Question

divide 600 into two parts such that 40% of one exceeds 60% of the Other by 120

Answer 1

Let the one part be 'x' and the other part be 600 - x
Therefore, according to the question.
(x × 40)/100 = {(600 - x) × 60}/100 + 120
On solving both L.H.S. and R.H.S. we get
2x/5 = {(600 - x)3}/5 + 120
2x/5 = (1800 - 3x)/5 + 120 
Taking L.C.M of (1800 - 3x)/5 + 120/1, and solving it, we get
2x/5 = (1800 - 3x + 600)/5
2x + 3x = 1800 + 600
5x = 2400
x = 2400/5
x = 480 
Therefore the first part is 480 and the other is 600 - 480 = 120
Answer.


Let us cross check the answer.
1st part = 480 and 2nd part = 120
40 % of the first part exceeds 60 % of the other by 120.
40 % of 480 = 60 % of 120 + 120
192 = 72 + 120
192 = 192
L.H.S. = R.H.S.
Hence proved.

Answer 2

Answer:

Step-by-step explanation:

40% of x - 60% of y = 120

40/100 ×x -60/100×y =120

2/5x - 3/5 y = 120

2x-3y =120×5. [x+y=600]

2x-3y= 600

Here both expressions are equal to 600.

2x-3y =600

(-) x+y = 600

Both 600's get cancelled

2x-3y= x+y

2x-x =y+3y

X=4y

ATP, replace y in x

2/5of 4y -3/5 of y = 120

8y-3y/5 = 120

5y/5 =120

y = 120

X=600-120

=48