Question

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, \( W_I \), of Section I is smaller than the average weight \( W_{II} \) of Section II. If the heaviest student, Deepak, of Section II is moved to Section I, and the lightest student, Poonam, of Section I is moved to Section II, then the average weights of the two sections are switched.
A: \( W_{II} - W_I = 1.0 \)
B: Moving Deepak from Section II to I (without any move to II) makes the average weights of the two sections equal.

• A. if the question can be answered using A alone but not B alone.
• B. if the question can be answered using B alone but not A alone.
• C. if the question can be answered using A and B together, but not using either A or B alone.
• D. if the question cannot be answered even using A and B together.

Hint:

In Data Sufficiency, check if each statement alone gives enough equations to solve for the unknown; if not, try combining.

Answer 1

The Correct Option is C

The problem involves understanding the average weight distribution of two sections within a class and how students' transfers affect these averages. Let's define the following:
 

• Let \( W_I \) be the average weight of Section I.
• Let \( W_{II} \) be the average weight of Section II.
• Total students in Class = 100, in Section I = 50, in Section II = 50.
• Average weight of Class = 45 kg implies total weight = 4500 kg.

Using the equations, we have: 
\[ 50W_I + 50W_{II} = 4500 \rightarrow (1) \]
Clue A: \( W_{II} - W_I = 1 \)
Substituting \( W_{II} = W_I + 1 \) in Eq. (1):
\[ 50W_I + 50(W_I + 1) = 4500 \]
\[ 100W_I + 50 = 4500 \]
\[ 100W_I = 4450 \rightarrow W_I = 44.5 \, \text{kg}, W_{II} = 45.5 \, \text{kg} \]
Clue B: When Deepak moves from Section II to I, the average weights become equal.
Let Deepak's weight be \( x \). For equilibrium:
Total weight after Deepak's move:
\[ \frac{50W_{II} - x}{49} = \frac{50W_I + x}{51} \]
Upon solving Clue B alone, we can't determine unique weights. Hence, Clue B alone is insufficient.
Now, using both clues:
With weights defined (44.5 kg and 45.5 kg from Clue A), verify:
\( W_I + 1 = W_{II} \) holds true.
Deepak's move confirms the Clue B fact with a valid x, verifying against given logic.
Therefore, using both Clue A and Clue B together, we accurately solve it:
Both Clues in combination solve but not individually, satisfying the answer: "if the question can be answered using A and B together, but not using either A or B alone."