Questions on Word Problems: Coins answered by real tutors!

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I do not know where to start on this one, please help me. John has $1.70 in nickels and dimes. He has 5 more dimes than nickels. With that said, how many nickels does he have?
-------------
N = # of nickels
D = # of dimes
D = N + 5 (he has 5 more dimes than nickels)
5N + 10D = 170 (they add to $1.70)
-----------
Now there are 2 eqns in 2 unknowns. My dog can solve those. (OK, he only did one, and it might have been luck.)
Sub for D in the 2nd eqn
5N + 10*(N+5) = 170
5N + 10N + 50 = 170
15N = 120
N = 8
D = N+5 = 13

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I PREFER D FOR DIMES & N FOR NICKLES MAKES IT EASIER TO REMEMBER WHAT'S WHAT.
D=9N
.10D=.05N+10.20
.10*9N=.05N+10.20
.9N=.05N+10.20
.9N-.05N=10.20
.85N=10.20
N=10.20/.85
N=12 NUMBER OF NICKLES.
D=9*12
D=108 NUMBER OF DIMES.
PROOF:
.10*108=.05*12+10.20
10.80=.60+10.20
10.80=10.80

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you have one hundred coins in your coin jar, consisting of nickels, dimes and quarters. the total value of the coins is &11.80 and you have 10 more nickels than dimes. How many of each coin do you have?
x=dimes
x +10= nickels
90-2x= quarters
-----------------
What you have is correct.
Then add:
10*x + 5*(x+10) + 25*(90-2x) = 1180
You can do it from here, I'm sure, but...
10x + 5x+50 + 2250-50x = 1180
-35x = -1120
x = 32 dimes (= $3.20)
42 = nickels (= $2.10)
26 = quarters (= $6.50)
100 coins------$11.80

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I'm presuming that you mean 1 penny and 5 pence coins, analogous to American penny and nickel. So let the NUMBER of penny coins be represented by p. Since each is worth one penny, the value of that number of coins is simply p as well. Let the number of P5 coins be represented by n (as in the American nickel). Since the value of each P5 coin is 5 pence, then the value of all the P5 coins would be 5n.

We are given that p = 3n + 1 (the number of P1 coins is one more than 3 times the number of P5 coins)

And we are given that p + 5n = 145 (the total amount inside is P145)

Rearranging:

Eq 1:
Eq 2: which can also be represented Eq 3:

Add Equation 1 to Equation 3:





So now we know that there are 18 P5 coins.

Since we know that the number of P1 coins is 3 times the number of P5 coins plus 1, we can say that there are P1 coins.

Check:

The value of 18 P5 coins is P90 (5*18=90) and the value of 55 P1 coins is P55 (1*55=55) and P55 + P90 = P145 which matches the given total value in the problem. Done.

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Otis Taylor has a box of coins that he uses when playing poker with his friends.
The box currently contains 44 coins, consisting of pennies, dimes, and quarters.
The number of pennies is equal to the number of dimes, and the total value is $4.37.
How many of each denomination of coin does he have in the box?
:
p = no. of pennies
d = no. of dimes
q = no. quarters
:
Write an equation for each statement:
:
"The box currently contains 44 coins, consisting of pennies, dimes, and quarters."
p + d + q = 44
:
"The number of pennies is equal to the number of dimes,"
p = d
:
" and the total value is $4.37."
.01p + .10d + .25q = 4.37
:
How many of each denomination of coin does he have in the box?
:
We want to eliminate q here.
Multiply the above equation by 4 and subtract from the total no. of coins eq:
1.0p + 1.0d + 1.0q = 44.00
.04p + .40d + 1.0q = 17.48
--------------------------- subtraction eliminates q
.96p + .60d + 0 = 26.52
:
Remember that p = d, therefore substitute d for p and find d
.96d + .60d + = 26.52
1.56d = 26.52
d =
d = 17 dimes
and
p = 17 pennies
:
Find Q:
17 + 17 + q = 44
34 + q = 44
q = 44 - 34
q = 10 quarters
:
:
Check solution in the $ equation
.01(17) + .10(17) + .25(10) =
.17 + 1.70 + 2.50 = 4.37
:
:
Was this enough depth for you?

