The dimensions of a certain machine are 48″ X 30″ X 52″. If the size of the machine is increased proportionately until the sum of its dimensions equals 156″, what will be the increase in the shortest side?

• as per the given condition let us say the proportionality factor is p.
  so the new dimensions are
  48p , 30p, 52p
  now sum of these will be 156
  48p+30p+52p = 156
  130p = 156
  p = 6/5
  
  so we have shortest new side = (6/5)*30 = 36
  increment is = 36-30 = 6 ans
• 10 years agoHelpfull: Yes(41) No(1)
• let the ratio
  48 : 30 : 52 = 130
  to make 130 to 156 then multiply both sides with 156/130
  so smallest dimession is 30*156/130=36
• 10 years agoHelpfull: Yes(5) No(0)
• just treat this as a ratio problem
  here sum of dimensons=48+30+52=130
  we increase dimension proportionally upto sum = 156
  so we hv to distribute 156-130=26 unit in dimensions
  so chnage in shortest side=(30/130)*26 = 6 ans
  
• 10 years agoHelpfull: Yes(1) No(0)
• SHORTEST SIDE I.E 30" sholud be increased by 7 ...
  proption will be 48 + 9,30+7,52+10
• 10 years agoHelpfull: Yes(0) No(1)
• ratio of dimensions is 48:30:52
  sum of dimension is 48+30+52=130
  required sum = 156
  remaining 156-130 = 26
  now 26 is divided in the ratio of its dimension which is 48:30:52
  the qus is ask about increasing in shortest dimension
  so 30*26/130 = 6 ans
• 10 years agoHelpfull: Yes(0) No(1)