Is this problem even possible?

Question

It seems like a really easy problem but I've been trying to get it for fifteen minutes and can't.

Basically, there's a fare for a boat trip. It's $39 for adults, $19 for children, and $5 for infants. There were 49 passengers who paid a total fare of $1475. There were also twice as many adult passengers as children. The question is how many infants were there?

(Btw, you're only supposed to use one variable)

My teacher said something crazy about purposely making impossible problems just for us to grow confident in our solving skills but this is ludicrous. Is this problem even possible?

Notice: By possible, I don't mean that there's no way to solve it, just that there's no answer that works.

Answer 1

If you let x equal the number of child passengers, you end up with an equation something like this:
(2x times $39) plus (x times $19) plus 49 - (2x + 1x) times $5 = $1475.

Answer 2

30 Adults @ $39 = $1170, 15 Children @ $19 = $285, 4 Infants @ $5 = $20 makes total of $1475.

Answer 3

You could solve this with only one variable as I will show last, but for explaination purposes, I will do it with:
a=# of adults
c=# of children
i=# of infants

$1475= 39a+19c+5i

So, we know that there are twice as many adults as children (don't count the infants):

a/2= c

We also know that there were a total of 49 passengers:
49= a + c + i
or
49 = a + (a/2) + i

therefore:
a=(2/3)*(49-i)

39a+19c+5i=1475

39a+19(a/2)+5i=1475

(97/2)a + 5i = 1475

(97/2)(2/3)(49-i) +5i =1475 Look! Only one variable...anyways...

82i/3 = 328/3

i=4

There were 4 infants on board.

Answer 4

OK...we'll write this algebraically:
x = # of adults
y = # of children
z = # of infants

x + y + z = 49 (# of passengers)
39x + 19y + 5z = 1475 (amount of fares)
x = 2y (twice as many adults as children

Step 1: Remove z.
z = 49 - x - y
39x + 19y + 5(49 - x - y) = 1475
39x + 19y - 5x - 5y + 245 = 1475
34x + 14y = 1230

Step 2: Substitute x = 2y for x.
34x + 14y = 1235
34(2y) + 14y = 1235
68y + 14y = 1235
82y = 1230
y = 15

x = 2y = 30
z = 49 - x - y = 49 - 30 - 15 = 4

There are 4 infants, 15 children, 30 adults.

Answer 5

the problem w/ only using one variable is not gonna work, because there's no information about the infants in terms of the other...... for example, information is given regarding adults and children, so u can write both in terms of one variable. but infants are alone, not in terms of anything really.

children = x
adults = 2x
infants = y

x+2x+y = 49
19x + 39(2x) + 5y = 1475

solve for y to change y in terms of x:
y = 49-3x

plug that into the second equation:
19x + 78x + 5(49-3x) = 1475
97x + 245 - 15x = 1475
82x = 1230
x = 15

so, there are 15 children. 2(15) = 30 adults. and plug in x=15 into that equation we had earlier to find out y :
y= 49 -3x = y = 4

so there are 4 infants.

Answer 6

Okay 30 adults, 15 children, and 4 infants

Now tell your teacher you got your answer off the internet

Answer 7

I agree with rahidz2003;

x + y + z = 49
39x + 19y + 5z = 1475
2y = x

You will get after solving these equations:

30 adults
15 kids
4 infants

Answer 8

A + C + I = 49
39A + 19C + 5I = 1475
2C = A

Solve the system of equations and you get

30 Adults
15 Children
4 Infants

Answer 9

let there be 'x' adults,'y' children and 'z' infants
the equations are
x+y+z=49
x=2y=>x-2y=0
39x+19y+5z=1475
now rewritin the 1st and the third equation without x
2y+y+z=49
or 3y+z=49..............(1)
78y+19y+5z=1475
or 97y+5z=1475.......(2)
15y+5z=245 (1)*5
subtracting
82y=1230
y=15
so x=30 and z=4
so there were 30 adults,15 children and 4 infants
subtracting

Answer 10

it's fairly straight forward, I could solve it in my head.
knowing there are two adults per child allows you to find the combination of price multiples nearest to but less than the total.
find that which is divisible by five