Percentage short Tricks for SSC CGL– Hello friends, once again I welcome all of you for another most important Math trick explanation. This Math trick will help you in the preparation of SSC CGL, RRB, RRB NTPC, UPSC & SSC GD Exams. Its name is Percentage short Tricks for SSC CGL
Percentage Tricks in Hindi
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=> To express X% as a fraction
X%=X/100
For example: 56%
56%=56/100=14/25
=> Suppose A=120
B=150
(Q1)->A is a what percent(%) of B ?
Solution: X%(let) of B=A
(X/100) ×150 =120
∴ X=(120×100)/150
=80% ans
(Q2)->B is what % of A ?
solution: X% of A=B
(X/100) ×120 =150
∴ X=(150×100)/120
=125% ans.
(Q3)->B is how % more than A ?
Solution: difference=150-120=30
∴ Required percentage = (30×100)/120
=25% ans.
(Q4)->A is how % less than B ?
Solution: difference=150-120=30
∴ Required percentage = (30×100)/150
=20% ans.
Percentage Problem Concept:
* Suppose X increases by 40% means
X×(100+40)/100
=(X×140)/100
* Suppose X decreases by 30% means
X×(100-30)/100
=(X×70)/100
(Q1)->If salary of A is 300. now 40% increases then new salary of A.
Solution:
(Q2)->If salary of A is 300. now 40% decreases then new salary of A.
Solution:
(Q3)->Salary of A is 120.now what percent of salary increases then new salary of A is 144 ?
Solution:
% increase=(24×100)/120
= 20% Ans
Note: Rs 24, Rs 120 पर increase हो रहा है तो percentage हमेशा initial amountपर calculate किया जाता है न कि increased वाले amount पर .
(Q4)->If salary of A increases 40% then new salary of A is 490.find out initial salary
Solution:
Now
(X×140)/100=490
∴ X=(490×100)/140
=350Ans
Common type of percentage problem that are frequently asked in various competitive exam.
(Q1)->If x% of Y% of 80 is the same as 25% of 900 then the value of xy is.
Solution: 80×(y/100)×(x/100)=900×(25/100)
4xy/500=25×9
∴xy=(25×9×500)/4
=28125 Ans
(Q2)->If 40% of 4/5 of 3/4 of a number is 48, then what is 1% of the same number?
Solution:Let the number be x
x×(3/4)×(4/5)×(40/100)=48
∴ x=(48×4×5×100)/(3×4×40)
=200
Now 1% of 200=200×(1/100)
=2 Ans.
(Q3)->What percent of 15 hours is 18 seconds?
Solution:x% of 15 hours = 18 seconds
x% of 15×60×60 seconds =18 seconds
15×60×60×(x/100)=18
∴ x=(18×100)/(15×60×60)
=1/30% Ans
(Q4)->What percent of 3.5 kg is 70 gms?
Solution:x% of 3.5 kg=70 gms
x%×3.5×1000 gms=70 gms
(x/100)×3.5×1000 gms=70gms
∴ x=(70×100)/3.5×1000
=2% Ans.
(Q5)->50% of a number when added to 50 is equal to the number.The number is.
Solution:Let the number be 100x
100x×(50/100)+50=100x
50x+50=100x
50x=50
∴ x=1
Number=100x
=100×1
=100 Ans
(Q6)->When 75 is added to 75% of a number,the answer is the number.Find 40% of that number.
Solution:Let the number be 100x
100x×75%+75=100x
=>100x ×(75/100)+75=100x
=>75x+75=100x
∴ x=3
Number=100x
=100×3
=300
Now 40% of 300=(300×40)/100
=120 Ans.
(Q7)->The number that is to be added to 10% of 320 to have the sum as 30% of 230 is.
Solution:Let the number to be added=x
320×(10/100)+x=(230×30)100
=>32+x=69
∴ x=37 Ans
(Q8)->If 60% of A=30% of B,B=40% of c,c=x% of A,then value of x is.
