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by Mani_mba » Sun Jul 27, 2008 4:57 am
5 is my answer to ur first question.
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by elimenda » Sun Jul 27, 2008 5:19 am
yes it is. can you explain me how u got there?
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by sudhir3127 » Sun Jul 27, 2008 5:23 am
My Answer is 5.

m+ n = Z+4
n = e + y

m + n = z + 4 + e + y
4 + y = -5
y = -9

e + z = 10

m + n = y + (e + z) + 4

m + n = -9 + (10) + 4

m + n = 5

Hope this is clear
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by elimenda » Sun Jul 27, 2008 5:58 am
yes it is, thanks!

i didn t get the format!
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please explain

by ektamatta » Sun Jul 27, 2008 8:06 pm
I didn't understsnd the format either. Can anyone please explain?
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by gmat_oct80 » Sun Jul 27, 2008 11:39 pm
1st question: Solution is

X+ 4 = 1
So X=-3
Y+4 = -5
So y = -9,
m= z+4,
e+z = 10,
e+y = n,
x+f = 2,
So f= 5
z+f = 5
So z= 0
so m= 4,
x+e = 7,
So e = 10
M + n = z+4+e+y = 0+4+10-9 = 5



2nd question:
question is same as 1st.
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by evansbd » Mon Jul 28, 2008 6:13 am
sudhir3127 wrote:My Answer is 5.

m+ n = Z+4
n = e + y

m + n = z + 4 + e + y
4 + y = -5
y = -9

e + z = 10

m + n = y + (e + z) + 4

m + n = -9 + (10) + 4

m + n = 5

Hope this is clear

Sudhir can you tell me how you got the bolded part?
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by b.kurland » Mon Jul 28, 2008 7:08 pm
i think the easiest way is to just solve for each necessary variable. a heavy reliance on algebra with this many variables could make the question more difficult/confusing than it ought to be.

x+4=1, so x=-3
x+e=7 : (-3)+e=7, so e=10
y+4=-5, so y=-9
y+e=n : (-9)+(10)=n, so n=1
z+e=10 : z+(10)=10, so z=0
z+4=m : (0)+4=m, so m=4

m+n=4+1=5

hope my elementary approach is helpful! :wink:
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