Hi, this blog is for all those IIFT call getters. IIFT has been focusing on economic issues and current affairs. For this you can go through a book "Basic Economics" by Thomas Sowell.
1.What is Economics?
Economics is the branch of knowledge that deals with use of scarce resource to satisfy unlimited human wants. We have to understand that human wants are unlimited in the sense that no amount of wealth and resources can fulfill all wants of a person. On the other hand resources are obviously limited. The basic subject matter of economics is to use these resources and produce goods and services that are capable of satisfying human wants to the maximum extent possible.
Samuelson and Nordhaus phrase it in following words:
“Economics is the study of how societies use scarce resources to produce valuable commodities and distribute them among different people.”
So it is a science related with optimization of use of resources for the fulfillment of human wants.
It must be noted that ‘economics is neutral among ends. It is concerned only with the use of scarce means’ (Robbins)
It means that the subject field includes the use of limited resources only, having alternative uses, to satisfy human wants. It does not include the notions of ‘good’ and ‘bad’. Anything that satisfies a human want (even tobacco, alcohol etc) and has limited availability comes under the subject field of economics. On the other hand, free air, though crucial for our existence in not a subject matter of economics, since it is not scarce.
Some goods are economic goods at one place or time, and non economic goods at other time. Water for example, is not an economic good for that living beside a river, but it is an economic good to those who are required to pay for it.
Economic goods are also known as commodities.
2. What is the subject matter of economics?
Many people think that economics is predominantly related to stock market. The truth is that stock market economics covers only a small portion of the scope of economics.
(A) From the definition of economics given above we can derive the subject matter easily. The definition implies that resources can be put to alternative uses, to produce different commodities. Among these possible alternative uses, a society needs to decide the first question – “What to produce?”
When everyone wants everything in unlimited quantity, a decision has to be taken about the priorities of production. These priorities are determined by the priority on the population in terms of their consuming preference. So our first basic question is related to the decision about the consumption, and hence it is related with Consumer Behaviour.
(B) After making the decision about what to produce among various possible groups of commodities, another question that comes to surface is – “How to produce?” The question arises from the fact that there might be more than one way to make a commodity. Decisions are to be taken about the production method among different available options
It has been said that “Teach a parrot to say -demand and supply; and you will have an economist.”
Some concepts are very fundamental to the understanding of economics. We try to grasp them one by one:
Utility: means capacity to satisfy any want of a person.
Measurement of Utility: There are two different schools of measurement of utility. One pioneered by Marshall suggested that utility of a product can be measured by an imaginary unit called “utils”. It is called the cardinal approach towards utility.
Another groups led by Hick and Allen held the opinion that there can not be a quantifiable measurement of utility, and we can only estimate relative utilities of different products and rank them higher or lower in relation to each other. The second approach is known as ordinal approach.
Marginal Utility: The term “marginal” is a unique modifier. It has similar meaning at all places where it is used i.e. something derived from the last unit. Marginal utility means utility derived from the last unit of consumption at the time of measurement. It varies with each unit of consumption. This concept assumes cardinal (quantifiability) approach to measurement of utility, and it is measured by increase in total utility by consumption of the last unit.
Marginal cost is the cost incurred on last unit of production at any point of time.
Marginal revenue is the sale proceeds of the last unit of sale at any point of time.
Marginal score of Saurav Ganguli is 31 on 24.jan 2007. Prior to the second match at between India and WI, marginal score was 98 (first match).I want to highlight the fact that 'marginal' is a continuously varying concept. The average, as we all know, is the cumulative score divided by number of innings played till date. It varies at a rate much less than the marginal one.
about Law of Increasing Returns last year. Are these two the same?
If not what exactly is the law of increasing returns and if they are the same then could you please explain the concept of marginal utility and the law of diminishing returns
Law of Diminishing Marginal Utility:
If we understand the concept of ‘marginal’ properly, there will be no difficulty in getting the meaning of Law of Diminishing Marginal Utility. The utility derived from last unit declines with each additional unit of consumption.
Example: If a person is thirsty, he will get great utility from the consumption of first glass of water, and less utility from the consumption of second. It will decline with each additional glass of water consumed. Soon it will come down to zero, or even less.
Law of returns to scale:
If all the factors (capital, labour, and land) engaged in production of a commodity are increased in a certain uniform ratio K, the level of production may increase by a level (A) more than K (B) equal to K (C) less than K.
In the first case, the production is under increasing returns to scale, I.e. the level of production rises faster than the level of factors of production, showing increased efficiency of production at higher level.
In the second case, it is a situation of constant returns to scale.
The third case shows the law of diminishing marginal returns to scale at work.
The last case is a general one since all production processes ultimately show diminishing returns to scale. The first case that is ‘increasing returns to scale’ is a specific situation in production not observed frequently.
