Caleb and Suzanne each have a pile of blue and white chips. Caleb determines in his pile, three out of every four chips is blue. Suzanne determines in her pile, one out of every five chips are blue. How many total chips does each student have if Caleb and Suzanne each have 6 blue chips?

Total number of chips Caleb have = 8

Total number of chips Suzanne have = 30

Step-by-step explanation:

In Caleb case -

Three out of every four chips is blue

⇒ If Caleb has total chips = 4

Then Blue chips = 3

⇒White chips = 4 - 3 = 1

Now,

If Caleb have 6 blue chips

⇒3 Blue chips = 4 total chips

  1 Blue chip = total chips

⇒6 blue chips = ×6 = 4×2 = 8 total chips

∴ we get

Number of chips become = 8

In Suzanne case -

One out of every five chips are blue.

⇒ If Suzanne has total chips = 5

Then Blue chips = 1

⇒White chips = 5 - 1 = 4

Now,

If Suzanne have 6 blue chips

⇒1 Blue chips = 5 total chips

⇒6 blue chips = 5×6 = 30 total chips

∴ we get

Number of chips become = 30

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!