Solution:A×(60/100)=B×(30/100)
=>3A/5=3B/10
=>3A/5=(3/10)×(2C/5)
=>3A/5=6C/50
=>3A/5=(6/50)(Ax/100)
∴ x=(50×100×3A)/6A×5
=500 Ans
Rough:B=C×(40/100)
=2C/5
C =A×(x/100)
=Ax/100
(Q9)->51% of a whole number is 714. 25% of that number is.
Solution:Let the whole number be 100x
100x×51%=714
=>100x×(51/100)=714
∴ x=14
whole number=100x
=100×14
=1400 Ans.
Now, 25% of 1400=(25/100)×1400
=350 Ans.
(Q10)->83¹/3% of Rs 90 is equal to 60% of ?
Solution:90×83¹/3%=x×60%
90×(250 /3×100)=x×(60/100)
∴ x=(250×90×100)/(3×100×60)
=125 Ans.
(Q11)->If 8% of x=4% of y,then 20% of x is.
Option:(a)10% of y (b)16%of y (c)40% of y (c)80% of y
Solution:x×(8/100)=y×(4/100)
=>8x=4y
∴ x=y/2
let,
20% of x = z% of y
(20/100)×x=y×(z/100)
20x=yz
20×y/2 =yz [x=y/2]
∴ z=10% Ans
Option (a) Ans
(Q12)->If 30% of A is added to 40% of B,the answer is 80% of B.what percentage of A is B ?
Solution:A×(30/100)+B×(40/100)=B×(80/100)
=>(3A/10)+(4B/10)=8B/10
=>3A+4B=8B
=>3A=4B
∴ A/B=4/3
Now, x% of A=B
(x/100)×A=B
xA=B×100
x=100B/A
=100×3/4
=75% Ans.
How to solve percentage problems quickly
(Q1) If Ravi salary is 25 % more than Raju’s salary the percentage by which Raju’s salary is less then Ravi salary.
Solution: BY TRICK
Required percentage=(25/100+25)×100)
=(25/125)×100
=20% Ans
(Q2)If A’s income is 50% less than that of B’s, then B’s income is what percent more than that of A?
Solution: BY TRICK
Required percentage:(50/100−50)×100
=(50/50)×100
=100% Ans
(Q3)Two numbers are more than the third number by 20% and 50% respectively. The first number is what percent of the second number?
Solution:Let third number=100
First number=120
Second number=150
∴ Required percentage=(120/150)×100
=80% Ans.
BY TRICK:
∴ Required percentage=(100+20/100+50)×100
=80% Ans.
(Q4)Two numbers are less than a third number by 30% and 37% respectively.The percent by which the second number is less than the first is.
Solution: Let the third number=100
First number=70
Second number=63
∴ Required percentage=(7/70)×100
=10% Ans.
(Q5)Two numbers are respectively 20% and 50% more than a third number. the ratio of the two numbers is.
Solution: Let the third number=100
First number=120
Second number=150
∴ Required ratio=120/150
=4/5
=4:5 Ans.
(Q6)25% of the annual salary of A is equal to 80% of the annual salary of the B.Monthly salary of B is 40% of the monthly salary of the C.Annual salary of C is Rs. 6 lac. what is the monthly salary of A?
Solution: monthly salary of C=600000/12
=Rs 50000
B’s monthly salary=(50000×40)/100
=Rs 20000
Annual salary of B=20000×12
=240000
Now,
Let Annual salary of A=X
( X×25)/100=(240000×80)/100
=>25X = 240000×80
∴ X=(240000×80)/25
=Rs 768000
monthly salary of A=768000/12
=Rs 64000 Ans.
(Q7)The monthly salaries of A and B together amount to Rs 40,000.A spends 85% of his salary and B, 95% of his salary.if now their saving is the same, then the salary(in Rs) of A is.
Solution: Let salary of A=X
salary of B=40000−X
Saving of A=15%
Saving of B=5%
Now according to question their saving are same so
( X×15)/100=(40000−X)×5/100
=>15X=200000−5X
=>20X=200000
∴ X=Rs 10000 Ans.