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P=D
P+D+Q=44
D+D+Q=44
2D+Q=44
Q=44-2D
.25Q+.10P+.01D=4.37
.25(44-2D)+.10D+.01D=4.37
11-.50D+.10D+.01D=4.37
-.39D=4.37-11
-.39D=-6.63
D=-6.63/-.39
D=17 NUMBER OF DIMES & PENNIES.
Q=44-2*17
Q=44-34
Q=10 NUMBER OF QUARTERS.
PROOF:
.25*10+.10*17+.01*17=4.37
2.50+1.70+.17=4.37
4.37=4.37

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Nana Nantambu found some coins while looking under her sofa pillows. There were equal numbers of nickels and quarters, and twice as many half dollars as quarters. If she found $2.60 in all, how many of each denomination of coin did she find?
.
Let x = number of nickels or number of quarters
from: "twice as many half dollars as quarters"
2x = number of half dollars
.
Because we know the total is $2.60:
2.60 = .05x + .25x + .50(2x)
2.60 = .05x + .25x + x
2.60 = 1.30x
2 = x (number of nickels:number of quarters)
.
Half dollars:
2x = 2(2) = 4
.
Solution:
2 nickels, 2 quarters, and 4 half dollars
.
SECOND QUESTION:
Otis Taylor has a box of coins that he uses when playing poker with his friends. The box currently contains 44 coins, consisting of pennies, dimes, and quarters. The number of pennies is equal to the number of dimes, and the total value is $4.37. How many of each denomination of coin does he have in the box?
.
Similar problem...
Let x = number of pennies or number of dimes
then, because we know there is a total of 44 coins:
"total coins" - "number of pennies" - "number of dimes" = "number of quarters"
44 - x - x = number of quarters
44 - 2x = number of quarters
.
Because the total was $4.37:
4.37 = .01x + .10x + .25(44 - 2x)
4.37 = .11x + .25(44 - 2x)
4.37 = .11x + (11 - .5x)
4.37 = .11x + 11 - .5x
4.37 = 11 - .39x
-6.63 = -.39x
17 = x (number of pennies and dimes)
.
quarters:
44 - 2x = 44 - 2(17) = 44 - 34 = 10 (quarters)
.
Solution:
17 pennies, 17 dimes, and 10 quarters

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There were equal numbers of nickels and quarters, and twice as many half dollars as quarters. If she found $2.60 in all, how many of each denomination of coin did she find?
--------------------
n - q + 0 = 0
0 -2q + h = 0
5n + 25q + 50h = 260
---------------------------
Use any method you know to get:
n = 2
q = 2
h = 4
============
Cheers,
Stan H.

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The box currently contains 44 coins, consisting of pennies, dimes, and quarters.
Quantity equation: p + d + q = 44
---------------------------------------------
The number of pennies is equal to the number of dimes,
Quantity equation: p - d = 0
------------------------
the total value is $4.37.
Value equation: p + 10d + 25q = 437 cents
---------------------------------------
How many of each denomination of coin does he have in the box?
--------
You have three equation with three variables:
p + d + q = 44
p - d + 0 = 0
p + 10d + 25q = 437
------------------------
Use any method you know to solve this system to get:
p = 17
d = 17
q = 10
==============
Cheers,
Stan H.

