nancy made 100 sandwiches.She sold them all for $100. one was for $5, other was for $2 and last was 10 cents.?

Question

how many of each sandwich did nancy make?

Answer 1

Let A be the number of $5 sandwiches sold
Let B be the number of $2 sandwiches sold
Let C be the number of $0.10 sandwiches sold

The important thing to realize is that you made $100 and also 100 sandwiches.
5A + 2B + 0.10C = 100
A + B + C = 100

Now this looks unsolvable because you have 3 variables and 2 equations... but let's see what we can do. Obviously C needs to be a multiple of 10, to make a whole dollar.

So the choices for C are 0, 10, 20, 30, ..., 100

In general:
C = 10*n, where n = 0, 1, 2, ..., 10

5A + 2B = 100-n
A + B = 100 - 10n
Multiplying the bottom by 5, we get
5A + 5B = 500 - 50n

Subtracting the first equation from the second you get:
3B = 500 - 50n - 100 + n
3B = 400 - 49n

Notice how we need a multiple of 3 when subtracting 49n from 400. Well, 400/3 has a remainder of 1, so 49n/3 needs a remainder of 1.

The only values of n that fit the pattern are 1, 4, 7, 10.

n = 1, C = 10:
3B = 400 - 49
3B = 351
B = 117, but this is too many.

n = 4, C = 40
3B = 400 - 4*49
3B = 400 - 196
3B = 204
B = 69, but this is already more than 100 sandwiches

n = 7, C = 70
3B = 400 - 7*49
3B = 400 - 343
3B = 57
B = 19
Then solving for A we have A = 11

Finally an answer!

Nancy sold:
11 sandwiches at $5 = $55
19 sandwiches at $2 = $38
70 sandwiches at $0.10 = $7

Double checking:
Total sandwiches (11 + 19 + 70) = 100
Total money made ($55 + $38 + $7) = $100

Answer 2

Alright,
Lets assign some variable names
Sandwich type 1 = S1
Sandwich type 2 = S2
Sandwich type 3 = S3

S1 + S2 + S3 = 100
Their prices should add up to $100 also

$5 * S1 + $2 * S2 + $.10 * S3 = $100

In order to solve this though, we need to make some sort of mathematical relationship between each variable and S1

S1 = 100 - (S2 +S3)
S3 = ($100 - $5 * S1 - $2 * S2)/.10 = 1000 - 50 * S1 - 20 * S2

Put these equations together

S1 = 100 - S2 - (1000 - 50*S1 - 20 * S2)
S1 = 100 - S2 - 1000 + 50 *S1 + 20*S2
S1 = -900 + 19* S2 + 50*S1
-49*S1 = -900 + 19 * S2
S2 = (-49 * S1 + 900)/19

Now, plug this into the equation
S3 = 1000 - 50 * S1 - 20 * S2
S3 = 1000 - 50 * S1 - 20 * ((-49 * S1 + 900)/19)
S3 = 1000 - 50 * S1 + (980 * S1)/19 -18000/19
S3 = 19000/19 - 950 * S1/19 + 980 * S1/19 - 18000/19
S3 = 1000/19 + 30 * S1/19


Now we know what each variable is equivalent to in terms of S1
so wemake the equations equivalent

$5 * S1 + $2 * S2 + $.10 * S3 = S1 + S2 + S3

5 * S1 + 2 * ((-49 * S1 + 900)/19) + .10 * (1000/19 + 30 * S1/19) = S1 + ((-49 * S1 + 900)/19) + (1000/19 + 30 * S1/19)

95 * S1 + 2(-49 * S1 + 900) + .10 * (1000 + 30 * S1) =19* S1 + ((-49 * S1 + 900)) + (1000 + 30 * S1)

95 * S1 - 98 * S1 + 1800 + 100 + 3 * S1 = 19* S1 + ((-49 * S1 + 900)) + (1000 + 30 * S1)

1900 = 1900

Multiple answers, (and they must be integers)

Answer 3

You have 2 equations with 3 unknowns,
5A+2B+0.1C = 100,
A+B+C=100

After eliminating A by subtracting the first equation from 5*the second equation, and then multiplying by 10, you get

30B + 49C = 4000.

C must be a multiple of 10 to obtain a whole dollar, so by trying multiples of 10, you find that C = 70, B = 19, and A = 11.

Answer 4

Could be many different combinations, but she still just made 100 sandwiches.

Could be 5 X 10 = 50.00
2 X 20 = 40.00
1 x 10 = 10.00 for a total of 100.00 dollars but the combination could vary.

Answer 5

42 for 2 bucks each and one half sandwich for a buck and one for5 bucks, and one for ten cents and one for 950

Answer 6

Need someone to do your homework for you eh??

Answer 7

1 for $5.00= $5.00
1 for $.10 cents .10
balance for

47.45 of 1 type each $2.00= 94.90

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$100.00