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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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04 Mar 2012, 15:12
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35% (medium)

Question Stats:

79% (02:10) correct 21% (02:27) wrong based on 232 sessions

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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

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04 Mar 2012, 15:49
1
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.

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

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04 Mar 2012, 16:05
1
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

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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Updated on: 11 Sep 2016, 20:37
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1
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)

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

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Originally posted by LogicGuru1 on 03 Jul 2016, 07:15.
Last edited by LogicGuru1 on 11 Sep 2016, 20:37, edited 1 time in total.
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Re: A certain bakery sells six different-sized wedding cakes  [#permalink]

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03 Jul 2016, 10:52
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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11 Sep 2016, 09:13
rohitrawat9990 wrote:
109.5= 24.5x6 - 15X
X=2.5

Sent from my iPhone using GMAT Club Forum mobile app

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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11 Sep 2016, 09:23
arturportugal wrote:
rohitrawat9990 wrote:
109.5= 24.5x6 - 15X
X=2.5

Sent from my iPhone using GMAT Club Forum mobile app

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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18 Oct 2016, 16:26
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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18 Oct 2016, 16:55
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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29 Oct 2017, 13:37
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
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Re: A certain bakery sells six different-sized wedding cakes  [#permalink]

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31 Oct 2017, 16:41
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

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Re: A certain bakery sells six different-sized wedding cakes &nbs [#permalink] 31 Oct 2017, 16:41
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