(Q8)Mohan donated 3% of his income to a charity and deposited 12% of the rest in the bank.if now he has Rs.12804, then his income was.
Solution:
Let income=100x
Now,
85.36x=12804
∴ x=150
income=100x
=100×150
=Rs 15000 Ans.
Rough:
(100x×3)/100=3x
rest=100x-3x=97x
deposit=12% of rest i.e 97x
(97x×12)/100
=11.64x
remaining=97x−11.64x=85.36x
(Q9)Mohan saves 14% of his salary while Shyam saves 22%. if both get the same salary and Shyam saves Rs. 1540, what is the saving of Mohan?
Solution: According to question Shyam saves 22% i.e equal to 1540
now,
22%=1540
∴ 1%=1540/22
∴100%=(1540/22)×100
=Rs 7000 salaries of Shyam
saving of ram=(7000×14)/100
=Rs.980 Ans.
IMPORTANT PERCENTAGE SHORTCUT TRICKS,CONCEPTS AND PROBLEM FOR COMPETITIVE EXAM
FOR EXAMPLE
(Q1)Due to a decrease in price per kg. of sugar by 25% by how much percent, a housewife should increase quantity so that total expenditure remains same.
Solution:Let initial price=Rs. 100
Now,
% Increase=(25/75)×100
=33.33% Ans
Rough:New price=100×(100-25)/100
=Rs75
Concept:[जब Question में कहा जाए कि Total Expenditure same है तो आपलोगों को Quantity always price का reverse लेना है.
Explanation:[पहले housewife Rs 100 में 75 kg sugar खरीदती थी जब price decrease किया तो Rs 75में 100kg sugar खरीदती है तो इस तरह से housewife 25 kg अधिक sugar खरीद पा रही है ]
BY TRICK:
% increase=(25/100−25)×100
=(25/75)×100
=33.33% Ans.
(Q2)Due to increase in price per kg of sugar by 25% by how much percent, a housewife should decrease quantity so that total expenditure remains same.
Solution:Let initial price=Rs100
Now,
%decrease= (25/125)×100
=20% Ans
Rough:New price=100×(100+25)/100
=Rs125
BY TRICK:
%Decrease=(25/100+25)×100
=(25/125)×100
=20% Ans.
(Q3)If the price of sugar rises from Rs 6 per kg to Rs 7.50 per kg, a person to have no increase in his expenditure on sugar, will have to reduce his consumption of sugar by:
Solution:
% Reduce=(1.50/7.50)×100
=20% Ans
(Q4)The length of the rectangle is increased by 60%.By what percent would the breadth be decreased to maintain the same area?
Solution:Area of rectangle=Length×Breadth
Let initial length=100
% Decrease=(60/160)×100
=37½% Ans.
Concept:Breadth, Length का reverse लिया गया है because Area is same.
BY TRICK:
% Decrease=(60/100+60)×100
=(60/160)×100
=37½% Ans.
(Q5)One side of a square is increased by 30%.To maintain the same area,the other side will have to be decreased by.
Solution:Area of square=Side×Side
Let initial length of side=100
Now,
% Decrease=(30/130)×100
=23¹/3% Ans.
BY TRICK:
% Decrease=(30/100+30)×100
=23¹/3% Ans.
(Q6)The height of a triangle is increased by 10% to retain the original area of the triangle,its corresponding base must be decreased by:
Solution:Area of Triangle=½×base×height
Let initial height=100
%Decrease=(10/110)×100
=9¹/11% Ans.
BY TRICK:
%Decrease=(10/100+10)×100
=(10/110)×100
=9¹/11% Ans
Next is percentage type-4. In this type, I have discussed a common type percentage problem that is frequently asked in Bank PO SSC Railway exam.
For Example: The price of sugar having gone down by 10%,sharad can buy 6.2Kg more for Rs.279.The difference between the original and the reduced price(per Kg) is:
And another example is:The price of sugar is decreased by 25%.Therefore a family increases its consumption so that the decrement in the expenditure of sugar is only 15%.if the consumption of sugar is 30Kg before the decrement,what is the consumption now.
