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A certain bakery sells six different-sized wedding cakes

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A certain bakery sells six different-sized wedding cakes [#permalink]

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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?
[Reveal] Spoiler: OA

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Re: Bakery [#permalink]

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New post 04 Mar 2012, 14:49
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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

Since the price of the largest cake (6th) is $24.50, then the price of the smallest cake (1st) is $(24.50-5x). Now, the prices of the cakes are evenly spaced so the sum of the prices is (average price)*(# of cakes)=(first+last)/2*(# of cakes). So, (24.50-5x+24.50)/2*6=109.50 --> x=2.5.

Answer: D.
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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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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?



Your result isn't the difference between any two cakes. Your result is the price of the first cake.
Let's say that x is the price of the first cake and y is the difference between the cakes, the formula will be: \(x + (x+y) + (x+2y) + (x+3y) + (x+4y) + (x+5y) = 109,5\).
Since you know that x is 12, and you also know that the price of the biggest cake is \(x+5y=24.5\), than just simply calculate the equation.

12 + 5y = 24.5; y = 2.5
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A certain bakery sells six different-sized wedding cakes [#permalink]

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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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Last edited by LogicGuru1 on 11 Sep 2016, 19:37, edited 1 time in total.

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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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New post 03 Jul 2016, 09:52
109.5= 24.5x6 - 15X
X=2.5


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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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New post 11 Sep 2016, 08:13
rohitrawat9990 wrote:
109.5= 24.5x6 - 15X
X=2.5


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How did you came up with 15X?

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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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New post 11 Sep 2016, 08:23
arturportugal wrote:
rohitrawat9990 wrote:
109.5= 24.5x6 - 15X
X=2.5


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How did you came up with 15X?


Let L6 represents the price of largest and L1 represents the smallest.

We are given that L6 = 24.5

L5 = 24.5-x ( each smaller sized is x less than its immediate larger)

L4 = 24.5 - 2x

L3 = 24.5 - 3x

L2 = 24.5 - 4x

L1 = 24.5 - 5x

Now, the sum of all prices = 109.5

So, L6 to L1, you will get

109.5= 24.5x6 - 15x

or x = 2.5
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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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New post 18 Oct 2016, 15:26
From the question we can derive the following equation:

(24.50-5x) + (24.50-4x)+... = 109.50

simplifies to... 24.50(6)-15x=109.50

Solve for x.

x=2.5

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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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New post 18 Oct 2016, 15:55
I first saw what 25*6 is just to give a ball park number and it was 150. I then realized that the numbers had to decrease by the "biggest" answer values.

I first started with D just b/c it ended in a .50. Once you start to go down the numbers you'll see that in order to end with a .50 (109.5) you need to have an amount taken away that ends in .50. B,C, and E are out. D worked.

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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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New post 29 Oct 2017, 12:37
Since the price of each cake is reducing by a constant value all the time, the prices will from Arithmetic Series
Average of 6 terms in Arithmetic is average of 3rd and 4th term. Thus now using options
Option C
4th term is 24.5 - 2*2 = 20.5
3rd term is 20.5-2 = 18.5
Average is 19.5
Total price of all 6 is 19.5*6 = 19*6+0.5*6 = 117
But since we need answer less than 117 (109.5) average value is less than 19.5.
Thus moving down to option D
4th term = 24.5-2*2.5 = 19.5
3rd term = 19.5 - 2.5 = 17
Average = 18.25
Total price = 18.25*6 = 18*6+0.25*6 = 108+1.5 = 109.5
Matching with question
Thus answer is option D
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Re: A certain bakery sells six different-sized wedding cakes [#permalink]

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New post 31 Oct 2017, 15:41
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


We can create the following equation:

(24.50) + (24.50 - x) + (24.50 - 2x) + (24.50 - 3x) + (24.50 - 4x) + (24.50 - 5x) = 109.50

6(24.50) - 15x = 109.50

147 - 15x = 109.5

37.5 = 15x

2.50 = x

Answer: D
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Re: A certain bakery sells six different-sized wedding cakes   [#permalink] 31 Oct 2017, 15:41
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