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Let x=number of nickels; 2x=twice the number of nickels; 2x-8 would be 8 fewer than twice the number of nickels
y=number of dimes; 2y=twice the number of dimes; 2y+2=2 more than twice the number of dimes
z=number of quarters
Now lets deal in pennies instead of $.
Value of the nickels=5x
Value of the dimes=10y
4 times the value of the nickels and dimes together=4(5x+10y)
Value of the quarters=25z
Now we are told the following:
x=2y+2 -------------------eq1
z=2x-8 ---------------------eq2
and
25z=4(5x+10y)+160-------------------------eq3
Lets rearrange and simplify the above equations so we can see what we are dealing with:
x-2y =2---------------------eq1
2x -z=8---------------------eq2
4x+8y-5z=-32-------------------eq3
Multiply eq1 by 4 and add it to
eq3 and we get:
8x-5z=-24----------------------------eq3a
Next multiply eq2 by 4 and subtract it from eq3a and we get
(8x-5z+-24) -(8x-4z=32) or
-z=-56
z=56-------------------------------------------number of quarters
substitute z=56 into eq2 and we get:
2x-56=8
2x=64
x=32-------------------------------------number of nickels
substitute x=32 into eq1 and we get:
32-2y=2
-2y=-30
y=15------------------------------------- number of dimes
CK
two more nickels than twice the number of dimes:
2*15+2=32
32=32
eight fewer quarters than twice the number of nickels:
32*2-8=56
56=56
value of the quarters was $1.60 greater than four times the value of the nickels and dimes together:
56*25=4(32*5+15*10)+160
1400=4(160+150)+160=(4*310)+160
1400=1240+160=1400
1400=1400
Hope this helps-----ptaylor

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Q+D=18
Q=18-D
.25Q+.10D=3.45
.25(18-D)+.10D=3.45
4.50-.25D+.10D=3.45
-.15D=3.45-4.50
-.15D=-1.05
D=-1.05/-.15
D=7 DIMES.
Q+7=18
Q=18-7
Q=11 QUARTERS.
PROOF:
.25*11+.10*7=3.45
2.75+.70=3.45
3.45=3.45

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.25q+.10d=4.50
q=d+4
.25(d+4)+.10d=4.50
.25d+1+.10d=4.50
.35d=4.50-1
.35d=3.50
d=3.50/.35
d=10 number of dimes.
q=10+4=14 number of quarters.
Proof:
.25*14+.10*10=4.50
3.50+1.00=4.50
4.50=4.50

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.25q+.10d=4.50
q=d+4
.25(d+4)+.10d=4.50
.25d+1+.10d=4.50
.35d=4.50-1
.35d=3.50
d=3.50/.35
d=10 number of dimes.
q=10+4=14 number of quarters.
Proof:
.25*14+.10*10=4.50
3.50+1.00=4.50
4.50=4.50

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.25q+.10d=4.50
q=d+4
.25(d+4)+.10d=4.50
.25d+1+.10d=4.50
.35d=4.50-1
.35d=3.50
d=3.50/.35
d=10 number of dimes.
q=10+4=14 number of quarters.
Proof:
.25*14+.10*10=4.50
3.50+1.00=4.50
4.50=4.50

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hank has a collection of dimes and quarters worth 4.50 he has 4 more quarters than dimes find the number of quarters
.
Let d = number of quarters
then
d+4 = number of quarters
.
.10d + .25(d+4) = 4.50
.10d + .25d + 1 = 4.50
.10d + .25d = 3.50
.35d = 3.50
d = 10
.
quarters:
d+4 = 10+4 = 14 (number of quarters)
.

