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Percentage and its applications
by Total Gadha - Wednesday, 4 April 2007, 07:31 PM
 

Your competence with percentages and its application should form a very important part of your armory of CAT 2010 preparation or, for that matter, preparation for any other MBA exam. The concept of percentage will be applied in not only your quant section but also your data interpretation (DI) section. Therefore, master this important concept and make it your habit to calculate percentages mentally.

 
The word "percentage" literally means "per hundred" or "for every hundred." Therefore, whenever you calculate something as a part of 100, that part is numerically termed as percentage.

In other words, percentage is a ratio whose second term is equal to 100. For example, 1:4 can be written as 25: 100 or 25%, 3: 8 can be written as 37.5: 100 or 37.5%, 3: 2 can be written as 150: 100 or 150%, and so on.

IMPORTANT CONCEPTS ASSOCIATED WITH PERCENTAGE

Basic formula of percentage:

percentage formula

Percentage of:

percentage of formula

Percentage increase/decrease:

Percentage increase/decrease when a quantity a increase/decreases to become another quantity b is
 percentage increase/ decrease

Percentage less than/greater than:

Have a look at the picture given below:

percentage greater than/ less than

You can see that Johnny is taller than Vicky. What will your answer be if I ask

(a)      By what percentage is Johnny taller than Vicky?

(b)      By what percentage is Vicky shorter than Johnny?

Answer:

percentage greater than/ less than

conversion of fractions into percentages

Therefore, to find the final quantity after a 20% increase, we can directly multiply the old quantity by a factor of 1.2 and get the new quantity. Similarly, for a 20% decrease, we can multiply the old quantity by 0.8 and get the new quantity. The factors to be multiplied for various percentage increase/decrease are given below:


multiplication factors for percentage increase/ decrease

The biggest advantage of using the factors is that for subsequent percentage increase/decrease, we just keep on multiplying the corresponding factors and get the final quantity.

Example:

1.       The performance bonus of a salesman increases by 10% in the first year, by 20% in the second year, and by 30% in the third year. What is the overall percentage increase in his bonus in 3 years?

Answers: Let the bonus at the start of the first year be Rs100.

Therefore, to find the final bonus we just multiply by factors.

Final bonus = 100 Ã— 1.1 × 1.2 × 1.3 = 171.6.

Therefore, overall percentage increase = 71.6%

 
NOTE: note that taking initial value of 100 makes the problem simpler; whatever increase we get is directly equal to the percentage increase.

2.       An amount was first reduced by 10% and then further reduced by 20% and Rs10 800 were left. What was the original amount?

Answer: Let the original amount be A.

Therefore A × 0.9 (factor for 10% decrease) × 0.8 (factor for 20% decrease) = 10 800

Or A = Rs15 000.

·        SOME SOLVED EXAMPLES

percentages solved problems

I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this in the CBT Club this week.

 

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Re: Percentage and its applications
by shiwa k - Monday, 28 May 2007, 05:53 PM
 

Hi tg,

Thanks a lot.. ur approach is too good.

 

Re: Percentage and its applications
by Tempo # 911 - Tuesday, 29 May 2007, 03:03 PM
 

Dear TG Sir,

In the watermelon question, it says 30g of water evaporated, which has reduced the percentage of water from 85% to 80% i.e, a reduction of 5%

5% of water is 30g.

85% of water would be 510g.

15% was solid matter i.e, 90g.

So, total 600g.

What is the fallacy in this question ?

 

Re: Percentage and its applications
by Total Gadha - Tuesday, 29 May 2007, 04:24 PM
  Hi Temperamental,

The 85% and 80% are on two different weights. Understand the difference. smile

Total Gadha
Re: Percentage and its applications
by ashish tyagi - Monday, 6 August 2007, 06:42 AM
  hi sir can u tell me how we will solve such type of questions like................

[1 ] A piece of equipment cost a certain factory Rs. 600,000. If it depreciates in value, 15% the first year, 13.5 % the next year, 12% the third year, and so on, what will be its value at the end of 10 years,
Re: Percentage and its applications
by ashish tyagi - Monday, 6 August 2007, 07:06 AM
  hi t g, percentage is really a very important chapter for ny m b a exam so pls provide  more gud concepts for it............., thnx gr8 work .........
Re: Percentage and its applications
by lavika gupta - Friday, 17 August 2007, 06:54 AM
 

hi tg.i have a doubt.please help.

