Question

DIVIDE 56 INTO TWO PARTS SUCH THAT THREE TIMES THE FIRST PART EXCEEDS ONE THIRD OF THE SECOND BY 48. THE PARTS ARE.

Answer 1

let the second part be x

so the first part is x/3+48

now x/3+48+x=56

(x+3x)/3=56-48

x+3x=8×3

4x=24

x=24/4=6

therefore the second part =x=6

and first part =x/3+48=6/3+48=2+48=50

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Answer 2

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