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# Tricky averages for Y6 pupils (aged 10 - 11)

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Four children weigh themselves in pairs and take the average mass of each pair. The average mass of the six pairs are: 47kg, 50kg, 51kg, 53kg, 54kg, 57kg.

Find the mass of the heaviest child.

I'm finding this hard to calculate the heaviest child. I know the heaviest two weigh 114kg and next heaviest two 108kg and the difference is 6kg, but can't work out a logical way to find the heaviest weight.

Any help would be greatly appreciated!

May 27, 2018
edited by Guest  May 27, 2018
edited by Guest  May 27, 2018
edited by Guest  May 27, 2018

#1
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Let the 4 kids be: A, B, C, D
A+B=2*47, A+C=2*50, A+D=2*51, B+C=2*53, B+D =2*54, C+D=2*57, solve for A, B, C, D

Solve the following system:
{A + B = 94 | (equation 1)
A + C = 100 | (equation 2)
B + C = 106 | (equation 3)

Subtract equation 1 from equation 2:
{A + B+0 C = 94 | (equation 1)
0 A - B + C = 6 | (equation 2)
0 A+B + C = 106 | (equation 3)

Add equation 2 to equation 3:
{A + B+0 C = 94 | (equation 1)
0 A - B + C = 6 | (equation 2)
0 A+0 B+2 C = 112 | (equation 3)

Divide equation 3 by 2:
{A + B+0 C = 94 | (equation 1)
0 A - B + C = 6 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Subtract equation 3 from equation 2:
{A + B+0 C = 94 | (equation 1)
0 A - B+0 C = -50 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Multiply equation 2 by -1:
{A + B+0 C = 94 | (equation 1)
0 A+B+0 C = 50 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Subtract equation 2 from equation 1:
{A+0 B+0 C = 44 | (equation 1)
0 A+B+0 C = 50 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Collect results: A = 44,   B = 50,   C = 56
C +  D =2 x 57
56 + D = 114
D = 114 - 56 = 58

A = 44 and B = 50 and C = 56 and D=58 - which is the heaviest child.

May 27, 2018
edited by Guest  May 27, 2018
edited by Guest  May 27, 2018