Quantitative:

Numbers

Ratio, Proportion, Variation

Averages,Mixtures and Alligation

Time,Speed and Distance

Time and Work

Percentages

Profit and Loss

Simple & Compound Interest

Fractions

Partnership

Progression

Permutation and Combination

Probability

Geometry and Mensuration

Geometry and Mensuration

test

Geometry and Mensuration

Simplification

Boats & Streams

Surds & Indices

Number System

Sets & Union

Logarithms

Numbers-Key Notes

Numbers-Key Notes-aptitude-questions-answers-basics

Numbers Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

Divisibility

1. A number is divisible by 2 if it is an even number.

2. A number is divisible by 3 if the sum of the digits is divisible by 3.

3. A number is divisible by 4 if the number formed by the last two digits is divisible by 4.

4. A number is divisible by 5 if the units digit is either 5 or 0.

5. A number is divisible by 6 if the number is divisible by both 2 and 3.

6. A number is divisible by 8 if the number formed by the last three digits is divisible by 8.

7. A number is divisible by 9 if the sum of the digits is divisible by 9.

8. A number is divisible by 10 if the units digit is 0.

9. A number is divisible by 11 if the difference of the sum of its digits at odd

places and the sum of its digits at even places, is divisible by 11.

Important formulas

i. ( a + b )( a - b ) = ( a 2 - b2 )

ii. ( a + b ) 2 = ( a2 + b2 + 2 ab )

iii. ( a - b )2 = ( a2 + b2 - 2 ab )

iv. ( a + b + c ) 2 = a2+ b2 + c2 + 2 ( ab + bc + ca )

v. ( a3 + b3 ) = ( a + b )( a2 - ab + b2 )

vi. ( a3 - b3 ) = ( a - b )( a2 + ab + b2)

vii. Sum of natural numbers from 1 to n      

                    viii. Sum of squares of first n natural numbers is =

                     ix. Sum of cubes of first n natural numbers is A

                                  x. HCF= (HCF of the numerators)/(LCM of the denominators)

xi. LCM= (LCM of the numerators)/HCF of the denominators

xii. Product of two numbers = Product of their H.C.F. and L.C.M

 

Note: When a number N is raised to any integral power m, the digit in the unit’s

place of the resulting value can be determined without actually evaluating the

power. The digits when raised to powers will give values in which the digits in

the unit’s place follow a cylindrical pattern. Following is the pattern to calculate

the digit in the unit’s place of any derived power.

HCF models:-

If N is a composite number such that N = ap . bq . cr ….. where a, b, c are prime factors of N and p,q,r ….. are positive integers, then

(a) The number of factors of N is given by the expression (p + 1) (q + 1) (r + 1)…

(b) It can be expressed as the product of two factors in 1/2 {(p + 1) (q + 1) (r + 1)…..} ways

(c) If N is a perfect square, it can be expressed

    (i)as a product of two DIFFERENT factors in 1/2 {(p + 1) (q + 1) (r + 1)…..   -1} ways

    (ii)as a product of two factors in 1/2 {(p + 1) (q + 1) (r + 1) ….+1} ways

(d) Sum of all factors of N = (ap+1 – 1 / a – 1) . (bq+1 – 1 / b – 1) . (cr+1 – 1 / c – 1)…..

(e) The number of co-primes of N (< N), Ø(N) = N(1 – 1/a) (1 – 1/b) (1 – 1/c) ….

(f) Sum of the numbers in (e) = N/2 . Ø(N)

(g) It can be expressed as a product of two factors in 2n-1, where ‘n’ is the number of different prime factors of the given number N.

Exercise Questions

1. 117 * 117 + 83 * 83 = ?

a) 20698

b) 20578

c) 21698

d) 21268

2. (1/4)3 + (3/4)3 + 3(1/4)(3/4)(1/4 + 3/4) =?

a) 1/64

b)27/64

c) 49/64

d)1

3. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as reminder. What is the smaller number ?

a) 240

b) 270

c) 295

d) 360

4. The 7th digit of (202)3 is

a) 2 b) 4c) 8 d) 6

5. H.C.F. of two numbers is 16. Which one of the following can never be their L.C.M

a) 32

b) 80

c) 64

d) 60

6. What is the remainder when 9 + 92 + 93 + .... + 98 is divided by 6?

a) 3b) 2 c) 0 d)5

7. The sum of the first 100 natural numbers is divisible by

a) 2, 4 and 8

b) 2 and 4

c)2 only

d)none of these

8. For what value of 'n' will the remainder of 351n and 352n be the same when divided by 7?

a) 2

b)3

c)6

d)4

9. Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

a) 144  b) 168c)192 d)none of these

10. Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.

a) 99840b) 99900 c)99960 d) 99990

Answer Key:

1.B; 2.D; 3.B; 4.C; 5.D; 6.C; 7.C; 8.B; 9.C; 10.C

Ratio,Proportion and Variation- Key PointsRatio,Proportion and Variation- Key Points-aptitude-questions-answers-basics

Ratio, Proportion, Variation Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Ratio:

1. Duplicate ratio of a : b is a2 : b2

2. Sub-duplicate ratio of a : b is sqrt of(a) : sqrt of (b)

3. Triplicate ratio of a : b is a3: b3

4. Sub-triplicate ratio of a : b is cuberoot of (a) : cuberoot of (b) 

If a/b = c/d; then a=bc/d

(i) (a+b)/b  = (c+d)/d

(ii) (a-b)/b = (c-d)/d

(iii) a/c = b/d

(iv) b/a = d/c

(v) (a+b)/(a-b) = (c+d)/(c-d)

If a/b=c/d; then simplest possible value of a=c, b=d

(vi) In the ratio a : b, if a > b, then  (a + x / b + x) < a / b (x > 0)

(vii) In the ratio a : b, if a < b, then  (a + x / b + x) > a / b (x > 0)

(viii) In the ratio a : b, if a = b, then  (a + x / b + x) = a / b (x > 0)

Proportion & Variation:

If a is directly proportional to b; then a=kb

If a is inversely proportional to b; then a=k/b

If a is directly proportional to b and inversely proportional to c, then a is directly proportional to b/c

=> a= kb/c

Exercise Questions

1. Ram, Sham and Suresh start business investing in the ratio 1/2 : 1/3: 1/6. The time for which each of them invested their money was in the ratio 8:6:12 respectively. If they get profit of Rs.18000 from the business, then how much share of profit will Ram get?a. Rs.4000b. Rs.6000     c. Rs.8000d. Rs. 10000

2. The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?a. 8 : 9b. 17 : 18c. 21 : 22   d. Cannot be determined

3. p,q and r are three positive numbers and Q=(p+q+r)/2; If (Q-p):(Q-q):(Q-r) = 2:5:7, then find the ratio of p,q and r ?a. 4:3:7b. 12:9:7  c.9:7:4    d. 4:3:2

4.A and B together have Rs. 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?a. Rs. 460  b. Rs. 484  c. Rs. 550  d. Rs. 664

5. Two numbers are respectively 20% and 50% more than a third number.The ratio of the two numbers isa. 2 : 5    b. 3 : 5    c. 4 : 5    d. 6 : 7

6. The ratio of the cost prices of two articles A and B is 4:5.The articles are sold at a profit with their selling prices being in the ratio 5:6.If the profit on article A is half of its cost price, find the ratio of the profits on the articles A and B?a. 7:10        b. 9:11        

c. 5: 9d. 10:11

7. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is the total amount?a. Rs. 14000  b. Rs. 15000  c. Rs. 20000  d. None of these

8. If Rs. 782 be divided into three parts, proportional to ½:2/3:3/4, then the first part is:a. Rs. 182  b. Rs. 190  c. Rs. 196  d. Rs. 204

9.A bag contains 50 paisa, 20 paisa and 10 paisa coins in the ratio 5:3:1.If the total amount in the bag is 640 Rs,find the difference in the amounts contributed by 50 paisa and 20 paisa coins.a. Rs.300  b. Rs.400  c. Rs.380  d. None of these

10. The speed of an engine is proportional to the square root of the number of wagons attached to it. Without any wagons attached to it the speed of the engine is 60km/hr. With 16 wagons attached to it the speed of the engine is 40km/hr; find the maximum number of wagons that can be attached so that the train moves.a. 144         b. 145         c. 142         d. 143

 

Answer Key:

1.c; 2.c; 3.b; 4.b; 5.c; 6.d; 7.a; 8.d; 9.c; 10.d

Averages - Key NotesAverages - Key Notes-aptitude-questions-answers-basics

Averages, Mixtures and Alligation Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews 

Average= (x1+x2+x3+......xn)/n where x1,x2,x3, are quantities.

  Weighted Average= (n1x1+n2x2+n3x3+.....nkxk) / n1+n2+n3+......nk

 where x1,x2,x3.... xk are the quality factorsn1,n2,n3,........nk are the quantity factors

Eg: If the average height of boys= 172cms and that of girls=154 cms, then find average height of the class with 18 boys and 12 girls?

