Question
Three friends, R, S and T shared toffee from a bowl. R took $1/3^{rd}$ of the toffees, but returned four to the bowl. S took $1/4^{th}$ 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 many toffees were originally there in the bowl?
The correct answer is
48
Toffee Fraction Word Problem Solution
This problem requires us to determine the initial quantity of toffees by reversing the actions described, starting from the final count.
Working Backward Solution Steps
- Final State: There are 17 toffees remaining in the bowl.
- Reverse T's Action: T returned 2 toffees. Before returning them, the count was $17 + 2 = 19$. T had taken half the available toffees ($N_T$) before returning 2. This means the $19$ toffees represent the half T didn't take plus the 2 returned. Let $N_T$ be the number of toffees before T took his share. T took $N_T / 2$. The amount remaining after T took his share was $N_T - N_T / 2 = N_T / 2$. After returning 2, the amount became $N_T / 2 + 2$. We know this equals 17.
So, $\frac{N_T}{2} + 2 = 17$.
Solving for $N_T$:
$\frac{N_T}{2} = 17 - 2$
$\frac{N_T}{2} = 15$
$N_T = 15 \times 2 = 30$. This means 30 toffees were present after S finished his actions. - Reverse S's Action: S returned 3 toffees. Before returning them, the count was $30 + 3 = 33$. S had taken $1/4^{th}$ of the available toffees ($N_S$) before returning 3. The amount left after S took his share was $N_S - \frac{N_S}{4} = \frac{3N_S}{4}$. After returning 3, the amount became $\frac{3N_S}{4} + 3$. We know this equals 30.
So, $\frac{3N_S}{4} + 3 = 30$.
Solving for $N_S$:
$\frac{3N_S}{4} = 30 - 3$
$\frac{3N_S}{4} = 27$
$N_S = 27 \times \frac{4}{3} = 9 \times 4 = 36$. This means 36 toffees were present after R finished his actions. - Reverse R's Action: R returned 4 toffees. Before returning them, the count was $36 + 4 = 40$. R had taken $1/3^{rd}$ of the original toffees ($N_0$) before returning 4. The amount left after R took his share was $N_0 - \frac{N_0}{3} = \frac{2N_0}{3}$. After returning 4, the amount became $\frac{2N_0}{3} + 4$. We know this equals 36.
So, $\frac{2N_0}{3} + 4 = 36$.
Solving for $N_0$:
$\frac{2N_0}{3} = 36 - 4$
$\frac{2N_0}{3} = 32$
$N_0 = 32 \times \frac{3}{2} = 16 \times 3 = 48$.
Original Toffee Calculation
The original number of toffees in the bowl was 48.
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Important Questions from Numerical Reasoning
- $P, Q, R, S, X$, and $Y$ are distinct single-digit whole numbers taking values from 0 to 9.
$PQ$ is a two-digit number with $Q$ being in the units place and $P$ in the tens place. Similarly, $RS$ is a two-digit number.
It is known that $PQ$ and $RS$ are consecutive numbers and
$(PQ)^2 + (RS)^2 = XYP$, with $XYP$ being a three-digit number.
The value of $Y$ is __________ - Let $p_1$ and $p_2$ denote two arbitrary prime numbers. Which one of the following statements is correct for all values of $p_1$ and $p_2$?
If $\oplus \div \odot = 2$, $\oplus \div \triangle = 3$, $\odot + \triangle = 5$, and $\Delta \times \otimes = 10$,
then the value of $(\otimes - \oplus)^2$ is:- The remainder when $98!$ is divided by $101$ is equal to ________
- A 'frabjous' number is defined as a 3 digit number with all digits odd, and no two adjacent digits being the same. For example, 137 is a frabjous number, while 133 is not. How many such frabjous numbers exist?
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