It looks like you want to analyze a problem related to the reduction in the price of oranges, where a customer can buy 4 more oranges for Rs. 1 due to the price drop. Let's break down the problem and solve it step-by-step.
Problem Statement
- Due to the reduction in the price of oranges, a customer can purchase 4 more oranges for Rs. 1 than before.
- We need to find the original price or the new price of the oranges.
Let’s define variables:
- Let the original price of one orange be Rs. xxx.
- Let the original number of oranges that can be bought for Rs. 1 be nnn.
Then:
n=1xn=\frac{1}{x}n=x1
After the price reduction:
- The price per orange becomes Rs. yyy (where y<xy<xy<x).
- The number of oranges that can be bought for Rs. 1 becomes n+4n+4n+4.
Thus:
n+4=1yn+4=\frac{1}{y}n+4=y1
Relationship between prices and quantities:
Since the problem only states the increase in quantity for Rs. 1, and not the exact prices, we can express yyy in terms of xxx and nnn. From the first equation:
n=1x ⟹ x=1nn=\frac{1}{x}\implies x=\frac{1}{n}n=x1⟹x=n1
From the second:
n+4=1y ⟹ y=1n+4n+4=\frac{1}{y}\implies y=\frac{1}{n+4}n+4=y1⟹y=n+41
Price reduction per orange:
Price reduction=x−y=1n−1n+4=(n+4)−nn(n+4)=4n(n+4)\text{Price reduction}=x-y=\frac{1}{n}-\frac{1}{n+4}=\frac{(n+4)-n}{n(n+4)}=\frac{4}{n(n+4)}Price reduction=x−y=n1−n+41=n(n+4)(n+4)−n=n(n+4)4
Summary:
- Original price per orange = 1n\frac{1}{n}n1
- New price per orange = 1n+4\frac{1}{n+4}n+41
- Price reduction per orange = 4n(n+4)\frac{4}{n(n+4)}n(n+4)4
Example:
If you want to find the prices numerically, you need additional information, such as the original price or the number of oranges bought initially for Rs.
- If you provide more details or specify what exactly you want to find (original price, new price, price reduction, etc.), I can help you solve it further!