(Q1)Due to increase in the price of sugar by 30% a man can buy 6kg less for Rs.520.then find out the original price per kg in Rs.?
Solution:Let original price=Rs 100
Now
Step-1
[Original price=Rs 100,New price=Rs130,Original quantity=130Kg,New Quantity=100Kg And Decreased quantity=30Kg]
Concept:According to Question A man can buy 6Kg less.But यहाँ पर 30Kg कम निकल रहा है,तो अब 30Kg को 6Kg बनाना होगा इसके लिए Quantity वाले part में 5 से Divide करना होगा|
[Concept in English:According to Question A man can buy 6Kg less.but here we got 30Kg less, so we will have to make 6kg, therefore, quantity divide by 5]
Step-2
price of 26 kg=Rs520
∴ Price of 1Kg=520/26
=Rs20 Ans.
Concept:[जब Expenditure same रहेगा तब इस process से solve करना है|इस Question में Expenditure same क्योंकि पहले Rs520 में जितना खरीदता था अब price increase होने से Rs520 में 6kg कम खरीद पा रहा है ]
BY TRICK:
Original price=(30×520)/(100+30)×6
=(30×520)/(130×6)
= Rs20 Ans.
(Q2)Due to a decrease in the price of sugar by 20% a man can buy 4Kg more for Rs80 then find the original price per Kg in Kg?
Solution:Let original price=Rs100
Step-1
[Original price=Rs 100,New price=Rs80,Original quantity=80Kg,New Quantity=100Kg And Increased quantity=20Kg]
Concept:According to Question A man can buy 4Kg more.But यहाँ पर 20Kg अधिक निकल रहा है,तो अब 20Kg को 4Kg बनाना होगा इसके लिए Quantity वाले part में 5 से Divide करना होगा|
[Concept in English:According to Question A man can buy 4Kg more.but here we got 20Kg more, so we will have to make 4kg, therefore, quantity divide by 5]
Step-2
the price of 16kg=Rs80
∴ price of 1Kg=80/16
=Rs 5 Ans.
BY TRICK:
Original price=(20×80)/(100-20)×4
=(20×80)/(80×4)
= Rs.5 Ans.
(Q3)A reduction of 21% in the price of wheat enables a person to buy 10.5kg more for Rs.100.what is the reduced price per Kg?
Solution:Let original price=Rs.100
Step-1
Step-2
Price of 50Kg=Rs.100
∴price of 1Kg=100/50
=Rs.2 Ans
BY TRICK:
Reduced price=(21×100)/(100×10.5)
=Rs.2 Ans
(Q4)The price of sugar having gone down by 10%,sharad can buy 6.2Kg more for Rs.279.The difference between the original and the reduced price(per Kg) is:
Solution:Let original price=Rs.100
Step-1:
Step-2:
(1)Original price:
Price of 90×0.62Kg=Rs.279
∴ price of 1Kg=279/(90×0.62)
=Rs.5/Kg Ans.
(2)Reduced price:
Price of 100×0.62Kg=Rs.279
∴ price of 1Kg=279/(100×0.62)
=Rs.4.5/Kg Ans
Difference=5-4.5
=Rs.0.5 Ans.
BY TRICK:
Original price=10×279/(100-10)×6.2
=10×279/(90×6.2)
=Rs.5/Kg
Reduced price=10×279/(100×6.2)
=Rs.4.5/Kg
IMPORTANT PERCENTAGE PROBLEM CONCEPTS AND SHORTCUT TRICKS FOR BANK PO SSC RAILWAY EXAM TYPE-4.1
(Q1)The price of sugar is increased by 30%.Therefore a family reduces its consumption so that the increment in the expenditure of sugar is only 10%.if the consumption of sugar is 26Kg before the increment,what is the consumption now.
