Read the information and answer the following question given below. The…

2025

Read the information and answer the following question given below.
The information is about the staffs (lab assistant and professor) in three colleges.
For A – total lab assistant and total professor is 82.
For B – total lab assistant is 4 more than total lab assistant in C.
For C – total lab assistant is 6 less than total professor
Note – total professor in A is 44 less than total lab assistant in B.
Total staffs in C is 24 more than that in A and total professor in al the college is 102.

Total professor in A is what percentage of total staff in B.

  1. A.

    9.09%

  2. B.

    11.11%

  3. C.

    12.5%

  4. D.

    30%

  5. E.

    25%

Attempted by 1 students.

Show answer & explanation

Correct answer: B

Concept

Translate each verbal clue of a data-interpretation caselet into one linear equation in the unknown head-counts, then solve the system step by step; a percentage "X is what percent of Y" is always computed as (X / Y) x 100, where Y is the FULL base named in the question.

Application

Let the lab assistants and professors be A_LA, A_P, B_LA, B_P, C_LA, C_P. Turn the clues into equations and solve in order:

  1. College A total: A_LA + A_P = 82.

  2. Total staff in C is 24 more than in A, so C_LA + C_P = 82 + 24 = 106.

  3. In C, lab assistants are 6 less than professors: C_LA = C_P - 6. Substituting into 106 gives 2*C_P - 6 = 106, so C_P = 56 and C_LA = 50.

  4. B's lab assistants are 4 more than C's: B_LA = 50 + 4 = 54.

  5. Professors in A are 44 less than lab assistants in B: A_P = 54 - 44 = 10 (and so A_LA = 82 - 10 = 72).

  6. Total professors across all colleges is 102: B_P = 102 - A_P - C_P = 102 - 10 - 56 = 36.

  7. Total staff in B = B_LA + B_P = 54 + 36 = 90.

  8. Required percentage = (A_P / total staff in B) x 100 = (10 / 90) x 100 = 11.11%.

Cross-check

Verify every clue with the solved values: A total = 72 + 10 = 82; C total = 50 + 56 = 106 = 82 + 24; C lab assistants 50 = 56 - 6; B lab assistants 54 = 50 + 4; A professors 10 = 54 - 44; professors 10 + 36 + 56 = 102. All constraints hold, confirming 10 out of 90, i.e. 11.11%.

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