In PTA fundraising activity, couples who came were charged $400. However if only one parent came, he was charged $250.00.
59 People attended the activity and $12,150 was raised. How many couples attended?
A CHALLENGE TO YOU
I have just been playing around with a harder version of this question.
Say you have all that info EXCEPT you do not know how many people attended.
In PTA fundraising activity, couples who came were charged $400. However if only one parent came, he was charged $250.00.
59 People attended the activity and $12,150 was raised. How many couples attended?
You want to know all the possibilities of how many singles and how many couples attended.
I mean all the possible pairs of numbers in order for EXACTLY $12,150 to be raised.
Can you do that ?
I found exactly 6 posibilities. Can you find ANY or all of those?
GOOD LUCK :))
There can be two equations with two variables: C = number of couples; S = number of singles.
One equation is for the number of people: 2C + S = 59
(59 is the total number of people, use '2C' because each couple contains two persons.)
The other equation holds the amount of money made: 400C + 250S = 12,150
Combining these two equations:
Keep the money equations : 400C + 250S = 12,150
Mutiply the number of people equation by 200: 400C + 200S = 11,800
Subtract down the columns: 50S = 350
Divide by 50: S = 7 (there were 7 single parents)
Since 2C + S = 59 ---> 2C + 7 = 59 ---> C = 26 (there were 26 couples)
In PTA fundraising activity, couples who came were charged $400. However if only one parent came, he was charged $250.00.
59 People attended the activity and $12,150 was raised. How many couples attended?
Let C=number of couples
Let S=number of singles
2C + S=59,and,
400C + 250S=12,150
Solve the following system:
{2 C+S = 59 | (equation 1)
400 C+250 S = 12150 | (equation 2)
Swap equation 1 with equation 2:
{400 C+250 S = 12150 | (equation 1)
2 C+S = 59 | (equation 2)
Subtract 1/200 × (equation 1) from equation 2:
{400 C+250 S = 12150 | (equation 1)
0 C-S/4 = (-7)/4 | (equation 2)
Divide equation 1 by 50:
{8 C+5 S = 243 | (equation 1)
0 C-S/4 = -7/4 | (equation 2)
Multiply equation 2 by -4:
{8 C+5 S = 243 | (equation 1)
0 C+S = 7 | (equation 2)
Subtract 5 × (equation 2) from equation 1:
{8 C+0 S = 208 | (equation 1)
0 C+S = 7 | (equation 2)
Divide equation 1 by 8:
{C+0 S = 26 | (equation 1)
0 C+S = 7 | (equation 2)
Collect results:
Answer: | C=26 and S=7
A CHALLENGE TO YOU
I have just been playing around with a harder version of this question.
Say you have all that info EXCEPT you do not know how many people attended.
In PTA fundraising activity, couples who came were charged $400. However if only one parent came, he was charged $250.00.
59 People attended the activity and $12,150 was raised. How many couples attended?
You want to know all the possibilities of how many singles and how many couples attended.
I mean all the possible pairs of numbers in order for EXACTLY $12,150 to be raised.
Can you do that ?
I found exactly 6 posibilities. Can you find ANY or all of those?
GOOD LUCK :))