Question:

The number of children in a camp is \(x\) and their average weight is \(20\) kg. If 5 children each weighing \(12\) kg join the camp or if 10 children each weighing \(21\) kg leave the camp, the average weight in both the cases remains the same. The value of \(x\) is:

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When two new averages are stated to be equal, express each scenario’s total and count, form the two average expressions, and equate them—cross multiplication quickly yields a linear equation in \(x\).
Updated On: Aug 14, 2025
  • \(18\)
  • \(16\)
  • \(15\)
  • \(14\)
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The Correct Option is C

Solution and Explanation

Step 1: Set up totals and averages.
Initial total weight \(= 20x\).
Case 1 (5 join, each \(12\) kg): total \(=20x+5\cdot12=20x+60\); number \(=x+5\); average \[ A_1=\frac{20x+60}{x+5}. \] Case 2 (10 leave, each \(21\) kg): total \(=20x-10\cdot21=20x-210\); number \(=x-10\); average \[ A_2=\frac{20x-210}{x-10}. \] Step 2: Equate the two averages and solve for \(x\).
\[ \frac{20x+60}{x+5}=\frac{20x-210}{x-10}. \] Cross-multiplying: \[ (20x+60)(x-10)=(20x-210)(x+5). \] Expanding: \[ 20x^2-140x-600=20x^2-110x-1050. \] Cancel \(20x^2\) and rearrange: \[ -140x+110x=-1050+600 \Rightarrow -30x=-450 \Rightarrow x=15. \] \[ \boxed{15} \]
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