**10 Important topics and concepts in Aptitude**

**1. Calendar:**

**Important Facts and Formulae:**

a)** Odd Days:** The number of days more than the complete number of weeks in a given period is a number of odd days during that period.

b) **Leap year:** Every year which is divisible by 4 is called a leap year. Thus each one of the year 1992, 1996, 2004, 2008, etc. is a leap year.

Every 4th century is a leap year but no other century is a leap year thus each one of 400, 800, 1200,1600,2000, etc is a leap year. None of the 1900,2010,2020,2100,etc is a leap year. A year which is not a leap year is called an Ordinary year.

c)An ordinary year has 365 days.

d) A leap year has 366 days.

e) **Counting of odd days:**

i) One ordinary year = 365 days =52 weeks+1 day, Therefore An ordinary year has 1 Odd day.

ii) One leap year = 366 days =52 weeks+2 days, Therefore a leap year has 2 Odd days.

iii) 100 years = 76 ordinary years+ 24 leap years

= [(76*52) weeks+76 days]+[(24*52)weeks+48 days]

= 5200 weeks+124days=[5217 weeks+5 days]

therefore 100 years contain 5 odd days.

iv) 200 years contain 10(1week+3days), i.e 3 odd days

v) 300 years contain 15 (2 weeks+ 1 day), i.e 1 odd day

vi) 400 years contain (20+1), i.e 3 weeks,

so 0 Odd days Similarly each one of 800,1200,1600, etc contains 0 odd days.

**Note: **(7n+m) odd days, where m less than or equal to 7 is equivalent to m odd days, thus, 8 odd days = 1 odd day etc.

f) Some codes o remember the months and weeks:

i) **Week**

Sunday – 1

Monday – 2

Tuesday – 3

Wednesday – 4

Thursday – 5

Friday – 6

Saturday – 0

ii) **Month**

Jan – 1 July – 0

Feb – 4 Aug – 3

Mar – 4 Sep – 6

Apr – 0 Oct – 1

May – 2 Nov – 4

June – 5 Dec – 6

**2. Boats and Streams:**

**Important Facts and Formulae:**

a) In water, the direction of the stream is called downstream.

b) Direction against the stream is called upstream.

3) The speed of boat in still water is U km/hr and

the speed of the stream is V km/hr then,

speed of downstream =U + V km/hr

speed of upstream = U – V km/hr.

**3. Time and Work:**

**Important Facts and Formulae:**

1. If A can do a piece of work in n days, then A’s 1 day work=1/n.

2. If A’s 1 day’s work=1/n, then A can finish the work in n days.

Ex: If A can do a piece of work in 4 days, then A’s 1 day’s work=1/4. If A’s 1 day’s work=1/5, then A can finish the work in 5 days

3. If A is thrice as good workman as B, then: Ratio of work done by A and B =3:1 Ratio of time taken by A and B to finish a work=1:3

4. Definition of variation: The change in two different variables follows some definite rule. It said that the two variables vary directly or inversely. Its notation is X/Y=k, where k is called constant. This variation is called direct variation. XY=k. This variation is called inverse variation.

5. Some pairs of variables:

i) The number of workers and their wages. If the number of workers increases, their total wages increase. If the number of days reduced, there will be less work. If the number of days is increased, there will be more work. Therefore, here we have direct proportion or direct variation.

ii) Number workers and days required to do a certain work is an example of inverse variation. If more men are employed, they will require fewer days and if there are less number of workers, more days are required.

iii) There is an inverse proportion between the daily hours of a work and the days required. If the number of hours is increased, less number of days are required and if the number of hours is reduced, more days are required.

6. Some important tips: More Men – Less Days and Conversely More Day-Less Men. More Men -More Work and Conversely More Work-More Men. More Days-More Work and Conversely More Work-More Days. The number of days required to complete the given work=Total work/One day’s work.

Since the total work is assumed to be one(unit), the number of days required to complete the given work would be reciprocal of one day’s work. Sometimes, the problems on time and work can be solved using the proportional rule ((man*days*hours)/work) in another situation.

7. If men are fixed, work is proportional to time. If work is fixed, then time is inversely proportional to men, therefore, (M1*T1/W1) = (M2*T2/W2).

