cm is common in between two , & 4cm and 4cm are uncovered in the two
  if their are 5 such strips then 4cm are uncommon in between them .
  and their are 3 overlapped strips means n-2 where n is no of strips oflength 3cm
  and their are 3 unoverlapped strips means n-2 of length 1 cm
  so total length is 4+(3*3)+(3*1)+4=20cm
  so if their are 55 strips then length is =
  4+(3*(55-2))+(1*(55-2))+4=220cm
• 12 years agoHelpfull: Yes(0) No(1)
• According to the question,
  the overlapping part must be 3 cm or in other words each time a new strip is added , the total width increases by 4cm.
  As there are total 55 strips,
  54 more strips were added to the initial 7cm strip.
  This gives us that there was total 54 times 4cm width was added to the 7cm strip width.
  therefore, width of such 55 strips= [7 + 54(4)]cm
  = 7+216 cm
  = 223 cm
  
• 12 years agoHelpfull: Yes(0) No(1)
• in general form, we can write the condition as
  length of n strips = 7n-3(n-1)
  {Checking :- put n=2 (i.e., when 2 strips kept together), we get length as 11cm.}
  Now,
  putting n=55, we will get our answer = 223.
• 8 years agoHelpfull: Yes(0) No(0)