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Three sisters $(R, S,$ and $T)$ received a total of $24$ toys during Christmas. The toys were initially divided among them in a certain proportion. Subsequently, $R$ gave some toys to $S$ which doubled the share of $S$. Then $S$ in turn gave some of her toys to $T$, which doubled $T’$s share. Next, some of $T’$s toys were given to $R$, which doubled the number of toys that $R$ currently had. As a result of all such exchanges, the three sisters were left with equal number of toys. How many toys did $R$ have originally?

  1. $8$
  2. $9$
  3. $11$
  4. $12$
in Numerical Ability by Boss (41.9k points)
edited by | 743 views
+1
Try to solve it in reverse direction.
+3
from equations it will not take more than 2 min

 

r   s    t

2(r-s)=8  , 2s-t=8, 2t-(r-s)=8

solving these , r=11, s=7 t=6
+3
Initially :- r:s:t
r-s : 2s : t
r-s : 2s-t : 2t
2(r-s) : 2s-t : 2t - (r-s)
Now, all have same number of toys = 24/3 = 8
2(r-s) = 2s-t = 2t -(r-s) = 8
Solving these, r=11, s=7, t=6

1 Answer

+16 votes
Best answer
We can proceed in reverse order:

$R\quad  S\quad  T$
$8\quad  8\quad   8$

and apply the given steps in reverse. We will get,

$R\quad  S\quad  T$
$4 \quad 8 \quad 12$
$4 \quad 14 \quad 6$
$11  \quad 7 \quad  6$

Correct Answer: $C$
by Active (2.1k points)
edited by
0
Nice
Answer:

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