**4. Profit and Loss:**

**Important Facts and Formulae:**

**Cost Price:** The price at which an article is purchased, is called its cost price, abbreviated as C.P.

**Selling Price:** The price at which an article is sold is called its selling price, abbreviated as S.P.

**Profit or Gain:** If S.P. Is greater than C.P. The seller is said to have a profit or gain.

**Loss:** If S.P. Is less than C.P., the seller is said to have incurred a loss.

__FORMULAE__

a). Gain=(S.P-C.P)

b). Loss=(C.P-S.P)

c). Loss or Gain is always reckoned on C.P.

d). Gain%=(gain*100)/C.P

e). Loss%=(loss*100)/C.P

f). S.P=[(100+gain%)/100]*C.P

g). S.P=[(100-loss%)/100]*C.P

h). C.P=(100*S.P)/(100+gain%)

i). C.P=(100*S.P)/(100-loss%)

j). If an article is sold at a gain of say,35%, then S.P=135% of C.P.

k). If an article is sold at a loss of say,35%, then S.P=65% of C.P.

l). Net selling price=Marked price-Discount

**5. Partnership:**

**Important Facts and Formulae:**

Definition: When two or more than two persons run a business jointly, they are called partners and the deal is known as Partnership.

The ratio of division of gains:

a) When the investments of all the partners are done at the same time, the gain or loss is distributed among the

partners in the ratio of their investments.

Suppose A and B invest Rs x and Rs y respectively for a year in a business, then at the end of the year:

(A’s share of profit):(B’s share of profit) = x:y

b). When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital*number of units of time). Now gain or loss is divided in the ratio of these capitals.

Suppose A invests Rs x for p months and B invests Rs y for q months, then (A’s share of profit):(B’s share of profit) = xp : yq

3.Working and sleeping partners: A partner who manages the business is known as working partner and the one who

simply invests the money is a sleeping partner.

__FORMULAE__

1.When investments of A and B are Rs x and Rs y for a year in a business, then at the end of the year

(A’s share of profit):(B’s share of profit)=x:y

2.When A invests Rs x for p months and B invests Rs y for q months, then A’s share profit : B’s share of profit = xp : yq

**Short cuts:**

1.In case of 3 A, B, C investments then individual share is to be found then

A=16000 , B=32,000 , C=40,000

Sol: A:B:C=16:32:40

=2:4:5` then individual share can be easily known.

2.If business mans A contributes for 5 months and B contributes for 9 months

then share of B in the total profit of Rs 26,8000, A = Rs 15000, B =Rs 12000

Sol: 15000*5 : 12000*9

25 : 36 for 36 parts = 268000*(36/61)

=Rs 158.16

**6. Percentages:**

**Important Facts and Formulae:**

By a certain percent, we mean that many hundredths. Thus x per cent means x hundredths written as x %. So x% = x/100.

#### EXAMPLE:

1.20% = 20/100 = 1/5 2.48% = 48/100 = 12/25

To express a/b as a percent, we have

### a/b=(a/b*100)%

Thus 1/4=(1/4*100)%=25%

0.6=6/10=(3/5*100)%=60%

If the price of commodity increases by r%, then the reduction in consumption so as not to increase the expenditure is given as follows

[r/(100+r)*100]%

If the price of commodity decreased by r%, then the increase in consumption so as not to decrease the expenditure is given as follows

[r/(100-r)*100]%

### RESULT ON POPULATION:

Let the population of a town be P now and suppose it increases at the rate R% per annum then,

i)Population after n years = P[1+(R/100)]nii) Population n years ago= P/[1+(R/100)]n

ii) Population n years ago= P/[1+(R/100)]n

### RESULT ON DEPRECIATION:

Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum then,

i)Value of machine after n years = P[1-(R/100)]n

ii)Value of machine after n years = P/[1-(R/100)]n

### IMPORTANT

i) If A is R% more than B, then B is less than A by [(R/(100+R))*100]%

ii)If A is R% less than B, then B is more than A by [(R/(100-R))*100]%

**7. Races and Games of Skill:**

**Important Facts and Formulae:**

**Races:** A contest of speed in running, riding, driving, sailing or rowing is called a Race.

**Race Course: **The ground or path on which contests are made is called a race course.

**Starting point: **The point at which a race begins is called starting point.

**Winning point or goal: **The point set to bound a race is called a winning point.

**Dead Heat Race: **If all the persons contesting a race reach the goal exactly at the same time, then the race is called a dead heat race.

**Start: **Suppose A and B are two contestants in a race. If before the start of the race, A is at the starting point and B is ahead of A by 12 meters.

Then we say that “A gives B a start 12 meters.

->To cover a race of 100metres in this case, A will have to cover 100m while B will have to cover 88m=(100-12)

->In a100m race ‘A can give B 12m’ or ‘A can give B a start of 12m’ or ‘A beats B by 12m’means that while A runs 100m B runs 88m.

**Games:** A game of 100m, means that the person among the contestants who scores 100 points first is the winner.

If A scores 100 points while B scores only 80 points then we say that ‘A can give B 20 points’.

