Holi Offer - Use Code HOLI24

Register Now

Quantitative Aptitude Tricks - PDF Download

Published on Thursday, July 30, 2015
Topics :
  1. Simplification
  2. Number Series 
  3. Percentage
  4. Profit and Loss
  5. Simple Interest and Compound Interest 
  6. Ratio and Proportion
  7. Time and Work
  8. Time Speed and Distance

#1 Simplification 

    Q1.

    812 ÷ 162 of 323 × √256 = 2?
    Sol :
    (23)12 ÷ (24)2 of (25)3 × 16 = 2
    236 ÷ 28 of  215 × 24 = 2?
    217 = 2?
    ? = 17

    Q2. 

    108 ÷ 36 of  1/4 + 2/× 31/4 = ?
    Sol :
    108 ÷ 9 + 2/5 × 13/4 = ?
    12+13/10
    ? = 133/10

    Q3.

    331/3% of 633 + 129 = 662/3% of = ?
    Sol :
    1/3 × 633 + 129 = 2/3 ×?
    ( 211+129 )×3/2 = ?
    ? = 340×3/2 = 170×3 = 510

    More Tricks on Simplification and Download PDF : Click Here

    #2 Number Series 

    Basic Concept Starts From Here : Click Here 

    Q1. 

    In each series only one number is wrong. Find out the Wrong number.
    • 5531, 5506,  5425, 5304, 5135, 4910, 4621 (IBPS PO 2012)
    Hint: -72, -92, -112


    • 1, 3, 10, 36, 152, 760, 4632 (IBPS PO 2012)
    Hint : ×1+2, ×2+4, ×3+6 ...

    • 4, 3, 9, 34, 96, 219, 435 (IBPS PO 2012)
    Hint : +13 -2 , +23 -2 , +33 -2, ...

    • 5, 7, 16, 57, 244, 1245, 7506 (Allahabad Bank PO 2010)
    Hint : ×1+12,×2+22
    • 2.5,3.5,6.5,15.5,41.25,126.75 (Allahabad Bank PO 2010)
    Hint : ×1/2+1/2, ×1+1 , ×3/2+3/2  ....

    #3 Percentage

    Basic Concepts Starts Here : Click Here

    Q1. 

    If the income of Ram is 10% more than that of Shayam's income. How much % Shyam's income is less than that of Ram's income ?
    Method I.
    By using formula
    less% = r/100+r  ×100 = 10/100+10  × 100
    = 10/110 ×100 = 9 1/11%
    Method II.
    quantitative aptitude questions with answers

    Q2.

     A man spends 40% on food, 20% on house rent, 12% on travel and 10% on education. After all these expenditure he saved Rs. 7200. Find the amount spent on travel ?
    Method I.
    Let total income x
    total expenditure 
    = x × (40%+20%+12%+10%)
    = x × 82%
    Total saving = x - x × 82%
    = x × 18%
    Then x × 18% = 7200
    x = 7200/18×100 = 40,000
    Expenditure on travel = 12% 
    x × 12% = 40,000×12/100 = Rs. 4800
    Method II.
    Total income = 100% - represent total
    quantitative aptitude tricks pdf download free
    100% -82% = 18% (saving)
    Expenditure on Travel = 7200/18×12 
    = 4800

    Q3. 

    When numerator of a fraction is increased by 10% and denominator decreased by 20% the resultant fraction becomes 5/8. Find the original fraction ?
    Method I.
    Let the original fraction be x/y then -
    quantitative aptitude tricks and tips
    Method II.
    Given Fraction = 5/8
    Original fraction = 5/8×80/110
    = 5/11 Ans.

    Q4. 

    If the length of a rectangle is increased by 20% and breath is decreased by 10%. Find the net% change in the area of that rectangle.
    Sol:
     net% change = x+y+ x×y/100
    (+20)×(-10)/100
    = +10-2
    =8
    Increase % = 8% Ans.

    Q5. 

    A reduction of 10% in the price of tea would enable and purchase to obtain 3 Kg. more for 2700 Rs. Find the reduced rate (new rate ) of tea ?
    Sol :
    10% 2700 = Rs. 270
    Rs. 270 is the rate of 3 kg. of tea
    1 kg of tea = Rs. 90/- kg,

    #4 Profit and Loss

    Basic Concept Starts Here : Click Here

    Statement

    A purchase an article at Rs 40 Rs. and sells it to B at rs. 50 and B sells its to C at Rs. 30
    short tricks for quantitative aptitude
    For A, Profit = 50-40 = 10
    For B, Loss = 50 -30 = 20

    For A, P =SP-CP
    For B, L= CP-SP

    For A, Percent Profit = Profit of A/CP of A×100
    For B, Percent loss = Loss of B/CP of B×100
    For A, 10/40×100 = 25%
    For B, 20/50×100 = 40%
    P% = P/CP×100
    L% = L/CP×100

    Q1. 

