Percentage Quantitative Aptitude FOR RRB

ssc preparation, important competetion questions, current affairs of india, latest question asked in competetion papers,


Percentage Quantitative Aptitude PDF
The word “percent” is derived from the latin words “per centum”, which means “per hundred”.
  • A percentage is a fraction with denominator hundred, It is denoted by the symbol %.
  • Numerator of the fraction is called the rate per cent.
VALUE OF PERCENTAGE:
Value of percentage always depends on the quantity to which it refers: Consider the statement, “65% of the students in this class are boys”. From the context, it is understood that boys from 65% of the total number of students in the class. To know the value of 65%, the value of the total number of student should be known. If the total number of student is 200, then,
The number of boys =130;
It can also be written as (200) × (0.65) =130.
Note that the expressions 6%, 63%, 72%, 155% etc. Do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.
To express the fraction equivalent to %:
Express the fraction with the denominator 100, then the numerator is the answer.


Example 1:
Express the fraction 11/12 into the per cent.
Solution:
11/12=(11/12×100)/100=(91 2/3)/100=912/3%
To express % equivalent to fraction:
a% =a/100
Example 2:
Express 45 5/6% into fraction.
Solution:
45 5/6% = (45 5/6)/100=275/(6×100)=11/24.
Example 3:
Rent of the house is increased from ` 7000 to `7700. Express the increase in price as a percentage of the original rent.
Solution:
Increase value = Rs 7700 – Rs 7000 = Rs 700
Increase % = (Increas value)/(Original value)×100= 700/7000×100=10
∴ Percentage rise = 10%
Example 4:
The cost of a bike last year was Rs19000. Its cost this year is Rs 17000. Find the per cent decrease in its cost.
Decrease % = (Decreas value)/(Original value) × 100
% decrease = (19000-17000)/19000×100
=2000/19000×100= 10.5%.
∴ Percentage decrease = 10.5%.
If A is x % if C and B is y % of C, then A is x/y × 100% of B.
Example 5:
A positive number is divided by 5 instead of being multiplied by 5. By what per cent is the result of the required correct value?
Solution:
Let the number be 1, then the correct answer = 5
The incorrect answer that was obtained =.
 The required % =  = 4%
If two numbers are respectively x% and y% more than a third number, then the first number is % of the second and the second is % of the first.
If two numbers are respectively x% and y% less than a third number, then the first number if % of the second and the second is % of the first.
x% of a quantity is taken by the first, y% of the remaining is taken by the second and    z% of the remaining is taken by third person. Now, if A is left in the fund, then the  initial amount
=(A×100×100×100)/((100-x)(100-y)(100-z)) in the beginning.
x% of a quantity is added. Again, y% of the increased quantity is added. Again z% of the increased quantity is added. Now it becomes A, then the initial amount
=(A×100×100×100)/((100+x)(100+y)(100+z))
Example 6:
3.5% income is taken as tax and 12.5% of the remaining is saved. This leaves Rs. 4,053 to spend. What is the income?
Solution:
By direct method,
Income = (4053×100×100)/((100-3.5)(100-12.5)) = Rs 4800.
If the price of a commodity increases by r%, then reduction in consumption, so as not to increase the expenditure is (r/(100+r)×100)%.
If the price of a commodity decreases by r%, then the increase in consumption, so as not to decrease the expenditure is (r/(100-r)×100)%.
Example 7:
          If the price of coal be raised by 20%, then find by how much a householder must reduce his consumption of this commodity so as not to increase his expenditure?
Solution:
          Reduction in consumption = (20/(100+20)×100)%
          = (20/(100+20)×100)% = 16.67%
POPULATION FORMULA
If the original population of a town is P, and the annual increase is r%, then the population after n years is P(1+r/100)^n and population before n years = P/(1+r/100)^n
If the annual decrease be r%, then the population after n years is P(1-r/100)^n and population before n years = P/(1+r/100)^n 
Example 8:
The population of a certain town increased at a certain rate per cent annum. Now it is 456976. Four years ago, it was 390625. What will it be 2 years hence?
Solution:
Suppose the population increases at r% per annum. Then, 390625 (1+r/100)^4 = 456976
           (1+r/100)^2 = √(456976/390625)= 676/625
            Population 2 years hence = 456976 (1+r/100)^2
            = 456976 × 676/625 = 494265 approximately. 
Example 9:
            The population of a city increase at the rate of 4% per annum. There is an additional annual increase of 1% in the population due to the influx of job seekers. Find percentage increase in the population after 2 years.
Solution:
The net annual increase = 5%
Let the initial population be 100.
Then, population after 2 years = 100×1.05×1.05 = 110.25
Therefore, % increase in population = (110.25-100) = 10.25%
If a number A is increased successively by x% followed by y% and then z%, then the final value of A will be A(1+x/100)(1+y/100)(1+z/100)
In case a given value decreases by an percentage then we will use negative sign before that.
First Increase and then decrease:
If the value is first increased by x% and then decreased by y% then there is (x-y-xy/100)% increase or decrease, according to the +ve or –ve sign respectively.
If the value is first increased by x% and then decreased by x% then there is only decrease which is equal to (x^2/100).
Example 10:
            A number is increased by 10% and then it is decreased by 10%. Find the net increase or decrease per cent.
Solution:
% change = (10×10)/100=1%
            i.e. 1% decrease.
Average percentage rate of change over a period.
            =((New Value-Old Value))/(Old Value)×100/n% where n = period.
The percentage error = (The Error)/(True Value)×100%
SUCCESSIVE INCREASE OR DECREASE
In the value is increased successively by x% and y% then the final increase is given     by (x+y+xy/100)%
In the value is decreased successively by x% and y% then the final decrease is given     by (-x-y-xy/100)%
Example 11:
            The price of a car is decreased by 10% and    20% in two successive years. What per cent    of price of a car is decreased after two      years?
Solution:
            Put x = -10 and y = -20, then
            -10-20+
             The price of the car decreases by 28%.
STUDENT AND MARKS
The percentage of passing marks in an examination is x%. If a candidate who scores y marks fails by z marks, then the maximum marks M = 100(y+z)/x
A candidate scoring x% in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more then the minimum required passing marks. Then the maximum marks M = 100(a+b)/x
In an examination x% and y% students respectively fail in two different subjects while z% students fail in both subjects then the % age of student who pass in both the subjects will be {100-(x + y – z)}%
Example 12:
            Vishal requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum he could have got?
Solution:
            If Vishal has 15 marks more, he could have scored 40% marks.
            Now, 15 marks more then 185 is 185+15 =     200
            Let the maximum marks be x, then 40% of x = 200
            ⇒ × x = 200  ⇒ x =500
            Thus, maximum marks = 500
Alternate method:
            Maximum marks =  (100(185+15))/40=(100×200)/40 = 500
Example 13:
           A candidate scores 15% and fails by 30 marks, while another candidate who scores 40% marks, gets 20 marks more then the minimum required marks to pass the pass the examination. Find the maximum marks of the examination.
Solution:
            By short cut method:
            Maximum marks = (100(30+20))/(40-15) =200

