for position C no of accepted applications from males is between 150 amp 200 while that of females i...........

Quantitative Aptitude(IBPS PO) Data Table

Directions (): Read the data carefully and answer the questions. Some data are missing which you have to calculate as per information provided in the questions.

NOTE:- A duplicate applicant is an applicant who has submitted additional (duplicate) application after submitting their original application. All application forms (original + duplicate) received from duplicate applicant were rejected. Remaining all application were accepted. None of the applicants applied for more than one post.

for position C, no. of accepted applications from males is between 150 & 200 while that of females is between 130 & 180. Which of the following can be a possible value (s) of average no. of duplicate applications submitted by duplicate applicants for position C?

A. 11 B. 5 C. 9 D. 13 E. 7

B, C & E

C & E

A & D

B & E

A, C & D

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Explanation:

All applicants who submitted more than one application, their original submission is also rejected.
FOR POSITION A:
63 applications were declared duplicate since these 63 applicants submitted more than one application
Total rejected applications = 63+63×4=315
Following the same, we get

Where, p = no. of duplicate applicants for B
q = average no. of duplicate applications from duplicate applicants for C
r = average no. of duplicate applications from duplicate applicants for D
s = total no. of applications received for D
t = no. of duplicate applicants for E
u = average no. of duplicate applications from duplicate applicants for E

let no. of accepted applications from males & females for C be x & y respectively
ATQ,
150 < x < 200
And, 130 < y < 180
To find, q = ?
Applications accepted from all (males + females) for C = 600 – 28(q + 1)
600 – 28(q + 1) = x + y ………………..(i)
Minimum possible value of ‘x + y’ = 151 + 131 = 282
Maximum possible value of ‘x + y’ = 199 + 179 = 378
So, 282 ≤ x + y ≤ 378 …………….(ii)
Using (i) & (ii)
282 ≤ 600 – 28(q + 1) ≤ 378
On solving above inequality,
222 ≤ 28 (q + 1) ≤ 318
9714≤q≤14514
Satisfying values of q = 7, 9