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OK
FIRST OF ALL, WE HAVE MORE UNKNOWNS IN THIS PROBLEM THAN WE HAVE EQUATIONS. IN THIS SITUATION, WE SOMETIMES HAVE MULTIPLE ANSWERS AND WE ALMOST ALWAYS WILL HAVE TO DO SOME TRIAL AND ERROR TO GET AT THE SOLUTIONS. HAVING SAID THAT, LETS SOLVE THE PROBLEM:
Let x=number of pennies
Let y=number of nickels
Let z=number of dimes
Let A=amount of money in the initial 222 coins
Now we are told that:
x+y+z=222-----------------------------------------------------eq1
And we know that (lets deal in pennies):
x+5y+10z=A----------------------------------------------------eq2
Now we are also told that:
"If the bank held 3 times as many dimes, half as many pennies, and 8 times as many nickels, it would contain twice as much money." This translates into the following equation (deal in pennies again):
(x/2)+5*8y+10*3z=2A multiply each term by 2 and simplify
x+80y+60z=4A---------------------------------------------------eq3
Next, multiply eq2 by 4 (we get: 4x+20y+40z=4A) and subtract eq3 from it and this yields:
4x+20y+40z=4A
-x-80y-60z=-4A
which equals:
3x-60y-20z=0------------------------------------------------------eq3a
Next, we multiply eq1 by 3 (we get: 3x+3y+3z=666) and subtract it from eq3a and this gives us:
-63y-23z=-666 multiply each term by -1
63y+23z=666 solve for z
z=(666-63y)/23-----------------------------------------------------eq3b
Eq3b gives us a relationship between the number of dimes and the number of nickels:
Now we know the following information about this problem which will help greatly in solving it:
(1) We cannot have negative coins--------our answers have to be positive numbers
(2) We cannot have fractions of coins-----our answers have to be whole numbers
Armed with the above information, by inspection, we can see from eq3b that y has to be less than or equal to 10, otherwise we start getting negative dimes. So, we have established that:
y<=10
Next, we will deal with the fractional aspect of this problem. We'll do this by calculating the value of z for the values of y between 1 and 10. Any whole numbers that we get are possible solutions:
y=1, z=26.22------------------------no good
y=2, z=23.5-------------------------no good
y=3, z=20.74------------------------no good
y=4, z=18---------------------------BINGO!!!!!!! possible answer
y=5, z=15.3-------------------------no good
y=6, z=12.5-------------------------no good
y=7, z=9.8--------------------------no good
y=8, z=7.04-------------------------no good
y=9, z=4.3--------------------------no good
y=10, z=1.57------------------------no good
So, out of all the possible values for y, we have only one possible solution:
y=4 and
z=18
Substitute y=4 and z=18 into eq1 and solve for x:
x+4+18=222 solve for x
x=200
So our answer is
x=200 pennies
y=4 nickels
z=18 dimes
CK
Original amount of money:
200*1+4*5+10*18=200+20+180=400
"If the bank held 3 times as many dimes, half as many pennies, and 8 times as many nickels, it would contain twice as much money." This translates into the following amount:
(200/2)*1+5*4*8+10*3*18=800
100+160+540=800
800=800
Niche problem!!!!
Hope this helps---ptaylor

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n=Nickels d=dimes so n=d+3 and n+d=25 so substituting n's value from the 1st equation into the 2nd equation we get (d+3)+d=25 or 2d+3=25 and subtracting 3 from each side we get 2d=22 so d=11 and since n=d(11)=3 then n=14
so there are 14 nickels and 11 dimes

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n=Nickels d=dimes so n=d+3 and n+d=25 so substituting n's value from the 1st equation into the 2nd equation we get (d+3)+d=25 or 2d+3=25 and subtracting 3 from each side we get 2d=22 so d=11 and since n=d(11)=3 then n=14
so there are 14 nickels and 11 dimes

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N = number of nickels.
D = number of dimes.
N = D + 3 (there are 3 more nickels than dimes)
D + N = 25 (total of dimes and nickels is 25).
-----
use D + N = 25
substitute D + 3 for N
equation becomes
D + D + 3 = 25
combine like terms:
2D + 3 = 25
subtract 3 from both sides:
2D = 22
divide both sides by 2:
D = 11
since N = D + 3, then N = 14
-----
answer is:
D = 11
N = 14
-----
she has 14 nickels and 11 dimes.
total number of coins is 25 (14+11=25)
number of nickels is 3 more than dimes (14-11=3)

