If the price of the eraser is reduced by 25% a person buy 2 more erase...
Let n erasers be available for a rupee
Reduced Price = (75/100 Ã - 1) = Re ¾
3/4 rupee fetch n erasers = 1 Rupee will fetch (n à - 4/3) erasers
Therefore, 4n/3 = n +2 ⇒ 4n = 3n +6 ⇒ n =6
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If the price of the eraser is reduced by 25% a person buy 2 more erase...
Given information:
- The price of the eraser is reduced by 25%.
- With this reduction in price, a person can buy 2 more erasers for a rupee.
To find:
- How many erasers are available for a rupee?
Assumptions:
- Let's assume that the original price of one eraser is x rupees.
Calculations:
1. Price reduction:
- The price of the eraser is reduced by 25%, which means the new price is 75% of the original price.
- The new price of one eraser is (75/100) * x = (3/4) * x rupees.
2. Number of erasers for a rupee:
- With the reduced price, a person can buy 2 more erasers for a rupee.
- This means that the person can buy (2 + 1) = 3 erasers for a rupee.
- Therefore, the new price of one eraser is (1/3) rupees.
3. Equating the two prices:
- The new price of one eraser is (3/4) * x rupees.
- The new price of one eraser is (1/3) rupees.
- Equating these two prices, we get:
(3/4) * x = (1/3)
Cross-multiplying, we get:
9x = 4
x = 4/9
Conclusion:
- The original price of one eraser is (4/9) rupees.
- Therefore, for a rupee, the number of erasers available is 1 / (4/9) = 9/4 = 2.25.
- Since we can't have a fractional number of erasers, the closest whole number is 2.
- Hence, the correct answer is option 'B': 6 erasers are available for a rupee.
If the price of the eraser is reduced by 25% a person buy 2 more erase...
Let n erasers be available for a rupee
Reduced Price = (75/100 Ã - 1) = Re ¾
3/4 rupee fetch n erasers = 1 Rupee will fetch (n à - 4/3) erasers
Therefore, 4n/3 = n +2 ⇒ 4n = 3n +6 ⇒ n =6