Solution:We know that
∴ X=286/13
=22Kg Ans
ROUGH:
Rs10 increased by 30% then,
=10×(100+30)/100
=Rs13
Rs260 increased by 10% then,
260×(100+10)/100
=(260×110)/100
=Rs286
BY TRICK:
Consumption=26×(100+10)/(100+30)
=26×110/130
=22Kg Ans.
(Q2)The price of sugar is decreased by 25%.Therefore a family increases its consumption so that the decrement in the expenditure of sugar is only 15%.if the consumption of sugar is 30Kg before the decrement,what is the consumption now.
Solution:
∴ X=255/7.50
=34Kg Ans.
ROUGH:
Rs300 decreased by 15% then
300×(100-15)/100
=(300×85)/100
=255
Rs10 decreased by 25% then,
10×(100-25)/100
=(10×75)/100
=7.50
BY TRICK:
consumption:30×(100-15)/100-25
=(30×85)/75
=34Kg Ans.
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.
Percentage Problems In Aptitude type-6 I have discussed a question based on examination and marks obtained. This type of problem frequently asked in Bank, SSC, Railways Exam.
FOR EXAMPLE:
(Q1) A candidate who scores 30% fails by 5 marks, while another candidate who scores 40% marks get 10 more then minimum pass marks. The minimum marks required to pass are.
Solution:Let total marks=100x
Now,
100x×(30/100)+5=100x×(40/100)-10
=>30x+5=40x-10
=>30x-40x=-10-5
=.-10x=-15
∴x=15/10
Total marks=100x
=100×(15/10)
=150
∴ minimum marks=100x×(30/100)+5
=100×(15/10)×(30/100)+5
=50 Ans
Concepts:[According to question A candidate who scores 30% fails by 5 marks.that means इस candidate को minimum pass marks के लिए 5 marks अधिक लना चाहिए था i.e 100x×(30/100)+5 marks लाना चाहिए था|
Now another candidate who scores 40% marks get 10 more than minimum passing marks.that means इस candidate को minimum pass marks के लिए 10 marks कम लाना चाहिए था i.e 100x×(40/100)-10 लाना चाहिए था|]
(Q2)In an examination, a candidate must secure 40% marks to pass. A candidate, who gets 220 marks fails by 20 marks. what are the maximum marks for the examination?
Solution:Let maximum marks=100x
now,
100x×(40/100) =220+20
40x=240
∴ x=6
maximum marks=100x
=100×6
600 Ans.
Concept:[220 marks लाया and 20 marks से fail हो गया that means उसे pass होने के लिये (220+20)marks लाना होगा]
(Q3)In an examination, there are three papers and a candidate has got 35% of the total to pass. in one paper ,he gets 62 out of 150 and in the second 35 out of 150. how much he get, out of 180, In the third paper to just qualify for a pass.
Solution: Let the marks required in the third paper= x.
then,
(62+35+x)=35% of(150+150+180)
=>97+x=(35/100)×480
∴ x=168-97
=71 Ans.
(Q4)In an examination, there are three subjects of 100 marks each. A student scores 60% in the first subject and 80% in the second subject. He scored 70% in aggregate. His percentage of marks in the third subjects is.
Solution: let percentage marks in third subjects=x
then,
60+80+x=70% of (100+100+100)
=>140+x=(70/100)×300
∴ x=70 i.e 70% Ans.
(Q5)In an examination,65% of the students passed in mathematics, 48% passed in physics and 30% passed in both. How much percent of students failed in both the subjects?
Solution:
Students passed only in mathematics=65%-30%=35%
Students passed only in physics=48%-30%=18%
students passed in both subjects=30%
total pass=35%+18%+30%
=83%
∴ total failed in both subjects=100%-83%
=17% Ans.
BY TRICK:
Percentage of students who passed in either mathematics or physics or both=(65+48-30)%=83%
∴ total failed in both subjects=100%-83%
=17% Ans.
(Q6)In an examination, 75% candidate who passed in English and 60% passed in mathematics. 25% failed in both the subjects. If 240 candidate passed in both.Find the total number of candidates.
