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Math problem?

I have this math problem, I can work it out by guessing and checking but need to use algebra.

Problem:

Moya had a total of $3.10 in her pocket in only 10- and 20- cent coins. If she has 19 coins altogether, how many of each coin does she have

Top 10 Answers
aznkid4321

Favorite Answer

a = 10 cent, b = 20 cent.

amount: cost total cost

a 10 10a

19-a 20 380 – 20a

10a – 20a + 380 = 310 <- sums of total costs = 3.10 dollars 70 = 10a, a=7, so b must = 12 haha i rule, i was just teaching this to my class at this tutoring place

0

Anonymous
12 20s and 7 10s?
0

Anonymous
let x = number of 10-cent coins, y = no. of 20-cent coints, since those are the unknowns which you want to find out.

set up the equations, according to the given:

a) x + y = 19 (total number of coins)

b) 0.10x + 0.20y = $3.10 (total money that Moya has)

to make things less confusing, multiply both sides of equation b) by 10.

new equation b): x + 2y = 31

now by using systems of equations:

x + y = 19

-(x + 2y = 31)

=> x + y = 19

-x – 2y = -31

==> -y = -12

y = 12: no. of 20-cent coins

refer to equation a) once more:

x + y = 19

x + 12 = 19

x = 7: no. of 10-cent coins

0

zanthus
12- 20 cent coins =2.40

7- 10 cent coins=.70

19 coins – $3.10

can’t help you with the algebra..I’m a savant 🙂

opps, I think the blond chick has a partial starting point:

let x be the number of 20 cent coins needed

ley y be the number of 10 cent coins needed

.20x+.10y= 3.10

x+y= 19

therfore x=19-y

solve

.20(19-y)+.10y=3.10

3.8- .20y +.10y=3.10

3.8 -.10y=3.10 (somehow aznkid above got here immediately

-.10y= 3.10-3.8

-.10y= -.70

y=7

OK, so we know that y, the number of 10 cent coins needed is 7!! Voila…Now that we know this, solving the rest is simple 🙂

From our original equation:

.20x + .10y =3.10

.20x+.10(7)= 3.10

.20x +.70 =3.10

.20x= 3.10-.70

.20x=2.40

x=12

Ok, So Algebraically, we know that the number of

10 cent pieces is 7 and the number of 20 cent pieces is 12

That’s algebra. I’m not sure what kind of space alien math aznkid is using, but it seems much faster for him :_) He jumped like 6 or 7 steps!!

It just goes to show that there are lots of ways to solve the same problem!!

0

James
7 and 12
0

p|nky
lets assume:

x = number of 10 cent coin

y = number of 20 cent coin

(1) 10x + 20y = 310

——————— : 10

x + 2y = 31

(2) x + y = 19

x = 19 – y

substitute x in (1) with (2) so:

(19 – y) + 2y = 31

19 + y = 31

y = 12

x = 19 – 12 = 7

0

Andrew D
.10x + .20y= 3.10

x + y= 19

Do the substitution method on the second equation and you get:

x = 19 – y

Now substitute it in the first equation.

.10(19 – y) + .20y = 3.10

Now solve this equation.

1.90 – .10y + .20y = 3.10 Distributive Property

-.10y + .20y = 1.20 Subtract 1.90 to both sides.

.10y = 1.20 Add left side

y = 12

Enter in Second equation:

12 + x = 19

x = 7

So your answers are:

12 twenty cent coins, and 7 ten cent coins

0

Anonymous
what exacltly is a 20 cent coin anyways????
0

Jonathan W
12 – .20

7 – .10

0

lablonde98
(20*x) + (10*y)

*=multiply

0

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