SOLUTION: Hi In a box there were some 50c coins and half as many 20cent coins. When 40 50c and 8 20cent coins were added.the value in the box would be 4 times as much as it was before. Wha

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Question 1208259: Hi
In a box there were some 50c coins and half as many 20cent coins. When 40 50c and 8 20cent coins were added.the value in the box would be 4 times as much as it was before.
What is the total value at first
What is the total of coins at the end.
This is word for word from textbook.
Found 3 solutions by math_tutor2020, ikleyn, MathTherapy:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answers:
total value at first = 720 cents = $7.20
total number of coins at the end = 66 coins
Extras: started with 6 coins worth 20cents each, and 12 coins worth 50cents each.

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Work Shown

x = starting number of 20cent coins
2x = starting number of 50cent coins
Note that x is half of 2x.

A = 20x = starting value of just the 20cent coins
B = 50*(2x) = 100x = starting value of just the 50cent coins
C = A+B = 20x+100x = 120x = value of original group of coins (before any extras are added). Values are in cents.

Now add 8 copies of a 20cent coin and 40 copies of the 50cent coin.
x bumps up to x+8 to represent the new number of 20cent coins.
2x bumps up to 2x+40 to represent the new number of 50cent coins.
D = 20*(x+8) = 20x+160 = new value of just the 20cent coins
E = 50*(2x+40) = 100x+2000 = new value of just the 50cent coins
F = D+E = (20x+160)+(100x+2000) = 120x+2160 = new total value in cents.

The new value (F) is 4 times as much as before (C)
new = 4*old
F = 4*C
120x+2160 = 4(120x)
120x+2160 = 480x
2160 = 480x-120x
2160 = 360x
x = 2160/360
x = 6
There are 6 coins worth 20 cents each in the box at first.
There are also 12 coins worth 50 cents each in the box at first (since 2x = 2*6 = 12)
C = total starting value = 6*20+12*50 = 720 cents = $7.20

Now add 8 copies of a 20cent coin and 40 copies of the 50cent coin.
6+8 = 14 copies of a 20cent coin
12+40 = 52 copies of a 50cent coin
The box now has 14+52 = 66 coins
F = new total value = 14*20+52*50 = 2880 cents = $28.80

The jump from C = 720 cents to F = 2880 cents is "times 4" since 720*4 = 2880
Or you could note that F/C = 2880/720 = 4.
We have confirmed the answers are correct.


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If you only cared about the total value, and not the coin counts, then you can follow this method:

m = total value of the original group of coins (before any extras added)
value in cents

Adding 40 coins worth 50cents each and 8 coins worth 20cents each will give an additional 40*50+8*20 = 2160 cents
m bumps up to m+2160 which represents the new total value.
This expression is 4 times that of the original value.

new = 4*old
m+2160 = 4m
4m-m = 2160
3m = 2160
m = 2160/3
m = 720 cents is the starting total value
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.
In a box there were some 50c coins and half as many 20cent coins.
When 40 50c and 8 20cent coins were added, the value in the box would be 4 times as much as it was before.
(a) What is the total value at first
(b) What is the total of coins at the end.
~~~~~~~~~~~~~~~~~~~~ 

The amount added is  40*(50c) + 8*(20c) = 2000c + 160c = 2160c.

Let x be the total value at first.


From the problem, 

    x + 2160 = 4x.


Hence,

    2160 = 3x,  x = 2160/3 = 720.


Thus, the total value at first was 720c.   It is the ANSWER to question (a).



To solve for (b), let T be the number of 20c coins at first.
Then the number of the 50c coins at first is 2T.


Write an equation for the total amount at first

    20T + 50(2T) = 720.


Simplify and find T

    20T + 100T = 720

    120T = 720

       T = 720/120 = 6.


Initially, there were  T + 2T = 6 + 2*6 = 6 + 12 = 18 coins.


At the end, there were 18 + 40 + 8 = 66 coins, in all.   It is the ANSWER to question (b).

Solved.

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

How this problem is worded in your textbook, it recalls me a lame horse with three legs.

It is not a right style writing Math problems.



Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
Hi
In a box there were some 50c coins and half as many 20cent coins. When 40 50c and 8 20cent coins were added.the value in the box would be 4 times as much as it was before.
What is the total value at first
What is the total of coins at the end.
This is word for word from textbook. 

Let original number of 20c coins be T
Then original number of 50c coins = 2T ( as many 20c as 50c, or # of 50c is TWICE the number of 20c)
Original value of 20c coins: $.2T
Original value of 50c coins: .5(2T) = $T
TRANSLATED, "When 40 50c and 8 20cent coins were added.the value in the box would be 4 times as much as
             it was before," gives us the following equation:
             T + 40(.5) + .2T + 8(.2) = 4(.2T)(T)
                   T + 20 + .2T + 1.6 = .8T2
                          1.2T + 21.6 = .8T2
                                    0 = .8T2 - 1.2T - 21.6
                               -- Multiplying by  to
                                                                                get rid of decimals 
                                    0 = 2T2 - 3T - 54
                                    0 = 2T2 - 12T + 9T - 54
                                    0 = 2T(T - 6) + 9(T - 6)
                                    0 = (T - 6)(2T + 9)
                                    0 = T - 6      OR     0 = 2T - 9
                                                   OR     9 = 2T
 
             Original number of 20c coins (T) = 6  OR   
             Original number of 50c coins: 2T = 2(6) = 12
             Original number of coins: 6 + 12 = 18
 
             Number of 20c and 50c coins ADDED: 40 + 8 = 48
             Number of 20c and 50c coins AT END: 18 + 48 = 66

             Original value of 20c coins: 6(.2) = $1.20
             Original value of 50c coins: 12(.5) = $6
             Original value of 20c and 50c coins: $1.20 + $6 = $7.20
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