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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:

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