Elitmus
Exam
Numerical Ability
Number System
abcde are 5 integer each have unique value and a>b>c>d>e>0 and c=9. so what is the value of a?
Read Solution (Total 11)
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- Ans =38
Soln-
given that (a+e)/2=7 & c=9
so--- a+e= 14
so passible combinations of a & e is (13,1),(12,2),(11,3).
Now
a>b>c>d>e>0
case-1-
13 > (12/11/10) > 9 > (8/7/6/5/4/3/2)>1>0
so ===> 3*7=21
case 2->
12 > (11/10) > 9 > (8/7/6/5/4/3) > 2 > 0
so ===> 2*6=12
case 3->
11 > 10 > 9 > (8/7/6/5/4) > 3 > 0
so===> 5
than - 21+12+5=38 - 10 years agoHelpfull: Yes(27) No(1)
- avgerage of a,e is 7(given) question is how many possible combination of abcde are???
sol:
a+e/2=7
a+e=14
ab9de
so the max value of e can be 13 and min value of a can be 1
so the combinations of 1b9d13 is 7*3=21
if a=2,e=12
then psbl sol 6*2=12
if a=3 e=11
then 5*1=5
so total combinations are 21+12+5=38 ans
- 10 years agoHelpfull: Yes(6) No(3)
- (a+e)/2=7 given
- 10 years agoHelpfull: Yes(3) No(2)
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- how many possible value of a?
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- mohit mathuriya : a,b,c,d,e can not exceed 9 i think.
- 10 years agoHelpfull: Yes(0) No(0)
- Can anyone please explain that how the average of a,e is 7?
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- can u please tell how u write avg of a &e=7
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bt its confirm ans is less than 45
bcz the avg of 5 digit no lies in middle
- 8 years agoHelpfull: Yes(0) No(1)
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