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Joe has a collection of nickels and dimes that is worth $5.65. If the number of
dimes were doubled and the number of nickels were increased by 8,the value of
the coins would be $10.45. How many dimes does he have?
:
n = no. of dimes
d = no. of nickels
:
Write an equation for each statement
:
"Joe has a collection of nickels and dimes that is worth $5.65."
.05n + .10d = 5.65
;
"If the number of dimes were doubled and the number of nickels were increased by 8,the value of the coins would be $10.45."
.05(n+8) + .10(2d) = 10.45
.05n + .40 + .20d = 10.45
.05n + .20d = 10.45 - .40
.05n + .20d = 10.05
:
Multiply the 1st equation by 2, subtract the above equation from it:
.10n + .20d = 11.30
.05n + .20d = 10.05
---------------------subtraction eliminates d
.05n = 1.25
n =
n = 25 nickels
:
How many dimes does he have?
:
Use the 1st equation to find d, substitute 25 for n
.05(25) + .10d = 5.65
1.25 + .10d = 5.65
.10d = 5.65 - 1.25
.10d = 4.40
d =
d = 44 dimes
;
;
Confirm our solutions, using the 2nd statement: 33 nickels and 88 dimes
.05(33) + .10(88) = 10.45
1.65 + 8.80 = 10.45

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Cannot have 103 coins of nickles & quarters that equal $1.25.
103*.05=$5.15 which is the minimum value of 103 coins consisting of nickles & quarters.

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Cannot have 103 coins of nickles & quarters that equal $1.25.
103*.05=$5.15 which is the minimum value of 103 coins consisting of nickles & quarters.

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I'll just work the problem for you. You'll have to
write it up using complete sentences and proper grammar,
and filling in your own explanations.

Let p = the number of pennies
Let n = the number of nickels
Let d = the number of dimes

We have the equation:



or multiplying through by 100,



Since half the coins are nickels, we have





 

We have the system:



Substituting  for  in the first equation







Divide through by 3:







Since the left side is even,
the right side must be even also,
and positive. Therefore  
must be even and positive, too, 
Therefore d must be odd and no more 
than 11.

Therefore there are 6 possibilities:

d = 1, 3, 5, 7, 9, or 11

If , then

 becomes





Substituting  and 
into








So one possibility is 

, , 

-------

If , then

 becomes





Substituting  and 
into








So another possibility is 

, , 

-----------------

If , then

 becomes





Substituting  and 
into








So another possibility is 

, , 

-------------

If , then

 becomes





Substituting  and 
into








So another possibility is 

, , 

-------------

If , then

 becomes





Substituting  and 
into








So another possibility is 

, , 

-------------

If , then

 becomes





Substituting  and 
into







So another possibility is 

, , 

-------------

So all the 6 possibilities are 

0 pennies, 11 nickels, 11 dimes, 22 coins, half of which are 11 nickels.
5 pennies, 14 nickels, 9 dimes, 28 coins, half of which are 14 nickels.
10 pennies, 17 nickels, 7 dimes, 34 coins, half of which are 17 nickels.
15 pennies, 20 nickels, 5 dimes, 40 coins, half of which are 20 nickels.
20 pennies, 23 nickels, 3 dimes, 46 coins, half of which are 23 nickels.
25 pennies, 26 nickels, 1 dimes, 52 coins, half of which are 26 nickels.

Edwin

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Let x=number of pennies
y=number of nickels
And z=number of dimes
(we'll deal in pennies)
x+5y+10z=165----------------------------eq1
y=x+z-----------------------------------------eq2
subtract eq2 from eq1 and we get:
6y+9z=165 subtract 9z from each side
6y=165-9z divide each side by 6
y=(165-9z)/6=(55-3z)/2-------------eq3
eq3 shows the relationship between number of nickels and number of dimes. Now, we know the following about this problem:
(1)----We cannot have fractions of coins; we must deal in whole numbers
(2)----And we cannot have negative coins; we must deal in positive numbers
By inspection, we see immediately that z, the number of dimes, cannot be greater than 18 lest we start having negative nickels. In fact, z cannot be greater than 16 otherwise we break the bank. Actually, if half the coins are nickels, z needs to be much less than 16. Lets start assuming values for z, starting at z=1 and see what happens:
z=1-----------y=52/2=26 now we substitute y=26 and z=1 into eq2:
26=x+1
x=25
CK
26 nickels, 1 dime and 25 pennies---------------half are nickels
and
26*5+1*10+25*1=165
130+10+25=165
165=165
Hope this helps----ptaylor

