Question

DIVIDE 150 INTO THREE PARTS SUCH THAT THE SECOND NUMBER IS FIVE-SIXTHS THE FIRST AND THE THIRD NUMBER IS FOUR-FIFTHS THE SECOND.


PLEASE ANSWER CLEARLY................

Answer 1

let x , y and 150 -x -y are three part of 150 .

a/c to question,

y = 5x/6

6y = 5x ------------------(1)

again,

(150-x-y) =4y/5

750 -5x -5y = 4y

750 -5x -9y = 0

5x + 9y = 750 ---------------(2)
solve this equation ,

put equation (1) in equation (2)

6y + 9y = 750
15y = 750
y = 50

put y = 50 in equation (1)

6 × 50 = 5x

x = 60

third number = 150 -x - y = 150 -60-50=40

hence, 60, 50, and 40 are three numbers

Answer 2

let x be the first term,

therefore,

second term = (5/6)x = 5x/6
third term = [(5/6)x]4/5 = 2x/3

therefore from given condition,

x + 5x/6 + 2x/3 = 150


making the denominators common,

6x/6 + 5x/6 + 4x/6 = 150
15x/6 = 150
15x = 150x6
x = 60

therefore,

first term = x = 60
second term = 5x/6 = 50
third term = [(5/6)x]4/5 = 40