Question
Download Solution PDFStudy the given table and answer the questions that follow.
The table represents the number of candidates appeared, qualified, and selected in a national level examination from four different schools M, N, O and P over the years 2012 to 2017.
Where A : Appeared; Q : Qualified ; S : Selected
Year |
M |
N |
O |
P |
||||||||
A |
Q |
S |
A |
Q |
S |
A |
Q |
S |
A |
Q |
S |
|
2012 |
7000 |
750 |
84 |
3000 |
400 |
34 |
2000 |
250 |
27 |
1500 |
160 |
23 |
2013 |
5500 |
500 |
55 |
3500 |
420 |
56 |
2500 |
340 |
36 |
2500 |
290 |
56 |
2014 |
6500 |
455 |
42 |
3800 |
460 |
66 |
3500 |
380 |
48 |
2700 |
480 |
67 |
2015 |
8500 |
854 |
89 |
4500 |
485 |
72 |
4500 |
560 |
57 |
3400 |
540 |
87 |
2016 |
9000 |
922 |
92 |
4690 |
678 |
86 |
5670 |
660 |
68 |
3700 |
650 |
89 |
2017 |
9200 |
942 |
94 |
6700 |
700 |
92 |
6780 |
680 |
89 |
4500 |
670 |
93 |
For which school is the average number of candidates selected over the years the maximum?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
The number of candidates selected for schools M, N, O, and P over the years 2012 to 2017.
Formula used:
Average = (Sum of selected candidates) ÷ (Number of years)
Calculation:
School M: 84 + 55 + 42 + 89 + 92 + 94 = 456
⇒ Average for M = 456 ÷ 6 = 76
School N: 34 + 56 + 66 + 72 + 86 + 92 = 406
⇒ Average for N = 406 ÷ 6 = 67.67
School O: 27 + 36 + 48 + 57 + 68 + 89 = 325
⇒ Average for O = 325 ÷ 6 = 54.17
School P: 23 + 56 + 67 + 87 + 89 + 93 = 415
⇒ Average for P = 415 ÷ 6 = 69.17
∴ The maximum average number of candidates selected is for School M with an average of 76.
Last updated on Jul 10, 2025
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