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# The price of a consumer good increased by p%. . .

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The price of a consumer good increased by p%. . .  [#permalink]

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Updated on: 07 Aug 2018, 06:09
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Question Stats:

67% (02:29) correct 33% (02:32) wrong based on 216 sessions

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The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$?

A. $$-2$$%
B. $$2$$%
C. $$22$$%
D. $$25$$%
E. Cannot be determined

Take a stab at this fresh question from e-GMAT. Post your analysis below.

Official Solution to be provided after receiving some good analyses.

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Originally posted by EgmatQuantExpert on 11 Nov 2016, 04:46.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:09, edited 1 time in total.
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Re: The price of a consumer good increased by p%. . .  [#permalink]

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11 Nov 2016, 06:09
Top Contributor
EgmatQuantExpert wrote:
The price of a consumer good increased by p% during 2012 and decreased by 12% during 2013. If no other change took place in the price of the good and the price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012, what was the value of p?

A. $$-2$$%
B. $$2$$%
C. $$22$$%
D. $$25$$%
E. Cannot be determined

Let $100 be the original price The price of a consumer good increased by p% during 2012 p% = p/100, so a p% INCREASE is the same a multiplying the original price by 1 + p/100 So, the new price = ($100)(1 + p/100)

The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
Simplify more: $110 = 88 + 0.88p Subtract 88 from both sides: 22 = 0.88p So, p = 22/0.88 = 25 Answer: RELATED VIDEO _________________ Test confidently with gmatprepnow.com Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4286 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 11:48 EgmatQuantExpert wrote: The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined Price's corresponding to year - 2011 = $$100$$ 2012 = $$100 + p$$ 2013 = $$\frac{88}{100}(100 + p)$$ Further , $$\frac{88}{100}(100 + p)$$ = $$110$$ Or, $$\frac{8}{100}(100 + p)$$ = $$10$$ Or, 800 + 8p = 1000 Or, 8p = 200 So, p = 25% Hence, answer will be (D) 25% _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Current Student Joined: 26 Jan 2016 Posts: 103 Location: United States GPA: 3.37 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 12:01 Lets start with a number for the original value. 100 is the easiest. So we're looking for a value of 110 at the end of 2013. Just by looking at the values we can get an idea of what to start testing. If we're increasing 100 by p% then decreasing it by 12% and the original value is still 10% higher we need a value much higher than 12. 25% is the easiest value to start with. 100+25%=125 125-12%=110 D Current Student Joined: 12 Aug 2015 Posts: 2627 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 12 Nov 2016, 19:32 For all the algebra loving people out there=> Let price at the beginning of 2012 be$x
so after the end of 2012=> x[1+p/100]
And finally at the end of 2013 => x[1+p/100][1-12/100]

As per question=> Price was simple 10 percent greater
Hence x[1+10/100] must be the final price.
Equating the two we get
=> x[110/100]=x[1+p/100][88/100]
=> 44p+4400=5500
=> 44p=1100
=> p=1100/44=> 100/4=> 25.
So p must be 25
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Re: The price of a consumer good increased by p%. . .  [#permalink]

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16 Nov 2016, 08:46
[size=150]Let Price = 100

[size=150]Increased by P% = 100(1+P/100)

Treat it like successive percents;

So a 12% decrease would mean 88% of (1+P/100) of 100

The key words are no other change took place. So there are no further successive percents, and the final price = 110

Therefore:

88/100 * (1+P/100) * 100 = 110

=> 8800 + 88P = 11,000
=> 88P = 2200
=> P = 2200/88
=> P = 25%
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Re: The price of a consumer good increased by p%. . .  [#permalink]

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Updated on: 18 Dec 2016, 21:12
Hey,

The given question can be solved in a number of ways. Let's look at two most common ways to solve this question. We will share two more ways of solving this question tomorrow.

Method 1
:

• Let us consider the price of the consumer good at the beginning of 2012 to be 100.
• Let us also assume the price to be “C” at the beginning of 2013, after an increase of p%.
• Since we know that with respect to the initial price, the price at the end of 2013 went up by 10%.
o Therefore, the price at the end of 2013 = $$100 + (10$$ % of $$100) = 110$$
• Now we can write -
o $$C – 12$$ % of $$C = 110$$
o $$C * (1 - \frac {12}{100}) = 110$$
o $$C = 110 * \frac {25}{22} = 125$$

Therefore, the price at the beginning of 2013 is 125 and we got this value after p% increase over the initial value.
Thus, we can write –
• $$100 + (p$$ % of $$100) = 125$$
• $$P = 25$$ %

Method 2
:

Conventional method –

• Let the price of consumer good at the beginning of 2012 be 100.
• After an increase of p%, the price at the beginning of 2013 will be –
o Price at the beginning of 2013 $$= 100 + (p$$ % of $$100) = 100 + p$$
Therefore the price at the beginning of 2013 is (100 + p)

