Question

Terry bought some gum and some candy. The number of packs of chewing gum was one more than the number of mints. The number of mints was three times the number of chocolate bars. If gum cost 6 cents a pack, mints cost 3 cents each, and chocolate bars cost 10 cents each, how many of each confection did he get for 80 cents?

Answer 1

He bought 13 pkgs of gum

He bought 12 mints

He bought 4 candy bars

Explanation:

Let +a+ = number of packages of gum Terry bought

Let +b+ = number of mints

Let +c+ = number of candy bars

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(1) +a+=+b+%2B+1+

(2) +b+=+3c+

(3) +24a+%2B+10b+%2B+35c+=+572+ ( in cents )

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There are 3 equations and 3 unknowns, so it's solvable

(2) +c+=+b%2F3+

Substitute (2) into (3)

(3) +24a+%2B+10b+%2B+35%2A%28+b%2F3%29+=+572+

Substitute (1) into (3)

(3) +24%2A%28+b+%2B+1+%29+%2B+10b+%2B+35%2A%28+b%2F3+%29+=+572+

(3) +24b+%2B+24+%2B+10b+%2B+%2835%2F3%29%2Ab+=+572+

Multiply both sides by +3+

(3) +3%2A34b+%2B+3%2A24+%2B+35b+=+3%2A572+

(3) +102b+%2B+72+%2B+35b+=+1716+

(3) +137b+=+1644+

(3) +b+=+12+

and

(1) +a+=+b+%2B+1+

(1) +a+=+12+%2B+1+

(1) +a+=+13+

and

(2) +c+=+b%2F3+

(2) +c+=+12%2F3+

(2) +c+=+4+

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He bought 13 pkgs of gum

He bought 12 mints

He bought 4 candy bars

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check:

(3) +24a+%2B+10b+%2B+35c+=+572+ ( in cents )

(3) +24%2A13+%2B+10%2A12+%2B+35%2A4+=+572+

(3) +312+%2B+120+%2B+140+=+572+

(3) +572+=+572+

OK

Answer 2

Mumber of mints=x
Number of packs of chewing gum=x+1
Number of chocolate bars=y

The number of mints was three times the number of chocolate bars; then
x=3y

We can suggest this system of equations:
x=3y
6(x+1)+3x+10y=80

we can solve this system of equations by substitution method.
6(3y+1)+3(3y)+10y=80
18y+6+9y+10y=80
18y+9y+10y=80-6
37y=74
y=74/37
y=2

x=3y
x=3*(2)=6

Number of mints=x=6
number of packs of chewing gum=x+1=6+1=7
number of chocolate bars=y=2

Answer: the number of mints is 6, the number of packs of chewing gum is 7, and the number of chocolate bars is 2.