SOLUTION: I am a parent and trying to help my daugther could you help!
1. Snookers Lumber can convert logs into either lumber or plywood. In a given day, the mill turns out three times as m
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Question 173678: I am a parent and trying to help my daugther could you help!
1. Snookers Lumber can convert logs into either lumber or plywood. In a given day, the mill turns out three times as many units of plywood as lumber. It makes a profit of $30 on a unit of lumber and $45 on a unit of plywood. How many of each unit must be produced and sold in order to make a profit of $17160?
Snooker Lumber must produce and sell__ units of lumber and__ units of plywood to make a profit of $17160.
2. A disc jockey must play 16 commercial spots during 1 hour of a radio show. Each commercial is either 30 seconds or 60 seconds long. If the total commercial time during 1 hour is 13 min, how many 30-second commercials were played that hour? How many 60-second commercials?
How many 30-second commercials were played that hour?
How many 60-second commercials were played that hour?
3. A student makes a $9.50 purchase at the bookstore with a $20 bill. The store has no bills and gives the change in quarters and fifty-cent pieces. There are 30 coins in all. How many of each kind are there?
How many quarters are there in the change?
How many fifty-cent pieces are there in the change?
4. The perimeter of a rectangle is 80.m. The length is 7 m more than twice the width. Find the dimensions.
What is the length? And Width?
5. Hockey teams receive 2 points when they win and 1 point when they tie. One season, a team won a championship with 60 points. They won 9 more games than they tied. How many wins and how many ties did the team have?
How many wins did the team have?
How many ties did the team have?
Found 2 solutions by Mathtut, gonzo:
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
these problems are linear equation problems
:
1.lets call the number of units of plywood and lumber p and L respectively
:
p=3L...eq 1
45(p)+30(L)=17160....eq 2
:
take p's value from eq 1 and plug it into all the p's in eq 2
:
45(3L)+30L=17160
:
135L+30L=17160
:
165L=17160
:
units of Lumber
:
units of plywood
:
2)lets call the 30 sec and 60 second spots t and s
:
t+s=16.......eq 1---->t=16-s
1/2t+1s=13...eq 2---->t+2s=26(multiplied all terms by 2 to get rid of fraction)
:
t+s=16.......eq 1
t+2s=26......eq 2 revised
:
lets subtract eq 1 from eq 2 so as to eliminate the t terms t-t=0
we are left with 2s-s=26-16
:
60 sec commercials
:
30 sec commercials
:
3)lets call the quarters and fifty cent pieces q and f respectively
:
q+f=30.....eq 1
.25q+.5f=11.50....eq 2 20-9.50=change=11.50
:
lets re write eq 1 . q=30-f now take q's value and plug it into eq 2
:
.25(30-f)+.5f=11.5
:
7.5-.25f+.5f=11.5
:
.25f=4
:
fifty cent pieces
:
quarters
:
4)remember perimeter equals 2 times the length plus 2 times the width
p=2L+2W where L and W are the length and width and p is the perimeter
:
2L+2W=80.....eq 1
L=2W+7.......eq 2
:
take L's value from eq 2 and plug it into eq 1
:
2(2W+7)+2W=80
:
4W+14+2W=80
:
6W=66
:
meters in width
:
meters in length
:
:
5)lets call the number of wins and ties , w and t respectively
:
2w+1t=60....eq 1
w=t+9......eq 2
:
take w's value from eq 2 and plug it into eq 1
:
2(t+9)+t=60
:
2t+18+t=60
:
3t=42
:
ties
:
wins
:
:hope that helps you out
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
question number 1:
-----
1. Snookers Lumber can convert logs into either lumber or plywood. In a given day, the mill turns out three times as many units of plywood as lumber. It makes a profit of $30 on a unit of lumber and $45 on a unit of plywood. How many of each unit must be produced and sold in order to make a profit of $17160?
Snooker Lumber must produce and sell__ units of lumber and__ units of plywood to make a profit of $17160.
-----
answer to question number 1:
-----
mill turns out 3 times as many units of plywood as lumber. (given)
let a = amount of plywood
let b = amount of lumber
equation is:
a = 3*b
-----
total profit is 17160
profit on one piece of plywood is 45 (given)
profit on one piece of lumber is 30 (given)
profit on b pieces of lumber is 30*b
profit on a pieces of plywood is 45*a
equation for total profit is:
30*b + 45*a = 17160
-----
you have 2 equations you need to solve simultaneously (same value for b and same value for a applies to both).
those equations are:
a = 3*b
30*b + 45*a = 17160
-----
since you know a = 3*b, you can substitute 3*b for a to get:
30*b + 45*(3*b) = 17160
this equation becomes:
30*a + 135*b = 17160
which becomes:
165*b = 17160
divide both sides of equation by 165 to get:
b = 17160/165 = 104
you have:
b = 104
you know that a = 3*b, so
a = 3*104 = 312
you now have:
b = 104
a = 312
-----
substitute in your profit equation to get:
30*b + 45*a = 17160
this becomes:
30*104 + 45*312 = 17160
which becomes:
3120 + 14040 = 17160
-----
both equations are satisfied (they are true when you substitute the values for a and b).
