Comprehension
These questions are based on the situation given below:
There are \( m \) blue vessels with known volumes \( V_1, V_2, \dots, V_m \), arranged in ascending order of volume, where \( v_1 > 0.5 \) litre, and \( V_m < 1 \) litre. Each of these is full of water initially. The water from each of these is emptied into a minimum number of empty white vessels, each having volume 1 litre. The water from a blue vessel is not emptied into a white vessel unless the white vessel has enough empty volume to hold all the water of the blue vessel. The number of white vessels required to empty all the blue vessels according to the above rules was \( n \).
Question: 1
Among the four values given below, which is the least upper bound on \( e \), where \( e \) is the total empty volume in the \( m \) white vessels at the end of the above process?
Among the four values given below, which is the least upper bound on \( e \), where \( e \) is the total empty volume in the \( m \) white vessels at the end of the above process?
Show Hint
Every blue vessel holds more than half a litre, so two cannot share one white vessel. Write the empty space in each white vessel and bound it using the smallest volume \( v_1 \).
Updated On: Nov 4, 2026
- \( m v_m \)
- \( m(1 - v_1) \)
- \( m v_1 \)
- \( m(1 - v_m) \)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: One blue vessel per white vessel:
Each \( v_i \ge v_1 \gt 0.5 \), so any two blue vessels together exceed 1 litre. Hence no white vessel can take two of them. So exactly one blue vessel goes into each white vessel, and \( n = m \).
Step 2: Empty volume:
The white vessel holding \( v_i \) has empty volume \( 1 - v_i \). Therefore \( e = \sum_{i=1}^{m} (1 - v_i) \).
Step 3: Upper bound:
The volumes are in ascending order, so \( v_i \ge v_1 \) and \( 1 - v_i \le 1 - v_1 \). Hence \( e \le m(1 - v_1) \). Equality holds when all the blue vessels have volume \( v_1 \), so no smaller value can be an upper bound.
Step 4: Check the other options:
\( m(1 - v_m) \) is a lower bound, not an upper bound. For \( m = 2,\ v = (0.6, 0.9) \): \( e = 0.4 + 0.1 = 0.5 \), but \( m(1 - v_m) = 0.2 \lt e \).
\( m v_1 \) and \( m v_m \) are upper bounds, but since \( v_1 \gt 0.5 \) we have \( 1 - v_1 \lt v_1 \le v_m \), so both are larger than \( m(1 - v_1) \) and are not the least.
Final Answer:
The correct option is (2): \( m(1 - v_1) \).
Each \( v_i \ge v_1 \gt 0.5 \), so any two blue vessels together exceed 1 litre. Hence no white vessel can take two of them. So exactly one blue vessel goes into each white vessel, and \( n = m \).
Step 2: Empty volume:
The white vessel holding \( v_i \) has empty volume \( 1 - v_i \). Therefore \( e = \sum_{i=1}^{m} (1 - v_i) \).
Step 3: Upper bound:
The volumes are in ascending order, so \( v_i \ge v_1 \) and \( 1 - v_i \le 1 - v_1 \). Hence \( e \le m(1 - v_1) \). Equality holds when all the blue vessels have volume \( v_1 \), so no smaller value can be an upper bound.
Step 4: Check the other options:
\( m(1 - v_m) \) is a lower bound, not an upper bound. For \( m = 2,\ v = (0.6, 0.9) \): \( e = 0.4 + 0.1 = 0.5 \), but \( m(1 - v_m) = 0.2 \lt e \).
\( m v_1 \) and \( m v_m \) are upper bounds, but since \( v_1 \gt 0.5 \) we have \( 1 - v_1 \lt v_1 \le v_m \), so both are larger than \( m(1 - v_1) \) and are not the least.
Final Answer:
The correct option is (2): \( m(1 - v_1) \).
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Question: 2
Let the number of white vessels needed be \( n_1 \) for the emptying process described above, if the volume of each white vessel is 2 litres. Among the following values, which is the least upperbound on \( n_1 \)?
Let the number of white vessels needed be \( n_1 \) for the emptying process described above, if the volume of each white vessel is 2 litres. Among the following values, which is the least upperbound on \( n_1 \)?
Show Hint
When calculating upper bounds for vessels or containers, consider the total capacity required and use the smallest integer greater than or equal to the calculated number of vessels.
Updated On: Oct 29, 2026
- \( m/4 \)
- smallest integer greater than or equal to \( n/2 \)
- \( n \)
- greatest integer less than or equal to \( (n/2) \)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
To find the least upper bound on \( n_1 \), we observe that the number of white vessels required is bounded by the smallest integer greater than or equal to \( n/2 \), considering that each white vessel has a volume of 2 litres and we are distributing the water evenly.
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