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Nadine has seven more nickels than Delano has dimes. If Delano gives Nadine four of his dimes, then Delano will have the same amount of money as Nadine. How much money do they have together? (Assume that Nadine has only nickels and Delano has only dimes.
:
Let x = no. of N's nickels
Let y = no. of D's dimes
:
"Nadine has seven more nickels than Delano has dimes."
x = y + 7
:
"If Delano gives Nadine four of his dimes, then Delano will have the same amount of money as Nadine."
.05x + .10(4) = .10(y-4)
.05x + .40 = .10y - .40
.05x = .10y - .40 - .40
.05x = .10y - .80
Substitute (y+7) for x
.05(y+7) = .10y - .80
.05y + .35 = .10y - .80
.35 + .80 = .10y - .05y
1.15 = .05y
y =
y = 23 dimes
and
x = 23 + 7 = 30 nickels
:
Check for equality
.05(30) + 4(.10) = .10(23-4)
1.50 + .40 = .10(19)
:
How much money do they have together?
23(.10) + 30(.05) =
2.30 + 1.50 = $3.80

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Warren has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 more nickels than dimes. How many quarters does Warren have?
.
Let d = times
then from: "He has 7 more nickels than dimes"
d+7 = nickels
and from: "Warren has 40 coins"
40-d -(d+7) = 33-2d = quarters
.
Since, he has a total of 4.05:
4.05 = .10d + .05(d+7) + .25(33-2d)
4.05 = .10d + .05d + .35 + 8.25 - .50d
4.05 = 8.60 - .35d
.35d = 4.55
d = 13 (number of dimes)
.
quarters:
33-2d = 33-2(13) = 33-26 = 7 (number of quarters)

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of 250 is not an integer (187.5). What gives here?

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The coins available are pennies, nickels, dimes, and quarters
Let the letters p,n,d,and q stand for these coins
Each of the following equals 40 cents
---------------------
40p
8n
4d
-------
35p + 1n
30p + 2n
25p + 3n
20p + 4n
15p + 5n
10p + 6n
5p + 7n
30p + 1d
20p + 2d
10p + 3d
15p + 1q
--------
25p + 1n + 1d
20p + 2n + 1d
15p + 3n + 1d
15p + 1n + 2d
10p + 4n + 1d
10p + 2n + 2d
10p + 1n + 1q
5p + 1n + 3d
5p + 3n + 2d
5p + 5n + 1d
5p + 2n + 1q
5p + 1d + 1q
--------
I count 26 ways to make 40 cents with these coins
(unless I missed some)

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x=# of 3cent stamps
3x-1=# of 8 cent stamps
(3x-1)-6=# of 15cent stamps
Total value is $2.47
.03x+.08(3x-1)+.15(3x-7)=2.47

Use the distributive property
.03x+.24x-.08+.45x-1.05=2.47
Combine like terms and simplify
(.03+.24+.45)x+(-.08-1.05)=2.47

(0.72)x-1.13=2.47
add 1.13 to both sides
.72x=3.6
divide both sides by .72
x=5
# of 3 cent stamps is 5
# of 8 cent stamps =3(5)-1=14
# of 15 cent stamps is 14-6=8
Let's double check our answer
.03(5)+.08(14)+.15(8)=.15+1.12+1.2=2.47

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Let J be the pennies in the first jar. Then are in the second jar.
You are told the total in both jars is





The first jar has 55 pennies, the second one has = pennies.
That's not a lot of pennies for two big jars :)

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3*.25=.75
2*.10=.20
5*.01=.05
.
Ed

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Total number of coins P5=
Total number of coins P10= , six more than P5 coins
Remember:
When total number of P5 coins multiplied by P5 + total number of P10 coins multplied by P10 equals P360.Putting into eqn:
, WORKING EQN
, distribute & combine similar terms then add/subtract

, divide P15 to isolate x
-------------------------> number of P5 coins
----------------------> number of P10 coins
Check via WORKING EQN,



