a+b+c=13
c+d+e=13
e+f+g=13
g+h+i=13
find the value of e.
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Answer / anjum kabir
A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
Value of E = 4
13*4 = 52
1+2+3+....+9 = 45
Then C+D+E must be equal to 7
or they must from 1 , 2 or 4.
1 and 2 for E cannot satisfy as because 1 and 2 cannot
come together in C+D+E
hence 1 and 4 and 2 and 4 should be together.
so you get two solutions and putting them any of the either
way you get E = 4. and everynumber gets satisfied with a
unique value.
| Is This Answer Correct ? | 57 Yes | 10 No |
Answer / vyas
the values from 1-9 can be substed for a-i....
a=6,b=5,c=2,d=7,e=4,f=8,g=1,h=3,i=9
| Is This Answer Correct ? | 35 Yes | 6 No |
Answer / manikandan
the alphabets a to i can be replaced by numbers 1-9
a=9,b=3,c=1,d=8,E=4,f=7,g=2,h=6,i=5
| Is This Answer Correct ? | 27 Yes | 8 No |
Answer / rohit parimal
a+b+c=13 - eq 1
c+d+e=13 - eq 2
e+f+g=13 - eq 3
g+h+i=13 - eq 4
we will add all the equations
a+b+c+c+d+e+e+f+g+g+h+i=52 - eq 5
a+b+c+d+e+f+g+h+i+(c+e+g)=52
as the numbers are from 1 to 9
the addition of nine numners from 1 to 9 is 45.
so
a+b+c+d+e+f+g+h+i=45 -eq 6
Putting the value from eq 6 in eq 5
45+c+e+g=52 - eq 7
c+e+g=7 - eq 8
eq 2 says c+d+e=13
Subtracting eq 8 from eq 2
c+d+e-c-e-g=13-7
d-g=6 - eq 9
when we put value 4 in g then the value of d should be 10.
Objction.
So value of g is 1 or 2.
eq 3 says e+f+g=13
When we put the vaue of g as 1 then f should be 8 and e
should be 4.
When we put the value of g as 2 then f should be 9 and e
should be 2.Objection (because no two alphabets can be
same.)
| Is This Answer Correct ? | 18 Yes | 3 No |
Answer / asp
a,b,c...e if rational no, infinite solutions possible
| Is This Answer Correct ? | 5 Yes | 6 No |
Answer / salma
c+d+e=13 or e+f+g=13
we can equally substitute the value of these
so...
e=13/3=4.3333333333333333333333333333333
| Is This Answer Correct ? | 6 Yes | 10 No |
Answer / sachin b. umbarkar
a+b+c=13
c+d+e=13
Thus, a+b+c=c+d+e
a+b=d+e .................(I)
e+f+g=13
g+h+i=13
Thus, g+h+i=e+f+g
h+i=e+g ...............(II)
Now add (I)and(II)
We get ,
a+b+h+i=d+g+2e
Thus,
e=(a+b+h+i-d-g)/2 .........Answer
| Is This Answer Correct ? | 1 Yes | 6 No |
Answer / kavitha
no unique solution but it can bnythng from 1-9
a=11
b=1
c=1
d=11
e=1
f=1
g=11
h=1
i=1
so at last e can be even 1
| Is This Answer Correct ? | 4 Yes | 11 No |
Answer / revathy
c=13-a-b
thus 13-a-b+d+e=13
e=a+b-d
This maynot have a unique solution
One such solution is:
if a=1, b=1, d=1
e=1
thus satisfying all the equations
| Is This Answer Correct ? | 6 Yes | 14 No |
Given that a,b,c,d,e each represent one of the digits between 1-9 and that the following multiplication holds a b c d e 4 ---------- e d c b a What digit does e represent a) 4 b) 6 c) 7 d) 8 e) none
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