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24 Apr 2026, 17:34
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
45% (02:39) correct
55%
(02:12)
wrong
based on 33
sessions
History
Date
Time
Result
Not Attempted Yet
Over 150 students were surveyed about how many internships they applied for, as shown in the table below:
Number of Internships ....... Percent Frequency
Applied for
...0 ............................................. 11 2/3%
....1 .............................................. 37 1/2%
...2 ............................................... 25%
...3 .............................................. 17 1/2%
...4 ............................................. 8 1/3%
At least how many students must have been surveyed?
(A) 120
(B) 150
(C) 180
(D) 200
(E) 240
Number of Internships ....... Percent Frequency
Applied for
...0 ............................................. 11 2/3%
....1 .............................................. 37 1/2%
...2 ............................................... 25%
...3 .............................................. 17 1/2%
...4 ............................................. 8 1/3%
At least how many students must have been surveyed?
(A) 120
(B) 150
(C) 180
(D) 200
(E) 240
26 Apr 2026, 20:54
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Let's create, a clean table:
So we take LCM of denominators:
60, 8, 4, 40, 12
Prime factors:
60 = 22 × 3 × 5
8 = 23
4 = 22
40 = 23 × 5
12 = 22 × 3
LCM = 23 × 3 × 5 = 120
So minimum N is 120.
Option: A
| Internships applied | Percent |
|---|---|
| 0 | 11 2/3% = 35/3% = 7/60 |
| 1 | 37 1/2% = 3/8 |
| 2 | 25% = 1/4 |
| 3 | 17 1/2% = 7/40 |
| 4 | 8 1/3% = 25/3% = 5/60 = 1/12 |
So we take LCM of denominators:
60, 8, 4, 40, 12
Prime factors:
60 = 22 × 3 × 5
8 = 23
4 = 22
40 = 23 × 5
12 = 22 × 3
LCM = 23 × 3 × 5 = 120
So minimum N is 120.
Option: A
02 May 2026, 03:33
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Over 150 students, so atleast 240.
AdarshSambare
