Ramesh, a local book vendor in Kozhikode, buys blank journals at a fixed price of Rs. 26 each. The journals are hand-illustrated by a street artist, and Ramesh increases the selling price of each subsequent journal by Rs. 3 compared to the previous one starting with Rs. 2. i.e., he sells the first journal for Rs. 2, the second for Rs. 5, the third for Rs. 8, and so on.
What is the minimum number of journals he must sell if he does not want to make a loss?
Note: Put the answer as an integer without any padded zeroes or decimal points. For example, if the answer is 100, then please put 100 as the answer and not 100.0 or 0100 or 00100.
Answer (1)
To determine the minimum number of journals Ramesh must sell to avoid a loss, we need to calculate both his costs and his revenues and set the revenue equal to or greater than the cost for him to break even or make a profit.
Step-by-Step Calculation:
Cost Calculation:
Ramesh buys each journal for Rs. 26.
If he sells n journals, his total cost C is: C = 26 n
Revenue Calculation:
He sells the first journal for Rs. 2, the second for Rs. 5, the third for Rs. 8, and so on. This sequence indicates an arithmetic series where:
The first term a 1 = 2
The common difference d = 3
The selling price for the n t h journal is given by: a n = 2 + ( n − 1 ) × 3 = 3 n − 1
The total revenue R when n journals are sold is the sum of the first n terms of this arithmetic series: R = 2 n × ( 2 + a n ) = 2 n × ( 2 + ( 3 n − 1 )) R = 2 n × ( 3 n + 1 )
Break-even Condition:
Ramesh does not want to make a loss, so R ≥ C .
Set up the inequality: 2 n ( 3 n + 1 ) ≥ 26 n
Simplifying gives: 3 n 2 + n ≥ 52 n 3 n 2 − 51 n ≥ 0
Factor out n : n ( 3 n − 51 ) ≥ 0
The solutions require solving the quadratic inequality, n ≥ 17 .
Thus, Ramesh needs to sell a minimum of 17 journals to ensure he does not incur a loss.
