Question 1182878: Five friends Adam, Bipin, Chen, Dana and Erica all take a maths test. Their
teacher decides not to tell them their individual marks but gives them the
following information:
-Adam, Bipin and Chen together scored 200 marks.
-Bipin, Chen and Dana together scored 215 marks.
-Chen, Dana and Erica together scored 224 marks.
-The two girls (Dana and Erica) scored more than Chen.
-The five of them together scored 350 marks and each of them scored a whole
number of marks.
What were their individual scores?
I can only solve Chen, which is 74. Then, I don't know how to solve.
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
You apparently have done some work on this, so I know you are at least trying to solve the problem yourself. So I will just give you one more hint that should make solving the rest of the problem easy.
I will guess that your work was something like this....
A+B+C+D+E = 350
A+B+C=200, so D+E=150
C+D+E=224, so A+B=126
A+B+D+E=276, so C=74
To finish the problem easily, use this:
C=74
C is less than D
C is less than E
D+E=150
D and E are both integers
That should tell you the scores for D and E; then the rest of the problem is easy.
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website! -Adam, Bipin and Chen together scored 200 marks.A+B+C = 200 -Bipin, Chen and Dana together scored 215 marks.B+C+D = 215 -Chen, Dana and Erica together scored 224 marks.C+D+E = 224 -The two girls (Dana and Erica) scored more than Chen.D > C and E > C -The five of them together scored 350 marks A+B+C+D+E=350
So we have this system of 4 equations in 5 unknowns,
which is "under-determined":
A+B+C = 200
B+C+D = 215
C+D+E = 224
A+B+C+D+E = 350
We can reduce it to a 3x3 system by letting
A+B=F and D+E=G in the 1st, 3rd, and 4th equations.
Then we have this 3x3 system:
F+C = 200
C+G = 224
F+C+G = 350
Subtracting the 1st eq. from the 3rd gives G = 150
Subtracting the 2nd eq. from the 3rd gives F = 126
Substitution of 126 for F in the 1st gives C = 74
So A+B = 126, D+E = 150, and C = 74.
Since we know that D and E are both greater than C=74,
let m be how much greater D is than C=74, and
let n be how much greater E is than C=74. then
D = 74+m
E = 74+n, where m and n are positive integers.
Substituting in
D+E = 150
(74+m)+(74+n) = 150
74+m+74+n = 150
148+m+n = 150
m+n = 2
The only positive integer solution to that is
m = 1 and n = 1.
So D = 74+m = 74+1 = 75
And E = 74+n = 74+1 = 75
We find B from B+C+D = 215
B+74+75 = 215
B+149 = 215
B = 66
We find A from A+B = 126
A+66 = 126
A = 60
So (A,B,C,D,E) = (60,66,74,75,75)
Edwin
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