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# Math: Number Theory - Percents

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Re: Math: Number Theory - Percents [#permalink]

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21 Jul 2010, 23:33
Compound Interest

$$Balance(final)=$$
$$=principal*(1+\frac{interest}{C})^{time*C}$$, where C = the number of times compounded annually.

If C=1, meaning that interest is compounded once a year, then the formula will be: $$Balance(final)=$$
$$principal*(1+interest)^{time}$$, where time is number of years.

Example: If $20,000 is invested at 12% annual interest, compounded quarterly, what is the balance after 2 year? Solution: $$Balance=20,000*(1+\frac{0.12}{4})^{2*4}=$$ $$=20,000*(1.03)^8=25,335.4$$ I dont understand why 0.12/4 ? where did we get no 4 ? thanks Math Expert Joined: 02 Sep 2009 Posts: 38985 Followers: 7748 Kudos [?]: 106424 [1] , given: 11626 Re: Math: Number Theory - Percents [#permalink] ### Show Tags 22 Jul 2010, 04:33 1 This post received KUDOS Expert's post xmagedo wrote: Compound Interest $$Balance(final)=$$ $$=principal*(1+\frac{interest}{C})^{time*C}$$, where C = the number of times compounded annually. If C=1, meaning that interest is compounded once a year, then the formula will be: $$Balance(final)=$$ $$principal*(1+interest)^{time}$$, where time is number of years. Example: If$20,000 is invested at 12% annual interest, compounded quarterly, what is the balance after 2 year?
Solution: $$Balance=20,000*(1+\frac{0.12}{4})^{2*4}=$$
$$=20,000*(1.03)^8=25,335.4$$

I dont understand why 0.12/4 ? where did we get no 4 ?

thanks

12% of annual interest is compounded quarterly, so it's compounded 4 times a year --> C=4.
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Re: Math: Number Theory - Percents [#permalink]

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22 Jul 2010, 07:09
thanks bunnel
But one more question, when it says compunded semiannually, does it mean half ? for example 6 ? do we take 3 ?
thanks
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Re: Math: Number Theory - Percents [#permalink]

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22 Jul 2010, 07:17
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Expert's post
xmagedo wrote:
thanks bunnel
But one more question, when it says compunded semiannually, does it mean half ? for example 6 ? do we take 3 ?
thanks

Compounded semiannually = Compounded twice a year.
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Re: Math: Number Theory - Percents [#permalink]

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02 Aug 2010, 20:59
The Probability material is really helpful thanks...
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Re: Math: Number Theory - Percents [#permalink]

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06 Aug 2010, 06:47
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Re: Math: Number Theory - Percents [#permalink]

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09 Oct 2010, 14:30
Generic questions based on percentages that can be remembered:

1. If the price(p) of an Item increases by x% then consumption(c) has to be decreased by 100 /(100+x) % to keep the expenditure(E) constant.

2. If two articles are sold at same price , and on first one the shopkeeper makes a profit of p% and on the other suffers a loss of p % , overall he suffers a loss. The loss is p*p /100 % ie., ( p square divided by 100).

Especially,I have seen the second type of question very common one in many competitive tests.

Note : Please let me know if you are interested in the reason behind each answer.

This is my first post in gmatclub.I thank all of the gmatclub members / moderators for providing such a wonderful environment.
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Re: Math: Number Theory - Percents [#permalink]

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23 Oct 2010, 06:32
Hey guys i am new in this forum... can you help me with these percentage problem?

1) The income of a broker remains unchanged though the rate of commission is increased from 4% to 5%. THe percentage slump in business is:

i)1 ii) 8 iii) 20 iv)80

2) p is 6 times as large as q. The percent that q is less than p is:

i) 83.33 ii) 16.5 iii) 90 iv) 60

3) In a market survey 20% voted for A and 60% voted for B. the remaining were uncertain. if the difference between who voted for B and those who were uncertain was 720 how many individuals were covered in the survey?

i) 3600 ii) 1440 iii)1800 iv) cant determine
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Re: Math: Number Theory - Percents [#permalink]

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23 Oct 2010, 07:21
syenaman wrote:
Generic questions based on percentages that can be remembered:

1. If the price(p) of an Item increases by x% then consumption(c) has to be decreased by 100 /(100+x) % to keep the expenditure(E) constant.