• After a decrease of 12%, the price at the end of 2013 will be –
o Value at the beginning of 2013 $$* (1 – \frac{12}{100}) = (100 + p) * \frac {22}{25}$$……………..(i)

• And we are also given that the overall increase in the price of consumer good is 10%.
• Therefore, the value at the end of 2013 = $$100 + (10$$% of $$100)$$ = 110………(ii)

• Equating equation (i) and (ii) we get –
o $$(100 + p) * \frac {22}{25} = 110$$
o $$100 + p = 125$$
Therefore, $$p = 25$$%

There are more innovative ways to solve this question. A few of them have not been discussed here. Can you all think of any other way to solve it? Would love to see a few other methods!

I will post two more ways to solve this question tomorrow. In the mean time, expecting some more responses with other ways to solve this question

Thanks,
Saquib
e-GMAT
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Originally posted by EgmatQuantExpert on 15 Dec 2016, 22:26.
Last edited by EgmatQuantExpert on 18 Dec 2016, 21:12, edited 1 time in total.
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Re: The price of a consumer good increased by p%. . .  [#permalink]

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16 Dec 2016, 03:11
1
joannaecohen wrote:
Lets start with a number for the original value. 100 is the easiest.

So we're looking for a value of 110 at the end of 2013.

Just by looking at the values we can get an idea of what to start testing. If we're increasing 100 by p% then decreasing it by 12% and the original value is still 10% higher we need a value much higher than 12. 25% is the easiest value to start with.

100+25%=125

125-12%=110

D

Hi,

Thanks for posting a different way of approaching this problem. In fact, the approach followed by you is very close to one of the innovative ways that we talked about in our official solution. The only difference being in your approach you have concluded that p should be much larger than 12. Going a step further, you can also conclude that p should be larger than 22% (12%+10%). Once you do so, you don't even need to pick any number. The only option which will satisfy it is D. 25%.

When we post our detailed solution using the two innovative methods tomorrow, we will explain how we can conclude that p should be greater than 22%.

Regards,
Saquib
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Re: The price of a consumer good increased by p%. . .  [#permalink]

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25 Dec 2016, 06:32
1
As discussed, let's look at one of the innovative ways of solving the above question. It is one of the quickest ways to solve a question that involves successive percentage increase/decrease on the same value. Please take a note of this approach and apply it on some GMAT questions to master it.

So, let's quickly look at this smart approach.

When a number is increased successively by two percentage, let's assume, $$a$$% and $$b$$%, the net increase in the value of the number can be expressed by the formula,

Net increase $$=a+b+\frac {ab}{100}$$

Le's take a simple example to understand. If we increase a number, let's say, X successively by 10% and 20% respectively, the net increase according to the above formula should be,

Net increase $$=10+20+\frac {10*20}{100}=10+20+\frac {200}{100} = 10+20+2 = 32$$%

Isn't that quick!! A nice method to keep in your arsenal to solve Percent question involving successive increase quickly.

One good thing about the above formula is that you can use it to calculate the net decrease in case of successive decrease too. All you need to do is in case of decrease represent the percent as negative. Easy isn't it . Let's see an application quickly.

If we decrease a number, let's say, X successively by 10% and 20% respectively, the net increase according to the above formula should be,

Net increase $$=(-10)+(-20)+\frac {(-10)*(-20)}{100}=-10-20+\frac {200}{100} = -10-20+2 = (-28)$$%

Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.

So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.

We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.

Regards,
Saquib
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The price of a consumer good increased by p%. . .  [#permalink]

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Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.

We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,

Net increase $$=p+(-12)+\frac {p*(-12)}{100}=p-12-\frac {3p}{25} = (\frac {22p}{25} - 12)$$%

It is given in the question that the net increase finally is $$10$$%. Hence, we can equate the two values.

$$(\frac {22p}{25} - 12)$$% = $$10$$%

or,
$$\frac {22p}{25} = 10+12 = 22$$%

or,
$$p = 25$$%

Now, with this understanding try to solve this question in an even better way. Give it a try and we will post the official solution in another innovative way soon.

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Originally posted by EgmatQuantExpert on 26 Dec 2016, 23:45.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:11, edited 1 time in total.
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Re: The price of a consumer good increased by p%. . .  [#permalink]

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27 Dec 2016, 06:56
EgmatQuantExpert wrote:
The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$?

A. $$-2$$%
B. $$2$$%
C. $$22$$%
D. $$25$$%
E. Cannot be determined

$$p - 12 - \frac{12p}{100} = 10$$

$$100p - 1200 -12p = 1000$$

$$88p = 2200$$
$$p = 25$$

Hence, the correct answer must be (D) 25

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Re: The price of a consumer good increased by p%. . .  [#permalink]

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