Q. If a merchant offers a discount of 40% on the marked price of his goods and thus ends up selling at cost price, what was the % mark up?

Re: Percentage and its applications
by lavika gupta - Friday, 17 August 2007, 07:00 AM
  hi tg.i think the answer to Q-2 under the topic-percentage increase/decrease is 135,rather than 108.
Re: Percentage and its applications
by tripti pathak - Saturday, 18 August 2007, 01:46 PM
  well i think the answer is

o.6 M.p= c.p
M.p/c.p =5/3
Mp-cp/cp=2/3
Mp is 66.67% of cp
Re: Percentage and its applications
by the underdog - Monday, 20 August 2007, 02:18 PM
 
Hiya TG,

Just a slight typographical error....it should be 180 x 0.75 = 135

Re: Percentage and its applications
by jaanaki sankar - Thursday, 22 November 2007, 11:57 PM
  hii
the discussion of chapters n methods are really good..
thanx a tonne.smile
Re: Percentage and its applications
by anurudh dubey - Saturday, 21 June 2008, 02:56 AM
 

of the adult population in NAGPUR ,45 % of men and 25 % of women are married .Wht percentage of the total population of adults is married (assume that no man marries more than one women and vice versa )

???

Re: Percentage and its applications
by TG Team - Saturday, 21 June 2008, 09:03 AM
  of the adult population in NAGPUR ,45 % of men and 25 % of women are married .Wht percentage of the total population of adults is married (assume that no man marries more than one women and vice versa )

Let total men be 'x'.
and women be 'y'.
so that 45x = 25y
=> y = 9x/5.
And total married population =45% of x + 25% of y =  9x/10
And total population =  x + y =  14x/5
=> Married population % = (9x/10)(5/14x)*100 = 450/14 = 32.14% smile
Re: Percentage and its applications
by Nikhil Goswami - Thursday, 3 July 2008, 07:42 PM
  The 3rd Variant in the question is missing
Assumption 1
Total Population = 200
Males = 100, Females = 100
Therefore, Male No.s = 45, Female No.s = 25
% of married Adults = (45+25)/200 = 70/200 = 35%

Assumption 2
Total Popultaion = 160
Males = 100, Females = 60
Therefore, Male No.s = 45, Female No.s = 15
% of married adults = (45+15)/160 = 60/160 = 37.5%

Hence u need a condition that what is the Ratio of Mens Population to Womens Population... hence Data Insufficient...
Re: Percentage and its applications
by mohit rastogi - Thursday, 17 July 2008, 10:34 PM
 

 Hi , Nikhil

 I think these assumptions are valid  in california onlytongueout.  i will stick to the fact that no. of married males = no. of married females.

Re: Percentage and its applications
by Saurabh Kumar - Monday, 21 July 2008, 10:19 PM
 

Re:

 

Thanks TG for making concept more clear but I would recomment to use some very tough problems which requires use of multiple concepts

Re: Percentage and its applications
by Abhishek Srivastava - Tuesday, 22 July 2008, 04:09 PM
 

I came across one of the question,plz help me to sort it out.

The radius of a circle is 21 cm,find the % of the numeric value of area to the numeric value of its radius.

Re: Percentage and its applications
by Abhishek Srivastava - Tuesday, 22 July 2008, 04:09 PM
 

I came across one of the question,plz help me to sort it out.

The radius of a circle is 21 cm,find the % of the numeric value of area to the numeric value of its radius.

Re: Percentage and its applications
by TG Team - Friday, 1 August 2008, 04:40 PM
  Hi Abhisheksmile
Area/Radius * 100 = (22*21*21*100)/(7*21) = 6600%.
or
Radius/Area * 100 = (21*7*100)/(22*21*21) = 100/66 = 1.5%
Re: Percentage and its applications
by Sibsankar Dasmahapatra - Tuesday, 16 September 2008, 12:27 PM
 

o.6 M.p= c.p
M.p/c.p =5/3
Mp-cp/cp=2/3
Mp is 66.67% of cp

...on the above question...i have a comment on the last part - > Mp is 66.67% of cp,,,as m.p/c.p=5/3 ; so m.p is 166.66% of c.p...that means m.p is 66.67% greater than c.p...

thanks

Sibu

Re: Percentage and its applications
by Rahul Sinha - Thursday, 18 September 2008, 11:38 AM
  Hi TG,
 
 
I am facing problem while solving below questions related to percentages.
 
