Quantitative Aptitude Tricks - PDF Download

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 + ²/₅ × 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 )׳/₂ = ?
? = 340׳/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 , ׳/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.

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

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 -

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

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 :

Q2. 

A sum of money double itself at compound interest in 15 years. In how many years will it become eight times.

Trick :

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:

#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:

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:

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 :

#8 Time, Speed and Distance

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