Ricky has some cookies. He ate 1/8 of his cookies and an additional 14 cookies on Friday. He then ate 3/10 of the remaining cookies and an additional 24 cookies on Saturday. He ate 40 cookies on Sunday and had 34 cookies left. How many cookies did Ricky have at first?

Soyoxxommm Jul 6, 2021

#1**+2 **

Suppose Rocky has X no. of cookies

Then cookies left after Friday

X- {(x/8)+14}

Cookies left after Saturday

[X-{(x/8) +14}] -[(3/10)(X-{(x/8)+14})+24]

Cookies left after Sunday

=[X-{(x/8) +14] - [(3/10)(X-{(x/8)+14})+24] -40....(1)

Since cookies left =34

Therefore equating the equation (1) with 34

(7/10){(7x/8) -14} -64 = 34

X=176

Thus Rocky has initially 176 cookies

apsiganocj Jul 6, 2021

#2**+2 **

$k - \left(\frac{1}{8} k + 14 + \frac{3}{10} \left(\frac{7}{8} k - 14 \right) + 24 + 40 \right)= 34$

$k - \frac{31k}{80} + \frac{21}{5} - 78 = 34$

$80k - 31k + 336 - 6240 = 2720$

$49k = 8624$

$k = \boxed{176}$

MathProblemSolver101 Jul 6, 2021

#3**+2 **

Ricky has some cookies. He ate 1/8 of his cookies and an additional 14 cookies on Friday. He then ate 3/10 of the remaining cookies and an additional 24 cookies on Saturday. He ate 40 cookies on Sunday and had 34 cookies left. How many cookies did Ricky have at first?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

x - [(1/8x + 14) + 3/10(x - 1/8x - 14) + 24 + 40] = 34

Total I Friday I Saturday I Sun I left

**x = 176 **

civonamzuk Jul 7, 2021