To solve this problem effectively, we can work backward step-by-step from the final quantity remaining (Backtracking Method). This avoids dealing with highly complex algebraic fractions.
Step 1: Analyzing the Third Customer Transaction
Let the number of oranges available before serving the third customer be R2.
The shopkeeper sells 51 of these remaining oranges plus 1 more. The remaining fraction of oranges after selling 51 is 54.
Since he also gives 1 additional orange away, the final number left is:
54×R2−1=3
Now, solve for R2:
54×R2=3+1
54×R2=4
R2=44×5=5 oranges
So, there were 5 oranges left before the third customer arrived.
Step 2: Analyzing the Second Customer Transaction
Let the number of oranges available before serving the second customer be R1.
The shopkeeper sells 31 of these oranges plus 1 more. The remaining fraction of oranges left over is 32.
Subtracting the extra 1 orange given away results in the 5 oranges we found in Step 1:
32×R1−1=5
Now, solve for R1:
32×R1=5+1
32×R1=6
R1=26×3
R1=3×3=9 oranges
So, there were 9 oranges left before the second customer arrived.
Step 3: Analyzing the First Customer Transaction
Let the initial total number of oranges the shopkeeper started with be X.
The shopkeeper sells half (21) of his initial stock plus 1 more. The remaining quantity left over is 21 of the stock.
Subtracting the extra 1 orange given away brings us to the 9 oranges found in Step 2:
21×X−1=9
Now, solve for the total stock X:
21×X=9+1
21×X=10
X=10×2=24 oranges
Final Answer
The total number of oranges the shopkeeper had initially is 24 oranges.
Therefore, the correct option is (b).
