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# help

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44.50=x/7+0.3x+x/10 is this the right equation for this problem and if so hoe do i solve it

Jeff is counting his small change and finds that he has collected \$44.50. He has one-seventh as many quarters as dimes, 0.3 times as many loonies as dimes, and one-tenth as many toonies as dimes.

Jun 27, 2017

#1
+1

Let the number of Dimes =D   P.S. This is "slang" for Canadian coin currency!!.

Quarters =1/7D

Loonies =3/10D

Toonies =1/10D

0.10D + 0.25(D/7) + 3D/10 + 2/10D =\$44.50

0.10D + 1/28D + 0.3D + 0.2D =\$44.50

0.6357142857...D =\$44.50    divide both sides by 0.6357142857

D = 70 Dimes

70/7 =10 Quarters

0.3 x 70 =21 Loonies

70 x 1/10 =7 Toonies. So that:

[70x.10] + [10 x 0.25] + [21 x \$1] + [7 x \$2]=\$44.50

Jun 27, 2017
edited by Guest  Jun 27, 2017
edited by Guest  Jun 27, 2017

#1
+1

Let the number of Dimes =D   P.S. This is "slang" for Canadian coin currency!!.

Quarters =1/7D

Loonies =3/10D

Toonies =1/10D

0.10D + 0.25(D/7) + 3D/10 + 2/10D =\$44.50

0.10D + 1/28D + 0.3D + 0.2D =\$44.50

0.6357142857...D =\$44.50    divide both sides by 0.6357142857

D = 70 Dimes

70/7 =10 Quarters

0.3 x 70 =21 Loonies

70 x 1/10 =7 Toonies. So that:

[70x.10] + [10 x 0.25] + [21 x \$1] + [7 x \$2]=\$44.50

Guest Jun 27, 2017
edited by Guest  Jun 27, 2017
edited by Guest  Jun 27, 2017