Question 86509: 1.A collection of 30 coins worth $5.50 consists of nickels, dimes, and quarters. There are twice as many dimes as nickels. How many quarters are there?
2.the school cafeteria sells milk at 25 cents per carton and salads at 45 cents each. One week the total sales for these items were $1,322.50. How many salads were sold that week?
3.At the homecoming football game, Senior Class Officers sold slices of pizza for $.75 each and hamburgers for $1.35 each. They sold 40 more slices of pizza than hamburgers, and sales totaled $292.50. How many slices of pizza did they sell?
4.For a recent job, a plumber earned $28/hour, and the plumber's apprentice earned $15/hour. The plumber worked 3 hours more than the apprentice. If together they were paid $213, how much did each earn? (Hint: First write an expression for the number of hours each worked on the job?
5.A plane whose air speed is 150 miles per hour flew from Abbot to Blair in 2 hours with a tail wind. On the return trip against the same wind, the plane was still 60 miles from Abbot after 2 hours. Find the wind speed and the distance between Abbot and Blair.
6.The cost of an adult ticket to a football game is $1.75. The cost of a student ticket is $1.25. Total receipts last week from ticket sales were $1,700. If the number of students tickets sold was twice the number of adult tickets, how many of each type were sold
7.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 1.A collection of 30 coins worth $5.50 consists of nickels, dimes, and quarters. There are twice as many dimes as nickels. How many quarters are there?
:
The number of each coin equation:
n + d + q = 30
:
The $ amt equation:
.05n + .10d + .25q = 5.50
:
"There are twice as many dimes as nickels" equation:
d = 2n
:
Substitute 2n for d in both equations
n + 2n + q = 30
and
.05n + .10(2n) + .25q = 5.50
Which is:
3n + q = 30
and
.25n + .25q = 5.50
:
Multiply the above equation by 4 and subtract it from 3n + q = 30
3n + 1q = 30
1n + 1q = 22
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2n + 0q = 8
n = 8/2
n = 4 nickels
:
Then we know that d = 2(4) = 8 dimes
:
Use the coins equation to find q:
4 + 8 + q = 30
q = 30 - 12
q = 18 quarters
:
Check solutions using the $ equation:
.05(4) + .10(8) + .25(18) =
.20 + .80 + 4.50 = 5.50
:
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2.the school cafeteria sells milk at 25 cents per carton and salads at 45 cents each. One week the total sales for these items were $1,322.50. How many salads were sold that week?
:
Seems like more information is needed for this one, the way it is:
It could be 1000 salads and 3490 milks
or 2000 salads and 1690 milks
or 3000 salads and 110 milks
:
:
3.At the homecoming football game, Senior Class Officers sold slices of pizza for $.75 each and hamburgers for $1.35 each. They sold 40 more slices of pizza than hamburgers, and sales totaled $292.50. How many slices of pizza did they sell?
:
Let x = no. of slices of pizza sold
Then
(x-40) = no. of hamburgers sold
:
.75x + 1.35(x-40) = 292.5
.75x + 1.35x - 54 = 292.5
2.10x = 292.5 + 54
2.10x = 346.5
x = 346.5/2.1
x = 165 slices of pizza
:
Check solution; 165-40 = 125 hamburgers
.75(165) + 1.35(125) =
123.75 + 168.75 = 292.50
:
:
4.For a recent job, a plumber earned $28/hour, and the plumber's apprentice earned $15/hour. The plumber worked 3 hours more than the apprentice. If together they were paid $213, how much did each earn? (Hint: First write an expression for the number of hours each worked on the job?
:
Let x = apprentice hrs
Then
(x+3) = plumber's hrs
:
15x + 28(x+3) = 213
15x + 28x + 84 = 213
43x = 213 - 84
x = 129/43
x = 3 hrs worked by the apprentice
3 + 3 = 6 hrs worked by the plumber
:
Check solution:
6(28) + 3(15) =
168 + 45 = 213
:
:
5.A plane whose air speed is 150 miles per hour flew from Abbot to Blair in 2 hours with a tail wind. On the return trip against the same wind, the plane was still 60 miles from Abbot after 2 hours. Find the wind speed and the distance between Abbot and Blair.
:
Let x = wind speed
Let y = one-way distance
:
With the wind equation: 2(150+x) = y
Against the wind equation: 2(150-x) + 60 = y
:
Rearrange the two equations to:
+2x - y = -300
-2x - y = -360
-------------- add
0x - 2y = -660
y = -660/-2
y = 330 mi is the distance
:
Find x:
2x - y = -300
2x - 330 = -300
2x = 330 - 300
x = 30/2
x = 15 mph is the wind
:
Check solutions:
2(150+15) =
2 * 165 = 330
:
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6.The cost of an adult ticket to a football game is $1.75. The cost of a student ticket is $1.25. Total receipts last week from ticket sales were $1,700. If the number of students tickets sold was twice the number of adult tickets, how many of each type were sold
:
Let x = no. of adult tickets
Then
2x = no. of student tickets
:
1.75x + 1.25(2x) = 1700
1.75x + 2.50x = 1700
4.25x = 1700
x = 1700/4.25
x = 400 adults, then 800 students
:
Check solutions:
1.75(400) + 1.25(800)
700 + 1000 = 1700
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