2. If two articles are sold at same price , and on first one the shopkeeper makes a profit of p% and on the other suffers a loss of p % , overall he suffers a loss. The loss is p*p /100 % ie., ( p square divided by 100).

Especially,I have seen the second type of question very common one in many competitive tests.

Note : Please let me know if you are interested in the reason behind each answer.

This is my first post in gmatclub.I thank all of the gmatclub members / moderators for providing such a wonderful environment.

Would love the explanation with some nice examples.
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Re: Math: Number Theory - Percents [#permalink]

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13 Nov 2010, 07:29
Thanks guys...found what i was looking for...you are wonderful..
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Re: Math: Number Theory - Percents [#permalink]

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16 Nov 2010, 05:52
Bunuel wrote:
PERCENTS

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, walker

--------------------------------------------------------

This topic is included in GMAT ToolKit App (iPhone/iPod Touch)

--------------------------------------------------------

Definition

A percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred"). It is often denoted using the percent sign, "%", or the abbreviation "pct". Since a percent is an amount per 100, percents can be represented as fractions with a denominator of 100. For example, 25% means 25 per 100, 25/100 and 350% means 350 per 100, 350/100.

• A percent can be represented as a decimal. The following relationship characterizes how percents and decimals interact. Percent Form / 100 = Decimal Form

For example: What is 2% represented as a decimal?
Percent Form / 100 = Decimal Form: 2%/100=0.02

Percent change

General formula for percent increase or decrease, (percent change):

$$Percent=\frac{Change}{Original}*100$$

Example: A company received $2 million in royalties on the first$10 million in sales and then $8 million in royalties on the next$100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next$100 million in sales?

Solution: Percent decrease can be calculated by the formula above:
$$Percent=\frac{Change}{Original}*100=$$
$$=\frac{\frac{2}{10}-\frac{8}{100}}{\frac{2}{10}}*100=60%$$, so the royalties decreased by 60%.

Simple Interest

Simple interest = principal * interest rate * time, where "principal" is the starting amount and "rate" is the interest rate at which the money grows per a given period of time (note: express the rate as a decimal in the formula). Time must be expressed in the same units used for time in the Rate.

Example: If $15,000 is invested at 10% simple annual interest, how much interest is earned after 9 months? Solution:$15,000*0.1*9/12 = $1125 Compound Interest $$Balance(final)=$$ $$=principal*(1+\frac{interest}{C})^{time*C}$$, where C = the number of times compounded annually. If C=1, meaning that interest is compounded once a year, then the formula will be: $$Balance(final)=$$ $$principal*(1+interest)^{time}$$, where time is number of years. Example: If$20,000 is invested at 12% annual interest, compounded quarterly, what is the balance after 2 year?
Solution: $$Balance=20,000*(1+\frac{0.12}{4})^{2*4}=$$
$$=20,000*(1.03)^8=25,335.4$$

Percentile

If someone's grade is in $$x_{th}$$ percentile of the $$n$$ grades, this means that $$x%$$ of people out of $$n$$ has the grades less than this person.

Example: Lena’s grade was in the 80th percentile out of 120 grades in her class. In another class of 200 students there were 24 grades higher than Lena’s. If nobody had Lena’s grade, then Lena was what percentile of the two classes combined?

Solution:
Being in 80th percentile out of 120 grades means Lena outscored $$120*0.8=96$$ classmates.

In another class she would outscored $$200-24=176$$ students.

So, in combined classes she outscored $$96+176=272$$. As there are total of $$120+200=320$$ students, so Lena is in $$\frac{272}{320}=0.85=85%$$, or in 85th percentile.

Official GMAC Books:

The Official Guide, 12th Edition: PS #10; PS #17; PS #19; PS #47; PS #55; PS #60; PS #64; PS #78; PS #92; PS #94; PS #109; PS #111; PS #115; PS #124; PS #128; PS #131; PS #151; PS #156; PS #166; PS #187; PS #193; PS #200; PS #202; PS #220; DS #2; DS #7; DS #21; DS #37; DS #48; DS #55; DS #61; DS #63; DS #78; DS #88; DS #92; DS #120; DS #138; DS #142; DS #143;

Generated from [GMAT ToolKit]

[Reveal] Spoiler:
Attachment:
Math_icon_percents.png

In the compound interest example, Please let me know how do we calculate 20000*(1.03)^8 . is there any simple way to find (1.03)^8.

thanx....
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Re: Math: Number Theory - Percents [#permalink]

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02 Aug 2011, 21:11
redwine01 wrote:
Hey guys i am new in this forum... can you help me with these percentage problem?