1) Hursh wanted to subtract 5 from anumber, Unfortunately he added  5 instead of subtracting.
   find the percentage change in the result.
 
2) Out of total production of iron from hematites 20% of the ore get wasted and out of the remaining iron
    only 25% is pure iron.If the pure iron obtained in a year from a mine of hematite was 80000kg then
    the quantity of hematite mined from that mine in the year is???
 
3) In an examination , 80% student passed in Physics, 70% in chemistry while 15% failed in both the subject.
    If 325 student passed in both the subjects.Find the total no. of students who appeared in the examination.
 
4) The ratio of Jim's salary for october to his salary for november was 15:1.333 and the ratio of the salary for
    November to that of December was 2:2.6666. The worker got 40rs more for December than for October and
    recieved a bonus constituting 40%  of the salary for 3 months. Find the bonus(Assume that the number of
    workdays is same in every month)
 
Please let me know how can i post questions for different topics.
 
 
Thank you
Rahul
Re: Percentage and its applications
by Sibsankar Dasmahapatra - Thursday, 18 September 2008, 04:01 PM
 

Hi Sinha..

             Even the questions are not asked to me ; but I put my explanations in below....

1) If the number is x ; then Hursh's wanted/right  value is x-5 , but the mistake/wrong value x+5.

so total change = (x+5)-(x-5)=10

so percentage change will be 10*100/(x-5) %...(let x=25 , then percentage change will be 10*100/20 %= 50 %....)

2) Do this problem by percentage change grph.....let  the quantity of hematite mined 100.

then 100 (by 20% of the ore get wasted ..that will be 80 )-----------80.........(25 % of 80 will be pure iron ; so it will be 20)--------20

If  pure iron obtained in a year from a mine of hematite was 80000kg then
    the quantity of hematite mined from that mine in the year is =100*80000/20=4,00,000 kg....(ans)

 

3) Let total no of student 100...and no of student pass for both =x% ;

then (80-x)+(70-x)+x+15 = 100 (if you draw the ven diagram then u can find this easily...)...

so x= 65 % ...now this 65 % of total student  =325

then no of total student = 100*325/65= 500 (ans...)

Hi Sinha

           let me think the  question.4...

Thanks

Sibu

Re: Percentage and its applications
by Tuhin Banerjee - Thursday, 18 September 2008, 05:23 PM
 

Hi Rahul is it the ratio of the  october salary  to his salary for november  15:1.333  or 1.5:1333 , please let me know?

Re: Percentage and its applications
by Interstellar Overdrive - Thursday, 18 September 2008, 06:53 PM
 

Hi All,

A method which I use to solve the problems of type 'Solved Example 1'.
 In this problem price increase by 10%, or we can say by '1/10'.
Therefore the decrease in consumption would be
1/(10+1) = 1/11 = 9.09%.

In other words if price( or any other value for that matter) increase by "a/b" then the subsequent decrease in cosumption would be "a/(b+a)".
After we get a fraction we need to convert it into a percentage using the table given by TG.


Ofcourse this would become a little complicated if the value of the fraction calculated is not found in the table.

Now say the value of decreases, then what would be the increase in consumption.
This can also be calculated as follows:
Suppose the value decreases by 37.5%, i.e. 3/8.
then the consumption would increse by '3/5' or 60%.

In other words if the value decreases by 'a/b', then the consumption would be increased by 'a/(b-a)'.
Notice that b will always be greater than a. This is so because the decrease can never be equal to or greater than 100%.


I had devised these formulae while studying percentages, so I thought it would be nice to share these with fellow TGites. Hope you liked these.

Thanks and Regards,
IO.

Re: Percentage and its applications
by Rahul Sinha - Monday, 22 September 2008, 01:58 PM
 

Sorry

my mistake....

its (1.5) : (1.333)

Re: Percentage and its applications
by Rahul Sinha - Monday, 22 September 2008, 02:04 PM
 

Thank you!!!

For question (1) i have answer with me as 33.33% as per book( Need to verify it again)

2) and 3) are correct clear to me.

For 3) can you please let me know how did you write the expression                                      (80-x)+(70-x)+x+15 = 100

I know it but just i wanted to make sure whether my thinking over that is correct or not.

