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10 Aug 2024, 19:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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(hard)
Question Stats:
54% (02:37) correct
46%
(02:25)
wrong
based on 84
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The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, will be
A. 2,222
B. 2,442
C. 2,592
D. 3,333
E. 2,834
A. 2,222
B. 2,442
C. 2,592
D. 3,333
E. 2,834
11 Aug 2024, 00:23
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Bunuel
The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, will be
A. 2,222
B. 2,442
C. 2,592
D. 3,333
E. 2,834
A. 2,222
B. 2,442
C. 2,592
D. 3,333
E. 2,834
Number of possible arrangements = 4! /2! = 12
Values =
1421
1412
1241
1214
1142
1124
4121
4112
4211
2411
2114
2141
Sum = 26664
Average = 26664/12 = 2222
Option A
11 Aug 2024, 22:55
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Available digits: 1 1 2 4
Total number of arrangements possible\(=\frac{4!}{2!}\) (Since arrangements within pair of 1's to be neglected)
No of integers when 4 in thousand place (4xxx): \(\frac{3!}{2!}=3\)
No of integers when 4 in hundredth place (x4xx): \(\frac{3!}{2!}=3\)
No of integers when 4 in tenth place (xx4x): \(\frac{3!}{2!}=3\)
No of integers when 4 in units place (xxx4): \(\frac{3!}{2!}=3\)
This is same for no of ways 2 can be arranged.
\(\text{Sum of 4 & 2: }\)
\((3*4000+3*400+3*40+3*4)+(3*2000+3*200+3*20+3*2)=3*6000+3*600+3*60+3*6=19998\)
No of integers when 1 in thousand place (1xxx): \(3!=6\). Similarly for all other places its 6 times.
\(\text{Sum of 1: }6*1000+6*100+6*60+6=6666\)
\(Sum=19998+6666=26,664\)
\(Average=\frac{26,664}{12}=2222\)
Answer is A
Total number of arrangements possible\(=\frac{4!}{2!}\) (Since arrangements within pair of 1's to be neglected)
No of integers when 4 in thousand place (4xxx): \(\frac{3!}{2!}=3\)
No of integers when 4 in hundredth place (x4xx): \(\frac{3!}{2!}=3\)
No of integers when 4 in tenth place (xx4x): \(\frac{3!}{2!}=3\)
No of integers when 4 in units place (xxx4): \(\frac{3!}{2!}=3\)
This is same for no of ways 2 can be arranged.
\(\text{Sum of 4 & 2: }\)
\((3*4000+3*400+3*40+3*4)+(3*2000+3*200+3*20+3*2)=3*6000+3*600+3*60+3*6=19998\)
No of integers when 1 in thousand place (1xxx): \(3!=6\). Similarly for all other places its 6 times.
\(\text{Sum of 1: }6*1000+6*100+6*60+6=6666\)
\(Sum=19998+6666=26,664\)
\(Average=\frac{26,664}{12}=2222\)
Answer is A
