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27 Jan 2026, 03:41
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
49% (02:21) correct
51%
(02:33)
wrong
based on 63
sessions
History
Date
Time
Result
Not Attempted Yet
A set of 44 positive numbers less than 5 has one mode. If 18 of the numbers are equal to 1.5 and the average of the 44 numbers is 2.0, how many integers could be the mode of the set?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
The set has 44 positive numbers, all less than 5.
Exactly 18 equal 1.5. The average is 2.0, so total sum = 44 × 2.0 = 88.
Sum of the 18 ones = 18 × 1.5 = 27, leaving sum of remaining 26 numbers = 88 - 27 = 61.
Condition for Mode :
The set has exactly one mode, meaning one value appears most frequently, including more than 18 times for 1.5.
Let that frequency be k > 18; it appears k times.
The other 26 - k numbers must have frequencies ≤ k - 1 ≤ 17 (since k ≥ 19).
Possible Integer Modes
Integers less than 5: 1, 2, 3, 4. For m (one of these) to be mode:
Needs ≥ 19 occurrences.
Sum contribution ≤ 61, so max occurrences: (61/m).
m=1: ≤61 (>19 possible), but using 1 gives 61-19=42, average is 6 - not possible since all numbers must be less than 5. (Discard)
m=2: ≤30 (>19 possible)
m=3: ≤20 (>19 possible)
m=4: ≤15 (<19 not possible)
Thus, 1, 2, or 3 work; 4 doesn't.
ANS = C
Exactly 18 equal 1.5. The average is 2.0, so total sum = 44 × 2.0 = 88.
Sum of the 18 ones = 18 × 1.5 = 27, leaving sum of remaining 26 numbers = 88 - 27 = 61.
Condition for Mode :
The set has exactly one mode, meaning one value appears most frequently, including more than 18 times for 1.5.
Let that frequency be k > 18; it appears k times.
The other 26 - k numbers must have frequencies ≤ k - 1 ≤ 17 (since k ≥ 19).
Possible Integer Modes
Integers less than 5: 1, 2, 3, 4. For m (one of these) to be mode:
Needs ≥ 19 occurrences.
Sum contribution ≤ 61, so max occurrences: (61/m).
m=1: ≤61 (>19 possible), but using 1 gives 61-19=42, average is 6 - not possible since all numbers must be less than 5. (Discard)
m=2: ≤30 (>19 possible)
m=3: ≤20 (>19 possible)
m=4: ≤15 (<19 not possible)
Thus, 1, 2, or 3 work; 4 doesn't.
ANS = C
29 Jan 2026, 00:32
Kudos
Bookmarks
kevincan
We have 1.5 present 18 times => Sum = 18 x 1.5 = 27
Sum of all 44 numbers = 2 x 44 = 88
=> Sum of the other 26 numbers = 88-27 = 61
To make an integer the single mode, it must occur at least 19 times.
Case 1: If 1 occurs 19 times => the other 7 terms will add up to 61 - 19 = 42 => average is 6 - not possible since all numebrs are less than 5
Case 2: If 2 occurs 19 times => the other 7 terms will add up to 61 - 38 = 23 => average is 3.28 - possible
Case 3: If 3 occurs 19 times => the other 7 terms will add up to 61 - 57 = 4 => average is 0.57 - possible
Case 4: If 4 occurs 19 times => the other 7 terms will add up to 61 - 76 = -15 => not possible since all numebrs are positive
No other cases are thus possible
Thus, there are only 2 integers possible (2 and 3)
Answer C
