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A certain candy manufacturer reduced the weight of Candy Bar M by 20

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A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 30 May 2011, 11:17
1
6
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

75% (01:03) correct 25% (01:02) wrong based on 246 sessions

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A certain candy manufacturer reduced the weight of Candy Bar M by 20 percent buy left the price unchanged. What was the resulting percent increase in the price per ounce of Candy Bar M?

a)5%
b)10
c)15%
d)20%
e)25%

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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 30 May 2011, 11:32
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Baten80 wrote:
A certain candy manufacturer reduced the weight of Candy Bar M by 20 percent buy left the price unchanged. What was the resulting percent increase in the price per ounce of Candy Bar M?

a)5%
b)10
c)15%
d)20%
e)25%


Before:
$1->1 oz
Price: 1/1 $/oz
After:
$1->0.8 oz
Price: 1/0.8 $/oz = 1.25 $/oz

Thus, the price increased per oz is 25cents.
25cents is 25% of $1

Ans: "E"
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 02 Jun 2011, 13:02
I used plugging in numbers, is that a good approach here.. I did get 25% percent though.. What level would this question be?
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 02 Jun 2011, 13:15
l0rrie wrote:
I used plugging in numbers, is that a good approach here.. I did get 25% percent though.. What level would this question be?


If you took less than 1.5 minutes and if you are more comfortable with PIN, then it's fine. However, I personally feel PIN is relatively more difficult/time taking for this problem. Level- 600-650 maybe.
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 02 Jun 2011, 13:26
I tried the algebra and got lost somewhere trying to set up the right equation.. That is really my weakness.. Plugging in numbers here worked pretty well for me I think I was done in around 1 minute but normally plugging in numbers is going like ^%$@%$#...So I guess I was just lucky this time..
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 04 Jun 2011, 08:35
Now i understand.

Let Previous price $100 for 100 ounce
Per ounce $ 1.
Now, 80 ounce for $100
so, per ounce 100/80=$1.25
Percentage increase 1.25 x 100 = 125
so 125-100 = 25%
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 06 Jun 2011, 22:45
I used Algebra and got 25%
let p be the price and w be the original weight
percent increase = ((p/0.8w - p/w)/(p/w))*100 = 25%
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 11 Jun 2011, 11:27
answer E

Suppose :

Earlier:
100 grams cost $100
so 1 gm costs $1

Now:
80 grams cost $100 (because weight has been reduced by 20%)
so 1 gm will cost 100/80 = $1.25

So change = 25 cents

% change = (25/100)*100 = 25%
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A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 07 Feb 2017, 05:16
Baten80 wrote:
A certain candy manufacturer reduced the weight of Candy Bar M by 20 percent buy left the price unchanged. What was the resulting percent increase in the price per ounce of Candy Bar M?

a)5%
b)10
c)15%
d)20%
e)25%


I got the answer through the algebraic method:

Assume:
For X ounces of choc manufacturer charges $Y
per ounce charge = $ Y/X
Later, for 80% of choc i.e. (4/5)X he charges the same i.e. $Y
here, per ounce choc costs $ 5Y/4X
resulting increase in price per ounce=\((5Y/4X - Y/X)= Y/4X\)
% INCREASE= \([(Y/4X)/(Y/X)]*100%= 25%\)
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 07 Feb 2017, 10:30
1
Baten80 wrote:
A certain candy manufacturer reduced the weight of Candy Bar M by 20 percent buy left the price unchanged. What was the resulting percent increase in the price per ounce of Candy Bar M?

a)5%
b)10
c)15%
d)20%
e)25%


(p/4)/(p/5)=5/4=125%
25% increase in price per ounce
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 09 Feb 2017, 17:30
Baten80 wrote:
A certain candy manufacturer reduced the weight of Candy Bar M by 20 percent buy left the price unchanged. What was the resulting percent increase in the price per ounce of Candy Bar M?

a)5%
b)10
c)15%
d)20%
e)25%


We are given that a certain candy manufacturer reduced the weight of Candy Bar M by 20 percent but left the price unchanged. If we let w = the original weight of the bar, the new weight is 0.8w = 4w/5. If we let p = the original price of the bar we know:

p/w = the original price per ounce of the bar

p/(4w/5) = 5p/4w = the new price per ounce of the bar

Finally we can determine the resulting percent increase in the price per bar due to the change:

[5p/4w - p/w]/[p/w] x 100

[5p/4w - 4p/4w]/[p/w] x 100

[p/4w]/[p/w] x 100

p/4w x w/p x 100 = 1/4 x 100 = 25%

Answer: E
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A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 09 Feb 2017, 23:11
JeffTargetTestPrep wrote:
Baten80 wrote:
A certain candy manufacturer reduced the weight of Candy Bar M by 20 percent buy left the price unchanged. What was the resulting percent increase in the price per ounce of Candy Bar M?

a)5%
b)10
c)15%
d)20%
e)25%


We are given that a certain candy manufacturer reduced the weight of Candy Bar M by 20 percent but left the price unchanged. If we let w = the original weight of the bar, the new weight is 0.8w = 4w/5. If we let p = the original price of the bar we know:

p/w = the original price per ounce of the bar

p/(4w/5) = 5p/4w = the new price per ounce of the bar

Finally we can determine the resulting percent increase in the price per bar due to the change:

[5p/4w - p/w]/[p/w] x 100

[5p/4w - 4p/4w]/[p/w] x 100

[p/4w]/[p/w] x 100

p/4w x w/p x 100 = 1/4 x 100 = 25%

Answer: E


Isn't that too much math?

How about just taking both variables as 100.

Original value = 100/100=1
New value=100/80=10/8

Change =10/8-1=>1/4

Percentage change =>1/4 *100 => 25

Hence E.


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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20  [#permalink]

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New post 30 Aug 2018, 18:11
Hello,

Well we can use this technique for couple of problems so its worth learning and understanding .

Say, Distance is constant, then Speed is inversely proportional to time

or we say
\(\frac{Sa}{Sb}\)= \(\frac{Tb}{Ta}\)

or

Speed Time Distance

40 m/hr 3hr 120

60m/hr 2hr 120



So i increased my speed by 50 % or by \(\frac{1}{2}\) , so time will reduce by 33% or by \(\frac{1}{3}\)

So generally increase of \(\frac{1}{x}\) leads to decrease of \(\frac{1}{x+1}\)


Comming back to our question . The price in both the cases is Constant or unchanged. So it must be of the above form.

Let initially the weight be w1 and price initially per gram is x1 and total price is p . Now after reducing the weight be w2 , price per gram is x2 and total price is p

we have x1*w1=p
now if w1 is decreased by 20 % that is by \(\frac{1}{5}\) ( note decrease is always on the form of \(\frac{1}{x+1}\) so \(\frac{1}{5}\)=\(\frac{1}{4+1}\) that means the corresponding price per ounce must rise by \(\frac{1}{x}\) which is\(\frac{1}{4}\) = 25% to keep the price p constant .


Hope i was able to put forward the method of indirect variation correctly.
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Re: A certain candy manufacturer reduced the weight of Candy Bar M by 20 &nbs [#permalink] 30 Aug 2018, 18:11
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