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<html><body><h2 id="content-id-8579401623-2"><b>Caselets Questions Overview</b></h2><p>Caselets 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 Caselets Questions.</p><h2 id="content-id-4196825307-3"><b> Comprehension On Caselets Questions</b></h2>Two boxes P and Q contain candies of four flavours: orange, mint, coffee and berry. Orange, mint and coffee candies are present in both boxes, while berry candies are present in only one of the two boxes. The number of orange candies in boxes P and Q is in the ratio 4:5, and in each box the number of coffee candies is 80% of the number of orange candies. In box P, the ratio of orange candies to mint candies is (K − 5):(K − 8), while in box Q, the ratio of orange candies to mint candies is (L + 5):(L − 1). The value of K is 25% less than the value of L. The ratio of the total number of mint candies in both boxes together to the total number of coffee candies in both boxes together is 11:12. The number of berry candies is equal to half of the total number of coffee candies in boxes P and Q together. If 16 berry candies are transferred from the box containing berry candies to the other box, the ratio of the total number of candies in boxes P and Q becomes 9:13. Furthermore, the initial number of mint candies in Box P was strictly greater than the number of berry candies transferred. Wherever a flavour is present, the number of candies of that flavour is a positive whole number, and K and L are positive integers.
<h2 id="content-id-5901643278-4"><b>Question On Caselets </b></h2>The cost prices of an orange, coffee, mint, and berry candy are in the ratio 2:3:4:5, and the total cost of all candies in Box Q is â&#130;¹44,160. Find the selling price of a single coffee candy if it is sold at a profit of (L + 12.5)%.
<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 : 100 Seconds </h3></div></div><h2 id="content-id-8693154072-5"><b> Caselets Questions Options</b></h2><ol><li><p>â&#130;¹169</p>
</li><li><p>â&#130;¹179</p>
</li><li><p>â&#130;¹159</p>
</li><li><p>â&#130;¹149</p>
</li><li><p>â&#130;¹139</p>
</li></ol> <h2 id="content-id-1658203974-6"><b> Caselets Questions Solution</b></h2><p>The correct answer is <strong>option 3</strong> i.e. <strong>â&#130;¹159</strong></p>

<p>Let orange candies in P and Q be:</p>

<p>P orange = 4n</p>

<p>Q orange = 5n</p>

<p>Coffee candies are 80% of orange candies:</p>

<p>P coffee = 4n × 80% = 16n/5</p>

<p>Q coffee = 5n × 80% = 4n</p>

<p>For whole-number values, let:</p>

<p>n = 5t</p>

<p>Therefore,</p>

<p>P orange = 20t</p>

<p>Q orange = 25t</p>

<p>P coffee = 16t</p>

<p>Q coffee = 20t</p>

<p>Total coffee = 16t + 20t = 36t</p>

<p>Given,</p>

<p>Mint : Coffee = 11 : 12</p>

<p>Total mint = 36t × 11/12</p>

<p>= 33t</p>

<p>Berry candies = 36t/2</p>

<p>= 18t</p>

<p>Total candies in both boxes:</p>

<p>= 20t + 25t + 16t + 20t + 33t + 18t</p>

<p>= 132t</p>

<p>After transferring 16 berry candies, the total number of candies remains 132t.</p>

