Home » Percentage Problems for Bank Exams – Speed Maths Tricks

Percentage Problems for Bank Exams – Speed Maths Tricks

Percentage Shortcuts for Bank Exams

The percentage is a fraction, whose denominator is 100. The term percent means for every 100

To solve the percentages problems we have to memorize some basic percentage values to do fast calculations in competitve exams.

  • 1/2 = 50%
  • 1/3 = 33.33%
  • ¼ = 25%
  • 1/5 = 20%
  • 1/6 =16.66%
  • 1/7 = 14.28%
  • 1/8 = 12.5%
  • 1/9 = 11.11%
  • 1/10 = 10%
  • 1/11 =9.09 %
  • 1/12 = 8.33%
  • 1/13 = 7.69%
  • 1/14 = 7%
  • 1/15 = 6.66%
  • 1/16 = 6.25%
  • 1/17 = 5.88%
  • 1/18 =5.55%
  • 1/19 = 5.26%
  • 1/20= 5%
  • 5%= 1/20 = 0.05
  • 10% =1/10 =0.1
  • 15% =3/20
  • 20% = 1/5
  • 25% = ¼
  • 30% =3/10
  • 40% =2/5
  • 50% =1/2
  • 55%=11/20
  • 60% = 3/5
  • 70% = 7/10
  • 75% = 3/4
  • 80% = 4/5
  • 90% = 9/10
  • 100%=1
  • 6 ¼ % = 1/6
  • 12 ½ % = 1/8
  • 16 2/3 % = 1/6
  • 33 1/3 % = 1/3
  • 66 2/3 % =2/3
  • 125% =5/4
  • 150% = 3/2

Percentage Change = [(final value – Initial Value)/Initial value]*100

If the price of item increases by R% Then the reduction in consumption as not to increase expenses is [R/ (100+R) ]* 100

If the price of the item decreases by R% Then the increase in consumption [R/(100-R)]*100.

Population:

After n years = P*[1+R/100]n

N years ago = P/ [(1+R/100)]n

Machine:

Value after n years = p * (1- R/100 )n

Value n years ago = P / (1- R/100)n

 If  A is R% more than B, Then B is less than A by [R/(100+R)] *100.

If A is R% less than B, Then B is more than A by [R/100-R ]* 100.

Smart Ways to Solve Percentages:

X is what % of more or less  than y?

Sol:  [(x-y) / y] * 100

 Eg: 72 is what % of 360?

(72/360)*100 = 20%

We have to think like this

10% of 360= 36 and then 20% of 360 =72

Eg: (503/ 800)*100

Sol: To solve this we have to think and segment  like this way.

50% of 800=400

25% of 800=200

10% of 800=80

1% of 800=8

Here we have 503= 400+80+23 i.e. 50%+10%+(1% *3) i.e. 63% (approx)

Eg: (1296/3600)*100

Sol: To solve this we have to think and segment  like this way.

50% of 3600 = 1800

25% of 3600 = 900

10% of 3600= 360

5% of 3600 = 180

1% of 3600 = 36

To get 1296 we have to add 900+360+36 i.e. 25% +10% +1% =36%.

Percentage Problems for Bank Exams – Speed Maths Tricks

Q.1)Nandini goes to a shop to buy a table costing Rs.2568. The rate of sales tax is 7%. She tells the shopkeeper to reduce the price of the table to such an extent that she has to pay Rs.2568 inclusive of sales tax. How much reduction is needed at the price of the table?

Solution :  Original price (100%) = Rs. 2568 sales tax = 7%

     she purchased including sales tax i.e 107% = Rs. 2568

                                                                     7%  = ?

=> (2568 * 7)/ 107  => 24 * 7

=> Rs. 168.

Q.2) If on selling 12 pencils any seller makes a profit equal to the selling of 4 pencil boxes.What is his percent profit?

Note: Use Shortcut formula wherever we can able to use. it saves time for solving.