Thank you,
Jojo

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Sandy has $3.75 in change. She has three more quarters than dimes, but twice as
many nickels as quarters. Express the given information as an algebraic
expression and simplify.
:
"Sandy has $3.75 in change."
.05n + .10d + .25q = 3.75
;
"She has three more quarters than dimes,"
q = d + 3
or
d = q - 3
:
" but twice as many nickels as quarters."
n = 2q
:
Substitute for d and n in the 1st equation:
.05(2q) + .10(q-3) + .25q = 3.75
.10q + .10q - .3 + .25q = 3.75
.45q = 3.75 + .30
.45q = 4.05
q =
q = 9 quarters
:
I'll let you find n and d

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-----------------------> EQN 1
In EQN 1 we get,
----------------------------> EQN 2
Also,
------------------> EQN 3
Subst. EQN 2 in EQN 3:



------>
number of Nickels (ANSWER)
For D, via EQN 2:
number of Dimes
Check via EQN 3:



Thank you,
Jojo

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Remember: ----------------------> EQN 1
where



Going back EQN 1,





LONGEST side (a)
2ND LONGEST (b)
SHORTEST (c)
Check via EQN 1:


Thank you,
Jojo

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Amanda has some pennies, nickels, dimes, and quarters in her piggy bank. There is more than $1.50 and less than $2.25. Using the clues, determine how much money Amanda has.
:
Write an equation for each statement:
:
"-there are 4 times as many nickels as quarters"
n = 4q
:
"-there are twice as many quarters as dimes"
q = 2d
:
"-there is an equal number of pennies and dimes"
p = d
:
Write a total$ equation within the given range
.01p + .05n + .10d + .25q = 2.00
:
try to get everything in terms of d:
In the 1st equation, substitute 2d or q (from the 2nd equation)
n = 4(2d)
n = 8d
:
Substitute:
.01d + .05(8d) + .10d + .25(2d) = 2.00
.01d + .40d + .10d + .50d = 2
1.01d = 2
d = obviously not a solution (has to be an integer)
:
Make a solution by multiplying 1.01 * 2, so the total is $2.02, then we have:
d =
d = 2 dimes
then
.01(2) + .40(2) + .10(2) + .50(2) = 2
.02 + .80 + .20 + 1.00 = $2.02
:
Find the number of each coin using d = 2:
p = d; 2 pennies
n = 8d: 16 nickels
d = 2 dimes
q = 2d; 4 quarters
;
:
Confirm solution from this:
.01(2) + .05(16) + .10(2) + .25(4) =
.02 + .80 + .20 + 1.00 = $2.02 is Amanda's total
:
There are probably other ways to do this. Did this make sense to you?

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OK, we'll let x=number of dimes
Then x=number of pennies, also
And 2x=number of quarters
And 4*2x=8x=number of nickels
Now we can set up our equation (or inequality) to solve:
(Lets deal in pennies to get rid of the decimals)
150<(x+10x+5*8x+25*2x)<225 simplifying
150<(x+10x+40x+50x)<225 collecting like terms, we have:
150<(101x)<225 divide each term by 101
1.48<(x)<2.23 since we cannot have a fraction of a coin, x has no choice but to be 2
For x=2, we have
2 pennies 2*1=2
2 dimes 2*10=20
4 quarters 4*25=100
16 nickels 16*5=80
TOTAL-------202 or $2.02 cents
And this satisfies the inequality
Hope this helps---ptaylor

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.05*3x+.10x=1.25
.15x+.10x=1.25
.25x=1.25
x=1.25/.25
x=5 10 cent pieces.
3*5=15 5 cent pieces.
Proof:
.05*15+.1*5=1.25
.75+50=1.25
1.25=1.25

