GATE2011 AG: GA-8

Question

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?

• A. $38$
• B. $31$
• C. $48$
• D. $41$

Comments

GO BACK…

Let the current toffees left at any point is C.

 T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left,

C/2 = 17 – 2

C = 30 toffees ( after R and S taken)

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S took 1/4th of what was left but returned three toffees to the bowl.

3C / 4  = 30 – 3

C = (27* 4) / 3

C = 36   (after R taken )

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R took 1/3rd of the toffees, but returned four to the bowl

2C / 3 = 36 – 4

C = (3*32) / 2

C = 48 Answer.

PS : Here all of the tree C’s are different. just showing the current moment left toffees.

 

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@Shaik Masthan sir pls explain this question 

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@pavansan, All answers on this question are explained very well and have unique approach.

I have already added my answer earlier itself. Nothing left simplify it further. I suggest you to just explore every answer. 

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if you analyse the options you will see there is on one option which is multiple of 3 so this can be another approch to solve this.

Answer 1

Let  $x$ be the number of toffees

Number of toffee taken by $R = \frac{x}{3}$

$R$ returned $4$ to bowl

So, number of toffees in bowl $=2(\frac{x}{3})+4$

$S$ takes $\frac{1}{4}$ of it

Number of toffees taken by $S=\frac{1}{4}(\frac{2x}{3}+4)$

$S$ returns $3$ to bowl

So, number of toffees in bowl $=\frac{3}{4}(\frac{2x}{3}+4)+3= \frac{x}{2}+6$

$T$ takes $\frac{1}{2}$ of it

Number of toffees taken by $T = \frac{1}{2}(\frac{x}{2}+6)$

$T$ returns $2$ to bowl

Number of toffees in bowl $=\frac{1}{2}(\frac{x}{2}+6) +2=\frac{x}{4}+5$

Now it is given $17$ toffees left in bowl

So, $\frac{x}{4}+5 =17$

$x=48$

Answer is $C$

Comments

@pooja , I think there is no need for last line.

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toffees can't be in fraction

therefor ans must be divisible by 3,4,2

ans :48

R:   16-4=12  left 36

S:    9-3=6 left 30

T:   15-2=13 left 17

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@pooja

Instead of taking 'x',if you had taken 12x(lcm of 3,4,2) as total no of toffees initially,calculation would have been easy.

Answer 2

Shortcut method

$\rightarrow$ R took $\frac{1}{3}^{rd}$ of the toffees

$\rightarrow$ So total toffees should be perfectly divisible by 3.

$\rightarrow$ From the given options we can see that only 48 is divisible by 3.

$\therefore$ Option C. $48$ is the correct answer.

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Detailed method

Start reading from last line and try to go back in time.

If the bowl had $17$ toffees left,how may toffees were originally there in the bowl?

$\rightarrow$ It means now we have $17$ toffees in the bowl.

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T took half of the remainder but returned two back into the bowl.

$\bigstar$  returned two back into the bowl.

$\rightarrow$ Out of the $17$ toffees $2$ toffees were given by T .

$\rightarrow$ So if T hadn't returned those $2$ toffees, then there would had been $17-2 =15$ toffees in the bowl.

$\bigstar$ T took half of the remainder

$\rightarrow$ These $15$ toffees are $\frac{1}{2}$ of the remainder which T didn't took and left in the bowl.

$\rightarrow$ So this means we had $15*2=30$ toffees before T took them from the bowl.

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S took $\frac{1}{4}^{th}$ of what was left but returned three toffees to the bowl.

$\bigstar$ returned three toffees to the bowl.

$\rightarrow$ Out of the $30$ toffees $3$ toffees were given by S.

$\rightarrow$ So if S hadn't returned those $3$ toffees, then there would had been $30-3=27$ toffees in the bowl.

$\bigstar$ S took $\frac{1}{4}^{th}$ of what was left

$\rightarrow$ These $27$ toffees are $\frac{3}{4}^{th}$ of what was left in the bowl. ($\because$ S took $\frac{1}{4}^{th}$ from the bowl.)

$\rightarrow$ So this means we had $27 * \frac{4}{3} =36$ toffees before S took them from the bowl.

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R took $\frac{1}{3}^{rd}$ of the toffees, but returned four to the bowl.

$\bigstar$ returned four to the bowl.

$\rightarrow$ Out of these $36$ toffees $4$ toffees were given by R.

$\rightarrow$ So if R hadn't returned those $4$ toffees,then there would had been $36-4=32$ toffees in the bowl.

$\bigstar$ R took $\frac{1}{3}^{rd}$ of the toffees

$\rightarrow$ These $32$ toffees are $\frac{2}{3}^{rd}$ of what was left in the bowl. ($\because$ R took $\frac{1}{3}^{rd}$ from the bowl.)

$\rightarrow$ So this means we had $32* \frac{3}{2} =48$ toffees before R took them from the bowl.

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$\therefore$ $48$ toffees were there in the bowl before any event happened.

So Option C. $48$ is the correct answer.

Comments

good approach sir :)

Answer 3

Let total no of toffees initially =  12x.(we have taken 12 i.e. lcm of 3,4,2 to make calculation easy)

After R's turn, total no of toffee left =  12x-4x + 4 = 8x + 4.

After 'S' has taken toffee,no of toffee left  =  (8x + 4) - (2x +1) + 3 = 6x +6

After 'T' has taken toffee, no of toffee left  = (6x + 6) - (3x +3) + 2 = 3x + 5.

Total toffee left = 3x + 5 = 17(given)

3x = 12 => 12x = 48

Hence total toffee initially were '12x' i.e. 48.

Comments

perfect use of LCM

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@Shashank Kumarhow did you choose 3,4 and 2 from question?

Answer 4

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

Comments

This is right sir,but it will take too much time solving this in real time,more error prone in high pressure situation.

Answer 5

Let it do it in reverse direction and in the end we will reach at total number of toffee. 

(17 - 2) * 2 = 30 (No. of toffee in the bowl before T picks).

(30 - 3) * (4/3) = 36 (No. of toffee in the bowl before S picks).

(36 - 4) * (3/2) = 48 (No. of toffee in the bowl before R picks).

So Option (C) is the answer. I hope this helps. :)

Answer 6

Answer Is solved with the guidance of Kapil Mittal Sir from GOClasses