    A person purchased an article for Rs. 80 and sold it for Rs. 100.Find his % profit.
    Sol:
    CP of the article = Rs. 80
    SP of the article = Rs. 100
    Profit of the person = 100-80 = Rs. 20
    % Profit of the person = Profit /CP×100
    %P = 20/80×100
    %P = 25%
    Trick:
    %P = 20/80×100 = 25%

    Q2. 

    A dishonest shopkeeper sells goods at his cost price but uses a weight of 900 gm for a kg. weight. Find his gain percent.
    Sol:
    The Cp of Shopkeeper = 900 gm
    The Sp of Shopkeeper
    = 1000 gm ( 1kg = 1000 gm )
    The profit of shopkeeper
    = 1000 -900 = 100 gm
    % profit shopkeeper 
    = Profit of shopkeeper/CP of shopkeeper×100
    %P = 100/900×100 = 111/9%

    Q3. 

    A person got 5% loss by selling an article for Rs. 1045. At what price should the article be sold to earn 5% profit ?
    Sol:
    Trick :
    New SP = 1045/95×105 = 1155

    Q4.

     A person sold an article at profit of 12%. If he had sold it Rs. 3.60 more, he would have gain 18%. What is the cost price ?
    Sol:
    Trick :
    CP = 3.60/6×100 = Rs. 60

    Q5. 

    If the CP of 12 articles is equal to the SP of 9 articles. Find the gain or loss.
    Sol : Let the CP of each article be Rs. 1
    Then CP of 9 articles = Rs. 9
    SP of 9 articles = Rs. 12
    Gain % = 3/9×100 = 331/3%

    # 5 Simple and Compound Interest

    Basic Concept Starts From Here : Click Here

    Q1. 

    At what rate of interest per annum will a sum double itself in 8 years ?
    Sol:
    Trick :
    quantitative aptitude techniques

    Q2. 

    A sum of money double itself at compound interest in 15 years. In how many years will it become eight times.
    Trick :
    tricks for solving quantitative aptitude questions pdf
    t2 = 45 years

    #6 Ratio and Proportion

    Q1.

    The ratio between the length and the breadth of a rectabgular field is 5:4 respectively. If the perimeter of that field is 360 meters. what is the breadth of that field in meters ?
    Sol :
    Perimeter = 2(5+4) = 18
    Mean value of 18 = 360
    Breadth = 360/18 × 4 = 80 meters

    Q2.

    A bag contains 50 P, 25 P and 10 P coins in the ratio 5:9:4 amounting to Rs. 206. Find the number of coins of each type.
    Sol:
    Let the number of 50P,25P and 10P coins be 5x,9x and 4x respectively
    5x/2+9x/4+4x/10 = 206
    50x + 45x + 8x = 4120
    103x = 4120
    x = 40
    No. of 50 P coins = 5×40 = 200
    No. of 25 P coins = 4×40 = 160
    No. of 10 P coins = 9×40 = 360

    Q3.

    A mixture contains alcohol and water in the ratio of 4:3. If 5 liters of water is added to the mixture the ratio becomes 4:5. Find the quantities of alcohol in the given mixture.
    Sol:
    Let the quantity of alcohal and water be 4x liters and 3x liters respectively.
    4x/3x+5 = 4/5
    8x =20
    x = 2.5

    Q4. 

    A:B = 5:9 and B:C = 4:7 Find A:B:C.
    Sol:
    tricks to solve quantitative aptitude

    #7 Time and Work

    Q1.

    A can finish a work in 24 days , B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days.The remaining work was done by A in ? (S.S.C.2003)
    Sol:
    shortcuts for solving aptitude questions pdf

    Q2. 

    X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last ? 
    (Bank PO,2004)
    Sol:
    short tricks for solving quantitative aptitude

    Q3.

    A work twice as fast as B. If B can complete a work in 12 days independently, the number of  days in which A and B can together finish the work is  ?
    Sol :
    quantitative aptitude solving techniques

    #8 Time, Speed and Distance

      Watch this detailed video lecture



    Download QT Tricks in PDF here

    Download Reasoning tricks PDF

    Download Quantitative Aptitude Made Easy eBook
    ebook store

    About us

    ramandeep singh

    Ramandeep Singh is a seasoned educator and banking exam expert at BankExamsToday. With a passion for simplifying complex concepts, he has been instrumental in helping numerous aspirants achieve their banking career goals. His expertise and dedication make him a trusted guide in the journey to banking success.

    • Follow me:
    Close Menu
    Close Menu