 

Thank you

Rahul

 

Re: Percentage and its applications
by Rahul Sinha - Monday, 22 September 2008, 02:19 PM
 

Hi TG's

Few questions:  i am having problem while solving it

1)If the length , breath and height of a cube are decreased, decreased  and increased by 5%, 5% and 20%
 respectively, then what will be the impact on the surface area of the cube(in % term)?

2)The minimum quantity of milk in liters(In whole numbers) that should be mixed in a mixture
  of 60ltr in which the initial ration of milk to water is 1:4 so that the resulting mixture
  has 15% milk is

Regards,

Rahul

Re: Percentage and its applications
by Sibsankar Dasmahapatra - Monday, 22 September 2008, 03:17 PM
 

Hi Rahul

     For your clarification of the equation (80-x)+(70-x)+x+15 = 100  (for your question no 3) ; I just give u an example...

let total student 10 ; out of 10 student 5 student pass in Physics , 4 pass in Math, 2 fail; if in this case 2 ( let x)  pass in both subject then pass only in Physics  5-2 =3; pass only in Math  3-2=1; pass in both subject =2 ( x) ;  fail 4 ; then your equation will be ==(5-2) + (3-2)+2 + 4 =10..

thanks

Re: Percentage and its applications
by Interstellar Overdrive - Monday, 22 September 2008, 05:03 PM
 

Soln

1) Let 'a' be the side of the cube..
The new length and breath would be '0.95a' and height would be '1.2a'.
Now initial surface area is 6(a^2).
New surface area wud be : 2[ (0.95a)*(0.95a) + (1.2a)*(0.95a) + (1.2a)(0.95a)]
= 2 [ 0.9025 (a^2) + 2.28 (a^2) ]
= 6.365 (a^2)

Therefor increase %age = 0.365/6 *100 = 6.0833%

2) In this problem, solution already has 20% milk. So by adding milk we will only increase the %age and not decrease it. I guess your question is wrong.

Thanks and Regards,
IO

Re: Percentage and its applications
by Surendran Chandravathanan - Thursday, 23 October 2008, 07:43 PM
 

Hi Rahul Sinha,

Is the answer for the 4th question you have posted (Salary & Bonus question) Rs. 295/- approx. ?????

Let me know whether it's correct.

Rgds,

Suren

Re: Percentage and its applications
by Navneet H - Saturday, 28 March 2009, 02:50 AM
  Consider the answer to the 3 question in this way:
If 80 percent have passed in phys tat means 20 have failed in it.
|| If 70 percent have passed in chem tat means 30 have failed in it. therefore totally 20+30-15=35 have failed in atleast one.
therefore 100-35=65 have passed in both.
65 %=325
100%=325/625*100
Re: Percentage and its applications
by Varun Agrawal - Tuesday, 14 July 2009, 06:12 PM
  @rahul

 Ques 4 :--  274 approx.. Bonus
Re: Percentage and its applications
by abhishek tripathi - Friday, 21 August 2009, 01:35 AM
  thank u once again 4 ur hard work sir 2 articles were very important :d one 2 change fractions into percentage and 2 one is percentage increase or decrease but sir tell me dnt u have any sections 4 di where we can learn few tips for faster approach in di i havent found any sections 4 di..do respond asap
abhisheksmile
Re: Percentage and its applications
by satyarth pandey - Monday, 31 August 2009, 03:39 PM
  hi need an solution to the following problem...

a student appears for 4 papers -Eng,Maths,phy,chem.Maximum marks for which are in the ratio of 1:1:2:2.His marks are in the ratio 4:8:13:15.if he got 80% of the total maximum marks in how many papers did he get more than 80%




Re: Percentage and its applications
by Hemant Ahire - Tuesday, 1 September 2009, 05:56 AM
  Ans is in 2 papers (in Maths and Chem) he got more than 80%. Please do let me know whether I m correct or wrong
Re: Percentage and its applications
by satyarth pandey - Tuesday, 1 September 2009, 12:46 PM
  yeah correct....
can you kindly elaborate....
Re: Percentage and its applications
by anuja kaushal - Tuesday, 1 September 2009, 06:45 PM
 

yes the ans z  in 2pprs..

one of d solutions can b lyk...

its given dt,d ratio for total marks z 1:1:2:2

for simplicity,lets assume the marks as 250,250,500  n 500.

