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Question
primary 6 | Maths
| Algebra
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Please help me to solve it
A+C=115
A+D=116
A+E=117
B+C=118
B+D=124
B+E=125
C+D=130
C+E=137
D+E=139
Total=1224
4A+4B+4C+4D+4E=1224
A+B+C+D+E=306
103+130+E=306 where A+B=103 and C+D=130
E= 73
A+118+139=306
A=49
49+B+130+73=306
B=54
49+54+C+139=306
C= 64
103+64+D+73=306
D=66
Therefore, third heaviest student = C =64kg.
Hope this helps.
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