cigar-word problem

Cuban cigar = $c
French .. = $f
Arabian .. = $a

c+f+a = 74.7

c = 3(f-0.7)/2 - 1
f = 2(c+1)/3 + 0.7
a = 3c/2 + 0.5

c+f+a = 74.7
c+ 2(c+1)/3 +0.7 + 3c/2 +0.5 = 74.7
c = $23

Put this in
c = 3(f-0.7)/2 - 1

f = $16.70

Option A

An Arabian cigar costs 50 cents more than 1.5 times the Cuban cigar. The three cigars together cost 74.70 dollars. What is the price of the French cigar?


Let the cost of a Cuban Cigar be $C and
Let the cost of a French Cigar be $F and
Let the cost of an Arabian Cigar be $A

A Cuban cigar would cost 1 dollar less than 1.5 times a French cigar,had the French cigar cost 0.7 dollar less than it does now. Implies
$C = (1.5*($F-0.7))-1

An Arabian cigar costs 50 cents more than 1.5 times the Cuban cigar. Implies
$A= (1.5*$C) + 0.5

The three cigars together cost 74.70 dollars.Implies
$C+$F+$A=$74.70

((1.5*($F-0.7))-1) + $F + (1.5*((1.5*($F-0.7))-1)) + 0.5 = 74.7
Implies $F = $16.70

Ans: A

Hi rijul,

How u r getting the second equation
f = 2(c+1)/3 + 0.7

Pls explain hw u arrive the above eqn.

rijul007 wrote:Cuban cigar = $c
French .. = $f
Arabian .. = $a

c+f+a = 74.7

c = 3(f-0.7)/2 - 1
f = 2(c+1)/3 + 0.7
a = 3c/2 + 0.5

c+f+a = 74.7
c+ 2(c+1)/3 +0.7 + 3c/2 +0.5 = 74.7
c = $23

Put this in
c = 3(f-0.7)/2 - 1

f = $16.70

Option A

kishokbabu wrote:Hi rijul,

How u r getting the second equation
f = 2(c+1)/3 + 0.7

Pls explain hw u arrive the above eqn.

rijul007 wrote:Cuban cigar = $c
French .. = $f
Arabian .. = $a

c+f+a = 74.7

c = 3(f-0.7)/2 - 1
f = 2(c+1)/3 + 0.7
a = 3c/2 + 0.5

c+f+a = 74.7
c+ 2(c+1)/3 +0.7 + 3c/2 +0.5 = 74.7
c = $23

Put this in
c = 3(f-0.7)/2 - 1

f = $16.70

Option A

@kishokbabu
I jst manipulated the expression
c = 3(f-0.7)/2 - 1


c = 3(f-0.7)/2 - 1

3(f-0.7)/2 = c+1
f-0.7 = 2(c+1)/3
f = 2(c+1)/3 + 0.7