A Simple Problem of Cherries

Let $c$ be the number of cherries in the bag. If Swarna eats half of the cherries, then the remainder is $c-\frac { 1 }{ 2 } c$ . If Madhur eats $\frac { 1 }{ 3 }$ of that, then the remainder is $\left( c-\frac { 1 }{ 2 } c \right) -\frac { 1 }{ 3 } \left( c-\frac { 1 }{ 2 } c \right)$ . If Bala eats $\frac{ 1 }{ 4 }$ of THAT, then the remainder is $\left[ \left( c-\frac { 1 }{ 2 } c \right) -\frac { 1 }{ 3 } \left( c-\frac { 1 }{ 2 } c \right) \right] -\frac { 1 }{ 4 } \left[ \left( c-\frac { 1 }{ 2 } c \right) -\frac { 1 }{ 3 } \left( c-\frac { 1 }{ 2 } c \right) \right]$ .

Since there are 6 cherries remaining, it can be said that $\left[ \left( c-\frac { 1 }{ 2 } c \right) -\frac { 1 }{ 3 } \left( c-\frac { 1 }{ 2 } c \right) \right] -\frac { 1 }{ 4 } \left[ \left( c-\frac { 1 }{ 2 } c \right) -\frac { 1 }{ 3 } \left( c-\frac { 1 }{ 2 } c \right) \right]=6$ , which when simplified gives $c=24$ .

Lots of algebra!

I ate only 4 cherries out of 24? Impossible!

Let $x$ be the original number of cherries. Swarna eats $\dfrac{1}{2}$ of them, so the remainder is $x-\dfrac{1}{2}x=\dfrac{1}{2}x$ . Madhur eats $\dfrac{1}{3}$ of the remainder, so the remainder is $\dfrac{1}{2}x-\dfrac{1}{3}\left(\dfrac{1}{2}x\right)=\dfrac{1}{3}x$ . Bala eats $\dfrac{1}{4}$ of the remaining cherries and $6$ cherries were left. So, $\dfrac{1}{3}x-\dfrac{1}{4}\left(\dfrac{1}{3}x\right)=\dfrac{1}{4}x=6$ . Hence, $x=4\times 6 =\boxed{24}$

assume number initially be 12, (LCM of 2, 3, 4)

Swarna ate 12/2 = 6 then remaining = 6

Madhur ate 6/3 = 2 remaining = 4

Bala ate 4/4 = 1 remaining = 3

when 3 cherries remaining, initially it was 12

so when 6 cherries remaining, initially it was 12/3 × 6 = 24

$6\div \frac { 3 }{ 4 } \div \frac { 2 }{ 3 } \div \frac { 1 }{ 2 } =\boxed{24}$