Percentage Questions Tricks Part 8th

Let’s continue with this chapter............

Concept 6^th   :- (Miscellaneous Questions based concept.)

Q26. Find a single equivalent increase, if a number is successively increased by 20%, 25% and 30%..

Note:- In such type of questions three successive increments are given, so first we solve first two of them then the result will be solved with the third.

Sol. 1^st Method :- First two increments are 20% and 25%.

= (+20) + (+25) + [{(+20) × (+25)}/100].

= 20 + 25 + (500/100).

= 45 + 5

= 50%.

Now two increments are 50% and 30%.

= (+50) + (+30) + [{(+50) × (+30)}/100].

= 50 + 30 + (1500/100).

= 80 + 15

= 95%.

Therefore the final equivalent increment = 95%.

2^nd Method :- let the number be 100.

1^st increment = 20%.

Now the number becomes [100 × {(100 + 20)/100] = 100 × (120/100) = 120.

Now the number =120.

2^nd increment = 25%.

Now the number becomes [120 × {(100 + 25)/100] = 120 × (125/100) = 150.

Now the number =150.

3^rd increment = 30%.

Now the number becomes [150 × {(100 + 30)/100] = 150 × (130/100) = 195.

Now the number = 195.

Total increment = 195 – 100 = 95.

Therefore the final equivalent increment = 95%.

Method 3^rd :- Direct Formula method

If there are three Successive x%, y% and z% (increases or decreases) then the resultant increase or decrease is given by :-

(± x) + (±y) + (±z) + [{(±x)×(±y)}/100] + [{(±y)×(±z)}/100] + [{(±z)×(±x)}/100] + [{(±x)×(±y)×(±z)}/(100×100)].

Now according to the question three successive increments are 20%, 25% and 30%.

= (+20) + (+25) + (+30) + {(+20)×(+25)/100} + {(+25)×(+30)/100} + {(+30)×(+20)/100} + {(+20)×(+25)×(+30)/(100×100)}.

= 20 + 25 + 30 + (500/100) + (750/100) + (600/100) + (15000/10000).

= 75 + 5 + 7.5 + 6 + 1.5 = 95

Therefore the final equivalent increment = 95%.

Q27. Find a single discount equivalent to a discount series of 10%, 15%, and 20%.

Sol. Analyse the given data :- we will consider negative (-) sign for discount.

First we will solve first two discounts 10% and 15%.

= (-10) + (-15) + [{(-10) × (-15)}/100].

= -10 – 15 + (150/100)

= - 25 + 1.5

= - 23.5%

= 23.5% discount.

Now discounts are 23.5% and 20%.

= (-23.5) + (-20) + [{(-23.5) × (-20)}/100].

= - 23.5 – 20 + (470/100)

= - 43.5 + 4.7

= - 38.8%

= 38.8% discount.

Hence the equivalent discount for the given series = 38.8%.

2^nd Method :- let the selling price be 100.

1^st discount = 10%.

Now the selling price becomes [100 × {(100 - 10)/100] = 100 × (90/100) = 90.

Now the selling price =90.

2^nd discount = 15%.

Now the selling price becomes [90 × {(100 - 15)/100] = 90 × (85/100) = 76.5.

Now the selling price =76.5.

3^rd discount = 20%.

Now the selling price becomes [76.5 × {(100 - 20)/100] = 76.5 × (80/100) = 61.2

Now the selling price = 61.2

Total discount = 100 – 61.2 = 38.8

Therefore the final equivalent discount = 38.8%.

(Note you may say that this is so long but dear this is the only way to explain the whole concept of the question to you now its upto you that how fast and short handedly you can solve these type of questions)

Hope you have enjoyed the base concept of this chapter till now. More Conceptual enjoyment will be continued for you soon, just keep visiting this blog regularly and writing me your related queries through comment or forum.

More types of questions will be followed soon on this blog........................