In an election 4000 votes were cast.One of the candidates got 40% votes.He was defeated by how m any votes?
a]2400  b]1600  c]800  d]600

Solution:
candidate who lost got 40% of 4000= 40/100 * 4000 = 1600
difference of votes = 2400-1600 = 800 votes

If the price of the kerosene was raised by 10%,find by how much percent a householder must reduce his consumption of kerosene so that not to increase his expenditure?

a]28%  b]20%  c]10.08%  d]10%

Raised price of kerosene is 110/100 of the original price
hence consumption should become 100/110 of the original consumption so as to keep the expenditure same
reduction = [1-100/10] of original consumption = 10/100 of original consumption= [10/100] * 100 = 9.09%

1]A towel was 50 cm broad and 100 cm long.When bleached,it was found to have lost 20 percent of its length and 10 percent breadth.The percentage decrease in area is
a] 28 %  b] 20%  c]  10.08%  d] 10%


original area = 100* 50
reduced length = 80% of 100 = 80
reduced breadth  = 90% of 50= 45
reduced area = 80*45
reduction in area = 500 - 3600 = 1400
% reduction = [400/5000] 100 = 28 %

 In an election between two candidates, A and B, A secured 56% of the votes and won by 48000 votes. Find the total number of votes polled if 20% of the votes were declared invalid.
a. 500000        b. 400000                    c. 600000        d. None

Valid Votes:
A got 56%  =>  B got 44%
Difference = 12% = 48000
So, 100% = 400000. These are valid votes.
But valid votes are only 80% of total votes.
So, 80% of total votes  = 400000  =>  total votes = 500000

A reduction of 10% in price of sugar enables a housewife to buy 5 kg more for Rs. 300/-. Find the reduced price per kg of sugar.
a. 5/-                b. 4.5/-                         c. 6/-                d. None

Total money = Rs. 300.
Saving of the lady = 10% of 300 = 30/-
With 30/- she bought 5 kg sugar => each kg costs Rs. 6/-

 From a 20lt solution of alt and water with 20% salt, 2lt of water is evaporated. Find the new % concentration of salt.
a. 20%             b. 23%                         c. 25%             d. None


In 20lt, salt = 20% => 4 lt.
New volume = 18 lt (2 lt evaporated)
So, new % = 4/18 X 100 = 22.22%

 In a list of weights of candidates appearing for police selections, the weight of A is marked as 58 kg instead of 46.4 kg. Find the percentage of correction required.
a. 30                b. 20                            c. 24                d. None



% correction = (58-46.4)/58 X 100 = 20%

A person spends 20% of his income on rent, 20% of the rest on food, 10% of the remaining on clothes and 10% on groceries. If he is left with Rs. 9520/- find his income.
a. 10000/-                    b. 15000/-                    c. 20000/-        d. None


Three successive decreases of 20%, 20% and 10% => 0.8 X 0.8 X 0.9 = 0.576
Again 10% decrease => 0.576 – 0.1 = 0.476.
So, 0.476 x = 9520 => x = 20000.

 A shopkeeper offers three successive discounts of 10%, 20% and 30% to a customer. If the actual price of the item is Rs. 10000, find the price the custome has to pay to the shopkeeper.
a. 5040/-                      b. 4000/-                      c. 6000/-          d. None
Total discount = 0.9 X 0.8 X 0.7 = 0.504 of actual price.
So, price = 0.504 X 10000 = 5040.

 If  10lt solution of water and alcohol containing 10% alcohol is to be made 20% alcohol solution, find the volume of alcohol to be added.
a. 1 lt                            b. 1.25 lt                      c. 1.5 lt             d. 2 lt


In 10 lt, alcohol is 10% = 1 lt.
Let x lt alcohol is added.
So, (1+x)/(10+x) = 20% = 1/5  =>  x = 1.25 lt.

A is twice B and B is 200% more than C. By what percent is A more than C?
a. 400                          b. 600                          c. 500              d. 200

A = 2B and B = 3C (ince 200% more)
A = 6C  =>  500 % more.

 In an examination, a student secures 40% and fails by 10 marks. If he scored 50%, he would pass by 15 marks. Find the minimum marks required to pass the exam.
a. 250                          b. 100                          c. 110              d. 125

50% of max marks – 40% of max marks = 25
 max marks = 250
Pass marks = 40% of max + 10 => 100 + 10 = 110.

If A is 20% taller than B, by what percent is B shorter than A?
a. 20%                         b. 25%                         c. 16.66%        d. None

A = 1.2 B
=> B = A/1.2
 => 0.8333A => 16.66%.
(OR) Decrease from 1.2 to 1 => 16.66%.

 

The population of a town increases at a rate of 10% for every year. If the present population is 12100, find the population two years ago.
a. 11000                      b. 9800                        c. 10000          d. 10120


1.1 X 1.1 X x = 12100 => x = 10000.

 A solution of salt and water contains 15% salt. If 30 lt water is evaporated from the solution the concentration becomes 20% salt. Find the original volume of the liquid before water evaporated.
a. 100 lt                        b. 120 lt                       c. 200 lt            d. None


let 'x' be the quantity of solution of salt and water
Salt content before evaporation of water= 15% of x = 0.15x (x = volume of solution)
Salt content after evaporation of water = 20% of [x-30] = 1/5  (since 30 lt evaporated)
equating both of them

(15/100) * x = (20/100) * (x-30)
3x/20 = 1/5 * (x - 30)
3x/20 - 1/5x = - 30/5
15x- 20x = - (6*20*5)
- 5x = -(6*20*5)
x = 120 litres

 If 240 lt of oil is poured into a tank, it is still 20% empty. How much more oil is to be poured to fill the tank?
a. 300 lt                        b. 60 lt                         c. 120 lt            d. None


22.20% empty
 => 80 % full = 240 lt
 => 20% = 60 lt

Finding increase in percentage difference by comparing two different values of percentages

A and B were hired for the same salary. A got two 40% hikes whereas B got a 90% hike. What is the percentage difference in the hikes thay got?
a. 16%                         b. 6%                           c. 10%             d. 8%

Solution
A => 1.4 X 1.4 = 1.96
B => 1.9     
1.96 -1.9=0.06
=> 6% difference.