2-DIMENSIONAL FIGURE AND AREA
If the sides of a triangle, square, rectangle, rhombus or radius of a circle are increased by a%, its area is increased by(a(a+200))/100 %
If the sides of a triangle, square, rectangle, rhombus or radius of a circle are decreased by a %
            Then its area is decreased by (a(200-a))/100%.
Example 14:
            If the radius of a circle is increased by 10%, what is the percentage increase in its area?
Solution:
Let R be the radius of circle.
Area of Circle, A =πR^2
Now, radius is increased by 10%
New radius, R’ = R + 10% of R = 1.1 R
New Area, A’ = π(1.1R)^2= 1.21 πR^2%
increase in area =(1.21πR^2-πR^2)/(πR^2 )×100=21%
Shortcut Method:
Radius is increases by 10%.
So, Area is increased by (10(10+200))/100 = 21%
If the both sides of rectangle are changed by x% and y% respectively, then % effect on area = x + y+xy/100 (+/- according to increase or decrease.
Example 15:
            If the length and width of a rectangular garden were each increased by 20%, then what would be the per cent increase in the area of the garden?
Solution:
By direct formula
% increase in area =(20 (20+200))/100=44%
If A’s income is r% more than that of B, then B’s income is less than that of A by (r/(100+r)×100)%
If A’s income is r% less than that of B, then B’s income is more than that of A by (r/(100-r)×100)%
Example 16:
            If A’s salary is 50% more than B’s, then by what percent B’s salary is less than A’s salary?
Soluti
Let B’s salary be Rs x
Then, A’s salary = x + 50% of x = 1.5x
B’s salary is less than A’s salary by ((1.5x-x)/1.5x×100)% = 100/3 = 33.33%
Shortcut method,
B’s salary is less than A’s salary by (50/(100+50)×100)%
=50/150×100% = 33.33%
Example 17:
            Ravi’s weight is 25% that of Meena’s and 40% that of Tara’s. What percentage of Tara’s weight is Meena’s weight.
Solution:
Let Meena’s weight be x kg and Tara’s weight be y kg. Then Ravi’s weight = 25% of Meena’s weight
= 25/100×x …..(i)
Also, Ravi’s weight = 40% of Tara’s weight
= 40/100×y …..(ii)
From (i) and (ii), we get
25/100×x=40/100×y
⇒ 25x = 40y
⇒ 5x = 8y ⇒ x = 8/5 y
Meena’s weight as the percentage of Tara’s weight
= x/y×100= ( 8/5 y)/y×100
= 8/5×100=160
Hence, Meena’s weight is 160% of Tara’s weight.
Example 18:
The monthly salaries of A and B together amount to `50,000. A spends 80% of his salary and B spends 70% of his salary. If now their saving are the same, then find the salaries of A and B.
Solution:
Let A’s salary by x, then B’s salary (50,000-x)
A spends 80% of his salary and saves 20%
B spends 70% of his salary and saves 30%
Given that
20% of x = 30% of (50,000-x)
20/100×x=30/100×(50,000-x)
50x/100=(30×50,000)/100
⇒ x = (30×50,000×100)/(100×50)=30,000
A’s salary Rs 30,000
B’s salary = Rs 50,000 – Rs 30,000 = Rs20,000

Comments

Popular From Month

Current Affairs 30th March 2017

India's wetland

Public Sector Banks/Private Sector Banks/Foreign Banks/Regional Rural Banks taglines and head name

Important Rivers Of India Part -2

भारत रत्न पुरस्कार विजेता व वर्ष

Indian History

Bank/Financial Department | Railway | SSC | UPSC | Defence | Andaman & Nicobar | Andhra Pradesh | Arunachal Pradesh | Assam | Bihar | Chandigarh | Chhattisgarh | Dama & Diu | Dadar & Nagar | Delhi | Goa | Gujarat | Haryana | Himachal Pradesh | J & K | Jharkhand | Karnataka | Kerala | Lakshadweep | Maharashtra | Mizoram | Madhya Pradesh | Manipur | Meghalaya | Nagaland | Odisha | Punjab | Puducherry | Rajasthan | Sikkim | Tamil Nadu | Telangana | Tripura | Uttar Pradesh | Uttarakhand| West Bengal | PSUs | Institute Jobs

Cabinet Ministers

Govt Jobs 2018 in India Updated on 31.03.2018