Sunday, December 27, 2009
Monday, November 23, 2009
LRDI Workshop 2
Questions 16-20:
The fitness school marathon attracted 16 entrants (including sports wiz) this year. Each of the five houses (IIM-A, IIM-B, IIM-C, IIM-L, and IIM-K) was represented by a team of 3 runners and the field was made up by the sports wiz, Aditya. The school houses were competing for the trophy. The number of points by each entrant would be equal to his finishing position.
The five houses were tied for the cup, their totals being equal, although no two entrants tied for the same position. In order to determine the order in which the houses would hold the cup (they agreed to hold it for 73 days each), they multiplied the finishers’ positions together in each house. The house with the smallest product, IIM-K, would hold the cup first and so on to the house with the largest product, IIM-A, which held it last. Unfortunately IIM-B and IIM-C houses were still tied, and had to be separated by the toss of coin.
Aditya later noted that no house had two finishers in consecutive positions, although IIM-B would have achieved this had he not managed to get in between two of their runners.
16. The possible position of Aditya is
a) 1 only b) 6 only c) 11 only d) 6 or 11 e) 16 only
17. The possible positions of IIM-C are
a) 4,6,15 b) 3, 10, 12 c) 3,5,12 d) 6, 11, 16 e) none of these
18. The product of IIM-K is
a) 136 b) 360 c) 120 d) 220 e) 128
19. The product of IIM-A is
a) 180 b) 320 c) 214 d) 455 e) 198
20. IIM-L’s positions are
a) 2,8,14 b) 2,9,14 c) 3,8,15 d) 1,8,13 e) 2,8,13
Questions 21 – 25:
A vending machine has five switches which when operated give soup, cold coffee, frappe, iced tea & lemon tea depending upon the switches that are turned on. The machine is such that each switch supplies two different drinks and each drink is supplied by two different switches. If two switches are turned on, the common drinks if any nullify each other and will not come out at all. To get the drink that one wants, one has to turn on the right combination of switches, put in the money, presses the delivery button, and out comes the drink.
Use the following information and answer the question that follow:
TURNING ON SWITCHES:
1. 1 and 3, we get cold coffee and frappe.
2. 1 and 2, we get soup and lemon tea.
3. 1 and 4, we get ice tea, soup, frappe and lemon tea.
Switch 1, 2, 3, 4 and 5 do not supply cold coffee, ice tea, soup, frappe and lemon tea respectively.
21. What drinks are supplied by turning on switches 2 and 3?
1. Frappe, cold coffee, ice tea 2. Lemon tea, ice tea, cold coffee
3. Soup, frappe 4. Frappe, cold coffee, soup, lemon tea
5. None of these
22. Which drink is supplied by turning on switches 2 and 4?
1. Soup, frappe, ice tea 2. Frappe, lemon tea
3. Frappe, ice tea 4. None of these
5. Cannot be determined
23. Soup is one of the drinks which will be supplied by turning on switches:
1. 1 & 3 2. 1 & 5 3. 2 & 4 4. 2 & 5 5. None of these
24. Which switches should be turned on to get just ice tea and soup?
1. 2 & 3 2. 1, 2, 3 & 4 3. 1, 3 & 4 4. 2, 3 & 4 5. None of these
25. What is produced by turning on switches 1, 2 & 3?
1. Cold coffee, soup, frappe, lemon tea
2. Cold coffee, ice tea, soup, lemon tea
3. Cold coffee, ice tea, soup, frappe
4. Cold coffee, soup
5. Cannot be determined
Questions 26 - 29:
In a competition, a person is allowed to shoot at 4 targets successively, followed by the next competitor. When all have finished one such round, the process is repeated. If a target is hit, the shooter is awarded 2 points. If he misses the target, the others are awarded one point each. The first person to get 60 points wins. In a contest between A = 60, B = 53, and C = 43, out of a total of 78 shots that were fired, only 43 hit the target.
26. Who was the first to shoot?
1. A 2. B 3. C 4. Cannot be determined
27. Who was the second to shoot?
1. A 2. B 3. C 4. Cannot be determined
28. How many targets did A hit?
1. 15 2. 19 3. 17 4. 21
29. How many targets did B miss?
1. 6 2. 10 3. 12 4. 14
Questions 30 - 31:
A tennis tournament of 100 games was conducted between two centers of Career Launcher (CP & KC). Each of the two centers, CP & KC selected their top 100 players. For both centers, the number of male in the teams was greater than 50. Number of female who played for KC center was more than that of CP center. In the tournament, male player of KC played only against male players of CP. However due to unequal number of female, some games were of mixed kind (female vs. male). CP won 60% of the games which were played between males, 33.33% of the games which were played between females and 66.67% of the mixed games. All the games were conclusive. In all CP won 51 games.
