Delicious brownies and cookies!

Let $C$ and $B$ be the cost of $1$ Cookie and $1$ Brownie respectively.

Money Calvin spent $=4B+3C$ and Money Lisa spent $=2B+16C$

Now, it is said that Lisa spent twice the amount that Calvin spent. Then, we have ---

$2\times$ (Amount Calvin spent) $=$ (Amount Lisa spent)

$\implies 2(4B+3C)=2B+16C$

$\implies 2(4B+3C)=2(B+8C)$

$\implies 4B+3C=B+8C \implies 3B=5C \implies \boxed{B=\frac{5}{3}C}$

* So, the cost of $1$ brownie is $\frac{5}{3}$ times the cost of a cookie. *

x is price of brownie and y is price of cookie. Next, set up the identity. 2(4x+3y)=2x+16y =>8x+6y=2x+16y =>6x=10y =>x/y=5/3

5/3 times

Calvin pays $4b+3c$ lily pays $2b+16c$ let, $b=kc$ $$ A/Q $2b+16c=2(4b+3c)$ $=> 2kc+16c=8kc+6c$ $=> k=\frac{5}{3}$

cost of brownie be x and cookie be y. A.T.Q, 2 (4x + 3y) = 2x + 6y i.e 8x + 6y = 2x + 16y i.e 6x = 10y
i.e x = 10/6 y i.e x = 5/3 y. So brownie's cost= 5/3 cookie,s cost .

4B + 3C = (2B + 16C)/2 4B + 3C = B + 8C 3B = 5C B = 5/3*C

4B+3C=X------------(1) & 2B+16C=2X-------(2) (1)-2*(2)+=>X=9/3PUTTING THIS VALU AT (1) 4B+3C=29/3C =>B=5/3C

Let b be the cost of 1 brownie and c be the cost of 1 cookie. Then as per give condition 2(4b+3c)=2b+16c => b=5c/3