<p>Given final ratio P : Q = 9 : 13</p>

<p>Final P = 132t × 9/22</p>

<p>= 54t</p>

<p>Final Q = 132t × 13/22</p>

<p>= 78t</p>

<p>Since 16 berry candies are transferred from Q to P:</p>

<p>Original P = 54t − 16</p>

<p>Original Q = 78t + 16</p>

<p>Now find mint candies in each box.</p>

<p>Mint in P:</p>

<p>= (54t − 16) − 20t − 16t</p>

<p>= 18t − 16</p>

<p>Mint in Q:</p>

<p>= (78t + 16) − 25t − 20t − 18t</p>

<p>= 15t + 16</p>

<p>For Box P:</p>

<p>Orange : Mint = (K − 5) : (K − 8)</p>

<p>Therefore,</p>

<p>20t/(18t − 16) = (K − 5)/(K − 8)</p>

<p>20t(K − 8) = (18t − 16)(K − 5)</p>

<p>20tK − 160t = 18tK − 90t − 16K + 80</p>

<p>K(2t + 16) = 70t + 80</p>

<p>K = (35t + 40)/(t + 8) ...(1)</p>

<p>For Box Q:</p>

<p>Orange : Mint = (L + 5) : (L − 1)</p>

<p>Therefore,</p>

<p>25t/(15t + 16) = (L + 5)/(L − 1)</p>

<p>25t(L − 1) = (15t + 16)(L + 5)</p>

<p>25tL − 25t = 15tL + 75t + 16L + 80</p>

<p>L(10t − 16) = 100t + 80</p>

<p>L = (50t + 40)/(5t − 8) ...(2)</p>

<p>K is 25% less than L:</p>

<p>K = 75% of L</p>

<p>K = 3L/4</p>

<p>Using (1) and (2):</p>

<p>(35t + 40)/(t + 8) = 3/4 × (50t + 40)/(5t − 8)</p>

<p>4(35t + 40)(5t − 8) = 3(50t + 40)(t + 8)</p>

<p>700t<sup>2</sup> − 320t − 1280 = 150t<sup>2</sup> + 1320t + 960</p>

<p>550t<sup>2</sup> − 1640t − 2240 = 0</p>

<p>55t<sup>2</sup> − 164t − 224 = 0</p>

<p>(t − 4)(55t + 56) = 0</p>

<p>Since t is positive:</p>

<p>t = 4</p>

<p>Now,</p>

<p>K = (35 × 4 + 40)/(4 + 8)</p>

<p>= 180/12</p>

<p>= 15</p>

<p>L = (50 × 4 + 40)/(5 × 4 − 8)</p>

<p>= 240/12</p>

<p>= 20</p>

<table border="1" cellspacing="0" style="width:300px">
	<tbody>
		<tr>
			<td>
			<p>Box</p>
			</td>
			<td>
			<p>Orange</p>
			</td>
			<td>
			<p>Mint</p>
			</td>
			<td>
			<p>Coffee</p>
			</td>
			<td>
			<p>Berry</p>
			</td>
			<td>
			<p>Total</p>
			</td>
		</tr>
		<tr>
			<td>
			<p>P</p>
			</td>
			<td>
			<p>80</p>
			</td>
			<td>
			<p>56</p>
			</td>
			<td>
			<p>64</p>
			</td>
			<td>
			<p>0</p>
			</td>
			<td>
			<p>200</p>
			</td>
		</tr>
		<tr>
			<td>
			<p>Q</p>
			</td>
			<td>
			<p>100</p>
			</td>
			<td>
			<p>76</p>
			</td>
			<td>
			<p>80</p>
			</td>
			<td>
			<p>72</p>
			</td>
			<td>
			<p>328</p>
			</td>
		</tr>
		<tr>
			<td>
			<p>Total</p>
			</td>
			<td>
			<p>180</p>
			</td>
			<td>
			<p>132</p>
			</td>
			<td>
			<p>144</p>
			</td>
			<td>
			<p>72</p>
			</td>
			<td>
			<p>528</p>
			</td>
		</tr>
	</tbody>
</table>

<p>Let CPs = 2P, 3P, 4P, 5P.</p>

<p>Box Q: Orange = 100, Coffee = 80, Mint = 76, Berry = 72</p>

<p>100 × 2P + 80 × 3P + 76 × 4P + 72 × 5P = 44160</p>

<p>200P + 240P + 304P + 360P = 44160</p>

<p>1104P = 44160</p>

<p>P = 40</p>

<p>Coffee CP = 3 × 40 = â&#130;¹120</p>

<p>L = 20</p>

<p>Profit = 20 + 12.5 = 32.5%</p>

<p>SP = â&#130;¹120 × 132.5% = â&#130;¹159</p>
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