Here n1 and n2 are no. of boys and girls (Quantity factor)x1 and x2 are average heights (Quality factor)

  Mixtures

For mixtures, Average, x'= (n1x1+n2x2)/ (n1+n2)                                                       

Alligation:

 

=> n1/n2 = (x2-x')/x'-x1

   

Exercise questions

 

1. The average wages of a worker during a fortnight comprising 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?

a)83   b) 92           

c)90             

d)97

 

2. The average of 5 quantities is 6. The average of 3 of them is 8. What is the average of the remaining two numbers?

a) 6.5      b) 4              c) 3               d) 3.5

 

3. The average temperature on Wednesday, Thursday and Friday was 250. The average temperature on Thursday, Friday and Saturday was 240. If the temperature on Saturday was 270, what was the temperature on Wednesday?

a) 240    b)210        

c) 270             

d) 300

 

4. The average age of a group of 12 students is 20years. If 4 more students join the group, the average age increases by 1 year. The average age of the new students is

a) 24               

b)26               

c)23               

d) 22

 

5. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students?

a) 57 kgs            

b) 56.8 kgs      

c)58.2 kgs  

d)52.2 kgs

 

6. The average of 5 quantities is 10 and the average of 3 of them is 9. What is the average of the remaining 2?

a) 11      

b) 12            

c) 11.5          

d) 12.5

 

7. The average age of a family of 5 members is 20 years. If the age of the youngest member be 10 years then what was the average age of the family at the time of the birth of the youngest member?

a) 13.5                

b) 14             

c)15                

d) 12.5

 

8. Average cost of 5 apples and 4 mangoes is Rs.36. The average cost of 7 apples and 8 mangoes is Rs.48. Find the total cost of 24 apples and 24 mangoes.

a) Rs.1044             

b) RS.2088          

c)Rs.720  

d) Rs.324

 

9. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:

a) Rs. 169.50              b) Rs. 170     c) Rs. 175.50            d) Rs. 18

 

10. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

a) 1/3              

b) 1/4                c) 1/5              d) 1/7

 

Answer Key

 

1.d; 2.c; 3.d; 4.a; 5.a; 6.c; 7.d; 8.b; 9.c; 10.c

Concepts and TheoryConcepts and Theory-aptitude-questions-answers-basics

 

Time, Speed and Distance Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Important Formula and Equations

1. Speed, Time and Distance:    Speed = Distance / time    Time  = distance /speed    Distance =speed*time2. km/hr to m/sec conversion:    x km/hr =[x*5/18] m/sec3. m/sec to km/hr conversion:    x m/sec =[x* 18/5] km/h4. If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a5. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then, the average speed during the whole journey is  [xy/(x+y)] km/hr

Key Note:

Caution average speed should not be calculated as average of different speeds, i.e., Ave. speed ≠(Sum of speed / No. of different Speed)There are two different cases when average speed is required.

Case I

When time remains constant and speed varies :If a man travels at the rate of x km/h for t hours and again at the rate of y km/h for another t hours, then for the whole journey, his average speed is given byAverage speed= Total distance/ Total time taken = (xt+yt)/(t+t) = (x+y)/2 kmph

Case II

When the distance covered remains same and the speeds vary :When a man covers a certain distance with a speed of x km/h and another equal distance at the rate of y km/h. then for the whole journey, the average speed is given by Average speed =2xy/(x+y) km/h.

Velocity :The speed of a moving body is called as its velocity. If the direction of motion is also taken into considerationVelocity = (Net displacement of the body)/(Time taken)

Relative speed:

a) Bodies moving in same directionWhen two bodies move in the same direction, then the difference of their speeds is called the relative speed of one with respect to the other.When two bodies move in the same direction, the distance between them increases (or decreases) at the rate of difference of their speeds.b) Bodies moving in opposite directionThe distance between two bodies moving towards each other will get reduced at the rate of their relative speed (i.e., sum of their speeds).Relative speed of one body with respect to other body is sum of their speeds.Increase or decrease in distance between them is the product of their relative speed and time.

Key notes to solve problems

When a moving body covers a certain distance at x km/h and another same distance at the speed of y km/h, then average speed of moving body during its entire journey will be [2xy/(x+y)]km/h

A man covers a certain distance at x km/h by car and the same distance at y km/h by bicycle. If the time taken by him for the whole journey by t hours, then Total distance covered by him is equal to 2txy/(x+y) km.A boy walks from his house at x km/h and reaches the school ' t1 ' minutes late. If he walks at y km/h he reaches ' t2 ' minutes earlier. Then, distance between the school and the house=  ((xy)/(y-x))* (t1+t2)/60 km

If a man walks with (x/y) of his usual speed he takes t hours more to cover a certain distance, then the time to cover the same distance when he walks with his usual speed, (xt)/(y-x)  hours.

If two persons A and B start at the same time in opposite directions from the points and after passing each other they complete the journeys in 'x ' and ' y ' hrs. respectively, then A's speed: B's

speed=speed= y: x

If the speed is (a/b) of the original speed, then the change in time taken to cover the same distance is given by Change in time = ((b/a)-1)*original time

Key notes to solve problems on Trains

The time taken by a train in passing a signal post or a telegraph pole or a man standing near a railway line = (Length of the train)/ (speed of the train)  The time taken by a train passing a railway bridge or a platform or a tunnel or a train at rest=

(x+y)/Speed  where, x = length of the train, y = length of the bridge or platform or standing train or tunnel

 

Time taken by faster train to pass the slower train in the same direction= (x+y)/(u-v); where, x = length of the first train ; y = length of the second train ; u = speed of the first train ; v = speed of the second train and u > v

Time taken by the trains in passing each other while moving in opposite direction =(x+y)/(u+v)

Time taken by the train to cross a man = x/(u-v) where, both are moving in the same direction and x= length of the train; u= speed of the train and v= speed of the man.

Time taken by the train to across a man running in the opposite direction= x/ (u+v)

If two trains start at the same time from two points A and B towards each other and after crossing, they take a and b hours in reaching B and A respectively. Then,  A's speed: B's speed= b: a

A train starts from a place at u km/h and another fast train starts from the same place after t hours at v km/h in the same direction. Find at what distance from the starting place both the trains will meet and also find the time of their meeting.Distance= uvt/(v-u) kmTime=ut/(v-u) hours

The distance between two places A and B is x km. A train starts from A to B at u km/h. One another train after t hours starts from B to A at v km/h. At what distance from A will both the train meet and also find the time of their meetingTime=(x-ut)/ (u+v) + t hoursDistance from A = u(((x-ut)/(u+v))+t) km

 Two trains starts simultaneously from the stations A and B towards each other at the rates of u and v km/h respectively. When they meet it is found that the second train had traveled x km more than the first. Then the distance between the two stations(i.e., between A and B)  is x(u+v)/ (v-u) km

Time and Work-Key Notes

Time and Work-Key Notes-aptitude-questions-answers-basics

Time and Work Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

Time & Work 

(M1D1HI)/ W1 = (M2D2H2)/W2                     Where M1 & M2 represents no of labourers; D1 & D2 represents no of days; H1 & H2 represents no of hours; W1&W2 represents work done.

If there are 2 persons A & B such that A can do work in ‘a’ days and B can do work in ‘b’days. Such that ‘a’ is a multiple of ‘b’, then, time taken by them to complete the work together = Bigger no/Sum of ratios                    

Eg: A can do work in 9 days, B can do work in 18 days. In how many days they will complete the work together.

Bigger no=18, Ratio=9:18=1:2

No of days = 18/(1 + 2)                 = 6 days

If ‘a’ is not a multiple of ‘b’, then time taken by A&B to complete the work together

=     (LCM)/(Sum of ratios)

                     Eg: A can do a piece of work 30 days. B can do work in 45 days. In how many days they will complete the work together?

LCM = 90, Ratio= 30:45=2:3No of days= 90/(2 + 3)  = 90/5 = 18

If there are 3 persons A, B & C whose time taken a,b,c days respectively, to complete a certain work. Time taken by them to complete the work =   LCM of (a, b, c)/(LCM/a + LCM/b + LCM/c)                Note: For 3 persons the common format is

Step1: Find the LCM

Step2: Find the individual share of work i.e. LCM/a, LCM/b, LCM/c.

Step3: Rest methods depend on the question i.e. follow the question patterns.     

Eg: A, B and C can do a work in 15,20,45 days respectively. In how many days they can complete the work together.

LCM=180

No of days= [180/ (180/15 + 180/20 + 180/45)

               = [180/ (12+9+4)]                                                  = [180/25]

               = 7.2 days 

Pipes & Cisterns 

If there are 2 pipes A & B such that A (inlet pipe) & B (outlet pipe). Such that A can fill tank in ‘a’ minutes and B can empty the tank in ‘b’ minutes , then time taken to fill the tank if both are operated together = Bigger no/Difference of ratios                                                     

Eg: A can fill tank in 9 minutes, B can empty the tank in 18 minutes.. In what time the tank be filled, if both pipes work simultaneously?

Bigger no=18, Ratio=9:18=1:2

Time taken to fill the tank = 18/(2 - 1)                                                                         = 18 minutes

If ‘a’ is not a multiple of ‘b’, then time taken by A&B to fill the tank.

=     (LCM)/(Difference of ratios)                     Eg.: An inlet pipe can fill the tank in 30 minutes. B an outlet pipe can empty the tank in 45 minutes. In what time the tank be filled if both pipes work simultaneously?

Time taken to fill the tank= LCM = 90= Ratio= 30:45=2:3= 90/(3 - 2)  = 90/1 = 90 minutes

If there are 3 pipes A, B & C, in which A, B are inlet pipes which takes a,b,minutes respectively to fill the tank and C an outlet pipe which takes c minutes to empty the tank

Time taken by them to fill the tank, if all of them are operated together.