-----
a = 3*b
312 = 3*104 (true)
-----
30*b + 45*a = 17160 (shown to be true just above).
-----
your answer to the question is:
the mill must produce and sell 312 units of plywood and 104 units of lumber to make a profit of 17160.
-----
question number 2:
-----
2. A disc jockey must play 16 commercial spots during 1 hour of a radio show. Each commercial is either 30 seconds or 60 seconds long. If the total commercial time during 1 hour is 13 min, how many 30-second commercials were played that hour? How many 60-second commercials?
-----
answer to question number 2:
-----
let a = number of 30 second commercial
let b = number of 60 second commercial
total time for commercials is 13 minutes * 60 seconds per minute = 780 seconds.
first equation for total time used is:
30*a + 60*b = 780
-----
total number of commercials is 16
second equation for total number of commercials used is:
a + b = 16
-----
this can also be solved by substitution same as in question number 1.
solve for a in the second equation:
a + b = 16
subtract b from both sides:
a = 16 - b
-----
substitute 16-b for a in the first equation:
30*a + 60*b = 780
becomes:
30*(16-b) + 60*b = 780
expand:
30*16 - 30*b + 60*b = 780
combine like terms:
30*16 + 30*b = 780
subtract 30*16 from both sides:
30*b = 780 - (30*16)
expand:
30*b = 780 - 480
simplify:
30*b = 300
divide both sides by 30:
b = 300/30
b = 10
-----
since b = 10, and a + b = 16, then:
a = 6
-----
you have:
a = 6
b = 10
-----
substitute for a and b in first equation:
30*a + 60*b = 780
30*6 + 60*10 = 780
180 + 600 = 780
780 = 780
-----
equations are good
answer to question number 2 is:
number of 30 second commercials is 6.
number of 60 second commercials is 10.
-----
question number 3:
-----
3. A student makes a $9.50 purchase at the bookstore with a $20 bill. The store has no bills and gives the change in quarters and fifty-cent pieces. There are 30 coins in all. How many of each kind are there?
How many quarters are there in the change?
How many fifty-cent pieces are there in the change?
-----
answer to question number 3:
-----
the change is 20.00 - 9.50 = 10.50
let a = # quarters
let b = # fifty cent pieces
-----
number of coins is 30:
a + b = 30
total change is 10.50
.25*a + .5*b = 10.5
-----
two equations are:
a + b = 30
.25*a + 5*b = 10.5
-----
we can use substitution as in questions 1 and 2 above, or we can solve as two simultaneous equations by manipulating the equations so one of the unknowns drops out allowing us to solve for the other unknown.
we'll do it the second way this time.
-----
two equations are:
a + b = 30
.25*a + .5*b = 10.5
-----
multiply the second equation by 4:
a + b = 30
a + 2b = 42
-----
subtract first equation from second equation to get:
a-a + 2b-b = 42-30
b = 12
-----
if b = 12, then a = 18 because a + b = 30
we have:
a = 18
b = 12
-----
substitute in the second equation to see if we get the right answer:
.25*a + .5*b = 10.5
.25*18 + .5*12 = 10.5
4.5 + 6 = 10.5
10.5 = 10.5
-----
values are good
answer to question number 3 is:
there are 12 fifty cent pieces in the change
-----
question number 4:
-----
4. The perimeter of a rectangle is 80.m. The length is 7 m more than twice the width. Find the dimensions.
What is the length? And Width?
-----
answer to question number 4:
-----
let a = length
let b = width
perimeter = 80 meters
length is 7 meters more than twice the width.
equation for that is:
a = 2*b + 7
formula for perimeter is:
2*a = 2*b = 80
substitute 2*b+7 for a to get:
2*(2*b+7) + 2*b = 80
simplify:
4*b + 14 + 2*b = 80
combine like terms:
6*b + 14 = 80
subtract 14 from both sides:
6*b = 66
b = 11
a = 2*b + 7
a = 2*11 + 7
a = 29
-----
substitute in perimeter equation to prove this is true.
2*a + 2*b = 80
2*29 + 2*11 = 80
58 + 22 = 80
80 = 80
equation is true
-----
answer to question # 4 is:
length is 29
width is 11
-----
question number 5:
-----
5. Hockey teams receive 2 points when they win and 1 point when they tie. One season, a team won a championship with 60 points. They won 9 more games than they tied. How many wins and how many ties did the team have?
-----
answer to question number 5:
-----
w = number of wins
t = number of ties
they won 9 more games than they tied:
w = t+9
2 points for a win
1 point for a tie
total points = 60:
2w + 1t = 60
-----
since w = t+9, substitute in total points equation to get:
2w + 1t = 60
2(t+9) + 1t = 60
2t + 18 + 1t = 60
3t + 18 = 60
3t = 42 (you subtract 18 from both sides)
t = 14 (you divide both sides by 3)
-----
w = 23 (w = t+9)
-----
you have:
w = 23
t = 14
substitute in equation for total points:
2w + 1t = 60
2*(23) + 1*(14) = 60
46 + 14 = 60
60 = 60
equeation is good.
answer to question number 5 is:
the team had 23 wins and 14 ties
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