Math Problem Statement

The average weight of a class of 100 students is 45 kg. The class consists of two sections, I and II, each with 50 students. The average weight, WI , of Section I is smaller than the average weight WII , of the Section II. If the heaviest student say Deepak, of section II is moved to Section I, and the lightest student, say Poonam, of Section I is moved to Section II, then the average weights of the two sections are switched, i.e., the average weight of Section I becomes WII and that of Section II becomes WI . What is the weight of Poonam?

Solution

Let’s solve this problem step-by-step.

Step 1: Setting up the Equations

We know:

  1. The total weight of all 100 students is 45×100=4500kg45 \times 100 = 4500 \, \text{kg}.
  2. Let the average weight of Section I be WIW_I and the average weight of Section II be WIIW_{II}.
  3. The number of students in each section is 50.

So, the total weight of Section I is 50×WI50 \times W_I, and the total weight of Section II is 50×WII50 \times W_{II}.

Since the sum of the weights of both sections is the total weight of the class: 50×WI+50×WII=450050 \times W_I + 50 \times W_{II} = 4500 Dividing both sides by 50: WI+WII=90W_I + W_{II} = 90

Step 2: Weight Swapping

When the heaviest student, Deepak, from Section II is moved to Section I, and the lightest student, Poonam, from Section I is moved to Section II, the average weights of the sections are switched. This implies:

  1. The new average weight of Section I becomes WIIW_{II}.
  2. The new average weight of Section II becomes WIW_I.

Let’s denote:

  • dd as Deepak's weight.
  • pp as Poonam's weight.

Step 3: Expressing the Weight Change in Each Section

For Section I:

  • Initial total weight = 50×WI50 \times W_I.
  • After Deepak joins and Poonam leaves, the new total weight becomes 50×WI+dp50 \times W_I + d - p.
  • The new average weight of Section I is WIIW_{II}, so: 50×WI+dp50=WII\frac{50 \times W_I + d - p}{50} = W_{II} Simplifying, we get: 50WI+dp=50WII50 W_I + d - p = 50 W_{II}

For Section II:

  • Initial total weight = 50×WII50 \times W_{II}.
  • After Poonam joins and Deepak leaves, the new total weight becomes 50×WII+pd50 \times W_{II} + p - d.
  • The new average weight of Section II is WIW_I, so: 50×WII+pd50=WI\frac{50 \times W_{II} + p - d}{50} = W_I Simplifying, we get: 50WII+pd=50WI50 W_{II} + p - d = 50 W_I

Step 4: Solving the Equations

Now we have two equations:

  1. 50WI+dp=50WII50 W_I + d - p = 50 W_{II}
  2. 50WII+pd=50WI50 W_{II} + p - d = 50 W_I

Adding these two equations, we get: (50WI+dp)+(50WII+pd)=50WII+50WI(50 W_I + d - p) + (50 W_{II} + p - d) = 50 W_{II} + 50 W_I 50WI+50WII=50WI+50WII50 W_I + 50 W_{II} = 50 W_I + 50 W_{II}

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Math Problem Analysis

Mathematical Concepts

Algebra
Averages
Systems of Equations

Formulas

Average weight = Total weight / Number of students
WI + WII = 90
(50 * WI + d - p) / 50 = WII
(50 * WII + p - d) / 50 = WI

Theorems

Basic properties of averages and algebraic equations

Suitable Grade Level

Grades 10-12