Percentage Problem Shortcut Tricks Bank PO,SSC,Railways Exam
Common Type Percentage Problem That Is Frequently Asked in BANK PO,SSC,Railways Exam
(Q1)Due to increase in the price of sugar by 20%.the consumption reduces by 30%.then find the percentage effect on total expenditure from it.
Solution:
%Effect=(16×100)/100
=16% Ans
CONCEPT:[यदि Price and Quantity/Consumption दोनों में change हो रहा हो तो Price and Quantity/Consumption को 100 मानना(let) है]
BY TRICK:
%Effect=(20-30-(20×30)/100)%
=-16%
=16% Decrease
Note:[-ve sign means decreases]
(Q2)The price of an article is reduced by 25% but the daily sale of the article is increased by 30%.the net effect on the daily sale receipts is.
Solution:
Net effect=(1×100)/40
=2½% Ans
BY TRICK:
Net effect=(-25+30-(25×30)/100)
=-2½% [-ve sign means decreases]
=2½% Ans
(Q3)If side of a square is increased by 10% then the percentage change in its area will be?
Solution:Area of square=side×side
%increase or change=(21×100)/100
=21% Ans
BY TRICK:
% change=(10+10+(10×10)/100)%
=(20+1)%
=21% Ans.
(Q4)If the radius of a circle is increased by 25% then area of circle will be increased by.
Solution:Area of circle=πr² [where r=radius]
% increase=(9×100)/16
=56.25% Ans.
BY TRICK:
%Increase=(25+25+(25×25)/100)%
=50+(25/4)%
=56.25% Ans.
(Q5)If radius is increased by 10% and height is decreased by 20% of cylinder. then find the percent change in volume of the cylinder.
Solution:Volume of cylinder=πr²h [where r=radius and h=height]
% change or increase=(32×100)/1000
=3.2% Ans
BY TRICK:
(10+10+(10×10)/100)%
now % change
=(21-20-(21×20)/100)%
=1-42/10
=-32/10
=-3.2%
i.e 3.2% decreases
(Q6)The percentage increases in the surface area of a cube when side is doubled .
Solution:Surface area of cube=6a²
% increase=(3×100)/1
=300% Ans
(Q7)The numerator of a fraction is increased by 20% and denominator is decreased by 20%.The value of the fraction becomes 4/5.The original fraction is.
Solution:Let original fraction =X/Y [where X=numerator and Y=denominator]
Now,
6X/4Y=4/5
=>30X=16Y
∴X/Y=16/30
=8/15 Ans.
(Q8)The price of an article was first increased by 10% and then again by 20%.If the last increased price be Rs 33,The original price was.
Solution:
% increase=(32×100)/100
=32% Ans.
Now,
Let original price=x
∴ x=(33×100)/132
=Rs 25 Ans.
BY TRICK:
% increase=(10+20+(10×20)100)%
= 32%
Now
Let original price=x
x×(100+32)/100=33
∴ x=(33×100)/132
=Rs 25 Ans.
(Q9)The cost of an article was Rs 75.The cost was first increased by 20% and later on it was reduced by 20%. The present cost of the article is.
Solution:
%Decrease=(1×100)/25
=4% Ans.
Now present cost of the article is:
Rough:
Rs 75 increased by 20%
i.e 75×(100+20)/100
=(75×120)/100
=Rs 90
Rs 75 Reduced by 20%
i.e 75×(100-20)/100
=(75×80)/100
=Rs 60
BY TRICK:
(20-20-(20×20)/100)%
=-4%=4% decrease
Now present cost of the article is:
(Q10)The difference between the value of the number increased by 20% and the value of the number decreased by 25% is 36. Find the number.
Solution:Let number =100x
now,
120x-75x=36
∴ 45x=36
∴ x=36/45
Number=100x=100×(36/45)
=80 Ans.
(Q11)A number is first decreased by 20%.The decreased number is then increased by 20%.The resulting number is less than the original number by 20.then the original number is.
Solution:
now,
100x-96x=20
4x=20
∴ x=5
original number=100x
=100×5
=500 Ans.
I have already shared a percentage problem type-1,type-2,type-3 and type-4 if you have not read that yet,you should do it the right way here is the link.
If you have any queries regarding this topic fell free to ask me in the comments below.
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Best examples and nicely explained
It is very helpful. Sir your way of explanation is very good.
Thanks rahul.