Quantitative Aptitude Shortcut tricks

Quantitative Aptitude Tricks
By Ramandeep Singh Page 1
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 )×3/2 = ?
? = 340×3/2 = 170×3 = 510
www.BankExamsToday.com

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.

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 and B together can complete a piece of work in 4 days. If A alone can complete
the same work in 12 days, in how many days can B alone complete that work
?(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 is thrice as good a workman as B and together is able to finish a job in 60 days
less than B. Working together, they can do it in ?
Sol :
#8 TIME, SPEED AND DISTANCE

CONCEPTS
1) There is a relationship between speed, distance and time:
Speed = Distance / Time OR
Distance = Speed* Time
2) Average Speed = 2xy / x+y
where x km/hr is a speed for certain distance and y km/hr is a speed at for same distance covered.
**** Remember that average speed is not just an average of two speeds i.e. x+y/2. It is equal to 2xy / x+y
3) Always remember that during solving questions units must be same. Units can be km/hr, m/sec etc.
**** Conversion of km/ hr to m/ sec and m/ sec to km/ hr
x km/ hr = (x* 5/18) m/sec i.e. u just need to multiply 5/18
Similarly, x m/sec = (x*18/5) km/sec
4) As we know, Speed = Distance/ Time. Now, if in questions Distance is constant then speed will be
inversely proportional to time i.e. if speed increases ,time taken will decrease and vice versa.
TIME AND DISTANCE PROBLEMS
Problem 1: A man covers a distance of 600m in 2min 30sec. What will be the speed in km/hr?
Solution: Speed =Distance / Time
⇒ Distance covered = 600m, Time taken = 2min 30sec = 150sec
Therefore, Speed= 600 / 150 = 4 m/sec
⇒ 4m/sec = (4*18/5) km/hr = 14.4 km/ hr.
Problem 2: A boy travelling from his home to school at 25 km/hr and came back at 4 km/hr. If whole
journey took 5 hours 48 min. Find the distance of home and school.

Solution: In this question, distance for both speed is constant.
⇒ Average speed = (2xy/ x+y) km/hr, where x and y are speeds
⇒ Average speed = (2*25*4)/ 25+4 =200/29 km/hr
Time = 5hours 48min= 29/5 hours
Now, Distance travelled = Average speed * Time
⇒ Distance Travelled = (200/29)*(29/5) = 40 km
Therefore distance of school from home = 40/2 = 20km.
Problem 3: Two men start from opposite ends A and B of a linear track respectively and meet at point
60m from A. If AB= 100m. What will be the ratio of speed of both men?
Solution: According to this question, time is constant. Therefore, speed is directly proportional to
distance.
Speed∝Distance
⇒ Ratio of distance covered by both men = 60:40 = 3:2
⇒ Therefore, Ratio of speeds of both men = 3:2
Problem 4: A car travels along four sides of a square at speeds of 200, 400, 600 and 800 km/hr. Find
average speed.
Solution: Let x km be the side of square and y km/hr be average speed
Using basic formula, Time = Total Distance / Average Speed
x/200 + x/400 + x/600 + x/800 = 4x/y ⇒ 25x/ 2400 = 4x/ y⇒ y= 384
⇒ Average speed = 384 km/hr

Quantitative Aptitude Shortcut tricks

More Related Content

What's hot (20)

Viewers also liked (16)

Similar to Quantitative Aptitude Shortcut tricks (20)

More from BankExamsToday.com (11)

Recently uploaded (20)

Quantitative Aptitude Shortcut tricks