Each one of ^@A^@ and ^@B^@ has some money. If ^@A^@ gives ^@$ 90^@ to ^@B^@ then ^@B^@ will have twice the money left with ^@A^@. But, if ^@B^@ gives ^@$ 30^@ to ^@A^@ then ^@A^@ will have thrice as much as is left with ^@B^@. How much money does each have?
Answer:
^@A^@ has ^@$ 186^@ and ^@B^@ has ^@$ 102.^@
- Let us assume ^@A^@ and ^@B^@ have ^@$ x ^@ and ^@$ y ^@ respectively.
- ^@A^@ gives ^@$ 90^@ to ^@B^@ and then ^@B^@ will have twice the money left with ^@A^@.
Money with ^@A^@ = $ ^@(x - 90)^@
Money with ^@B^@ = $ ^@(y + 90)^@ @^\begin{aligned} \therefore \space & 2(x - 90) = (y + 90) \\ \implies & 2x - 180 = y + 90 \\ \implies & 2x - y = 270 && \ldots (1) \end{aligned}@^ - ^@B^@ gives ^@$ 30^@ to ^@A^@ and then ^@A^@ will have thrice as much as is left with ^@B^@.
Money with ^@A^@ = $ ^@(x + 30)^@
Money with ^@B^@ = $ ^@(y - 30)^@ @^\begin{aligned} \therefore \space & x + 30 = 3(y - 30) \\ \implies & x + 30 = 3y - 90 \\ \implies & x - 3y = -120 && \ldots (2) \end{aligned}@^ - On multiplying ^@(1)^@ by ^@3^@ we get @^\begin{aligned} & 6x - 3y = 810 && \ldots (3)\end{aligned}@^Now, subtracting ^@(2)^@ from ^@(3),^@ we get @^\begin{aligned} & 5x = 930 \\ \implies & x = 186 \end{aligned}@^
- Now, substituting ^@x = 186^@ in ^@(1),^@ we get@^ \begin{aligned} & 2(186) - y = 270 \\ \implies & y = 372 - 270 = 102 \end{aligned}@^
- Hence, ^@A^@ has ^@$ 186^@ and ^@B^@ has ^@$ 102.^@
