SOLUTION: Hi Mr Thomas spent $1972 on cars and dolls. He bought 3 times as many cars as dolls. Each doll cost $10 more than each car. She paid $476 more for the cars than the dolls. What is

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Question 1208177: Hi
Mr Thomas spent $1972 on cars and dolls. He bought 3 times as many cars as dolls. Each doll cost $10 more than each car. She paid $476 more for the cars than the dolls. What is the cost of a doll.

Found 5 solutions by ikleyn, greenestamps, mccravyedwin, josgarithmetic, MathTherapy:
Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.
Mr Thomas spent $1972 on cars and dolls. He bought 3 times as many cars as dolls.
Each doll cost $10 more than each car. She paid $476 more for the cars than the dolls.
What is the cost of a doll.
~~~~~~~~~~~~~~~~~~~


How Mr. Thomas suddenly became "She" ?


        If it was an "artificial intelligence", who created this problem,
        hence, this artificial intelligence has a defective logic inside, which should be fixed.

        In order for do not scare people around.


        Another version is that a professor who composed this problem was drunk.


        Third version is that the problem's creator does not read what he/she writes and posts.


        Appropriate problem to submit it for  shNobel prize competition.


//////////////////////////


From the problem, we know that the total cost was $1972, and the cost for the cars 
was $476 more than for the dolls.


Hence, the cost for the cars  was  1972%2F2 + 476%2F2 = 1224 dollars,

   and the cost for the dolls was  1972%2F2 - 476%2F2 =  748 dollars.


Let the number of the dolls be x.  
Then the number of cars is 3x.


The price for one doll (each doll) is  748%2Fx.

The price for one car  (each car) is  1224%2F%283x%29 = 408%2Fx.


From the problem, we have this equation for the price difference

    748%2Fx - 408%2Fx = 10  dollars.


Simplify and find x

    %28748-408%29%2Fx = 10

    340%2Fx = 10

    x = 340%2F10 = 34.


Hence, the cost of a doll is 748%2F34 = 22 dollars.    ANSWER

Solved.

Simply,  easy,  fresh and funny.         And educative.



Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


He bought 3 times as many cars as dolls:
let x = # of dolls
then 3x = # of dolls

Each doll cost $10 more than each car:
let y = cost of each car
then y+10 = cost of each doll

total cost of the cars: (3x)(y) = 3xy
total cost of the dolls: (x)(y+10)=xy+10x

The total cost was $1972; the cost of the cars was $476 more than the cost of the dolls:
3xy%2B%28xy%2B10x%29=4xy%2B10x=1972
3xy-%28xy%2B10x%29=2xy-10x=476

Eliminate the xy term from the two equations:
4xy%2B10x=1972
4xy-20x=952
30x=1020
x=34

He bought x=34 dolls and 3x=102 cars

Substitute x=34 in one of the equations to solve for y:
2xy-10x=68y-340=476
68y=816
y=12

The cost of each car was y = $12; the cost of each doll was y+10 = $22

ANSWER: $22

CHECK:
cost of 34 dolls and 102 cars: 34($22)+102($12) = $748+$1224 = $1972
difference between cost of cars and cost of dolls: $1224-$748 = $476


Answer by mccravyedwin(406) About Me  (Show Source):
You can put this solution on YOUR website!
Take advantage of available technology. 
Use a separate letter for every unknown, 
which makes setting up a problem easy.

Let the number of cars = c
Let the cost of a car = x
Let the number of dolls = d
Let the cost of a doll = y

Mr. Thomas spent $1972 on cars and dolls. 

cx + dy = 1972

He bought 3 times as many cars as dolls. 

c = 3d

Each doll cost $10 more than each car. 

y = x + 10

She paid $476 more for the cars than the dolls. 

Cx = Dy + 476

What is the cost of a doll?

Go to this site: 

https://www.emathhelp.net/en/calculators/algebra-2/system-of-equations-solver/?

Where it says "Enter a system of equations, type in 

cx + dy = 1972, c = 3d, y = x + 10, cx = dy + 476

Click "calculate".  

Scroll down and read 

c = 102, d = 34, x = 12, y = 22

Cost of a doll = $y = $22

Edwin

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
                 PRICE          QTY         COST

CARS              p              3d          3pd

DOLLS             p+10           d          d(p+10)

Total                                        1972

"She paid $476 more for the cars than the dolls."
3pd-d%28p%2B10%29=476


Very interesting to notice the sum of two expressions and then the difference of
those same two expressions.
system%283pd%2Bd%28p%2B10%29=1972%2C3pd-d%28p%2B10%29=476%29
-
3pd%2Bpd%2B10d=1972
4pd%2B10d=1972
2pd%2B5d=986
-
3pd-pd-10d=476
2pd-10d=476
pd-5d=238

system: system%282pd%2B5d=986%2Cpd-5d=238%29

Easier if kept in the form system%282pd%2B5d=986%2C2pd-10d=476%29 because now easy to eliminate for pd
to solve for d.


15d=510
highlight%28d=34%29----------This is how many dolls bought.

pd-5d=238
pd=238%2B5d
p=%28238%2B5d%29%2Fd
p=%28238%2B5%2A34%29%2F34
p=12---------car price
Doll is $10 more than a car, so p%2B10=highlight%2822%29. Price of a doll




-----------------------

d=34 so p=12.
check:
3pd+d(p+10)=1972 ?
3*12*34+34(12+10)
1224+748
1972
checks correctly.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
Hi
Mr Thomas spent $1972 on cars and dolls. He bought 3 times as many cars as dolls. Each doll cost $10 more than each car. She paid $476 more for the cars than the dolls. What is the cost of a doll.

Let cost of ALL dolls purchased be x
Since $476 MORE was spent on cars than dolls, then amount spent on cars was x + 476
Since a total of $1,972 was spent, we get: x + x + 476 = 1,972
                                                    2x = 1,972 - 476
                                                    2x = 1,496
                              Cost of ALL dolls, or matrix%281%2C5%2C+x%2C+%22=%22%2C+%221%2C496%22%2F2%2C+%22=%22%2C+%22%24748%22%29
                           So, cost of ALL cars = 1,972 - 748 = $1,224

Let price of each doll, be D
Then price of each car = D - 10

Number of dolls purchased: 748%2FD
Number of cars purchased: %221%2C224%22%2F%28D+-+10%29
Since he bought 3 times as many cars as dolls, we get the following
NUMBER-OF-ITEMS equation:   matrix%281%2C3%2C+%221%2C224%22%2F%28D++-++10%29%2C+%22=%22%2C+3%28748%2FD%29%29
                           
                             matrix%281%2C3%2C+102%2F%28D++-++10%29%2C+%22=%22%2C+187%2FD%29 --- Canceling GCF, 12, in numerators
                                102D = 187(D - 10) --- Cross-multiplying
                                102D = 187D - 1,870
                         102D - 187D = - 1,870
                               - 85D = - 1,870
            Cost of each doll, or