The Quick Meals Diner served 411 dinners. A child's plate cost $3.20 and an adult's plate cost $6.60. A total of $2,005.40 was collected. How many of each type of plate was served?

Question

The Quick Meals Diner served 411 dinners. A child's plate cost $3.20 and an adult's plate cost $6.60. A total of $2,005.40 was collected. How many of each type of plate was served?

Round answers to the nearest whole person.

___ child plates were served.

___ adult plates were served.

Answer 1

Hello, Jake,

We have two unknowns so we need at least two equations. Line 1 tells us the total dinners (D) is 411, made up of plates for child (Pc) and adult (Pa). Line 2 tells us that the total value of 2005.40 is made up of sales from child and adult plates at their respective prices. Line 3 puts what we know into the line 2 equation. Line 4 takes line 1 and rearranges it to give us a value for Pc. Substitute that value in line 3 to yield line 5. Now we have only one variable in line 5, so solve for Pa in lines 6 and 7. Pc is 411 - Pa. All is entered in lines 8 and 9. Check the results by multiplying the meals times their prices to see if they add up to the original figures.

Bob

1. D = Pc + Pa = 411
2. Total $ = C*Pc + A*Pa = 20040
3. 2005.40 = 3.20Pc + 6.60Pa
4. Pc = 411 - Pa
5. 2005.40 = 3.20*(411 - Pa) + 6.60Pa
6. 2005.40 = 1315.20 - 3.2Pa + 6.60Pa
7. 690.20 = 3.5Pa
8. Pa = 203 and Pc = 208

Answer 2

Let's call the number of child's plates C and the number of adult's plates A.

If 411 dinners were served, then it follows that C+A = 411

Since there are two unknowns here (C and A) we need another equation to solve this.

If the total amount of money collected was $2005.40 then we need to represent this as a combination of child's and adult plates based on the cost of each plate.

Therefore, 2005.40=3.20C + 6.60A

Now we have two equations so we can solve. This can be done by either elimination or substitution. Let's do substitution.

In the first equation, we can subtract C so that A=411-C

We can now substitute this for the A in the second equation as follows:

2005.40 = 3.20C + 6.60(411-C)

Now we can expand this so we have 2005.40 = 3.2C + 2712.6 -6.6C

If we subtract the 2712.6 from both sides and combine the like terms on the right hand side, we now have:

-707.2 = -3.4C

Dividing both sides by -3.4 gives us C=208

Now we can plug this into the first equation 208 + A = 411 which gives us A = 203

So in summary, there were 208 child's plates and 203 adult plates served.