Sol :   [4/(12-4)]*100 = [4/8]*100 = 50%.


Q.3) If the two numbers are respectively 20% and 50% of a third number, what is the percentage of the first number to the second number?

Sol: Let the 3rd  number be 100.

Then, the first and second numbers will be 20 and 50, respectively.

Required % = [ 20/50]*100 =40.

Q.4) Three successive discounts of 10%, 12% and 15% amount to a single discount of

Ans:  use this formula [x+y+ x *y/100]%. Here , discount means decreasing value so

we have to use as follows : (-x) + (-y) + (-x * -y)/100.

Step 1:

-10 – 12 +120/100 = -22+6/5 = -22 +1.2 =20.8

Step 2:

-20.8 – 15 + (-20.8 * -15)/100 = – 35.8 + ( 312/100) = -35.8 +3.12 = 32.68



5) In a vessel of fruits, there are 25% of apples, 40% of  oranges and the rest are mangoes. If there are 2000 fruits in the vessel. how many mangoes are there in the vessel?

Sol:   Total fruits = 100% = 2000

         Mangoes = Total Fruits – (apples + Oranges) = 100% – (25% + 40%)

                          = 100 – 65 = 35%.

Given 100%  = 2000

            35% = ?

=> (35 * 2000)/100 = 700.

 

6) Usha Deposits, Rs.500 in the postal savings account. If this is 20% of her monthly income. What is her monthly income?

Solution:

Here, Monthly income  = 100%

From the problem , we get            20%   =  500.

                                     Then              100% = ?

By cross multiplication,

     (100* 500 )/20  = 2500.

 

7) In an Examination, a candidate who scores 25% of the maximum marks fails by 60% marks. Another candidate who secures 42% of the maximum marks gets 8 marks more than  necessary for passing. Find the maximum marks and the percentage necessary for passing.

Sol:   m/4  + 60  = (42m/ 100) -8.

   => 42m/100 – m/4 =  60 + 8.

   => 42m – 25m  = 68 * 100

=> m = 400.

Minimum marks for passing = 25% of 400 +60 = 160.

% necessary for passing = (pass marks/ total marks ) * 100

                                            =( 160/400 ) * 100 = 40%

8) If both length and breadth of a rectangle are increased by 50% each What would be the net increase in its area?

Solution:

We have the shortcut formula to find the net percentage.
a + b + ab/100 i.e. 50 + 50 + 2500/100 = 125

9) Because of the power shortage, a factory has to work only for 80% of the time it used to . By what percent must the output per worker increase. If the total output is to be the same as before?

Solution: ( Use this, If the price of the item decreases by R% Then the increase in consumption [R/(100-R)]*100.)

Decreased 20%,  [20/100 -20]* 100 = 1/4 *100 = 25%

 [20/100 -20]* 100 = 1/4 *100 = 25%

10) A and B respectively secure 30% and 20% fewer marks than C. A’s Marks are What Percent of B’s Marks?

Solution:

 Let us assume C = 100 marks

Then, A = 70  and B =80

Required percentage  = (70/ 80) * 100  = 87  1/2.

11) If the population of the village of 10,000 people increases by 10% every year for two years and then decreases by 10% every year for the next two years. What would be the population of the village four years from now?

Solution:  (Use this formula 

Population Increase for n years:

After n years = P*[1+R/100]n      

After 2 years = 10000(1+10/100)^2  = 110 * 110 = 12100.

Population Decrease for n years:

After n years = P*[1  – R/100]n

After next 2 years = 12100(1 – 10/100)^2  = 90 * 90  *1.21= 9801.

12) When a discount of 20% offered by  a  trader, his sales increased by 80%. What was the effect on his receipts?

Solution:

Formula =  – a + b  – ab/100 

 Net percentage   =  – discount% + increment %  – (discount%) (increment%)/ 100

          = -20 + 80 -1600/100 = -20 + 80-6 = +44% that means 44% increased.

 

 

error: Content is protected !!