Average, Mixture and Alligation Question 3 - CAT 2023
There are three persons A, B and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by
kg. If the weight of E is 12 kg more than that of D, then the value of
is:
Step-by-Step Explanation
Let the initial total weight of persons A, B, and C combined be W.
- Scenario 1 (D joins):
The room now contains 4 people. The new average is x less than the original three-person average:
(W / 3) − ((W + D) / 4) = x ⟹ (4W − 3W − 3D) / 12 = x ⟹ W − 3D = 12x --- (Equation 1)
- Scenario 2 (E joins instead of D):
The room contains 4 people. The new average is 2x more than the original three-person average:
((W + E) / 4) − (W / 3) = 2x ⟹ (3W + 3E − 4W) / 12 = 2x ⟹ 3E − W = 24x --- (Equation 2)
Add Equation 1 and Equation 2 together to eliminate W:
W − 3D + 3E − W = 12x + 24x
3E − 3D = 36x ⟹ E − D = 12x
We are given that the weight difference between E and D is 12 kg (E − D = 12):
12 = 12x ⟹ x = 1
Correct Option: C
💡 FundaMakers Tip:
Use the Global Deviation Method to solve this faster. When group sizes stay identical, the change in the total sum equals the difference between the incoming entities.
Shift from case 1 to case 2 changes the average by x − (−2x) = 3x.
Across 4 people, the total weight shift is 4 × 3x = 12x.
This total shift must be exactly the difference between the two people (E − D = 12). Equating 12x = 12 gives x = 1 instantly without writing out any long balance equations.