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<html><body><h2 id="content-id-8493670521-2"><b>MIxtures And Alligations Questions Overview</b></h2><p>MIxtures And Alligations Questions have a good weightage in the Ssc Chsl Exam and the type of question asked in Ssc Chsl exam is similar to the question mentioned below. It has been solved and explained by KD Live Experts under the guidance of Neetu Mam and they have tried to elaborate the concept used in MIxtures And Alligations Questions.</p><h2 id="content-id-2568349017-3"><b>Question On MIxtures And Alligations </b></h2>Two vessels, Vessel A and Vessel B, contain mixtures of milk and water. Vessel A contains V liters of a mixture with a milk-to-water ratio of 3 : 2. Vessel B contains a mixture with a milk-to-water ratio of 5 : 3, and its total volume is (V + 40) liters. When the entire contents of Vessel A and Vessel B are mixed together in a large container, the total volume of the new mixture is 360 liters. The proportion of milk in Vessel A is less than the proportion of milk in Vessel B.

Match the numerical value:


	
		
			Column I
			Column II
		
	
	
		
			(A) 60 liters of pure milk is added to the initial mixture of Vessel A. Find the new percentage concentration of milk in Vessel A.
			(P) If one root of the quadratic equation x2 − kx + 250 = 0 is 10, then find the value of the other root.
		
		
			(B) A shopkeeper takes out 40 liters from Vessel B's initial mixture and adds the same volume of pure water. He then sells this new mixture at the cost price per liter of the original Vessel B mixture. What is his profit percentage?
			(Q) 61.25%
		
		
			(C) A third mixture is formed by mixing Vessel A and Vessel B in the ratio 1 : 1. Find the percentage concentration of milk in this third mixture.
			(R) The positive root of the quadratic equation x2 − 77x + 76 = 0.
		
	



<div class="row show--dificulty--bar" style="margin-top:8px;"><div class="col-sm-6 show--dificulty--bar" style="display:inline-flex !important;border-right:0.1px solid #eee;"><h3 style="font-size:12px !important;">Difficulty : Easy </h3></div><div class="col-sm-6 show--dificulty--bar" style="display:inline-flex;"><h3 style="font-size:12px !important;"> Average Time : 86 Seconds </h3></div></div><h2 id="content-id-0593124867-4"><b> MIxtures And Alligations Questions Options</b></h2><ol><li><p>A → Q, B → R, and C → P</p>
</li><li><p>Only B → Q, C → R</p>
</li><li><p>Only B → P, C → Q</p>
</li><li><p>A → R, B → P, and C → Q</p>
</li><li><p>A → P, B → R, and C → Q</p>
</li></ol> <h2 id="content-id-7659213804-5"><b> MIxtures And Alligations Questions Solution</b></h2><p>The correct answer is <strong>option 3 </strong>i.e. <strong>Only B → P, C → Q.</strong></p>

<p>Let the volume of Vessel A be V liters.</p>

<p>Volume of Vessel B = (V + 40) liters</p>

<p>Given that the total volume after mixing is 360 liters,</p>

<p>V + (V + 40) = 360</p>

<p>2V + 40 = 360</p>

<p>2V = 320</p>

<p>V = 160 liters</p>

<p>Therefore,</p>

<p>Vessel A = 160 liters</p>

<p>Vessel B = 200 liters</p>

<p>Initial quantities in Vessel A</p>

<p>Milk : Water = 3 : 2</p>

<p>Milk = (3/5) × 160 = 96 liters</p>

<p>Water = (2/5) × 160 = 64 liters</p>

<p>Initial quantities in Vessel B</p>

<p>Milk : Water = 5 : 3</p>

<p>Milk = (5/8) × 200 = 125 liters</p>

<p>Water = (3/8) × 200 = 75 liters</p>

<p>(A) 60 liters of pure milk is added to Vessel A</p>

<p>Initial milk = 96 liters</p>

<p>New milk = 96 + 60 = 156 liters</p>

<p>New volume = 160 + 60 = 220 liters</p>

<p>Percentage concentration of milk = (156/220) × 100 = 70.91%</p>

<p>(B) A shopkeeper takes out 40 liters from Vessel B and adds the same volume of water</p>

<p>Milk removed = (5/8) × 40 = 25 liters</p>

<p>Water removed = (3/8) × 40 = 15 liters</p>

<p>Remaining milk = 125 − 25 = 100 liters</p>

<p>Remaining water = 75 − 15 = 60 liters</p>

<p>After adding 40 liters of water:</p>

<p>Milk = 100 liters</p>

<p>Water = 60 + 40 = 100 liters</p>

<p>Total volume = 200 liters</p>

<p>Assume the cost of pure milk = â&#130;¹1 per liter and water is free.</p>

<p>Cost price per liter of original Vessel B mixture = 5/8</p>

<p>Cost price per liter of modified mixture = 100/200 = 1/2</p>

<p>Profit % = [(5/8 − 1/2)/(1/2)] × 100 = (1/8 ÷ 1/2) × 100 = 25%</p>

<p>(C) Vessel A and Vessel B are mixed in the ratio 1 : 1</p>

<p>Milk percentage in Vessel A = 60%</p>

<p>Milk percentage in Vessel B = 62.5%</p>

<p>Required milk percentage = (60 + 62.5)/2 = 61.25%</p>

<p>Column II</p>

<p>(P) If one root of the quadratic equation x<sup>2</sup> − kx + 250 = 0 is 10, find the other root.</p>

<p>Product of roots = 250</p>

<p>Other root = 250/10 = 25</p>

<p>Therefore,</p>

<p>(P) = 25</p>

<p>(Q) 61.25%</p>

<p>Therefore,</p>

<p>(Q) = 61.25%</p>

<p>(R) Positive root of x<sup>2</sup> − 77x + 76 = 0</p>

<p>x<sup>2 </sup>− 77x + 76 = (x − 1)(x − 76)</p>

<p>Roots are 1 and 76</p>

<p>Positive root = 76</p>

<p>Hence,</p>

<p>B → P</p>

<p>C → Q</p>

<p>A → No Match<br>
 </p>


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