Solution:Let x candidates passed in both subjects.
percentage of students who passed in either English or mathematics or both=(75+60-x)%
=>75%=(135-x)%
∴ x=60%
Now,
60%=240
∴ 1%=240/60
∴ 100%=(240/60)×100
=400 candidate Ans.
Concept:[25% failed in both subjects it means only 75% candidates passed in either English or mathematics or both]
(Q7)In an examination A got 25% marks more than B, B got 10% less than C and C got 25% more than D. if D got 320 marks out of 500, the marks obtained by A.
Solution:
(Q8)Three sets of 40,50 and 60 students appeared for an examination and the pass percentage was 100,90 and 80 respectively. The pass percentage of the whole set is.
Solution:pass percentage of the whole set=(40×100+90×50+60×80)/(40+50+60)
=88²/3% Ans.
Percentage Aptitude Shortcut Tricks type-8. In this, I have discussed a question based on Income,Expenditure, and salary. This type of problem frequently asked in Bank,SSC,Railways Exam.
(Q1)Mohan gives 35% of the money he had to his wife and gave 50% of the money he had to his sons. The remaining amount of 11,250 he kept for himself. what was the total amount of money Mohan had?
Solution:Let total amount of money Mohan had=100x
Now,
15x=11250
∴ x=11250/15
=750
Total amount of money Mohan had=100x
=100×750
=75000 Ans.
Rough:
Total money gives by Mohan to his wife+son=35%+50%=85%
Now,
100x×(85/100)
=85%
remaining=100x-85x=15x
(Q2)Mr. Nand spent 20% of his monthly income on food and 15% on children’s education, 40% of the remaining he spent on entertainment and transport together and 30% on medical.He is left with an amount of 8775 after all these expenditures. What is Mr. Nand monthly income?
Solution:Let monthly income of Mr. Nand=100x
19.5x=8775
∴ x=8775/19.5
= 450
monthly income of Mr. Nand=100x
=100×450
=45000 Ans.
Rough:
Total spent on Food+Education=20%+15%=35%
i.e 100x(35/100)
=35x
Remaining=100x-35x=65x
Total spent on entertainment+Transport+Medical=70%
i.e 65x(70/100)
=45.5x
Remaining=65x-45.5x=19.5x
(Q3)Mohan had a certain amount with him. He spent 20% of that to buy an article and 5% of the Remaining on Transport, then he gifted Rs 120. If he is left with Rs 1400, the amount he spent on transport is.
Solution:Let certain amount=100x
76x-120=1400
=>76x=1520
∴ x=1520/76
=20
Transport=4x
=4×20
=Rs 80 Ans.
(Q4)Aman spends 40% of the amount he received from his father on Hostel expenses, 20% on Books and Stationary and 50% of the remaining on Transport. He saves Rs 450 which is half the Remaining amount after spending on Hostel expenses,Books etc. and Transport. How much money did he get from his father?
Solution:Let Aman get money from his father=100x
10x=450
∴ x=45
Aman get money from his father=100x
=100×45
=Rs 4500 Ans.
Rough:
Total spent on Hostel+(Books+Stationary)=40%+20%=60%
i.e 100x×(60/100)
=60x
Remaining=100x-60x=40x
50% remaining on transport
i.e 40x×(50/100)
=20x
Remaining=40x-20x=20x
(Q5)A man spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%. His saving is increased by.
Solution:Let income=100x
Expenditure(i.e spend) is increased by 10% than new Expenditure
=75x×(100+10)/100
=75x×(110/100)
=82.5x
New saving=120x-82.5x=37.5x
Increased saving=37.5x-25x=12.50x
% increase=(12.50x×100)/25x
=50% Ans
(Q6)The monthly salaries of A and B together amount to Rs 40000. A spends 85% of his salary and B, 95% of his salary. If now their saving is the same, then the salary of A.
Solution:Let salary of A=100x
Salary of B=40000-100x
A spends 85% of his salary that means A saved 15% salary
∴ Saving of A=100x(15/100)
=15x
B spends 95% of his salary that means B saved 5% salary
∴ Saving of B=(40000-100x)×(5/100)
=(40000-100x)/20
According to question their saving are same so,
15x=(40000-100x)/20
300x=40000-100x
∴ x=40000/400
= 100
Now,
Salary of A=100x
=100×100
=Rs 10000 Ans.
