Weighing Pumpkins- Need Equations- Algebra 1 Level...?

Question

The Zunino family weighed their 5 pumpkins 3 at a time. The following weights were recorded: 3006, 3035, 3093, 3122, 3151, 3180, 3209, 3238, 3267, 3325. Find the weight of each pumpkin.

Please help me write the equations to solve this problem. Thanks!

Answer 1

First, I am going to assign all the weights to the proper pumpkins and then find the weight of each pumpkin.

Let's say a, b, c, d, and e represent the weight of each pumpkin so that a <= b <= c <=d <= e

We know:
a+b+c = 3006
a+b+d = 3035

That means
d - c = 3035 - 3006 = 29

At the other end, we know:
c+d+e = 3325
b+d+e = 3267

That means
c - b = 3325 - 3267 = 58

Putting these two together we have
d - c + c - b = 29 + 58
d - b = 87

Now, let's try to use what we have gathered so far:
a+c+d = a + c + b + 87 = 3006 + 87 = 3093
b+c+e = d - 87 + c + e = 3325 - 87 = 3238

So there are four more weights remaining. These are the weights of (a+b+e), (a+c+e), (a+d+e), (b+c+d)
We know (a+b+e) = (a+c+e) - 58
And (a+c+e) = (a+d+e) - 29

And looking at the weights, we can only have
a+d+e = 3209
a+c+e = (a+d+e) - 29 = 3209 - 29 = 3180
a+b+e = (a+c+e) - 58 = 3180 - 58 = 3122

And that leaves
b+c+d = 3151

Now, let's try to find each weight
b+c+d = 3151
c-58+c+c+29 = 3151
3c-29 = 3151
3c = 3151+29
3c = 3180
c = 3180/3
c = 1060

b = c - 58 = 1060 - 58 = 1002
a = 3006 - b - c = 3006 - 1002 - 1060 = 944
d = 29 + c = 29 + 1060 = 1089
e = 3325 - d - c = 3325 - 1089 - 1060 = 1176

So here you have it
944, 1002, 1060, 1089, 1176

I hope this helped

Kia

Answer 2

This 10 weights correspond to C(5 taken by 3)
a,b,c,d,e
The ten equations are
a+b+c=3006
a+b+d=3035
a+b+e=3093
a+c+d=3122
a+c+e =3151
a+d+e =3180
b+c+d =3209
b+c+e =3238
b+d+e 3267

c+d+e =3325

Answer 3

♣ for 5 unknown values we’ve got 10 equations!
But how variables are combined we do not know;
Each pumpkin participates 6 times in 10 acts of weighing;
Each weighing result being uj, we get
s3 =∑uj {j=1,10} =31626;
total weight of 5 pumpkins is
P = s3/6 = ∑pj {j=1,5} =5271, where pj is weight of each pumpkin,
being p1 ♥ on the other hand if we weighed by 2 pumpkins at a time we would get results vj as follows:
☺P-u1= 5271-3006=2265 =v1;
☺P-u2= 5271-3035=2236 =v2;
☻P-u3= 5271-3093=2178 =v3;
☻P-u4= 5271-3122=2149 =v4;
☻P-u5= 5271-3151=2120 =v5;
☻P-u6= 5271-3180=2091 =v6;
☻P-u7= 5271-3209=2062 =v7;
☻P-u8= 5271-3238=2033 =v8;
☺P-u9= 5271-3267=2004 =v9;
☺P-u10= 5271-3325=1946 =v10;
{lines marked as ☻ are illustrative and can be omitted!}
♦{this is check #1, and not for your teacher:
Each pumpkin participates 4 times in 10 acts of weighing;
Each weighing result being vj, we get
s2 =∑vj {j=1,10} =21084;
total weight of 5 pumpkins is
P = s2/4 = ∑pj {j=1,5} =5271; check done!};
♥ now we notice:
3 lightest pumpkins weight u1=3006;
2 lightest pumpkins weight v10=1946;
hence p3=u1-v10 =3006 -1946 =1060; ▬▬
♦{this is check #2:
3 heaviest pumpkins weigh u10=3325;
2 heaviest pumpkins weigh v1=2265;
hence p3=u10-v1 =3325 -2265 =1060; check done!};
♠ here we conclude:
1)there must be the lightest pumpkin
p1=v9-p3 =2004-1060 =944; ▬▬
2)there must be the heaviest pumpkin
p5=v2-p3 =2236-1060 =1176; ▬▬
3)three lightest pumpkins weight
3006= p1+p2+p3 =
= 944+p2+1060 =2004+p2, hence
p2= 3006-2004=1002; ▬▬
4)three heaviest pumpkins weight
3325= p3+p4+p5 =
= 1060 +p4 +1176 =2236+p4, hence
p4= 3325-2236=1089; ▬▬
▬ result:
p1=944; p2=1002; p3=1060; p4-1089; p5=1176;