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?

_________________

Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.

FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.