Being simple is the key my friend .
You should not try to be fancy when a simple approach is right in front of your eyes.
This is essentially an algebra problem with two unknown variables.
Let the price of FIRST (ALSO THE SMALLEST) CAKE be Y
Then every other cake will become {Y+X}, {Y+X+X} ......and so on
BASED ON THE INFO, YOU WILL GET:-
First cake price = y
Second cake price = y+x
Third cake price= y+x+x
Fourth cake price = y+x+x+x
Fifth cake price= y+x+x+x+x
sixth cake price = y+x+x+x+x+x
SIXTH CAKE IS LARGEST
y+5x= 24.50
Total price of all cakes =(FIRST $ + SECOND $ + THIRD $....... + SIXTH $) ====> 6y+15x = 109.40
Now you got two unique equation for two unknown variable
WHATS THE PRPBLEM THEN :- SOLVE IT
y+5x=24.50 (Multiply by 3) ...........Eq1
6y+15x=109.40 ............. Eq2
Subtract Eq1 from Eq2
y==>12
put value of y in Eq 1
12+5x=24.50
5x= 12.50
x=\(\frac{12.50}{5}\) ===>2.5 (OPTION D)
ANSWER IS D
ENJOY
enigma123 wrote:
A certain bakery sells six different-sized wedding cakes. Each cake costs x dollars more than the next one below it in size, and the price of the largest cake is $24.50. If the sum of the prices of the six different cakes is $109.50, what is the value of x?
(A) 1.50
(B) 1.75
(C) 2.00
(D) 2.50
(E) 3.00
This is how I am trying to solve this, but got stuck.
Total price = 109.5
Price of each cake i.e. average = \(\frac{109.50}{6}\) = \(18.25\)
Price of the smallest cake = x
Price of the largest cake = 24.5
As this is the evenly spaced set [Price difference of x between any 2 cakes] average of x and 24.5 will be equal to the average 18.25 i.e.
\(\frac{x+24.5}{2}\) = 18.25
x = 12
I am stuck after this. can you please help?
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