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05 Aug 2024, 15:45
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C
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Difficulty:
95%
(hard)
Question Stats:
26% (02:55) correct
74%
(03:07)
wrong
based on 47
sessions
History
Date
Time
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Not Attempted Yet
Ram, a merchant bought 120 articles, each at a price of Rs. 60. He sold exactly y of these articles to Shyam at a profit of y% and the remaining articles at a profit of (100 − y)% to his friend Jayesh. What is the minimum profit (in Rs) that Ram could've made in this transaction?
A. 3,250
B. 3,460
C. 3,570
D. 3,600
E. 3,650
A. 3,250
B. 3,460
C. 3,570
D. 3,600
E. 3,650
30 Sep 2024, 10:06
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Cost Price = \(120 \times 60\)
\(SP_{S} = y \times 60 \times \Big( 1 + \dfrac{y}{100} \Big)\)
\(SP_{J} = (120 - y) \times 60 \times \Big( 1 + \dfrac{100 - y}{100} \Big) = (120 - y) \times 60 \times \Big( 2 - \dfrac{y}{100} \Big)\)
\(\text{Profit} = SP - CP = y \times 60 \times \Big( 1 + \dfrac{y}{100} \Big) + (120 - y) \times 60 \times \Big( 2 - \dfrac{y}{100} \Big) - 120 \times 60 \qquad \to (1)\)
\(= 60y + (60\times \dfrac{y^2}{100}) +(120\times 60 \times 2) - (120 \times 60 \times \dfrac{y}{100}) - (2\times 60\times y) + (60 \times \dfrac{y^2}{100}) - (120\times 60)\)
\(= 1.2 y^2 - 132 y + 7200 = y^2 - 110y +6000\) (This is the final equation which is after simplifying the equation (1))
For a quadratic equation, \(ax^2 + bx + c\) the minimum value occur at \(x = -\frac{b}{2a} \)
Therefore, minimum value is \(y = -\dfrac{(-110)}{2\times 1} = 55\)
Subtitute, \(y = 55\) at equation (1):
Profit = \(55 \times 60 \times \Big( 1 + \dfrac{55}{100} \Big) + (120 - 55) \times 60 \times \Big( 2 - \dfrac{55}{100} \Big) - 120 \times 60 = 5115 + 5655 - 7200 = 3570\)
ANS C
\(SP_{S} = y \times 60 \times \Big( 1 + \dfrac{y}{100} \Big)\)
\(SP_{J} = (120 - y) \times 60 \times \Big( 1 + \dfrac{100 - y}{100} \Big) = (120 - y) \times 60 \times \Big( 2 - \dfrac{y}{100} \Big)\)
\(\text{Profit} = SP - CP = y \times 60 \times \Big( 1 + \dfrac{y}{100} \Big) + (120 - y) \times 60 \times \Big( 2 - \dfrac{y}{100} \Big) - 120 \times 60 \qquad \to (1)\)
\(= 60y + (60\times \dfrac{y^2}{100}) +(120\times 60 \times 2) - (120 \times 60 \times \dfrac{y}{100}) - (2\times 60\times y) + (60 \times \dfrac{y^2}{100}) - (120\times 60)\)
\(= 1.2 y^2 - 132 y + 7200 = y^2 - 110y +6000\) (This is the final equation which is after simplifying the equation (1))
For a quadratic equation, \(ax^2 + bx + c\) the minimum value occur at \(x = -\frac{b}{2a} \)
Therefore, minimum value is \(y = -\dfrac{(-110)}{2\times 1} = 55\)
Subtitute, \(y = 55\) at equation (1):
Profit = \(55 \times 60 \times \Big( 1 + \dfrac{55}{100} \Big) + (120 - 55) \times 60 \times \Big( 2 - \dfrac{55}{100} \Big) - 120 \times 60 = 5115 + 5655 - 7200 = 3570\)
ANS C
22 Jan 2025, 21:03
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