= LCM of abc/ (LCM/a + LCM/b - LCM/c)

Eg: A, B two inlet pipes takes 15,18 minutes to fill the tank and C an oulet pipe takes 45 minutes to empty the tank respectively. In what time the tank be filled if all of them are operated together?

LCM=90No of days= [90/(90/15 + 90/18 - 90/45)               =  [90/(6+5-2)]                                   =  [90/9]               = 10 minutes

Note: In case of division of money with respect to share of each person’s work then share of A = bc/ab+bc+ac

In case of division of money with respect to share of each person’s work then share of B = ac/ab+bc+acIn case of division of money with respect to share of each person’s work then share of C = ab/ab+bc+ac

Same as Share of A:(LCM/a)/ (LCM/a + LCM/b + LCM/c)              Share of B:(LCM/b)/ (LCM/a + LCM/b + LCM/c)              Share of C:(LCM/c)/ (LCM/a + LCM/b + LCM/c)

                                 Eg: A,B,C can do a work in 15,20,45 days respectively. They get Rs 500 for their work. What is the share of A?

LCM = 180

Share of A = (LCM/a x Total amount)/LCM/a + LCM/b + LCM/c

                =  (180/15)/(180/15 +180/20 + 180/45)

                = (12/25) * 500

                =  Rs.240 

Exercise questions 

1. Two workers A and B manufactured a batch of identical parts. A worked for 2 hours and B worked for 5 hours and they did half the job. Then they worked together for another 3 hours and

they had to do (1/20)th of the job. How much time does B take to complete the job, if he worked alone?A) 24 hours B) 12 hours C) 15 hours D) 30 hours

2. Pipe A can fill a tank in 'a' hours. On account of a leak at the bottom of the tank it takes thrice as long to fill the tank. How long will the leak at the bottom of the tank take to empty a full tank, when pipe A is kept closed?A) (3/2)a hours             B) (2/3)a              C) (4/3)a               D) (3/4)a

3. A and B working together can finish a job in T days. If A works alone and completes the job, he will take T + 5 days. If B works alone and completes the same job, he will take T + 45 days. What is T?A) 25  B) 60           C) 15            D) None of these

4. A man can do a piece of work in 60 hours. If he takes his son with him and both work together then the work is finished in 40 hours. How long will the son take to do the same job, if he worked alone on the job?A) 0 hours B) 120 hours C) 24 hours   D) None of these

5. A, B and C can do a work in 5 days, 10 days and 15 days respectively. They started together to do the work but after 2 days A and B left. C did the remaining work (in days)A) 1    B) 3             C) 5             D) 4

6. X alone can do a piece of work in 15 days and Y alone can do it in 10 days. X and Y undertook to do it for Rs.720. With the help of Z they finished it in 5 days. How much is paid to Z?A) Rs.360 B) Rs.120 C) Rs.240 D) Rs.300

7. Ram starts working on a job and works on it for 12 days and completes 40% of the work. To help him complete the work, he employs Ravi and together they work for another 12 days and the work gets completed. How much more efficient is Ram than Ravi?A)50%B) 200%  C) 60%            D)100%

8. A red light flashes 3 times per minute and a green light flashes 5 times in two minutes at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?A) 30 B) 24              C) 20              D) 60

9. A and B can do a piece of work in 21 and 24 days respectively. They started the work together and after some days A leaves the work and B completes the remaining work in 9 days. After how many days did A leave?A) 5    B) 7                C) 8                D) 6

10. Ram, who is half as efficient as Krish, will take 24 days to complete a work if he worked alone. If Ram and Krish worked together, how long will they take to complete the work?A) 16 days B) 12 days  C) 8 days  D) 18 days

Answer Key

1.C; 2.A; 3.C; 4.B;5.D; 6.B; 7.D; 8.A; 9.B; 10.C

Concepts and TheoryConcepts and Theory-aptitude-questions-answers-basics

Percentages Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Percentages 

By a certain percent, we mean that many hundredths. Thus, x percent means x hundredths, written as x%.

To express x% as a fraction, we have x%=x/100.Thus, 20%= 20/100= 1/5To express a/b as a percent, we have, a/b= (a/b)*100%.Thus, 1/4= (1/4)*100%= 25%.

1. If A is R% more than B, then B is less than A by R/ (100+R)*1002. If A is R% less than B, then B is more than A by R/(100-R)*1003. If the price of a commodity increases by R%, then reduction in consumption, not to increase the expenditure is: R/(100+R)*1004. If the price of a commodity decreases by R%, then the increase in consumption, not to decrease the expenditure is: R/(100-R)*100

Results on population:

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

1. Population after n years= p (1+(R/100))n

2. Population n years ago= P/(1+(R/100))n

3. If a number is increased by x% and thereafter reduced by x%, then the number will be reduced by x2/100 percent

4. If a number is reduced by x% and there after increased by x% then the number will be reduced by x2/100 percent

5. If in an examination, in which the minimum pass percentage is x%, a candidate secures y marks and falls by z marks, then the total number of marks in this examination will be 100*(y+z)/x

6. If in an examination x% and y% candidates respectively fail in two different subjects while z% candidates fail in both the subjects, then the percentage of candidates who pass in both the subjects will be [100-(x+y+z)]%

Tips:

1. If an object's price is increased or decreased by x% and the other factor is decreased by y% then the net effect is given by:Net Effect= [x+y+xy/100]%

2. If the net effect is nil, ie, there is no loss or no gain, then the above formula becomes: y=100x/100+x

3. If the price of an article is successively increased by x%,y% and z% then single equivalent increase in the price will be [x+y+z+{xy+yz+zx}/(100)+xyz/1002]%

4. If after spending p1% first, then p2% from the remaining and so on, B is the balance amount, then the total (original) amount is given by:Total amount= B*100*100...../ (100-p2).....

Population formula: 1)If the population increases by x% during the first year, by y% during the second year, by z% during the third year, the population after three years will be:P(1+x/100)(1+y/100)(1+z/100)

Concepts and TheoryConcepts and Theory-aptitude-questions-answers-basics

Profit and Loss Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Important formula and Equations

Gain= SP-CPLoss= CP-SPGain Percentage= (Gain*100)/ CPLoss Percentage= (Loss*100)/ CPSelling Price=((100+Gain %)/100)*CP or ((100-Loss%)/100)*CPCost Price= (100*SP)/(100+Gain%) or (100*SP)/(100-Loss%)

When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by:

Loss %= (Common loss ans gain %)2/10 = (x/10)2

If a trader professes to sell his goods at cost price, but uses false weights, then

Gain%=((Error)/(True value)-(Error))*100%

Key Notes

When an article is sold at a profit of x%. If it would be sold for Rs.n less, there would be a loss of y%, then the cost price of the article CP=(n*100)/(x+y)

A man sells an article at a gain of x%. If it would have been sold for Rs.n more, there would have a profit of y%, then CP= (n*100)/(y-x)

A person brought two articles for Rs.n. On selling one article at x% profit and other at y% profit, he get the same selling price of each, thenCP of first article= Rs. (100+y)n/200+x+y

CP of second article= Rs. (100+x)n/ 200+x+y

When m articles are brought for Rs.n and n articles are sold for Rs.m and m>n, then profit%= ((m2-n2)*100)/n2

If A sells an article to B at a profit of r1 %, B sells it to C at a profit of r2 % and C sells it to D at a profit of r3 %, then, cost price of D= Cost Price of A (1+r1/100)(1+r2/100)(1+r3/100)

If A sells an article to B at a loss of r1 %, B sells it to C at a loss of r2 % and C sells it to D at a loss of r3 %, then, cost price of D= Cost Price of A (1-r1/100)(1-r2/100)(1-r3/100)

A dealer purchases a certain number of articles at x articles for a rupee and the same number at y articles for a rupee. He mixes them together and sells at z articles for a rupee.Then his gain or loss %=([2xy- 1]/z(x+y))*100; according to positive or negative sign.

If P1 is rate gain w.r.t. selling price S1 and P2 is rate gain w.r.t. selling price S2Then CP=(100/P1-P2)* difference between selling prices

If P1 is rate gain w.r.t. selling price S1 and P2 is rate loss w.r.t. selling price S2Then CP= (100/ P1+P2) * difference between selling prices

When a man sells two things at the same price each and in this process his loss on first thing is x% and gain on second thing is x%, then in such a type question, there is always a loss andLoss= 2*SP/((100/x)2 -1)

When a man buys two things on equal price each and in those things one is sold on the profit of x% and another is sold on the loss of x%, then there is no loss or no gain percent.A sells an article at a profit of r1 % to B and B again sells it to C at a profit of r2 %. If C pays Rs. P to B, then CP of the article forA= Rs. 100*100*P/(100+r1)(100+r2)

When a shopkeeper on selling an article for Rs.n, gains as much percent as the cost price of

it,then CP of the article                                 

If there is loss in place of profit,

then CP of the article=  

If two articles are sold at the same price (i.e., the selling prices are equal) and the magnitude of percentage of profit x on one article is the same as the magnitude of percentage of loss x on the second article, then there is an overall loss and the percentage of loss is x2/100.

If a shopkeeper claims to sell the goods at cost price and gives x units less than the actual weight, then the profit percentage made by the shopkeeper is [x / actual weight – x] x 100.

