What will be the least possible number of plants , if three pieces of timber 42m, 49m, 63m long have to be divided into planks of the same length 

The equal length under the worst case scenario is the hcf of the length of the planks.

HCF(42,49,63)= 7

Therefore, each plank length is of 7 m.

Now, no. Of planks from 42 m one = 42/7 = 6

From 49 m= 49/7 = 7

From 63 m = 63/7 = 9

So, total no. Of planks required = 6+7+9 = 22
  • 6
IF U WANT THE LEAST U WILL TAKE HCF 
hcf of the numbers 
42 , 49 , 63 will be = 7 

one plank will be = 7m 

plank a = 42m/7m = 6 
plank b = 49m/7m = 7 
plank c = 63m/7m = 9 

total number of planks will be 6 + 7 + 9 = 22
  • 27
IF U WANT THE LEAST U WILL TAKE HCF , hcf of the numbers 42 , 49 , 63 will be = 7 one plank will be = 7m plank a = 42m/7m = 6 plank b = 49m/7m = 7 plank c = 63m/7m = 9 total number of planks will be 6 + 7 + 9 = 22
  • 6
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