TOP 10 Shortcuts for Percentage Problems - SBI Clerk / IBPS Exams 2017:
Dear Readers, Here we have given the TOP 10 Shortcuts for Percentage Problems for SBI Clerk / IBPS Exams 2017, candidates those who are preparing for the upcoming SBI Clerk/IBPS Exams 2017 can make use of it.
Fraction and Percentage Equivalents:
Two number are in the ratio x:y if the first number is increased by a% and the second number is increased by b%, then
a) Percentage increase in the sum of the number = (ax+by/x+y)×100
b) Percentage change in difference of the numbers = (ax-by/x-y) × 100
Sum original numbers = x+y. --> sum of new numbers = x + ax/100 + y + by/100
Increase in sum = ax/100 + by/100 = ax+by/100 --> increase = ax+by/x+y×100
1) Sukumar’s expenditure and savings are in the ratio 5:3. If his income increases by 12% and expenditure by 15%, then how much percent do his savings increase?
a) 3% b) 5% c) 7% d) 27%
Savings = Income – Expenditure
Increase in savings = (18×12-5×15) / 8-5
Assume that there are two persons A and B and A’s salary is X% more than B’s salary.
Then B’s salary is lesser than A’s salary by X/(100+X) x 100%
Then B’s salary is lesser than A’s salary by X/(100+X) x 100%
Variant of the above formula:
If A’s salary is X% lesser than B’s salary,
then B’s salary will be more than A’s salary by X/(100-X) x 100%
If A’s salary is X% lesser than B’s salary,
then B’s salary will be more than A’s salary by X/(100-X) x 100%
1) Ram’s salary is 30% more than Renu’s salary, by how much percent is Renu’s salary less than Ram?
From the question you know that Ram’s salary is 30% more than that of Renu.
Therefore our value of X (to use in formula) = 30
Now, Renu’s formula is lesser than that of Renu by X/(100+X) x 100%
= 30/(100 + 30) x 100%
= 30/130 x 100%
1) Raman`s salary was decreased by 50% and subsequently increased by 50%. How much percent does he lose?
a) 45% b) 50% c) 75% d) 25%
Let the original salary = Rs.100
( x2/100 ) % or 100 × −50/100 × 50/100 = -25 % or Decrease = 25%
1) The salary of a person was reduced by 10%. By what percent should his reduced salary be raised so as to bring it at par with his original salary?
a) 12.1% b) 10% c) 5% 6) 11.11%
Let the original salary be Rs.100. New salary = Rs.90.
Percentage increase in salary = ( (Old – New/New) × 100) %
= ( (100 – 90/90) × 100) % = ( 100/ 9 ) %
1) Shaba’s mathematics test had 75 problems i.e.10 arithmetic, 30 algebra and 35 geometry problems. Although she had answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly. She did not pass the test because she got less than 60% of the problems right. How many more questions she would have to answer correctly to earn 60% of the passing grade?
a)2 b) 6 c) 5 d) 7
Number of questions attempted correctly = (70% of 10 + 40% of 30 + 60% of 35)
70 /100 × 10 + 40/ 100 × 30 + 60 /100 × 35 = 7 + 12+21= 40
questions to be answered correctly for 60% grade=60% of 75 = 45
therefore required number of questions= (45 - 40) = 5.
1) The population of a town is 1, 76,400. If it increases at the rate of 5% per annum, what will be its population 2 years hence? What was it 2 years ago?
a) 170000 b) 160000 c) 180000 e) None of this
Population Increase for n year = p × (1 + R/100)n
=176400 × (100 + 5)2/100
= 176400 × ( 105/100) 2
= 176400 × ( (21/ 20) × (21 /20)) = 194481.
Population 2 years ago = (174600/ (1+ 5/100)2 ) = ( 174600/ (105 /100)2 ) = 160000.
If A’s salary is 20% less then B`s salary, by how much percent is B's salary more than A`s ?
a) 25% b) 30% c) 45% d) None of this
Required percentage = (x/(100 – x)) × 100
= (20/100 −20) × 100 = 25%.
In a box of fruits, there are 25% of apples, 40% of orange and the rest are mangoes. If there are 2000 fruits in the box. How many mangoes are there in the box?
a) 600 b) 700 c) 500 d) 300
Total fruits = 100% = 2000
Mangoes = total fruits – (apples + oranges) = 100%-(25%+40%)
= 100-65 = 35%
Given 100% = 2000
-->(35 *2000)/100 = 700.
1) When a discount of 20% offered by a trader, his sales increased by 80%. What was the effect on his receipts?
a) 55% b) 44% c) 45% d) None
Shortcut formula,
-a + b –ab/100
Net percentage = - discount % + Increment% - (discount%) (increment%) /100
=20 +80-1600/100 = -20 +80-16 = 44% increased
If the present population of a city is P and there is a increment of R1%, R2% ,R3% in first, second and third year respectively, then
Population of city after 3 yr=P(1+R1/100)(1+R2/100)(1+R3/100)
1) Population of a city in 2004 was 1000000. If in 2005 there is an increment of 15 % , in 2006 there is a decrement of 35 % and in 2007 there is an increment of 45 %, then find the population of city at the end of the year 2007.
Population of city after 3 yr=P(1+R1/100)(1+R2/100)(1+R3/100)
1) Population of a city in 2004 was 1000000. If in 2005 there is an increment of 15 % , in 2006 there is a decrement of 35 % and in 2007 there is an increment of 45 %, then find the population of city at the end of the year 2007.
a) 1083875 b) 1125685 c) 8546945 d) None
Solution :
Solution :
Required population = P (1 + R1/100)(1 - R2/100)(1 + R3/100)
= P (1 + 15/100)(1 - 35/100)(1 + 45/100)
= 1083875
= 1083875
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