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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.
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    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
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    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 -
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    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
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    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 :
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    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 :
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    #8 Time, Speed and Distance

      Watch this detailed video lecture



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    Ramandeep Singh, your guide to banking and insurance exams. With 14 years of experience and 5000+ selections, Ramandeep understands the path to success, having transitioned himself from Dena Bank and SBI. He's passionate about helping you achieve your banking and insurance dreams.

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