agn d ratio of obtained marks z 4:8:13:15

dis ratio can b writn as 120:240:390:450(multipling d ratio term 4:8:13:15 by 30)

ie;the obtained marks are 120/250,240/250,390/500 n 450/500

which means only 2 of them,240/250 n 450/500 r abv 80%.

this solution also satisfies the condition that he got 80% of marks in total(as the total marks r 1500(250+250+500+500)n marks obtained r 1200(120+240+390+450)).

hence d ans dt he gets abv 80%marks in two sbjcts...smile

Re: Percentage and its applications
by satyarth pandey - Tuesday, 1 September 2009, 08:02 PM
  thanx for the solution  smile
got the right ans..was doing a calc mistk sad 
my way of doing :  x=t/6 and y=0.02t
now if u wnt to find the % of say 1st paper then : 4y/x*100 will give u d ans
                                   
Re: Percentage and its applications
by anuja kaushal - Wednesday, 2 September 2009, 12:38 AM
  bt i cnt understand ur solution,means these x n ystand for wt?n wt z d formula used...
Re: Percentage and its applications
by satyarth pandey - Wednesday, 2 September 2009, 02:31 PM
  the complete thing goes like this.....
x is the common factor in the ratio 1:1:2:2
y is the common factor in the ratio 4:8:13:15
now total marks : 1*x+1*X+2*X+2*x =t :: x=t/6
also given : 4y+8y+13y+15y=0.8t :: y=0.02t

now to calculate percent of a subject : marks obtained / total marks
                                      in 1st case  : 4y/x  = 4*0.02*6= .48
                                          2nd case : 8y/x  = 8*0.02*6= .96
                                          3rd case  : 13y/2x = 13*0.02*3= .78
                                           4th case :  15y/2x= 15*0.02*3=.90        


Re: Percentage and its applications
by anuja kaushal - Wednesday, 2 September 2009, 02:49 PM
 

ok...nw i understood...

n thts a very good approach indeed...

Re: Percentage and its applications
by abhishek rai - Monday, 10 May 2010, 07:32 AM
  Thnx TG and also, Interstellar Overdrive.....
Re: Percentage and its applications
by afroz m - Saturday, 22 May 2010, 07:32 PM
  Hi TG sir,
can i not get the complete lesson on each topic without joining the TG CBT, these seem to be incomplete lessons on Quant?? Please suggest.

Re: Percentage and its applications
by Pravin Vaidya - Monday, 24 May 2010, 01:27 AM
 

I feel there is a typo error in the explanation given for percentage increase/decrease.
It has been highlighted below in BOLD style.

--------------------------------------------------------------------------------Percentage increase/decrease when a quantity a increase/decreases to become another quantity b is

Where , [(a-b)/a]*100 when a<b (decrease)  should be
        [(a-b)/a]*100 when b<a (decrease)

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

Re: Percentage and its applications
by vishnu rajelekshmmi - Friday, 24 September 2010, 09:05 PM
  how x+y=14/5x????it sholud b 18/5x
Re: Percentage and its applications
by Bhanu Singh - Wednesday, 10 November 2010, 12:14 AM
  try to use simple words please.else it makes difficult to understand.. like.. 25% dearer.. as mentioned in one ques.. it make things bit confusing ..
Re: Percentage and its applications
by mansi goel - Saturday, 29 October 2011, 09:36 AM
  hello Sir..i have a slight problem in the answers to the given ques..these are from Percentages-

1.In a village consisting of p persons, x% can read and write. Of the males alone, y% , and of the females alone z% can read n write. Find the number of males in the village in terms of p,x,y,z and z<y.
options-
a).p(x-z)/(y+x-z) b).p(x-z)/(y+x-2z) c).p(y-x)/(x-z) d). p(x-z)/(y-z)

2.Australia scored a total of x runs in 50 overs. India tied the scores in 20% less overs. If India's average run rate had been 33.33% higher the scores would hav been tied 10 overs earlier. Find how many runs were scored by Australia.
options-
250,240,200,CBD

3.In an election, the total turnout was 80% out of which 16% of the total voters were declared invalid.Find which of the following can be the percentage of the votes got by the winner of the election if the candidate who came second got 20% of the total voters on the voting list.(there were only three contestants, only one winner, and the total number of voters on the voters' list= 20,000)
options-
1.44.8% 2.46.6% 3.48% 4. NOT

