Question:

A mess contractor can either serve 450 students with the meal that he prepares or can cater to 270 cops with the same meal. If 300 students have already eaten in the mess, how many cops can be fed with the remaining meal?

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In “same meal for two groups” problems, convert everything to one unit (e.g., student-meals), then use ratios to switch groups.
Updated On: Jul 15, 2026
  • 20
  • 45
  • 90
  • 180
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The Correct Option is C

Approach Solution - 1

Treat the total prepared meal as a fixed resource.
Capacity equivalence: $450$ student-meals $=$ $270$ cop-meals, so one cop consumes $450/270=5/3$ times as much as one student.
Students already fed: $300$ student-meals; remaining resource $=450-300=150$ student-meals.
Convert to cops: $150 \div (5/3)=150 \times \dfrac{3}{5}=90$ cops.
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Approach Solution -2

A fixed quantity of prepared food can serve either 450 students or 270 cops, and 300 students have already eaten. We need to know how many cops the leftover food can feed. Since a cop's portion is larger than a student's, we can test each option by checking whether it uses up exactly the remaining food.

  1. 20: Feeding 300 students uses \( \frac{300}{450}=\frac{2}{3} \) of the food, leaving \( \frac{1}{3} \). Feeding only 20 cops would use just \( \frac{20}{270} \) of the food, far less than the \( \frac{1}{3} \) remaining, so this option is too small.
  2. 45: Feeding 45 cops would use \( \frac{45}{270}=\frac{1}{6} \) of the total food, but \( \frac{1}{3} \) remains after the students ate, so this option leaves food unaccounted for and is too small.
  3. 90: Feeding 90 cops uses \( \frac{90}{270}=\frac{1}{3} \) of the total food, which exactly matches the \( \frac{1}{3} \) remaining after 300 students ate.
  4. 180: Feeding 180 cops would use \( \frac{180}{270}=\frac{2}{3} \) of the total food, but only \( \frac{1}{3} \) remains after the students ate, so this option needs more food than is left.

Since 300 students use up two-thirds of the prepared food, exactly one-third remains, and that one-third is precisely enough to feed 90 cops.

Therefore, the correct answer is 90.

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Approach Solution -3

Assume the prepared food equals 1350 units of a common measure, chosen so it splits evenly both ways: a student's portion is then \( 1350/450=3 \) units and a cop's portion is \( 1350/270=5 \) units. Feeding 300 students uses \( 300 \times 3=900 \) units. We can check each option by seeing whether adding its unit requirement to the 900 already used comes to exactly 1350.

  1. 20: Feeding 20 cops needs \( 20 \times 5=100 \) units, bringing the total used to \( 900+100=1000 \), well short of 1350.
  2. 45: Feeding 45 cops needs \( 45 \times 5=225 \) units, bringing the total to \( 900+225=1125 \), still short of 1350.
  3. 90: Feeding 90 cops needs \( 90 \times 5=450 \) units, bringing the total to \( 900+450=1350 \), using up the food exactly.
  4. 180: Feeding 180 cops needs \( 180 \times 5=900 \) units, bringing the total to \( 900+900=1800 \), well beyond the 1350 units available.

Only feeding 90 cops uses up exactly the 450 units of food left over after 300 students have eaten.

Therefore, the correct answer is 90.

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