Latest Elitmus Aptitude Question SOLUTION: If whole numbers a>b>c>d>e>0 and avg. of a and e is 7 and b and d is 5 then how many such combinations possible

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(#M40151506) ELITMUS Category question Keep an EYE

If whole numbers a>b>c>d>e>0 and avg. of a and e is 7 and b and d is 5 then how many such combinations possible

Asked In Elitmus ankit (9 years ago)

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Devendra Marghade  (9 years ago)

14

a+e=2*7=14 and b+d=2*5=10
For a to e in descending order (a,e)=(8,6) (9,5) (10,4) (11,3) (12,2) (13,1) and
(b,d)=(6,4) (7,3) (8,2) (9,1)
Now with (a,e)=(11,3) possible (b,d)=(6,4), So (a,b,c,d,e)=(11,6, c, 4,3) with c=5 i.e. 1 combination
With (a,e)=(12,2) possible (b,d)=(6,4) (7,3), So (a,b,c,d,e)=(12,6, c, 4,2) with c=5 i.e. 1 combination
and (12,7, c, 3, 2) with c=4,5,6 i.e. 3 combinations
With (a,e)=(13,1) possible (b,d)=(6,4) (7,3) (8,2), So (a,b,c,d,e)=(13,6,c,4,1) with c=5 i.e 1 combination &
(13,7,c,3,1) with c=4,5,6 i.e. 3 combinations and
(13,8,c,2,1) with c=3,4,5,6,7 i.e. 5 combinations
So total combinations 1+1+3+1+3+5=14

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Devendra Marghade  (9 years ago)

14 combinations

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Lavanya Rathnam  (9 years ago)

what is the sylabus for elitmus tell me subjcts ..... is their ajny progrm lange........

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ravi bhushan  (9 years ago)

3 4 5 6 17 2 3 4 7 19 1 2 3 8 21
2 4 5 6 18 2 3 5 7 18 1 2 4 8 20
1 4 5 6 19 2 3 6 7 17 1 2 5 8 19
1 3 4 7 20 1 2 6 8 18
1 3 5 7 19 1 2 7 8 17 so answer is.... 14
1 3 6 7 18

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vishwa  (8 years ago)

actually a+e avg is 7 which is the edge value by basing on this we can directly say that a+e=7*2=14 so 14 combinations will occur

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