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16 Jun 2018, 00:13
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (02:40) correct
47%
(02:04)
wrong
based on 377
sessions
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A set of 1522 numbers has a single mode. If the mode repeats 12 times and if no number occurs fewer than 2 times, what is the range of the number of unique numbers that could be in the set?
A. 128
B. 139
C. 552
D. 617
E. 756
A. 128
B. 139
C. 552
D. 617
E. 756
16 Jun 2018, 03:48
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Given: Single mode and it occurs 12 times. 1522 - 12 = 1510 numbers whose mode we don't know.
To find the RANGE(max-min) of number of unique numbers, we'll take two cases:
Case 1: Where the number of unique numbers is Maximum:
Since we already know that one number is occurring 12 times, we'll keep that aside and look at the remaining 1510 numbers.
1510/2 = 755.
Divided by 2 because every number is repeated atleast twice as stated in the question.
So there are 755 numbers that occurs twice in this series.
(1,1,2,2,3,3,4,4,..... 755 numbers like this, and, 123,123,123,123(12 times))
So total unique numbers = 755 + 1 = 756.
Case 2: For the Unique numbers to be minimum.
1510/11 = 137 + 3 remainder
Divided by 11 because we have a single mode that occurs 12 times. So the other numbers can repeat 11 times. And the remainder 3 can be of a same number.
So the total unique numbers in this set is 137 + 1 + 1 = 139
Range = 756 -139 = 617
OA - D.
To find the RANGE(max-min) of number of unique numbers, we'll take two cases:
Case 1: Where the number of unique numbers is Maximum:
Since we already know that one number is occurring 12 times, we'll keep that aside and look at the remaining 1510 numbers.
1510/2 = 755.
Divided by 2 because every number is repeated atleast twice as stated in the question.
So there are 755 numbers that occurs twice in this series.
(1,1,2,2,3,3,4,4,..... 755 numbers like this, and, 123,123,123,123(12 times))
So total unique numbers = 755 + 1 = 756.
Case 2: For the Unique numbers to be minimum.
1510/11 = 137 + 3 remainder
Divided by 11 because we have a single mode that occurs 12 times. So the other numbers can repeat 11 times. And the remainder 3 can be of a same number.
So the total unique numbers in this set is 137 + 1 + 1 = 139
Range = 756 -139 = 617
OA - D.
General Discussion
19 Sep 2019, 06:08
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Bunuel
set has 1522 numbers, and a single mode of 12
no number occurs fewer than 2 times: each number occurs more than or equal to two times
range: MAX unique numbers in the set - MIN unique numbers in the set
max: 1 number occurs 12 times, so max unique possible is 1522-12=1510/2=755+1=756
min: 1 number occurs 12 times, so min unique possible is 1522-12=1510/11=137…~138+1=139
range: 756-139=617
Answer (D)