Percentage Shortcut Tricks Bank PO SSC Railway Exams type-9.
In type-9, I have discussed a question based on Depreciation and population increase. This type of percentage problem frequently asked in Bank, SSC, Railways Exam.
(Q1)The population of a town increases by 5% every year. If the present population is 9261, The population 3 Years ago was.
Solution: F.Q=I.Q×M.F
Where I.Q=Initial Quantity
F.Q=Final Quantity
M.F=Multiplying Factor
Now
F.Q=I.Q×M.F
9261=I.Q×(105/100)×(105/100)×(105/100)
I.Q=(9261×100×100×100)/(105×105×105)
=8000 Ans
CONCEPT:[इस Type के Question को Solve करते Time आपलोगो को सिर्फ इन बातों पर ध्यान रखना है Question में After n Years का value निकालने के लिए कह रहा है या n Years Ago का | (1)यदि Question में After n Years का value निकालने के लिए कह रहा हो तो Present value जो Question में दिया होगा वह Initial Quantity होगा| (2) यदि Question में n Years Ago का value निकालने के लिए कह रहा हो तो Present value जो Question में दिया होगा वह Final Quantity होगा|]
(Q2)The present population of a city is 180000. If it increases at the rate of 10%per annum, its population after 2 years will be.
Solution: F.Q=I.Q×M.F
F.Q=180000×(110/100)×(110/100)
=217800 Ans.
(Q3)The present price of a scooter is 7290. If its value decreases every year by 10%, then its value after 3 years back was.
Solution: F.Q=I.Q×M.F
7290=I.Q×(90/100)×(90/100)×(90/100)
∴ I.Q=(7290×100×100×100)/(90×90×90)
=10000 Ans.
(Q4)In a factory, the production of cycles rose to 48,400 from 40,000 in 2 years. The rate of growth per annum is.
Solution: Let rate of growth=x%
F.Q=I.Q×M.F
48,400=40000×((100+x)/100)×(100+x)/100
48,400/4=(100+x)²
12100=(100+x)²
(110)²=(100+x)²
100+x=110
∴ x=10% Ans
(Q5)The population of a village was 9800. In a year with the increase in the population of males by 8% and that of females by 5%, the population of the village become 10458.What was the number of males in the village before increase?
Solution:Let the number of males=x
∴ number of females=9800-x
x×(108/100)+(9800-x)×(105/100)=10458
=>108x+9800×105-105x=1045800
=>3x=1045800-1029000
=16800
∴ x=16800/3
=5600 Ans.
Percentage Shortcut Tricks Bank PO SSC Railway Exams type-9.1.
(Q1)A large watermelon weighs 20kg with 96% of its weight being water. It is allowed to stand in the sun and some of the water evaporates so that only 95% of its weight is water.Its reduced weight will be:
Solution:
Weight of fresh watermelon=20Kg
the weight of fresh pulp=20×(4/100)
=0.8Kg
Now,
Concept:[watermelon में से water सूखेगा न कि pulp. pulp तो always same ही रहेगा चाहे Fresh हो या Dry ]
so,
5%=0.8Kg
1%=0.8/5 Kg
∴ 100%=(0.8/5)×100 Kg
=16 Kg Ans.
(Q2)Fresh grapes contain 80% water while Dry grapes contain 10% water. If the weight of dry grapes is 500Kg then what is its total weight when it is fresh.
Solution:
Weight of Dry grapes=500Kg
the weight of Dry pulp=500×(90/100)
=450Kg
Now,
Concept:[Grapes में से water सूखेगा न कि pulp. pulp तो always same ही रहेगा चाहे Fresh हो या Dry ]
so,
20%=450Kg
1%=450/20 Kg
∴ 100%=(450/20)×100 Kg
=2250 Kg Ans.
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