**8. Puzzles:**

**Important Facts and Formulae:**

Puzzles are dealt in a detailed manner with certain solutions. Different puzzles are gathered from ShakuntalaDevi’s puzzle books. Keeping in mind certain puzzles for Infosys some reasoning problems are also dealt. Puzzle name at the top of each problem will give a brief idea regarding the mode of application.

__Selecting a candidate:__

For an advertisement of six local posts, twelve persons applied for the job.Can you tell in how many different ways the selection can be made?

Solution: 6^12

__Set of Bat and Ball:__

When I wanted to buy a bat and ball, the shopkeeper said they would together cost Rs.3.75.But I did not want to buy a ball.The shopkeeper said that bat would cost 75paise more than the ball.What were the cost of the bat and the ball?

Soluton:

Given that bat and ball together cost Rs.3.75 = 375paise

Let the cost of the ball alone be x.

Given cost of the bat is 75p greater than cost of the ball.

So cost of the bat = x+75

x+x+75 = 375

2x = 375 – 75

2x = 300

x = 150p

Hence cost of the ball = Rs.1.50

=>Cost of the bat = 1.50 + 75 = Rs.2.25

__Playing Children__

A group of boys and girls are playing.15 boys leave.There remain 2 girls for each boy.Then 45 girls leave.There remain 5 boys for each girl.How many boys were in the orginal group?

Solution:

Let B and G represent no.of boys and girls in the original group respectively.

G ———> 2

B-15 ———-> 1

G/B-15 = 2/1

i.e., 2 girls are left for 15 boys who are alone.

G-45 ——————–>1

B-15 ————————->5

5 boys are left out when 15 girls are alone.

=>G/B-15=2/1 ——————————– (1)

=>G-45/B-15 =1/5 —————————- (2)

(1) & (2) =>

G = 2B-30

5G – 225 = B – 15

5 ( 2B – 30 ) = B – 15 + 225

10B = B – 15 + 225 + 150

9B = 360

B = 40

(1)=> G/40-15 = 2

G=50 girls.

**9. Odd Man Out and Series:**

**Important Facts and Formulae:**

In any type of problems, a set of numbers is given in such a way that each one except one satiesfies a particular definite property. The one which does not satisfy that characteristic is to be taken out.

*Some important properties of numbers are given below :*

a). Prime Number Series:

Example: 2,3,5,7,11,……………..

b).Even Number Series

Example: 2,4,6,8,10,12,…………

c).Odd Number Series:

Example: 1,3,5,7,9,11,……………….

d).Perfect Squares:

Example: 1,4,9,16,25,…….

e).Perfect Cubes:

Example: 1,8,27,64,125,……

f).Multiples of Number Series:

Example: 3,6,9,12,15,……. are multiples of 3

g).Numbers in Arthimetic Progression(A.P):

Example: 13,11,9,7…….

h).Numbers in G.P:

Example: 48,12,3,…..

*SOME MORE PROPERTIES:*

a). If any series starts with 0,3,….., generally the relation will be (n2-1).

b). If any series starts with 0,2,….., generally the relation will be (n2-n).

c). If any series starts with 0,6,….., generally the relation will be (n3-n).

d). If 36 is found in the series then the series will be in n2 relation.

e). If 35 is found in the series then the series will be in n2-1 relation.

f). If 37 is found in the series then the series will be in n2+1 relation.

g). If 125 is found in the series then the series will be in n3 relation.

h). If 124 is found in the series then the series will be in n3-1 relation.

i). If 126 is found in the series then the series will be in n3+1 relation.

j). If 20,30 found in the series then the series will be in n2-n relation.

k). If 60,120,210,……….. is found as series then the series will be in n3-n relation.

l). If 222,………… is found then the relation is n3+n

m). If 21,31,………. is series then the relation is n2-n+1.

n). If 19,29,………. is series then the relation is n2-n-1.

o). If series starts with 0,3,………… the series will be in n2-1 relation.

**10. Trains:**

**Important Facts and Formulae:**

a). The time taken by a train x meters long in passing a signal post or a pole or a standing man = time taken by the train to cover x meters.

b). Time taken by a train x meters long in passing a stationary object of length y meters = the time taken by the train to cover x+y meters.

c). Suppose two trains or two bodies are moving in the same direction at u kmph and v kmph such that u > v then their relative speed is u-v kmph.

d). If two trains of length x km and y km are moving in opposite diredtions at u kmph and vmph, then the time taken by the train to cross each other = (x+y)/(u+v) hr.

e). Suppose two trains or two bdies are moving in opposite direction at u kmph and v kmph then, their relative speed = (u+v) kmph.

f). If two train starts at the same time from 2 points A & B towards each other and after crossing they take a & b hours in reaching B & A respectively then A’s speed : B’s speed = (b^1/2 : a^1/2 ).