1. ## story problem

STORY PROBLEM-
AT THE COOKIE COUNTER THERE WERE 6 MORE CHOCOLATE MINT COOKIES THAN WHITE FROSTED COOKIES, BUT THE SUM OF THE MINT AND FROSTED WAS EQUAL TO THE TOTAL OF PEANUT BUTTER COOKIES. THEY SAW EXACTLY TWICE AS MANY PEANUT BUTTER COOKIES AS OATMEAL COOKIES. THERE WERE ON THIRD AS MANY PEANUT BUTTER AS CHOCOLATE CHIP.THERE WERE 36 CHOCOLATE CHIP COOKIES.

HOW MANY CHOCOLATE CHIP_____
HOW MANY CHOCATATE MINT______
OATMEAL_______
PEANUT BUTTER_________
WHITE FROSTED__________
TOTAL ___________________

2. Originally Posted by mhughes21
STORY PROBLEM-
AT THE COOKIE COUNTER THERE WERE 6 MORE CHOCOLATE MINT COOKIES THAN WHITE FROSTED COOKIES, BUT THE SUM OF THE MINT AND FROSTED WAS EQUAL TO THE TOTAL OF PEANUT BUTTER COOKIES. THEY SAW EXACTLY TWICE AS MANY PEANUT BUTTER COOKIES AS OATMEAL COOKIES. THERE WERE ON THIRD AS MANY PEANUT BUTTER AS CHOCOLATE CHIP.THERE WERE 36 CHOCOLATE CHIP COOKIES.

HOW MANY CHOCOLATE CHIP_____
HOW MANY CHOCATATE MINT______
OATMEAL_______
PEANUT BUTTER_________
WHITE FROSTED__________
TOTAL ___________________
Turn off your caps Lock key.

You can figure out what the symbols mean:

x = y + 6

x + y = z

z = 2w

z = 12

Solve for x, y and w.

M = F +6 -------------(1)

BUT THE SUM OF THE MINT AND FROSTED WAS EQUAL TO THE TOTAL OF PEANUT BUTTER COOKIES.
M +F = P --------------------(2)

THEY SAW EXACTLY TWICE AS MANY PEANUT BUTTER COOKIES AS OATMEAL COOKIES.
P = 2(Ot) -----------------(3)

THERE WERE ON THIRD AS MANY PEANUT BUTTER AS CHOCOLATE CHIP.
P = (1/3)C -------------------(4)

THERE WERE 36 CHOCOLATE CHIP COOKIES.
C = 36 --------------------(5)

So, combining Eq.(5) and Eq.(4),
P = (1/3)(36) = 12 peanut butter cookies

Substitute that into Eq.(3),
12 = 2(Ot)
Ot = 12/2 = 6 oatmeal cookies.

Substitute the P = 12 into Eq.(2),
M +F = 12
M = 12 -F -------(6)

M from Eq.(1) = M from Eq.(6),
F +6 = 12 -F
F +F = 12 -6
2F = 6
F = 6/2 = 3 white frosted cookies

Hence, M = F +6 = 3 +6 = 9 choc mint cookies