4. 10% of mexico's population migrated to South Asia, 10% of the remaining migrated to America and 10 % of the rest migrated to Australia. If the female population , which ws left in mexico, remained only 3,64,500, find the population of mexico b4 migratn and its effects if it is givn that b4 the migratn female population ws half the male population and this ratio did nt change after the migration.
options-
1.10,00,000 2. 12,00,000 3. 15,00,000 4. 16,00,000 5. 12,50,000

please post the approach as well....
thank you!
Re: Percentage and its applications
by Mohit Sharma - Saturday, 21 April 2012, 08:25 PM
  Hi sir,
 Gr8 wrk again.
Please help with the following prob
Ques) In an examination, Mohit obtained 20% more than Sushant but 10% less than Rajesh. If the marks of Sushant is 1080, find the percentage marks obtained by Rajesh if the full marks is 2000.

Please explain sir.
Re: Percentage and its applications
by arsh arora - Saturday, 21 April 2012, 10:42 PM
 

hi mohit..ans) 72%

exp;-sushant=1080 implies mohitz=1296 now using percentage knowledge we know 10% less means 11.11% more which is 1/9 in frction..thereby 1/9 * 1296=144 to be added to 1296 gives 1440 rajeshz marks!!!!!...so 72 %...hope its clear...

Arsh Arora

Re: Percentage and its applications
by Mohit Sharma - Sunday, 22 April 2012, 06:40 PM
  Thanks arsh... Got the point smile
Re: Percentage and its applications
by Prerna Golani - Friday, 11 May 2012, 04:51 PM
  Solution to question No1
let the number of persons be p
let the number of males be m,then mumber of females is (p-m).
No of persons that can RW be px/100.
No of males that can RW be my/100.
No of females that can RW be (p-m)*z/100.
therefore
no of persons that can RW=No of males hat can RW+No of females that can RW
=>px/100=my/100+(p-m)*z/100;
=>px=my+(p-m)z
=>p(x-z)=m(y-z)
=>m=p(x-z)/y-z

correct option is d

solution to question 4
Let the original population be x.
10% migration means population 10/100*x=1/10x;
people left in mexico=x-1/10x=9/10x=0.9x
Similarily Population after 10% migrated to America = 0.9x * 0.9 = 0.81x
Population after 10% of the rest migrated to Australia = 0.81x * 0.9 = 0.729x
Let number of males be m and females be f
f=1/2m=>m=2f--------------(1)

m+f=total population left out=0.729x;
put value of m in above equation
3f=0.729x
now f= 364500
therefore x= 15,00,000
option 3 is correct
Ratio & Proportion
by achal singhal - Tuesday, 26 June 2012, 02:00 PM
 

Hi TG,

Request you to please write on Ratio & Proportions. It would be of great help.

Thanks

Re: Percentage and its applications
by Baskar N - Friday, 14 June 2013, 09:58 AM
  sir
can i take sushants marks as 80% of mohitz..if i take like that i will get mohitzs mark as 1350...pls explain
Re: Percentage and its applications
by maverick 3118 - Monday, 22 July 2013, 09:54 PM
  please explain...the proces for 4th question?the one about salary and bonus !

Re: Percentage and its applications
by maverick 3118 - Monday, 22 July 2013, 10:04 PM
  How to solve the problem of depreciation?Like the problem: an equipment
depreciating 15% in first year,13.5 %in second,12 %in third,depreciation at the end of 10 years?

Re: Percentage and its applications
by maverick 3118 - Monday, 22 July 2013, 10:18 PM
  'd' is the correct reply.M=p(x-z)/y-z

Re: Percentage and its applications
by maverick 3118 - Tuesday, 23 July 2013, 07:11 AM
  Answer to 2nd ques: 200??

Re: Percentage and its applications
by maverick 3118 - Tuesday, 23 July 2013, 07:55 AM
  ans 4>1500000

Re: Percentage and its applications
by steffy varghese - Friday, 26 July 2013, 02:07 PM
  dear tg sir
i have a doubt in a problem reharding percentage
if the price of the petrol is increased by 25% and kevin intends to spend only 15% on petrol,then by what % must he reduce the quantity of petrol purchased?
a.6.67%
b.10%
c.8%
d.none of these
(can u pls tel me how to solve these kind of sums in an easier way)