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Sam found a number of nickels, dimes and quarters. He found 3 more dimes than
nickels but twice as many quarter as dimes. The total value of the coins was
$5.05. How many coins of each type did Sam find?
;
Let:
n = no. of nickels
d = no. of dime
q = no. of quarters
:
Write an equation for each statement:
;
"He found 3 more dimes than nickels"
d = 3+n
or
n = (d-3)
:
" but twice as many quarter as dimes."
q = 2d
:
"The total value of the coins was $5.05."
.05n + 10d + .25q = 5.05
:
How many coins of each type did Sam find?
:
Substitute for q and n in the total$ equation
.05(d-3) + .10d + .25(2d) = 5.05
.05d - .15 + .10d + .50d = 5.05
.05d + .10d + .50d = 5.05 + .15
.65d = 5.20
d =
d = 8 dimes
then
n = 8 - 3
n = 5 nickels
and
q = 2*8
q = 16 quarter
:
8 + 5 + 16 = 29 coins total
;
:
Check solution in the total$ equation
.05(5) + .10(8) + .25(16) =
.25 + .80 + 4.00 = 5.05; confirms our solutions

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FIVE MORE THAN THREE-FOURTHS OF A NUMBER c IS 26. FIND THE NUMBER
.
Let N = the number
.
From:"FIVE MORE THAN THREE-FOURTHS OF A NUMBER c IS 26." we get:
5 + (3/4)N = 26
Multiplying both sides by 4:
20 + 3N = 104
3N = 84
N = 28

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Mary has $3.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and five more nickels than dimes, how many coins of each type does she have?
.
Let q = number of quarters
then from "she has twice as many dimes as quarters"
2q = number of dimes
and from "five more nickels than dimes"
2q+5 = number of nickels
.
Since she has a total of $3.00:
.25q + .10(2q) + .05(2q+5) = 3.00
.25q + .20q + .10q + .25 = 3.00
.55q + .25 = 3.00
.55q = 2.75
q = 2.75/.55
q = 5 (number of quarters)
.
number of dimes:
2q = 2(5) = 10 (number of dimes)
.
number of nickels:
2q+5 = 2(5)+5 = 10 + 5 = 15 (number of nickels)

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Warren has 40 coins; nickels dimes and quarters worth $4.05. He has seven more nickels than dimes. How many quarters does he have?
:
Write 3 equations, one for each statement:
:
"Warren has 40 coins;"
n + d + q = 40; eq1
:
" nickels dimes and quarters worth $4.05."
.05n + .10d + .25q = 4.05; eq2
;
"He has seven more nickels than dimes."
n = d + 7
n - d = 7; eq3
:
Add equations 1 and 3:
n + d + q = 40
n - d + 0 = 7
------------------adding eliminates d
2n + q = 47
q = (47-2n)
:
Multiply eq2 by 10, subtract eq1
.5n + 1d + 2.5q = 40.5
1n + 1d + 1q = 40
------------------------subtraction eliminates d again
-.5n + 0 + 1.5q = .5
-.5n + 1.5q = .5
:
Substitute (47-2n) for q in the above equation, find n
-.5n + 1.5(47-2n) = .5
-.5n + 70.5 - 3n = .5
-.5n - .3n = .5 - 70.5
-3.5n = -70
n =
n = +20 nickels
:
Find d;
n - d = 7
20 - d = 7
d = 13 dimes
;
Obviously, that leaves us with 7 quarters.
;
You should check these solutions using eq2:

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I will set it up and you finish.
5 times 10 times 25 = combination of coins

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working in cents

let n=number of coins

"The average value of all the pennies, nickels, dimes and quarters in Bob's pocket is 20 cents" __ 20n

"If he had one more quarter, the average would be 21 cents" __ 20n+25=21(n+1)

20n+25=21n+21 __ 4=n

4 coins worth 80 cents means 3 quarters and a nickel
__ no pennies or dimes

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"Susan has saved $85.75 in quarters and loonies" __ .25Q+1L=85.75 __ 25Q+100L=8575

"She has one quarter more than three-fourths the number of loonies" __ Q=.75L+1 __ 100Q=75L+100



***SOLUTION HERE***


multiplying 1st eqn by 4 __ 100Q+400L=34300 __ substituting __ (75L+100)+400L=34300

475L=34200 __ L=72

substituting __ .25Q+(72)=85.75 __ .25Q=13.75 __ Q=55