In the above case, the error percentage = [x / actual weight] x 100

If two articles are bought for the same price (i.e., the cost prices are equal) and one is sold at a profit of p1% and the second is sold at a profit of p2%, then the overall percentage of profit is ((p1 + p2 )/2) x 100

If the selling price of m articles is equal to cost price of n articles, where m > n, then profit percentage is ((m – n )/m)x 100.If m < n, then loss percentage = ((n – m)/m) x 100.

Discount

Discount% = Discount / Marked price * 100%

An article sold at selling price(SP1) at a loss of x% is to be sold at selling price(SP2) to gain y%, then SP2 = SP1(100 + y)/ (100-x)

If selling an object for Rs.x a person loses a certain sum and selling for Rs.y he gains the same amount, CP is given by CP = (x+y)/2.

When the price of an article is reduced by p% a man can buy x quantity of the article for Rs.y thenreduced price = 1/x ( y * p / 100) per unit.original price = reduced price * 100 / (100 - p).

If the MP (marked price) of an article above CP is M% and after allowing a discount of d%, the gain is g%,Then M% = d+g * 100% / 100 - d, and if there is a loss of l%, then M% = d-l * 100% / 100-d.

A person sells goods at a profit of x%. Had he sold it for Rs. X more, y% would have been gained. Then CP is given by Rs. X *100 / y-x.

A person sells goods at a loss of x%. Had he sold it for Rs. X more, he would have gained y% . Then CP is given by Rs. X * 100/ y+x.

When there are two successive profits of x% and y% the net gain% is given by: Net gain = [ x + y + { xy / 100 }]%.

When there are two successive losses of x% and y% the net loss% is given by: Net loss = [ - x - y + { xy / 100 }]%.10)

When there is a gain of x% and a loss of y% the net effect is given by: Net effect = [ x - y - { xy / 100}]%.

l. If d1, d2, d3….. are percentages of successive discounts on a marked price MP, then the selling price SP = MP (1 – d1/100) (1 – d2/100) (1 – d3/100)

2. If d1, d2, d3…. are the percentages of successive discounts offered, then the effective discount is d% = 100[1- (1 – d1/100) (1 – d2/100) (1 – d3/100)…]

3. If x and y are two successive discount percentages, then it is equivalent to a single discount percentage of x + y – xy/100.

Simple and Compound Interest- Key NotesSimple and Compound Interest- Key Notes-aptitude-questions-answers-basics

Simple and Compound Interest Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Important formula and equations

Principal: The money borrowed or lent out for a certain period is called the principal or the sum.Interest: Extra money paid for using other's money is called interest.Simple Interest (SI): If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Let Principal= P, Rate= R% per annum (p.a) and Time= T years. Then (i) Simple Interest= (P*R*T)/100(ii) P = (100*SI)/(R*T); R= (100*SI)/(P*T) and T= (100*SI)/(P*R)

Key notes on Simple Interest

A sum of money becomes n times itself in T years at simple interest, then the rate of interest is Rate= 100(n-l)%               T If a sum of money becomes n times in T years at SI then it will be m times of itself in ..... yearsRequired time= (m-l)*T years                          (n-l) If SI on a sum of money is 1/xth of the principal and the time T is equal to the rate percent R, then

Rate= Time=A certain sum is at SI at a certain rate for T years. And if it had been put at R1 % higher rate, then it would fetch Rs.x more, then the Principal= x*100                 T*R1The annual payment that will discharge a debt of Rs.P due in T years at the arte of interest R% per annum is Annual payment =100P                               100T+RT(T-1)                   2                                                                                   Let the rate of interest for first 1 years is r1% per annum, for the next t2 years is r2 % per annum and for the period beyond that is r3 %. Suppose all together the simple interest for t3 years is Rs.I. Then Principal=100*I                                                                            t1r1+t2r2+(t3-t1-t2)r3The simple interest on a certain sum of money at r1 % per annum for t1 years is Rs.m. The interest on the same sum for t2 years at r2 % per annum is n.Then the sum= (m-n)*100                         r1t1-r2t2

Key notes on Compound interest

Compound Interest: (Amount - Principal)Amount= P* (1+R/100)n When the interest is compounded K times a year, Amount= P( 1 + R / K*100)kt

When the interest is paid half yearly, say at r%per annum compound interest, then the amount after t years is given by: P( 1 + R / 2*100)2t Similarly, if the interest is paid quarterly, say at r% per annum compound interest, then the amount due after t years is given by: P( 1 + r / 4 * 100)4t

Under the method of equated instalments, the value of each instalment is the same.

Equal Annual Instalment under(a) Simple Interest, x = 2P(100 + nr)                                     n[200 + (n - 1)r](b) Compound Interest, x = Pr / 100[1 – (100/100 + r) n ]

Exercise questions

1.A father left a will of Rs.35 lakhs between his two daughters aged 8.5 and 16 such that they may get equal amounts when each of them reach the age of 21 years. The original amount of Rs.35 lakhs has been instructed to be invested at 10% p.a. simple interest. How much did the elder daughter get at the time of the will?A) Rs.17.5 lakhs B)Rs.21 lakhs C) Rs.15 lakhs D) Rs. 20 lakhs

2.What will Rs.1500 amount to in three years if it is invested in 20% p.a. compound interest, interest being compounded annually?A) 2400 B) 2592 C) 2678 D)2540

3. If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to treble if invested at the same rate of interest in a scheme where interest is computed using simple interest method?A) 9 years B) 22 years C) 18 years D)33 years

4. The population of a town was 3600 three years back. It is 4800 right now. What will be the population three years down the line, if the rate of growth of population has been constant over the years and has been compounding annually?

A) 6000 B) 6400 C) 7200 D)9600

5. A man invests Rs.5000 for 3 years at 5% p.a. compound interest reckoned yearly. Income tax at the rate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of the third year.A) 5624.32 B)5630.50 C)5788.125 D)5627.20

6. The difference between the compound interest and the simple interest on a certain sum at 12% p.a. for two years is Rs.90. What will be the value of the amount at the end of 3 years?A) 9000 B) 6250 C) 8530.80 D)8780.80

7. Vijay invested Rs.50,000 partly at 10% and partly at 15%. His total income after a year was Rs.7000. How much did he invest at the rate of 10%?A) Rs.40,000 B)Rs.40,000 C)Rs.12,000 D)Rs.20,000

8. A sum of money invested for a certain number of years at 8% p.a. simple interest grows to Rs.180. The same sum of money invested for the same number of years at 4% p.a. simple interest grows to Rs.120 only. For how many years was the sum invested?A) 25 years B) 40 years C) 33 years and 4 months D)Cannot be determined

9. How long will it take for a sum of money to grow from Rs.1250 to Rs.10,000, if it is invested at 12.5% p.a simple interest?A) 8 years B) 64 years C) 72 years D)56 years

10. Rs.5887 is divided between Shyam and Ram, such that Shyam's share at the end of 9 years is equal to Ram's share at the end of 11years, compounded annually at the rate of 5%. Find the share of Shyam.A) 2088 

B) 2000 C) 3087 D)None of these

Answer Key

1.B; 2.B; 3.B; 4.B; 5.A; 6.D; 7.B; 8.A; 9.D; 10.C

FractionsFractions-aptitude-questions-answers-basics

Fractions Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Simple or Vulgar Fraction

A number expressed with numerator and denominator. Say I have 3 of 10 apples then I will express it as 3/10. The total is written below a horizontal or diagonal line, and the number of parts comprising the fraction (numerator) is written above. Such fractions are called vulgar fractions or simple fractions. Eg:[ 3/4 ]

Decimal Fraction

Expressing the fraction in decimal values (denominator a power of 10) is called decimal fraction. 1/2 is expressed as 0.5 in decimal fraction. Eg:[ 0.45773 ]

Converting a decimal to vulgar fraction:

Step 1: Calculate the total numbers after decimal point.

Step 2: Remove the decimal point from the number.

Step 3: Put 1 under the denominator and annex it with "0" as many as the total in step a.

Step 4: Reduce the fraction to its lowest terms.Example: Consider 0.44

Step 1: Total number after decimal point is 2

Step 2 and 3: 44/100 

Step 4: Reducing it to lowest terms : 44/100 = 22/50 = 11/25

Converting a recurring decimal to vulgar fraction

A decimal with recurring value is called recurring decimal. E.g: 2/9 will give 0.22222222...... where 2 is recurring number.

Method:

Step 1: Separate the recurring number from the decimal fraction.

Step 2: Annex denominator with "9" as many times as the length of the recurring number.

Step 3: Reduce the fraction to its lowest terms.

Example: Consider 0.2323232323

Step 1: The recurring number is 23

Step 2: 23/99 [the number 23 is of length 2 so we have added two nines] 

Step 3: Reducing it to lowest terms : 23/99 [it can not be reduced further].

Mixed Recurring to Fractions:

If N= 0.abcbcbc…. Then N = abc - a / 990 = Repeated & non-repeated digits - Non repeated digits / As many 9's as repeated digits followed by as many zero as non - repeated digits

Eg: 0.25757..... = 257 - 2 / 990 = 255 / 990 = 17 / 60.

PartnershipPartnership-aptitude-questions-answers-basics

Partnership Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Important formula and equation

Partnership:

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

Ratio of Division of Gains:

I.When investments of all the partners are for 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 .

II. When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x 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 .

Working and Sleeping Partners:

A partner who manages the the business is known as a working partner and the one who simply invests the money is a sleeping partner .

Kind of partners :There are two kinds of partners.

