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Three friends, $R, S$ and $T$ shared toffee from a bowl. $R$ took $\frac{1}{3}^{\text{rd}}$ of the toffees, but returned four to the bowl. $S$ took $\frac{1}{4}^{\text{th}}$ of what was left but returned three toffees to the bowl. $T$ took half of the remainder but returned two back into the bowl. If the bowl had $17$ toffees left, how may toffees were originally there in the bowl?

  1. $38$
  2. $31$
  3. $48$
  4. $41$
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let there are X toffee in the bowl.

R takes ${\frac{1}{3}}^{rd}$ of the toffee ===> ${\frac{x}{3}}^{}$ toffees with R and remaining = ${\frac{2x}{3}}^{}$

But R returns 4 to the bowl ==> R took ${\frac{x}{3}}^{}-4$ toffees and remaining = ${\frac{2x}{3}}^{}+4$


S takes ${\frac{1}{4}}^{th}$ of the toffee from remaining ===> ${\frac{1}{4}}^{}({\frac{2x}{3}}^{}+4)$ toffees with S and remaining = ${\frac{3}{4}}^{}({\frac{2x}{3}}^{}+4)$ = $\frac{x}{2} + 3$

But S returns 3 to the bowl ==> S took ${\frac{1}{4}}^{}({\frac{2x}{3}}^{}+4)-3$ toffees and remaining = $(\frac{x}{2} + 3+3)$ = $\frac{x}{2} + 6$


T takes ${\frac{1}{2}}^{th}$ of the toffee from remaining ===> ${\frac{1}{2}}^{}({\frac{x}{2} + 6})$ toffees with T and remaining = ${\frac{1}{2}}^{}({\frac{x}{2} + 6})$

But T returns 2 to the bowl ==> T took ${\frac{1}{2}}^{}({\frac{x}{2} + 6})-2$ toffees and remaining = ${\frac{1}{2}}^{}({\frac{x}{2} + 6})+2$


Finally left in Bowl = ${\frac{1}{2}}^{}({\frac{x}{2} + 6})+2$ = $({\frac{x}{4} + 3}+2)$ = $({\frac{x}{4} + 5})$

 $({\frac{x}{4} + 5})$ = 17

 $({\frac{x}{4}})$ = 12

x=48
Answer:

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