4. Divide 60 into two parts, such that 3 times the smaller part may exceed 100 by as much as
9 times the bigger falls s

2 thoughts on “<br />4. Divide 60 into two parts, such that 3 times the smaller part may exceed 100 by as much as<br />9 times the bigger falls s”

1. Step-by-step explanation:

   Given:–

   The number = 60

   To find:–

   Divide 60 into two parts, such that 3 times the smaller part may exceed 100 by as much as

   9 times the bigger falls short of 200.

   Solution :–

   Given number = 60

   Let the smaller part be a

   Then the bigger part = 60-a

   Given condition is

   3 times the smaller part may exceed 100 by as much as 9 times the bigger falls short of 200.

   =>3a= [200-9(60-a)]+100

   =>3a = 200 -540+9a +100

   =>3a = 300-540 +9a

   => 3 a = -240 +9a

   =>3 a-9a = -240

   => -6a = -240

   =>a = -240/-6

   =>a = 40

   => 60-a

   = 60-40

   =20

   Answer:–

   The two parts for the given number 60 are 40 and 20

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2. Answer:

   Step-by-step explanation:

   Divide 60 into two parts such that 3 times the smaller part may exceed 100 by as much as 9 times the bigger falls short of 200.

   Solution –

   Let the two parts be x and ( 60 – x ) respectively .

   Here x is the smaller part and ( 60 – x ) is the larger part.

   Now, according to the question,

   100 – 3x = 200 – 9 ( 60 – x )

   100 – 3x = 200 – 540 + 9x

   12x = 640 – 200 = 440

   x = 36.67 approximately .

   So, the parts are 36.67 and 23.33 respectively .

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