+0

# In PTA fundraising activity, couples who came were charged \$400. However if only one parent came, he was charged \$250.00.

0
387
6
+76

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?

Mashirio  Nov 22, 2015

#3
+91780
+5

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 :))

Melody  Nov 22, 2015
edited by Melody  Nov 22, 2015
Sort:

#1
+17711
+5

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)

geno3141  Nov 22, 2015
#2
+5

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:

Guest Nov 22, 2015
#3
+91780
+5

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 :))

Melody  Nov 22, 2015
edited by Melody  Nov 22, 2015
#5
+5

How about: 1, 6, 11, 16, 21, 26 couples, and:

47, 39, 31, 23, 15, 7 singles.

Guest Nov 22, 2015
#6
+91780
0

That sounds good to me. Who is this mysterious guest ??     LOL

I'm thinking Bertie but then I have been known to be wrong  :D

Melody  Nov 23, 2015
edited by Melody  Nov 23, 2015

### 30 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details