Question

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 short of 200.
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Answer 1

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 .

Answer 2

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