1.Working or active partner :When a partner devotes his time for the business in addition to invest his money, he is called a working partner. With mutual agreement, the active partners get some fixed percentage of profit as working allowance.

2.Sleeping or non - active partnerA partner who simply invests money, but doest not attend to the business is called a sleeping partner.

Kinds of Partnership :

1.  Simple partnershipIf the capitals of several partners are invested for the same period. It is called a simple partnership.

2. Compound or complex partnership :If the capitals of the partners are invested for different intervals of time, the partnership is called compound or complex.

Key notes

If the capitals of two partners are invested for the same duration of period and let A 1 and A 2 be their investments and total profit is Rs. P, then share of the partners in the profits are (A1*P)/(A1+A2)  and Rs. (A2*P)/A1+A2           If the capitals of two partners be Rs.A1 and A2 for the periods t1 and t2 respectively and the profit be Rs.P, then shares of the partners in the profit are

(A1*t1*p)/A1t1+A2t2 and Rs. (A2*t2*P)/A1t1+A2t2

ProgressionProgression-aptitude-questions-answers-basics

Progression Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews 

Important Equations and Formula

Sum of first n natural numbers= n(n+1)/2Sum of the squares of first n natural numbers= (n(n+1)(2n+1))/6Sum of the cubes of first n natural numbers= [n (n+1)/2]2

Sum of first n natural odd numbers= n2

Average = Sum of items/ Number of items

Arithmetic Progression (AP): An AP is of the form a, a+d, a+2d, a+3d,..... where a is called the 'first term' and d is called the 'common difference'.nth term of an AP; tn=a+(n-1)dSum of the first n terms of an AP; Sn= n/2 [2a+(n-1)d] or Sn= n/2 (first term+last term)

Geometrical Progression (GP):A GP is of the form a, ar, ar2, ar3......... where a is called the 'first term' and r is called the 'common ratio'.nth term of a GP; tn= arn-1

Sum of the first n terms in a GP; Sn= a(1-rn)/1-rSum of infinite series of progression; S= a/(1-r)Geometric mean of two number a and b is given as GM= sqrt(ab)

Harmonic Progression (HP)

If a1,a2,a3,...................an are in AP, then 1/a1, 1/a2, 1/a3, ........1/an, are in HPnth term of this HP, tn =1/(1/a1+(n-1)(a1-a2/a1a2) ) nth term of this HP from the end, tn=1/ (1/a1-(n-1)(a1-a2/a1a2))

If a and b are two non-zero numbers and H is harmonic mean of a and b then a, H, b from HP and then H=2ab/(a+b)

Arithmetico-Geometric series

A series having terms a, (a+d)r, (a+2d)r2,...... etc is an Arithmetico-Geometric series where a is the first term, d is the common difference of the Arithmetic part of the series and r is the common ratio of the Geometric part of the series.

The nth term tn= [a+(n-1)d]rn-1

The sum of the series to n terms isSn= a/1-r+ dr (1-rn-1)/ (1-r)2 - [a+(n-1)d]rn/ 1-rThe sum to infinity, S= a/ 1-r + dr (1-rn-1)/(1-r)2 ; r<1

Exponential Series

ex = 1+x/1!+x2/2!+x3/3!+..........   (e is an irrational number)coefficient of xn= 1/n!; Tn+1=xx/n!e-x = 1-x/1!+x2/2!- x3/3!+..........

Logarithmic Series

loge (1+x)= x-x2/2+x3/3+x4/4+........  (-1<x 1)

loge (1-x)= -x-x2/2-x3/3- x4/4 -........  (-1x< 1)

loge (1+x)/(1-x)= 2-(x+x3/3+x5/5+.........) (-1<x 1)

Permutation and Combination- Key NotesPermutation and Combination- Key Notes-aptitude-questions-answers-basics

Permutation and Combination Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews 

nPr =  n!/ (n-r)! 

nCr =  n!/r!(n-r)! 

nCr = nC(n-r)nCr =  nPr/r!

0!=11!=1nPn=n!nP1=nnC1=nnCn=1

Exercise questions

1. How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively?A)5!     B)4!            C)6! - 2!       D)6! / 2!

2. There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?A) 2 * 19!             B)18! * 18                 C) 19! * 18               D)2 * 18!

3. There are 12 yes or no questions. How many ways can these be answered?A) 1024  B) 2048            C) 4096  D)144

4. How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?A) 256    B) 12            C) 81            D) None of these

5. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?A) 9        B)26              

C)126             D) 3920

6. How many numbers are there between 100 and 1000 such that at least one of their digits is 6?A) 648    B) 258           C) 654            D)252

7. How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters?A) 510                  B) 105           C) 10P5 D) 10C5

8. In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?A) 9!/4    B) 9!/(4!*5!)C) 4!*5!D) None of these

9. In how many ways can 15 people be seated around two round tables with seating capacities of 7 and 8 people?A) 15!/(8!)            B) 7!*8!            C) (15C8)*6!*7!            D)2*(15C7)*6!*7!

10. If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?A) 24 B)31 C) 32 D)30

11. How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?A) 16C7 * 7!  B) 12C4 * 4C3 * 7!  C) 12C3 * 4C4  D) 12C4 * 4C3

12. In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?A) 5C3   B) 5P3           C) 53            D)35

13. How many number of times will the digit '7' be written when listing the integers from 1 to 1000?A) 271    B) 300            C) 252           D)304

14. There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done isA) 5B) 21C) 33 D) 60

15. What is the value of 1*1! + 2*2! + 3!*3! + ............ n*n!, where n! means n factorial or n(n-A(n-2)...1A) n(n-A(n-A!))    B) (n+A!)/(n(n-A))     C) (n+A! - n!)  D) (n + A! - 1!)

 

Answers

1.B; 2.D; 3.C; 4.C; 5.C; 6.D; 7.A; 8.C; 9.C; 10.C; 11.B; 12.D; 13.B; 14.B; 15.D

Concepts and TheoryConcepts and Theory-aptitude-questions-answers-basics

Probability Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Probability is a measure of how likely a particular outcome to an event is to happen.It ranges from 0 to 1. A probability of 0 means that the outcome cannot happen. A probability of 1 means that the outcome will definitely happen. And in between 0 and 1 means that the outcome may happen.

Example with a coin.

When a coin is tossed the outcome (or event) can be heads or tails. What is the probability it is tails?Since each outcome, heads or tails, is equally likely we can say that the probability of each is 0.5.p(coin toss is tails)= 1/2

Basic rule of probability

More generally we can say that where there are n equally likely outcomes then the probability of each of these possibilities will be 1/n. So we can say thatp(outcome)= (number of ways it can happen)/(total number of possible outcomes)

This is the basis of all probability questions in the GMAT

Example with die

What is the probability of rolling a 6 when you throw a 6 sided die.Each number from 1 to 6 is equally likely to be thrown and only one of those outcomes is a 6 so using the general rule we can say thatp(throw a six)= 1/6

Example with cards

If you pick a card at random from a deck of cards what is the probability that it is an ace?There are 52 cards in a pack and those there are 4 aces sop (an ace)= 4/52 - 1/13

The probability two outcomes for independent events both occur can be found by multiplying their probabilities.p(A and B)- p(A)*p(B)

Example with coins

What is the probability of throwing two heads in a row when tossing a coin?This is the same as asking what the probability that the first coin tossed will be head AND the second coin tossed will be a head.So the probability that of tossing two heads in a row is 1/4

Example with a jar

A jar contains 2 red balls and 4 green balls. What is the probability that two balls selected at random from the jar are both green?Each ball is equally likely to be selected from the jar so we can work out the probability of the first ball selected being green.Here is where we need to be careful, once we have taken 1 green ball out of the jar, the jar contains only 3 green balls ans 2 red balls so Now we can say that,So the probability that of picking out two green balls is 2/5.

Total Probability Formula.

P(A) = P(AjH1)P(H1) + ¢ ¢ ¢ + P(AjHn)P(Hn):

Total probabilityEvents H1;H2; : : : ;Hn form a partition of the sample space S if(i) They are mutually exclusive (Hi ¢ Hj = ;; i 6= j) and(ii) Their union is the sample space S; i=1 Hi=S;The events H1; : : : ;Hn are usually called hypotheses and from their definition follows that P(H1) +¢ ¢ ¢ + P(Hn) = 1 (= P(S)):

Let the event of interest A happens under any of the hypotheses Hi with a known (conditional) probability P(AjHi): Assume, in addition, that the probabilities of hypotheses H1; : : : ;Hn are known. Then P(A) can be calculated using the total probability formula.

Total Probability Formula:P(A) = P(AjH1)P(H1) + ¢ ¢ ¢ + P(AjHn)P(Hn):The probability of A is the weighted average of the conditional probabilities P(AjHi) with weights P(Hi):

Bayes Formula:

Let the event of interest A happens under any of hypothesesHi with a known (conditional) probability P(AjHi): Assume, in addition, that the probabilities of hypotheses H1; : : : ;Hn are known (prior probabilities). Then the conditional (posterior) probability of the hypothesis Hi; i = 1; 2; : : : ; n, given that event A happened, isP(HijA) = P(AjHi)P(Hi) P(A) ; where P(A) = P(AjH1)P(H1) + ¢ ¢ ¢ + P(AjHn)P(Hn):

Assume that out of N coins in a box, one has heads at both sides. Such “two-headed” coin can be purchased in Spencer stores. Assume that a coin is selected at random from the box, and without inspecting it, flipped k times. All k times the coin landed up heads. What is the probability that two headed coin was selected?

