How do I solve this money problem for math?

Question

There are 100 eggs for sale. All of the eggs are one of three types. One type for 6 dollars each, one for 3 dollars each and one for a dime each. I used 100 dollars and bought all those 100 eggs. How may of each type did I buy?

There are 100 eggs for sale. All of the eggs are one of three types. Type A sells for 6 dollars each, Type B sells for 3 dollars each and Type C for a dime each. I used 100 dollars and bought ALL 100 eggs. How many of each type did I buy?

Answer 1

Correct answer .

6$x 1 Eggs = 6$
3$x 29Eggs = 87$
0.1$x70 Eggs= 7$

100Eggs= 100$

1Eggs of ( 6$) =6$
29Eggs 0f (3$)=87$
70Eggs of (0.1$)=7$

Now make the Total ( 1+29+70) =!00Eggs & Total of $ (6$+87$+7$) =100$

This is the answer 100$ you can buy 100Eggs

Answer 2

Let x be number of type A,y be number of type B,z be number of type C

We have the equations

x + y + z = 100

6x+3y+ 0.10z = 100

This is a set of two equations in three unknowns that
has to be solved for positive integer values of x,y,z
It can be done even though one usualy needs as many equations as there are unknowns to solve a system of equations

Multiply the second equation by 10. Then we have the set

60x + 30y + z = 1000
x + y + z = 100

By subtracting bottom equation from top we get

59x + 29y = 900

There exist algorithms to solve for x and y in integers but
since I don't know one I will attempt to solve it anyway.

x + 58x +29y = 29(31) +1

I divided 900 by 29 and wrote it as divisor times quotient plus remainder

From this 58x + 29y = 29(31) + (1 - x)

Now divide both sides by 29

2x + y = 31 + (1-x)/29

The right side will be an integer if we let

1-x = -29t where t is an integer to be determinded

x = 29t + 1

Substitute into the equation

2x + y = 31 + (1-x)/29

58t + 2 + y = 31 - t

Then y = 29 -59t

z = 100 - (x + y)

= 70 +30t

So we have x = 29t + 1
y = 29 - 59t
z = 70 + 30t

Lets examine the expressions for x and y

If t is a positive integer then y is a negative integer
We can't have a negative number of eggs

If t is negative then x is negative

But if t = 0 then x =1, y = 29, z = 70

1 Type A,29 Type B, and 70 Type C

Answer 3

Although the equations mentioned above are correct, you cannot substitute to solve because we are talking about two different things, dollar value and quantity.

if we substitute we get 450.1c = $100, but this cannot be correct.

Lets try a different way.

take each dollar value to determine how many make $100

$value(quantity) = $100

6(16.6) = $100
3(33.3) = $100
.1(1000) =$100

Now it's a balancing act. Think of a hanging scale where you take from one side to balance the other.

100 eggs = $100

If we take $3(30) we have 70(quantity) as a remainder which can be replaced with the .10 eggs to get $90 + $7 = $97

But we still need the $6 quantity so lets try substitution by subtraction:

$3(29) + $6(1) + .10(70) = $100


**Update

Good solution from Kevin below except:

y = 29 + (59 - 59x)/29
y = 29 - 59/29 (x - 1)

should be

y = 29 + (59 - 59x)/29
y = 29 - 59/29 (1 - x)

Answer 4

ALL answers are WRONG.....i am sorry but there are 3 unknowns and only 2 valid equations...you need more information or the questions is incorrectly worded...10 eggs of each type is the WRONG answer. If you actually bought 10 of each kind of egg, yes you did spend $100 but you only put 30 eggs in total - and the question said you bought 100 eggs for $100. Please fix the question.

Answer 5

Let
x = eggs worth $ 6.00 each
y = eggs worth $ 3.00 each
z = eggs worth $ 0.10 each

Thus,
x + y + z = 100
6x + 3y + 0.1z = 100

Multiply -0.1 to the 1st equation
-0.1x - 0.1y - 0.1z = -10
6x + 3y + 0.1z = 100

Add the 2 equations:
5.9x + 2.9y = 90

Multiply 10:
59x + 29y = 900

Solve for y:
29y = 900 - 59x
y = (900 - 59x)/29
y = (841 + 59 - 59x)/29
y = 29 + (59 - 59x)/29
y = 29 - 59/29 (x - 1)
Since y has to be a whole number, x - 1 must be a multiple of 29. In this problem, x can only be 1, because x = 2 will make y a negative value. Thus, x = 1, and y= 29, leaving z = 70.

Therefore, you bought
1 egg worth $ 6 (or $ 6)
29 eggs worth $ 3 (or $87)
70 eggs worth $ 0.10 (or $7)
Total:
100 eggs = $ 100

^_^

Answer 6

1, 29,70 is a correct answer it fulfills all requirements. all of you that said 10 each say it with me "would you like fries with that" or maybe you are more of a "paper or plastic" kind of person. the question clearly states that you buy 100 eggs. come on if you buy 10 eggs of 3 types you get 30 eggs. i have a minor in mathematics but could have done this in jr high. Or maybe saying " you can upsize that for a quarter" has ruined you brain. have a nice day.

Answer 7

yes finally one person (fire) knows there need to be 100 eggs. but the question is weird. you mentioned all of the eggs are one of three types..so does it mean all 100 eggs cost either all $6, all $3, or all a dime. but i think you meant there are three types of eggs among the 100 eggs right?

Answer 8

we'll call the $6 eggs type a, the $3 eggs type b and the $0.10 eggs type c

there are two equations
a eggs + b eggs + c eggs = 100 eggs

and

6a + 3b + 0.10c = $100.00

now use substitution to solve

you'll need to take partial derivatives for this one maybe

Answer 9

10 of each

Answer 10

Yes it would be 10 of each