Question 1208259
.
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.
~~~~~~~~~~~~~~~~~~~~


<pre>
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 <U>ANSWER</U> 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 <U>ANSWER</U> to question (b).
</pre>

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.