## Percentage Type 4 Shortcut Tricks

At first we need to follow some traditional rule then we go through the shortcut tricks. Percentage Type 4 Shortcut Tricks is based on, to find the number of valid votes that the other candidate got, Percentage is a very important chapter and it uses most chapters for calculation. some problems are based on Relation between Percentage or x and y.

This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

when we say 100 percent in mathematical notation we write 100% so 35 % means 35 per 100 and 65% means 65 per 100 it is a proportion per hundred and it is used to find marks, profit percent or loss percent of a particular product. it is also used in find out the depreciation at the rate percent per annul.

we can expressed ration number as percent, Here is some different percent related problems are given with shortcut tricks below.

### Example 1 :

Sanjay got 88 marks in Hindi, 81 marks in Science, 74 marks in Maths, 68 marks in Social Science and 57 marks in History. The maximum marks of each subject is 100. How much overall percentage of marks did he get ?
Percentage = 368 x 100 / 500 = 73.6.

### Example 2 :

In a class test, it is required to get 45% marks to pass. Joy got 618 marks and failed by 57 marks. What is the maximum marks in class test ?
Joy got 618 marks and failed by 57 marks
X x 45% = 618 + 57
X x 45% = 675
X = 675 x 100 / 45 = 1500.

### Example 3 :

A engineering student has to secure 60% marks to pass. He gets 70 and fails by 50 marks. Find the maximum marks.
He gets 70 and fails by 50
So , 60% = 70 + 50
60% = 120
100% = 120 x 100 / 60 = 200 marks
So , the maximum marks is 200.

### Example 4 :

The average marks of Rahim in 9 subject is 68. His average marks in 8 subjects except Math is 65. How many marks did he get in Math ?
( 68 x 9 ) – ( 65 x 8 ) = ?
612 – 520 = 92.

### Example 5 :

Ranjan got 82 marks in Math, 78 marks in Physics, 65 marks in Computer, 68 marks in English, The maximum marks of each subject are 85. How much overall percetage of marks did ranjan get ?
Total marks get in all subject ( 82 + 78 + 65 + 68 ) = 293
Maximum marks ( 85 x 4 ) = 340
Percentage = 293 x 100 / 340 = 86.17.

### Example 6 :

A student scores 25% & failed by 35 marks while another student who scores 65% get 45 marks more than minimum required marks to pass. Find the maximum marks in the exam ?
25% – 35 = 65% + 45
65% – 25% = 45 + 35
40% = 80
100% = 80 x 100 / 40 = 200 marks.

### Example 7 :

If N is equals to 20% of M and P is equals to 30% of N, then which one of the following equals to 40% of P ?
N = 20% of M = ( 20 / 100 ) x M = 0.2 M.
P = 30% of N = ( 30 / 100 ) x N = 0.3 N = 0.3 x 0.2 M.
So here 40% of P = ( 40 / 100 ) x P = ( 0.4 )( 0.3 )( 0.2 M )
= 0.024 M.

### Example 8 :

In an examination it is required to get 57% of the aggregate marks to pass. A student gets 237 marks and is declared failed by 7% marks. What are the maximum aggregate marks a student can get ?
57% = 237 + 7%
57% – 7% = 237
50% = 237
100% = 237 x 100 / 50 = 474.

### Example 9 :

A student has obtain 34% of the total marks to pass in paper. He got 113 and failed by 40 marks. The maximum marks are :
Let the maximum number is X
Then, 34% of X = 113 + 40
34 / 100 x X = 153
X =153 x 100 / 34
X = 15300 / 34 = 450.

### Example 10 :

Rajah has to score 60% to pass exam. He scores 225 marks & failed by 15%. Find the maximum marks of exam.