Denote with Ak the event that randomly selected coin lands heads up k times. The hypotheses areH1-the coin is two headed, and H2 the coin is fair. It is easy to see that P(H1) = 1=N and P(H2) = (N ¡ 1)=N.The conditional probabilities are P(AkjH1) = 1 for any k, and P(AkjH2) = 1=2k:By total probability formula,P(Ak) = (2k + N- 1)/2kN and

P(H1jAk) = 2k/(2k + N- 1) 

Conditional Probability

Problem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test:

Analysis: This problem describes a conditional probability since it asks us to find the probability that the second test was passed given that the first test was passed. In the last lesson, the notation for conditional probability was used in the statement of Multiplication Rule 2.

Multiplication Rule 2:When two events, A and B, are dependent, the probability of both occurring is P(A and B)= P(A).P(B|A)The formula for the Conditional probability of an event can be derived from Multiplication on Rule 2 as follows.

Step1: P(A and B)=P(A).P(B|A) start with multiplication rule 2.Step2: P(A and B)/P(A) = (P(A).P(B|A))/P(A) Divide both sides of the equation by P(A)Step3: P(A and B)/P(A) = P(B|A) Cancel P(A)s on right-hand side of the equationStep4: P(A and B)/P(A) = We have derived the formula for conditional probability

Problem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test?Solution: P(Second|First) = P(First and Second)/P(First)= = 0.25/0.42 = 0.60 = 60%

Geometry and Mensuration-Key PointsGeometry and Mensuration-Key Points-aptitude-questions-answers-basics

 

Geometry and Mensuration Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

Geometry and Mensuration

Mensuration: Mensuration is the branch of mathematics which deals with the study of Geometric shapes , Their area , Volume and different parameters in geometric objects.

Some important mensuration formulas are:

1.Area of rectangle (A) = length(l) * Breath(b);

 

2.Perimeter of a rectangle (P) = 2 * (Length(l) + Breath (b))

 

3.Area of a square (A) = Length (l) * Length (l)

 

4.Perimeter of a square (P) = 4 * Length (l)

 

5.Area of a parallelogram(A) = Length(l) * Height(h)

  

 

6.Perimeter of a parallelogram (P) = 2 * (length(l) + Breadth(b))

 

7.Area of a triangle (A) = (Base(b) * Height(b)) / 2

  

And for a triangle with sides measuring “a” , “b” and “c” , Perimeter = a+b+c

and s = semi perimeter = perimeter / 2 = (a+b+c)/2

And also . Area of triangle,

  

This formulas is also knows as “Hero’s formula”.

 

8.Area of triangle(A)

  = 

 

9. Area of isosceles triangle = 

Where , a = length of two equal side , b= length of base of isosceles triangle.

 

10.Area of trapezium (A) =(a+b)/2

Where , “a” and “b” are the length of parallel sides.

 

11.Perimeter of a trapezium (P) = sum of all sides

 

12.Area f rhombus (A) = Product of diagonals / 2

 

13.Perimeter of a rhombus (P) = 4 * l

where l = length of a side

 

14.Area of quadrilateral (A) = 1/2 * Diagonal * (Sum of offsets)

 

15. Area of a Kite (A) = 1/2 * product of it’s diagonals

 

16. Perimeter of a Kite (A) = 2 * Sum on non-adjacent sides

 

17. Area of a Circle (A)  =

  , where , r= radius of the circle

 

18. Circumference of a Circle  =

r= radius of circle

d= diameter of circle

 

19. Total surface area of

cuboid = 2(lb+bh+lh)

where , l= length , b=breadth , h=height

 

20. Total surface area of

cuboid =  6l2

where , l= length

 

21. length of diagonal of cuboid =

 

 

 

22. length of diagonal of cube = 

 

23. Volume of cuboid = l * b * h

 

24. Volume of cube = l * l* l

 

25. Area of base of a cone =

 

 

26. Curved surface area of a cone =C

Where , r = radius of base , l = slanting height of cone

 

27. Total surface area of a cone = 

 

28. Volume of right circular cone = 

Where , r = radius of base of cone , h= height of the cone (perpendicular to base)

 

29. Surface area of triangular prism = (P * height) + (2 * area of triangle)

Where , p = perimeter of base

 

30. Surface area of polygonal prism = (Perimeter of base * height ) + (Area of polygonalbase * 2)

 

31. Lateral surface area of prism = Perimeter of base * height

 

32. Volume of Triangular prism = Area of the triangular base * height

 

33. Curved surface area of a cylinder =

 

Where , r = radius of base , h = height of cylinder

 

34. Total surface area of a cylinder = 

 

35. Volume of a cylinder =

 

36. Surface area of sphere = 

where , r= radius of sphere , d= diameter of sphere

 

37. Volume of a sphere =

 

 

38. Volume of hollow cylinder =

 

where , R = radius of cylinder , r= radius of hollow , h = height of cylinder

 

39. Surface area of a right square pyramid =

 

Where , a = length of base , b= length of equal side ;

of the isosceles triangle forming the slanting face.

 

40. Volume of a right square pyramid =

 

 

41. Area of a regular hexagon =

 

 

42. area of equilateral triangle = 

 

43. Curved surface area of a Frustums = 

 

44. Total surface area of a Frustums =  

 

45. Curved surface area of a Hemisphere = 

 

46. Total surface area of a Hemisphere =

 

47. Volume of a Hemisphere = 

 

48. Area of sector of a circle = 

where , \theta = measure of angle of the sector , r= radius of the sector

 

Exercise questions

1. What is the area of an equilateral triangle of side 16cm ?a) 243 cm2b) 64 3 cm2

c) 363 cm2d)323 cm2

2. Consider the following figure. <A=x+30; <D= x-40. Find <B?a. 125ob. 55oc. 155od. 122o

3. Find the area of a square, the product of whose diagonals is 66 cm2a) 30 cm2b) 33 cm2c) 36 cm2d) 42 cm2

4. A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1 cubic centimeter cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?a) 9b) 61c) 98d) 54

5. Find the area of a trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.a) 225 cm2b)275 cm2c) 285 cm2d)315 cm2

6. Examine the figure. <ADB=25o; Find <OBC:a. 115ob. 25oc. 50od. 65o

7. The sector of a circle has radius of 21 cm and central angle 1350. Find its perimeter.a) 91.5 cmb)93.5 cmc) 94.5 cmd) 92.5 cm

8. The volumes of two cones are in the ratio of 1 :10 and the radii of the cones are in the ratio of 1 : 2, what is the ratio of their vertical heights?a) 2 : 5b) 1 : 5

c) 3 : 54) 4 : 5

Answer Key

1.c; 2.b; 3.b; 4.d; 5.c; 6.a; 7.a; 8.a

SimplificationSimplification-aptitude-questions-answers-basics

 

Simplification Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

I. 'BODMAS' Rule: This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of a given expression. Here, 'B' stands for 'Bracket', 'O' for 'of', 'D' for Division', 'M' for 'multiplication', 'A' for 'Addition' and 'S' for 'Subtraction'. Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order (),{} and []. After removing the brackets, we must use the following operations strictly in the order:(i) of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction.

Eg: (-5)(4)(2)(-1/2)(3/4)=?-5*4*2*-1/2*3/4= 15

 

II. Modulus of a Real Number: Modulus of a real number a is defined as

a|= a, if a>0 or|a|= -a, if a<0Thus, |5|= 5 and |-5|= -(-5)=5

 

III. Virnaculum (or Bar): When an expression contains Virnaculam, before applying the 'BODMAS' rule, we simplify the expression under the Virnaculum.

 

Exercise Questions

1.((469 + 174)2 - (469 - 174)2)/(469 x 174) = ?a. 2   b. 4   c. 295   d. 643Answer: Option b

Explanation: Given exp. = ((a + b)2 - (a - b)2)/ab =4ab/ab = 4                                                    

 

2. David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?a. 19   b. 28   c. 30   d. 37

Answer: Option c

Explanation: Suppose their paths cross after x minutes. Then, 11 + 57x = 51 - 63x => 120x = 40

x=1/3

Number of floors covered by David in (1/3) min. = (1/3)*57= 19

So, their paths cross at (11 +19) i.e., 30th floor.

 

3. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:a. 22   b. 23   c. 24   d. 26

Answer: Option dExplanation: Let the number of hens be x and the number of cows be y.

Then, x + y = 48 .... (i)

and 2x + 4y = 140 =>   x + 2y = 70 .... (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

Therefore, The required answer = 26.

 

4. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?a. 160   b. 175   c. 180   d. 195

Answer: Option B

Explanation: Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 * 8 * 4) = 160.

Therefore, 160 * 2.40 + x * 3.20 = 432

=> 3.20x = 432 - 384 = 48

   => x = 15.

Hence, total hours of work = (160 + 15) = 175.

