There is a rectangular Garden whose length and width are 60m X 20m.There is a walkway of uniform width around garden. Area of walkway is 516m^2. Find width of walkway
i.1
ii.2
iii.3
iv.4
PROVIDE SOLUTION ALSO...

• ANSWER : III (3m)
  
  SOLUTION :
  
  Area of Garden = 60*20 = 1200 sq m.
  Total Area of garden and walkway = 1200+516= 1716 sq m.
  Let assume width of walk way be x
  So (60+2x)(20+2x) = 1716
  Or x = 3, -43
  3 is correct answer
  Since 3 is positive.
• 13 years agoHelpfull: Yes(65) No(6)
• why 60+2x and 20+2x ?
  
• 10 years agoHelpfull: Yes(12) No(1)
• Area of Garden = 60*20 = 1200 sq m.
  Total Area of garden and walkway = 1200+516= 1716 sq m.
  
  It means dimensions of final rectangular area are 66*26.
  
  so width of walkway = 3 m ... option (iii)
• 13 years agoHelpfull: Yes(9) No(11)
• Answer:3m
  Area of Garden = 60*20 = 1200 sq m.
  Total Area of garden and walkway = 1200+516= 1716sq m.
  Let us assume that width of the walk way be x
  therefore(60+2x)(20+2x) = 1716
  x = 3, -43
  since width can't be negative
  3 is correct answer
• 13 years agoHelpfull: Yes(8) No(4)
• let the width of rectangle be x
  so the length & breath is increased by 2x(draw on a paper
  then only u'll understand)
  so new total area along with walkway is (60+2x)*(20+2x)
  so (60+2x)*(20+2x)-60*20=516
  =>(60+2x)*(20+2x)=1716
  =>(30+x)*(10+x)=429=33*13
  =>x=3
  which is da width of gateway
• 8 years agoHelpfull: Yes(7) No(0)
• when options are given and final area is known, there is no need to go for complete solution.
  Area is 1716 sq m.
  Only possible if digit at unit place of width and breadth is 6 .
  only possible when walkway is having uniform width of 3 mtrs.
• 13 years agoHelpfull: Yes(2) No(13)
• PLEASE CAN U SIMPLY YOUR ANSWER MORE..............
• 13 years agoHelpfull: Yes(1) No(5)
• area of walkway=516=sum of areas of new rectangles formed by walkway.
  =>(60+2x)*x*2+20*x*2=516
  => x^2+40x-129=0
  => x=3
• 13 years agoHelpfull: Yes(1) No(4)
• there is a formula for these type ques.if the fence is around the rectangle then area of fence width (w) is 2w(l+b+2w)
• 9 years agoHelpfull: Yes(1) No(0)
• let, x be the pathway length and width
  area of garden =60*20=1200m^2
  area of pathway = 516m^2 (given)
  now, the total area i.e, area of garden + pathway = 1200+516= 1716
  this total area has length of 60+2x and width of 20+2x(if u draw a clear diagram u can easily understand)
  therefore, (60+2x)(20+2x)=1716;
  by solving we get x^2 +40x-129=0
  x^2+43x-3x-129=0 (129=43*3)
  x(x+43)-3(x+43)=0
  (x-3)(x+43)=0
  x=3 or x=-43
  length cannot be negative so answer is 3
  so answer is 3
• 8 years agoHelpfull: Yes(1) No(0)
• thanks now i understand the concept...........
• 13 years agoHelpfull: Yes(0) No(4)
• (2*60w) + (2*20w) + 4w^2 = 516
  120w + 40w + 4w^2 = 516
  160w + 4w^2 = 516
  (w-3)(w+43) = 0
  w = 3
  so option iii) is correct
• 8 years agoHelpfull: Yes(0) No(0)
• let whole width of garden=60+2x
  let whole length of garden=20+2x
  where x=common width of gateway
  thus,
  (60+2x)(20+2x)-1200=516
  =>1200+120x+40x+4x^2-1200=516
  4x^2+160x-516=0
  =>x=3
• 7 years agoHelpfull: Yes(0) No(0)
• Total Area is=(60*20)+516=1716 m^2.
  (60+2x)(20+2x)=1716
  => (30+x) (10+x)=429
  if we put 3 then it satisfy the above equation.
• 5 years agoHelpfull: Yes(0) No(0)