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DIVIDE 56 INTO TWO PARTS SUCH THAT THREE TIMES THE FIRST PART EXCEEDS ONE THIRD OF THE SECOND BY 48. THE PARTS ARE.
Answers (1)
Assume two parts are x and y.
3x = y/3 + 48
9x -y=144...(1)
x +y = 56....(2)
Equation (1) + Equation (2)
10 x = 144 + 56
x= 200/10 = 20
y = 56 - 20 = 36
So such two number are 20 and 36.
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