(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.