Finding percentage increase,Given the change is the quantity/number over a span of time

The population of a town doubled every 5 years from 1960 to 1975. What is the percentage increase in population in this period?
a. 800                          b. 400              c. 700              d. 600

Solution :From 1960 to 1975, in 15 years population doubled every 5 yrs => three times
So, 2 X 2 X 2 = 8 times => 700% more.

In a test of 80 questions, Jyothsna answered 75% of the first 60 questions correctly. What % of the remaining questions she has to answer correctly so that she can secure an overall percentage of 80 in the test?
a. 80%                         b. 90%             c. 85%             D. 95%                                              
Solution
[(75% X 60) + (x% X 20)] / 80 = 80%  =>  x = 95.   (since required is 80%)
(OR)  60 out of 80 is 3/4. So,  (3/4 X 75)  +  (1/4 X x)  = 80  =>  x =95.

Given the reminder of a number when it is divided by a certain divisor.Finding the reminder if the same number is divided by another divisor

A number when divided by 32 gives a the reminder 29.This number when divided by 8 will leave the reminder
a] 3 b] 5 c] 7 d] 29                                                                                                                     Ans 5

let the number be x,
When it is divided by 32,let the quotient be k and reminder 29
32) x (k
      ---
      29

therefore x = 32k  + 29
                  = [(8 * 4 * k) + (8*6) - 5]
5 is the reminder

Finding percentage increase in area of a circle when the radius is increased by a certain percent

1] If radius of a circle is increased by 100%,then the area is increased by
a] 100%  b] 200%  c]300%  d]400%                                                                                  

Solution
let the original radius be R cm

New radius = {(100+100)/100}R cm = (200/100) R = 2R cm

original area = pi * R^2  cm^2

New area =  pi * (2R)^2  cm^2

change in area = pi *4* R^2 -  pi  * R^2 = 3 * pi * R^2

% increase in area =  3 * pi * R^2 * [1/ pi  * R^2 ] * 100 =300 %                       
                          
Practice questions

2] If radius of a circle is increased by 6%,then the area is increased by
a] 12.36%  b] 12.35%  c]12.34%  d]12.33%         

*3] If radius of a circle is doubled,how many times of the area of original circle will it become now?
a] 100%  b] 200%  c]300%  d]400%       
 
4] If radius of a circle is increased by 1%,what is the increased percent in the area 
a] 1%  b] 2%  c]1.1%  d]2.01%         

    

Solving for unknown numbers

1]Three fourth of a number is 60 more than its one third.The number is
a] 84  b]108  c]144  d]none

Solution
Let the number be 'x'
(3/4) x = 60+(1/3) x
(3/4) x - (1/3) x = 60
(9x-4x)/12 = 60
5x = 12*60
x = 144



2]If one seventh of a number exceeds its eleventh part by 100.The number is
a] 770  b]1100 c]1825  d]1925


3]A number exceeds its four seventh by 18. What is the number ?
a]42   b]24  c]144  d]none


4]A number exceeds its one third by 50. What is the number?
a] 70  b]75  c]144  d]none

5]A number exceeds its three fifth by 14. What is the number?
a] 70  b]35c]144  d]45

  Friends,Math-o-Phobics and Apti enthusiasts..this is a blog dedicated to Mathematical Aptitude.
I have always been weak in Mathematics or should i say i suffered from matho-phobia that completely ruined my otherwise excellent acedemic records starting from my class 7th Karnataka State Public Exams upto my grad exams.Whatever explainations i may give,at the end of the day am just weak in MATH,thats all!
  Being a capricorn,the perfection seeker i am,entire life it gnawed me that i sucked at math! No matter how much i tried i couldnt get out of that phobia and nor did my folks helped me in getting rid of it.It resulted in just pass marks in the plus two and ruined my entire life coz i couldnt go according to my nerdy plans of getting into the course i wanted and the college i wanted! As a result,i hated my college life,my course and ended up in a career i didnt want.Life became an entire mess.i blame everything on my MATH Result even today and will keep doing it till i die.I had math in Grad too which i made sure to pass with good marks out of fear of not getting graduation! That's a different story altogether.
  Many things changed but the yearning to master the math always remained.And hence i kept it as one of the MANY GOALS of my life[capricorns suck i know!].I was hell bound to learn it and master it,no matter how long it took.My perseverance for 8 years has paid and finally am out of Math-o-phobia coz today i am fairly good at aptitude as i easily crack it in the entrance exams and all.
  This blog is for those people who feel mathematical aptitude is an iron nut.Its just a concoction of questions with solutions and practice questions.Solve them and feel confident for entrance exams as there is no escaping this section in any entrance exam at any level.Its a necessary evil.Why not make it a friend by giving some time to it once in a while?
Moral of the story=Most of us suffer from Math-o-phobia,consciously subconsciously.Elders need to pay attention towards it.Given proper guidance and practice time,we can come out of it..!

Thanks for stopping by
Happy math-ing!
Komal :)