1) The income of a broker remains unchanged though the rate of commission is increased from 4% to 5%. THe percentage slump in business is:

i)1 ii) 8 iii) 20 iv)80

2) p is 6 times as large as q. The percent that q is less than p is:

i) 83.33 ii) 16.5 iii) 90 iv) 60

3) In a market survey 20% voted for A and 60% voted for B. the remaining were uncertain. if the difference between who voted for B and those who were uncertain was 720 how many individuals were covered in the survey?

i) 3600 ii) 1440 iii)1800 iv) cant determine

Hi Guys!

Thank you.
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Re: Math: Number Theory - Percents [#permalink]

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10 Aug 2011, 04:06
1)iii 20
Lets say initial income=x, amended income=y then 4x/100=5y/100 ie x/y=5/4
If income declines from 5 to 4, its 20% decrease.

2)i 83.33
p=6q ie p/q=6/1 therefore p-q/p=6-1/6=5/6=83.33%

3)iii 1800
Votes for A=20%, B=60%, uncertain(U)=20% and B-U=60%-20%=720 ie 40%=720 ie 100%=1800
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Re: Math: Number Theory - Percents [#permalink]

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12 Feb 2012, 01:07
Bunuel wrote:
gurpreetsingh wrote:
Bunuel wrote:
PERCENTS

Example: Lena’s grade was in the 80th percentile out of 120 grades in her class. In another class of 200 students there were 24 grades higher than Lena’s. If nobody had Lena’s grade, then Lena was what percentile of the two classes combined?

Solution:
Being in 80th percentile out of 120 grades means Lena outscored $$120*0.8=96$$ classmates.

In another class she would outscored $$200-24=176$$ students.

So, in combined classes she outscored $$96+176=272$$. As there are total of $$120+200=320$$ students, so Lena is in $$\frac{272}{320}=0.85=85%$$, or in 85th percentile.

[Reveal] Spoiler:
Attachment:
Math_icon_percents.png

In another class she would outscored $$200-24=176$$ students.
I think it should be 200-24-1 = 175 as 24 were higher than Lena , thus 24+1 are lower than her, we need to count her as well

The point here is that Lena herself is not in the other class. So in another class she outscored 200-24=176 not 175.

Hope it's clear.

But the question does not say Lena is not in other class. So, how to interpret this, based on the provided solutions?
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Re: Math: Number Theory - Percents [#permalink]

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12 Feb 2012, 02:23
"Lena’s grade was in the 80th percentile out of 120 grades in HER class. In ANOTHER class ..." So another class is not Lena's class.
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Re: Math: Number Theory - Percents [#permalink]

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12 Feb 2012, 04:19
Bunuel wrote:
"Lena’s grade was in the 80th percentile out of 120 grades in HER class. In ANOTHER class ..." So another class is not Lena's class.

OK. Thanks.
I observed that, but I looked at those two statements "independently", because we are not allowed to make assumptions unless sufficient supporting material is available.
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Re: Math: Number Theory - Percents [#permalink]

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12 Feb 2012, 04:23
Chembeti wrote:
Bunuel wrote:
"Lena’s grade was in the 80th percentile out of 120 grades in HER class. In ANOTHER class ..." So another class is not Lena's class.

OK. Thanks.
I observed that, but I looked at those two statements "independently", because we are not allowed to make assumptions unless sufficient supporting material is available.

It's a PS question, so sentences are not independent like the statements in DS. Check similar GMAT Prep problem: amy-s-grade-was-90th-percentile-of-the-80-grades-for-her-98164.html

Hope it helps.
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Re: Math: Number Theory - Percents [#permalink]

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12 Feb 2012, 04:50
Bunuel wrote:
Chembeti wrote:
Bunuel wrote:
"Lena’s grade was in the 80th percentile out of 120 grades in HER class. In ANOTHER class ..." So another class is not Lena's class.

