Basically, a percent is a simple
ratio times 100. CAT will ask you to calculate percent changes, and here you have to be very careful with order, i.e., what’s the starting number and what’s the ending number. IMPORTANT:
in a percent change, the starting number is always 100%. Thus, we can say:
Can you quickly calculate 15% of 42 in your head? In this lesson, we’ll examine a fast way to perform this calculation and others.
The technique I’ll demonstrate is based on the fact that it’s incredibly easy to find 10% of any value, and 1% of any value.
Finding 10% of a value
You probably already know that the “trick” is to move the decimal point one space to the left. Some examples:
10% of 5.2 = 0.52
10% of 4,321 = 432.1
10% of 837,160 = 83,716
Finding 1% of a value
To find 1% of a value, just move the decimal point two spaces to the left. Some examples:
1% of 5.2 = 0.052
1% of 4,321 = 43.21
1% of 837,160 = 8,371.6
Once we know how to find 10% and 1%, we can apply some number sense to quickly find other percents.
Finding 5% of a value
If we can find 10% of y, then 5% of y will equal half of 10% of y.
For example, since 10% of 240 = 24, we know that 5% of 240 is half of 24. In other words, 5% of 240 = 12
Likewise, since 10% of 3.6 = 0.36, we know that 5% of 3.6 = 0.18 (i.e., half of 0.36)
Finding 15% of a value
Now that we’re experts at finding 10% and 5% in our heads, we can combine percents. For example, let’s find 15% of 260
To mentally perform this calculation, we need to recognize that 15% = 10% + 5%.
10% of 260 = 26
And 5% of 260 = 13
So, 15% of 260 = 26 + 13 = 39
Here’s another one: 15% of 42
10% of 42 = 4.2
5% of 42 = 2.1
So, 15% of 42 = 4.2 + 2.1 = 6.3
Combining percents
Now that we’ve found 15% by combining 10% and 5%, we can use the same approach to find other percents. We need only take the required percent and break it into sums of 10% and/or 1%.
Let’s begin by finding 2% of 0.31
To perform this calculation, we’ll use the fact 2% = 1% + 1%
1% of 0.31 = 0.0031
1% of 0.31 = 0.0031
So, 2% of 0.31 = 0.0031 + 0.0031 = 0.0062
Aside: This technique can prevent test-takers from making careless errors. When students find 2% of 0.31 by calculating (0.02)(0.31), there’s a chance that they’ll misplace the decimal point in the final answer. When we find 1% by moving the decimal point 2 spaces to the left, such mistakes are less likely.
Okay, now try 21% of 210
Here, we’ll use the fact 21% = 10% + 10% + 1%
10% of 210 = 21
10% of 210 = 21
1% of 210 = 2.1
So, 21% of 210 = 21 + 21 + 2.1 = 44.1
Last one: 55% of 70
One approach is the recognize that 55% = 10% + 10% + 10% + 10% + 10% + 5%, however a faster approach is to recognize that 55% = 50% + 5%.
50% of 70 = 35
5% of 70 = 3.5
So, 55% of 70 = 35 + 3.5 = 38.5
This technique takes only a few minutes to master, and it can save you valuable time on test day.
Now try the following calculations in your head (scroll down the page to find the answers):
a) 15% of 90
b) 11% of 170
c) 3% of 11,000
d) 115% of 82
Answer key
a) 13.5
b) 18.7
c) 330
d) 94.3
MUST KNOW SHORTCUT FORMULA’s :
Percentage is a fraction whose denominator is always 100. x percentage is represented by x%.
To express x% as a fraction :
We know
x% = x/100
Thus 10% = 10/100 (means 10 parts out of 100 parts)
= 1/10 (means 1 part out of 10 parts)
To express x/y as a percentage :
We know that x/y = (x/y× 100 )
Thus 1/4 = ( 1/4 ×100 )% = 25%
and 0.8 = ( 8/10 ×100 )% = 80%
If the price of a commodity increases by R%, then reduction in consumption 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 as not to decrease the expenditure is –
[ R/(100-R)×100 ] %
Result on Population :
Let the population of a town be P now and suppose increases the rate of R% per annum, then :
Population after n years = P ( 1+ R/100 )n
Population n years ago = P /( 1+ R/100 )n
Result on Depreciation :
Let the present value of a machine be P. Suppose depreciates at the rate of R% per annum Then :
1. Value of the machine after n Years
= P ( 1- R/100 )n
2. Value of the machine 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 ]%
Net % change = x + y + xy/100
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