 

5. One-third of Rahul's savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?a. Rs. 30,000    b.  Rs. 50,000  c. Rs. 60,000   d. Rs. 90,000

Answer: Option C

Explanation: Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 - x) respectively. Then,

(1/3)x= (1/2)(150000-x)

x/3+x/2=75000

5x/6=75000

x=75000*6/5= 90000

Therefore, Savings in Public Provident Fund = Rs. (150000 - 90000) = Rs. 60000

 

6. A sum of Rs. 1360 has been divided among A, B and C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. B's share is:a. Rs.120   b. Rs. 160   c. Rs.240   d. Rs.300

Answer: Option C

Explanation: Let C's share = Rs. x

Then, B's share = Rs. x/4; A's share = Rs.2/3 * x/4= Rs. x/6

Therefore x/6 + x/4 + x = 1360

  17x/12 = 1360

x= 1360*12/17= Rs.960

Hence, B's share= Rs.960/4= Rs.240

 

7. If a-b= 3 and a2+b2=29, find the value of aba. 10   b. 12   c. 15   d. 18

Answer: Option A

Explanation: 2ab = (a2 + b2) - (a - b)2

= 29 - 9 = 20

=> ab = 10.

 

8. If 45-[28-{37-(15-*)}]= 58, then * is equal to:

a. -29   b. -19   

c. 19   d. 29

Answer: Option c

Explanation: 45-[28-{37-(15-*)}]= 58 =>    45-[28-{37-15+*}]=58

45-[28-37+15-*]=58 =>   45[43-37-*]=58

45-[6-*]=58 => 45-6+*=58

39+*=58 => *=58-39

= 19

Boats and StreamsBoats and Streams-aptitude-questions-answers-basics

 

Boats and Streams  Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

Boats and Streams 

The water of a stream, usually, keeps flowing at a certain speed, in a particular direction. This speed is called the current of the stream. A boat develops speed because of its engine power. The speed with which it travels when there is no current is called speed of boat in still water. When the boat moves in the direction of the current is said to be with the stream/ current or downstream. When the boat moves in the direction opposite to that of the current, it is said to be against the stream is called upstream.

Eg:-If the speed of a boat in still water is ‘u’km/hr and the speed of the stream is ‘v’km/hr then:

* Speed downstream=(u+v)km/hr

* Speed upstream = (u-v)km/hr

If the speed downstream is u km/hr and the speed upstream is v km/hr, then: 

* Speed of boat in still water = ½(u+v)km/hr

* Speed (Rate) of stream = ½ (u-v)km/hr 

Examples 

a) A man can row a boat 12 km/h with the stream and 8km/h against the stream.

Find his speed in still water. 

a) 2km/hr b) 4km/hr c) 8km/hr d) 10km/hr

Solution: Speed of boat in still water = ½(u+v) km/hr = ½ (12+8)=10km/hr

b) A man can row a boat 27km/h with the stream and 11km/h against the stream.

Find speed of stream 

a) 2km/hr b)4km/hr c)8km/hr d)10km/hr

Solution: Speed (Rate) of stream = ½ (u-v) km/hr = ½ (27-11)=8km/hr

c) A boat running downstream covers a distance of 16km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water? 

a)4km/hr b)6km/hr c)8km/hr d) None of these 

Rate of downstream=(16/2 ) kmph=8kmphRate of upstream =(16/4) kmph=4kmphTherefore Speed in still water=1/2(8+4) kmph=6kmph 

Note: If ratio of downstream and upstream speeds of a boat is ‘a:b.’

Then ratio of time taken= b:a

Speed of stream=a-b/a+b *Speed in still water

Speed in still water =a+b/a-b *Speed of stream

Exercise Questions

1. A man rows downstream 32 km and 14km upstream. If he takes 6 hours to cover each distance, then the velocity (in kmph) of the current is:

a)1/2 b)1 c)1and ½ d)2

Solution: Rate downstream=(32/6)kmph; Rate upstream=(14/6)kmphVelocity of current=1/2(32/6-14/6)kmph=3/2kmph=1.5kmph

2. In one hour, a boat goes 11km along the stream and 5km against the stream.

The speed of the boat in still water (in km/hr)is:

a)3 b)5 

c)8 d)9

Solution: Speed in still water=1/2(11+5)kmph=8kmph

3. Speed of a boat in still water is 16km/h. If it can travel 20km downstream in the same time as it can travel 12 km upstream, the rate of stream is. 

a)1km/h b)2km/h c)4km/h d)5km/h

Solution: Speed downstream: Speed upstream=20:12=5:3

Speed of current=5-3/5+3*16=4km/h

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Surds and IndicesSurds and Indices-aptitude-questions-answers-basics

Surds and Indices Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

I. Laws of Indices:

i. am * an = am+n

ii. am/an = am-n

iii. (am)n =amn

iv. (ab)n = anbn

v. (a/b)n = an/bn

vi. a0= 1

 

II. Surds: Let a be a rational number and n be a positive integer such that a1/n = n√a is irrational. Then, na is called a surd of order n.

 

III. Laws of Surds:

     

1. To find √ (a + √b) write it in the form m + n + 2√mn, such that m + n = a and 4mn = b, then √ (a + √b) = ±(√m + √n)

2. (√a.√a.√a….∞) = a3. If (√a + √a + √a……..∞) = p, then p (p – 1) = a.4. If a + √b = c + √d, then a = c and b = d.

 

Examples:1. Simplify: (i) (81)3/4        (ii) (1/64)-5/6        (iii) (256)-1/4

Solution:(i) (81)3/4 =(34)3/4 =33=27.            (ii) (1/64)-5/6 = 645/6 = (26)5/6= 25 = 32        

            (iii) (256)-1/4 =( 1/256)1/4 = [( 1/4)4]1/4 =1/4

 

2. If x=3+2√2, then the value of ( √x- (1/√x)) is:........Solution:

Exercise Questions

1.The value of (√8)1/3 is:a.2   b. 4   c. 2   d. 8

Answer: Option c.(√8)1/3 = (81/2)1/3= 81/6 = (23)1/6= 21/2= √2.

 

2. The value of 51/4 * (125)0.25 is:a. √5       b.5√5       c.5       d.25Answer: Option c50.25 * (53)0.25 = 51 = 5.

 

3. The value of (32/243)-4/5 is:a. 4/9       b. 9/4       c. 16/81       d. 81/16

Answer: Option d.(32/243)-4/5 = (243/32)4/5 = [(3/2)5]4/5 = 81/16

 

4. (1/216)-2/3 ÷ (1/27)-4/3 = ?a. 3/4       b. 2/3       c. 4/9       d. 1/8Answer: Option c.(1/216)-2/3 ÷ (1/27)-4/3 = 2162/3 ÷ 274/3 = (63)2/3 ÷ (33)4/3 = 4/9

 

5. (2n+4 -2.2n)/(2.2n+3) = 2-3 is equal to:a. 2n+1       b. -2n+1 + 1/8        c. 9/8 - 2n       d. 1Answer: Option d.

(2n+4 -2.2n)/2.2n+3  + 1/23= 7/8 + 1/8 = 1  

   

6. If 5√5 * 53 ÷ 5-3/2 = 5a+2 , the value of a is:a. 4       b. 5       c.6       d. 8

Answer: Option a53/2 * 53 ÷ 5-3/2 = 5a+2

53/2 + 3 + 3/2 = 5a+2

3/2 + 3 + 3/2 = a+2a+2=6; a=4

 

7. If √2n =64, then the value of n is:a. 2       b. 4       c. 6       d. 12

Answer: Option d√2n =64 => 2n/2 = 64= 26

n/2=6; n=12

 

8.The simplified form of (x7/2 /x5/2).√y3 /√y )is :a.x2/y       b. x3/y2           c. x6/y3       d. xyAnswer: Option d

(x7/2 /x5/2 ). (√y3 /√y) = x7/2 -5/2. y3/2 - 1/2 = xy

 

Number System- KeynotesNumber System- Keynotes-aptitude-questions-answers-basics

Number System Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews 

Number Systems

 

When we consider a number in a decimal system we can divide it into units, tens, hundreds, one tenth, one hundredth etc. For example the numeral 572.65 can be written as (5*102) + (7*101) + (2*100) + (6*10-1) + (5*10-2).

We say that “10” is the base of the number system.

 

Base

The number which decides the place value of a symbol or a digit in a number.  Alternatively, it is the number of distinct symbols that are used in that system. The base should be a positive integer other than 1. If N is any integer, r is the base of the system and a0, a1, a2… an be the digits required to present N, then

N= anrn+an-1rn-1+ ……. +a1r+a0, where 0≤ai≤r-1

Eg: (i) (143)5 = 1*52 + 4*51 + 3*50 = 48

(ii) (1101)2 = 1*23 + 1*22 + 0*21 + 1*20 = 13.

Note:  The subscript indicates the base. In the above examples 5 and 2 are bases. We can also represent fractions in other bases. For example (0.572)8 = 5*1/8 + 7*1/82 + 2*1/83.

The following table lists some number systems along with their base and symbols.

A=10, B=11, C=12, D=13, E=14, F=15, some books denote ten as “E” and eleven as “e”.

The conversion of a number from one base to the other and the arithmetic operations involving bases other than 10 are discussed in worked out examples.

We need to remember the elementary rules while adding binary numbers.

1.Convert (216.42)8 into base 10.

Sol. (216.45)8 = 2*82 +1*81 + 6*80 + 4*8-1 + 2*8-2

= 128+8+6+ ½ + 1/32 = (142.53125)10

 

2.Convert (1101.11)2 into base 10.

Sol. (1101.11)2 = 1*23 + 1*22 + 0*21 + 1*20 + 1*2-1 + 1*2-2

= 8+4+1+ ½ + ¼ = (13.75)10

 

3.Convert (456)10 into base 8.