OK. Thanks.
I observed that, but I looked at those two statements "independently", because we are not allowed to make assumptions unless sufficient supporting material is available.

It's a PS question, so sentences are not independent like the statements in DS. Check similar GMAT Prep problem: amy-s-grade-was-90th-percentile-of-the-80-grades-for-her-98164.html

Hope it helps.

Yes, it helped me lot, Bunnel. I thank you very much. Now, I realized the difference between PS && DS. (shame on me )
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Re: Math: Number Theory - Percents [#permalink]

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05 Mar 2012, 01:35
Bunuel wrote:
gurpreetsingh wrote:
Bunuel wrote:
PERCENTS

Example: Lena’s grade was in the 80th percentile out of 120 grades in her class. In another class of 200 students there were 24 grades higher than Lena’s. If nobody had Lena’s grade, then Lena was what percentile of the two classes combined?

Solution:
Being in 80th percentile out of 120 grades means Lena outscored $$120*0.8=96$$ classmates.

In another class she would outscored $$200-24=176$$ students.

So, in combined classes she outscored $$96+176=272$$. As there are total of $$120+200=320$$ students, so Lena is in $$\frac{272}{320}=0.85=85%$$, or in 85th percentile.

[Reveal] Spoiler:
Attachment:
Math_icon_percents.png

In another class she would outscored $$200-24=176$$ students.
I think it should be 200-24-1 = 175 as 24 were higher than Lena , thus 24+1 are lower than her, we need to count her as well

The point here is that Lena herself is not in the other class. So in another class she outscored 200-24=176 not 175.

Hope it's clear.

This kind of clear thinking is a little awe-inspiring, especially when my brain's already turned into oatmeal from 2 months of prepping!
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Re: Math: Number Theory - Percents [#permalink]

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27 Oct 2012, 02:09
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The formula for percentage is the following and it should be easy to use:

Study it below carefully before looking at the examples

$$\frac{is}{of}= \frac{%}{100}$$

OR

$$\frac{part}{whole} = \frac{%}{100}$$

We will take examples to illustrate.Let us start with the formula on the left

An important thing to remember: Cross multiply

It means to multiply the numerator of one fraction by the denominator of the other fraction

Examples #1:

25 % of 200 is____

In this problem, of = 200, is = ?, and % = 25

We get:

is/200 = 25/100

Since is in an unknown, you can replace it by y to make the problem more familiar

y/200 = 25/100

Cross multiply to get y × 100 = 200 × 25

y × 100 = 5000

Divide 5000 by 100 to get y

Since 5000/100 = 50, y = 50

So, 25 % of 200 is 50

Examples #2:

What number is 2% of 50 ?

This is just another way of saying 2% of 50 is___

So, set up the proportion as example #1:

is/50 = 2/100

Replace is by y and cross multiply to get:

y × 100 = 50 × 2

y × 100 = 100

Since 1 × 100 = 100, y = 1

Therefore, 1 is 2 % of 50

Examples #3:

24% of___ is 36

This time, notice that is = 36, but of is missing

After you set up the formula, you get:

36/of = 24/100

Replace of by y and cross multiply to get:

36/y = 24/100

y × 24 = 36 × 100

y × 24 = 3600

Divide 3600 by 24 to get y

3600/24 = 150, y = 150

Therefore, 24 % of 150 is 36

Now, we will take examples to illustrate how to use the formula for percentage on the right

Examples #4:

To use the other formula that says part and whole, just remember the following:

The number after of is always the whole

The number after is is always the part

If I say 25 % of___ is 60, we know that the whole is missing and part = 60

Your proportion will will like this:

60/whole = 25/100

After cross multiplying, we get:

whole × 25 = 60 × 100

whole × 25 = 6000

Divide 6000 by 25 to get whole

6000/25 = 240, so whole = 240

Therefore, 25 % of 240 is 60

Examples #5:

___% of 45 is 9

Here whole = 45 and part = 9, but % is missing

We get:

9/45 = %/100

Replacing % by x and cross multiplying gives:

9 × 100 = 45 × x

900 = 45 × x

Divide 900 by 45 to get x

900/45 = 20, so x = 20

Here we go!. I hope these formula for percentage were helpful.
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Re: Math: Number Theory - Percents   [#permalink] 27 Oct 2012, 02:09

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