4.Convert (27)10 into base 2.

Thus (27)10  =(11011)2

 

5.Find (1100101)2 + (110)2

 

 

6.Find (101110)2 + (111011)2

Sets and Union- KeynotesSets and Union- Keynotes-aptitude-questions-answers-basics

 

Sets and Union Aptitude basics, practice questions, answers and explanations Prepare for companies tests and interviews

 

A set can be defined as a collection of things that are brought together because they obey a certain rule. These 'things' may be anything you like: numbers, people, shapes, cities, bits of text ..., literally anything. The key fact about the 'rule' they all obey is that it must be well-defined. In other words, it enables us to say for sure whether or not a given 'thing' belongs to the collection. If the 'things' we're talking about are English words, for example, a well-defined rule might be: '... has 5 or more letters'. A rule which is not well-defined (and therefore couldn't be used to define a set) might be: '... is hard to spell'

Elements

A 'thing' that belongs to a given set is called an element of that set.For example: Henry VIII is an element of the set of Kings of England

Notation

Curly brackets {...... } are used to stand for the phrase 'the set of ...'. These braces can be used in various ways.

For example: We may list the elements of a set: { − 3, − 2, − 1,0,1,2,3}.

We may describe the elements of a set: { integers between − 3 and 3 inclusive}.

We may use an identifier (the letter x for example) to represent a typical element, a | symbol to stand for the phrase 'such that', and then the rule or rules that the identifier must obey:

{x | x is an integer and | x | < 4} or {x|x ϵ Z, |x| <4 }

The last way of writing a set - called set comprehension notation - can be generalized as:

x | P(x), where P(x) is a statement (technically a propositional function) about x and the set is the collection of all elements x for which P is true.

The symbol  ϵ  is used as follows:

ϵ means 'is an element of ...'. For example: dog ϵ {quadrupeds}

ɇ means 'is not an element of ...'. For example:

Washigton DC ɇ {European capital cities}

A set can be finite: {British citizens}or infinite: {7, 14, 21, 28, 35, …. }.

Sets will usually be denoted using upper case letters: A, B, ...

Elements will usually be denoted using lower case letters: x, y, ...

Some Special Sets

1.Universal Set

The set of all the 'things' currently under discussion is called the universal set (or sometimes, simply the universe). It is denoted by U. The universal set doesn’t contain everything in the whole universe. On the contrary, it restricts us to just those things that are relevant at a particular time. For example, if in a given situation we’re talking about numeric values – quantities, sizes, times, weights, or whatever – the universal set will be a suitable set of numbers (see below). In another context, the universal set may be {alphabetic characters} or {all living people}, etc.

2.Empty set

The set containing no elements at all is called the null set, or empty set. It is denoted by a pair of empty braces: { } or by the symbol f. It may seem odd to define a set that contains no elements. Bear in mind, however, that one may be looking for solutions to a problem where it isn't clear at the outset whether or not such solutions even exist. If it turns out that there isn't a solution, then the set of solutions is empty.

For example:

If U = {words in the English language} then {words with more than 50 letters}= f .

If U = {whole numbers} then {x|x2 = 10} = f .

Operations on the empty set

Operations performed on the empty set (as a set of things to be operated upon) can also be confusing. (Such operations are nullary operations.) For example, the sum of the elements of the empty set is zero, but the product of the elements of the empty set is one (see empty product). This may seem odd, since there are no elements of the empty set, so how could it matter whether they are added or multiplied (since “they” do not exist)? Ultimately, the results of these operations say more about the operation in question than about the empty set. For instance, notice that zero is the identity element for addition, and one is the identity element for multiplication.

3.Equality

Two sets A and B are said to be equal if and only if they have exactly the same elements. In this case, we simply write:

A = B

Note two further facts about equal sets:

The order in which elements are listed does not matter.

If an element is listed more than once, any repeat occurrences are ignored.

So, for example, the following sets are all equal:

{1, 2, 3} = {3, 2, 1} = {1, 1, 2, 3, 2, 2}

(You may wonder why one would ever come to write a set like {1, 1, 2, 3, 2, 2}. You may recall that when we defined the empty set we noted that there may be no solutions to a particular problem - hence the need for an empty set. Well, here we may be trying several different approaches to solving a problem, some of which in fact lead us to the same solution. When we come to consider the distinct solutions, however, any such repetitions would be ignored.)

4.Subsets

If all the elements of a set A are also elements of a set B, then we say that A is a subset of B, and we write: A ⊆ B

For example: If  T = {2, 4, 6, 8, 10} and E = {even integers}, then T ⊆ E

If A = {alphanumeric characters} and P = {printable characters}, then A ⊆ P

If Q = {quadrilaterals} and F = {plane figures bounded by four straight lines}, then Q ⊆ F

Notice that A ⊆ B does not imply that B must necessarily contain extra elements that are not in A; the two sets could be equal – as indeed Q and F are above. However, if, in addition, B does contain at least one element that isn’t in A, then we say that A is a proper subset of B. In such a case we would write: A ⊂ B

In the examples above:

E contains 12, 14, ... , so T ⊂ E

P contains $, ;, &, ..., so A ⊂ P

But Q and F are just different ways of saying the same thing, so Q = F.

The use of ⊂ and ⊆ is clearly analogous to the use of < and ≤ when comparing two numbers.

Note: Every set is a subset of the universal set, and the empty set is a subset of every set.

5.Disjoint

Two sets are said to be disjoint if they have no elements in common.

For example: If A = {even numbers} and B = {1, 3, 5, 11, 19}, then A and B are disjoint.

Operations on Sets

1.Intersection

The intersection of two sets A and B, written A ∩ B, is the set of elements that are in A and in B.

(Note that in symbolic logic, a similar symbol,^, is used to connect two logical propositions with the AND operator.)

For example, if A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, then A ∩ B = {2, 4}.

We can say, then, that we have combined two sets to form a third set using the operation of intersection.

2.Union

In a similar way we can define the union of two sets as follows:

The union of two sets A and B, written A ∪ B, is the set of elements that are in A or in B (or both).

(Again, in logic a similar symbol,V, is used to connect two propositions with the OR operator.)

So, for example, {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}.

You'll see, then, that in order to get into the intersection, an element must answer 'Yes' to both questions, whereas to get into the union, either answer may be 'Yes'.

The ∪ symbol looks like the first letter of 'Union' and like a cup that will hold a lot of items. The ∩ symbol looks like a spilled cup that won't hold a lot of items, or possibly the letter 'n', for intersection. Take care not to confuse the two.

3.Difference

The difference of two sets A and B (also known as the set-theoretic difference of A and B, or the relative complement of B in A) is the set of elements that are in A but not in B.

This is written A - B, or sometimes A \ B.

For example, if A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, then A - B = {1, 3}.

4.Complement

The set of elements that are not in a set A is called the complement of A. It is written A′ (or sometimes AC, or Â). Clearly, this is the set of elements that answer 'No' to the question Are you in A?.

For example, if U = N and A = {odd numbers}, then A′ = {even numbers}.

Notice the spelling of the word complement: its literal meaning is 'a complementary item or items'; in other words, 'that which completes'. So if we already have the elements of A, the complement of A is the set that completes the universal set.

5.Cardinality

The cardinality of a finite set A, written | A | (sometimes #(A) or n(A)), is the number of (distinct) elements in A. So, for example:

If A = {lower case letters of the alphabet}, | A | = 26.

Some special sets of numbers

Several sets are used so often, they are given special symbols.

1.The natural numbers

The 'counting' numbers (or whole numbers) starting at 1, are called the natural numbers. This set is sometimes denoted by N. So N = {0, 1, 2, 3, ...}.

Note that, when we write this set by hand, we can't write in bold type so we write an N in blackboard bold font: N

2.Integers

All whole numbers, positive, negative and zero form the set of integers. It is sometimes denoted by Z. So Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}

In blackboard bold, it looks like this: Z

3.Real numbers

If we expand the set of integers to include all decimal numbers, we form the set of real numbers. The set of reals is sometimes denoted by R.

A real number may have a finite number of digits after the decimal point (e.g. 3.625), or an infinite number of decimal digits. In the case of an infinite number of digits, these digits may:

recur; e.g. 8.127127127...

... or they may not recur; e.g. 3.141592653...

In blackboard bold: R

4.Rational numbers

Those real numbers whose decimal digits are finite in number, or which recur, are called rational numbers. The set of rationals is sometimes denoted by the letter Q.

A rational number can always be written as exact fraction p/q; where p and q are integers. If q equals 1, the fraction is just the integer p. Note that q may NOT equal zero as the value is then undefined.

For example: 0.5, -17, 2/17, 82.01, 3.282828... are all rational numbers.

In blackboard bold: Q

5.Irrational numbers

If a number can't be represented exactly by a fraction p/q, it is said to be irrational.

Examples include: √2, √3.

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1.If x,a and m are any three numbers connected by the relation:

m=ax     (a>0, a≠1), then,

“x” is defined as the logarithm of “m” to the base “a” and is written as:

x= loga m

2.Some important results:

(a) m =a log am

(b) x =log a (ax)

(c) log a 1 = 0

3.Some important theorems:

(a) log a (mn) = log a m + log a n

(b) log a (m/n) = log a m – log a n

(c) log a (mn) = n. log a m

(d) log a m = (log b m) / (log b a) ……. Change of base theorem

(e) log a a  = 1

(f) log a b * log b a = 1