30. How many male players were from CP center?
1. 62 2. 64 3. 54 4. 55 5. None of these
31. How many mixed games were played?
1. 7 2. 8 3. 9 4.10 5. None of these
The fitness school marathon attracted 16 entrants (including sports wiz) this year. Each of the five houses (IIM-A, IIM-B, IIM-C, IIM-L, and IIM-K) was represented by a team of 3 runners and the field was made up by the sports wiz, Aditya. The school houses were competing for the trophy. The number of points by each entrant would be equal to his finishing position.
The five houses were tied for the cup, their totals being equal, although no two entrants tied for the same position. In order to determine the order in which the houses would hold the cup (they agreed to hold it for 73 days each), they multiplied the finishers’ positions together in each house. The house with the smallest product, IIM-K, would hold the cup first and so on to the house with the largest product, IIM-A, which held it last. Unfortunately IIM-B and IIM-C houses were still tied, and had to be separated by the toss of coin.
Aditya later noted that no house had two finishers in consecutive positions, although IIM-B would have achieved this had he not managed to get in between two of their runners.
16. The possible position of Aditya is
a) 1 only b) 6 only c) 11 only d) 6 or 11 e) 16 only
17. The possible positions of IIM-C are
a) 4,6,15 b) 3, 10, 12 c) 3,5,12 d) 6, 11, 16 e) none of these
18. The product of IIM-K is
a) 136 b) 360 c) 120 d) 220 e) 128
19. The product of IIM-A is
a) 180 b) 320 c) 214 d) 455 e) 198
20. IIM-L’s positions are
a) 2,8,14 b) 2,9,14 c) 3,8,15 d) 1,8,13 e) 2,8,13
Questions 21 – 25:
A vending machine has five switches which when operated give soup, cold coffee, frappe, iced tea & lemon tea depending upon the switches that are turned on. The machine is such that each switch supplies two different drinks and each drink is supplied by two different switches. If two switches are turned on, the common drinks if any nullify each other and will not come out at all. To get the drink that one wants, one has to turn on the right combination of switches, put in the money, presses the delivery button, and out comes the drink.
Use the following information and answer the question that follow:
TURNING ON SWITCHES:
1. 1 and 3, we get cold coffee and frappe.
2. 1 and 2, we get soup and lemon tea.
3. 1 and 4, we get ice tea, soup, frappe and lemon tea.
Switch 1, 2, 3, 4 and 5 do not supply cold coffee, ice tea, soup, frappe and lemon tea respectively.
21. What drinks are supplied by turning on switches 2 and 3?
1. Frappe, cold coffee, ice tea 2. Lemon tea, ice tea, cold coffee
3. Soup, frappe 4. Frappe, cold coffee, soup, lemon tea
5. None of these
22. Which drink is supplied by turning on switches 2 and 4?
1. Soup, frappe, ice tea 2. Frappe, lemon tea
3. Frappe, ice tea 4. None of these
5. Cannot be determined
23. Soup is one of the drinks which will be supplied by turning on switches:
1. 1 & 3 2. 1 & 5 3. 2 & 4 4. 2 & 5 5. None of these
24. Which switches should be turned on to get just ice tea and soup?
1. 2 & 3 2. 1, 2, 3 & 4 3. 1, 3 & 4 4. 2, 3 & 4 5. None of these
25. What is produced by turning on switches 1, 2 & 3?
1. Cold coffee, soup, frappe, lemon tea
2. Cold coffee, ice tea, soup, lemon tea
3. Cold coffee, ice tea, soup, frappe
4. Cold coffee, soup
5. Cannot be determined
Questions 26 - 29:
In a competition, a person is allowed to shoot at 4 targets successively, followed by the next competitor. When all have finished one such round, the process is repeated. If a target is hit, the shooter is awarded 2 points. If he misses the target, the others are awarded one point each. The first person to get 60 points wins. In a contest between A = 60, B = 53, and C = 43, out of a total of 78 shots that were fired, only 43 hit the target.
26. Who was the first to shoot?
1. A 2. B 3. C 4. Cannot be determined
27. Who was the second to shoot?
1. A 2. B 3. C 4. Cannot be determined
28. How many targets did A hit?
1. 15 2. 19 3. 17 4. 21
29. How many targets did B miss?
1. 6 2. 10 3. 12 4. 14
Questions 30 - 31:
A tennis tournament of 100 games was conducted between two centers of Career Launcher (CP & KC). Each of the two centers, CP & KC selected their top 100 players. For both centers, the number of male in the teams was greater than 50. Number of female who played for KC center was more than that of CP center. In the tournament, male player of KC played only against male players of CP. However due to unequal number of female, some games were of mixed kind (female vs. male). CP won 60% of the games which were played between males, 33.33% of the games which were played between females and 66.67% of the mixed games. All the games were conclusive. In all CP won 51 games.
30. How many male players were from CP center?
1. 62 2. 64 3. 54 4. 55 5. None of these
31. How many mixed games were played?
1. 7 2. 8 3. 9 4.10 5. None of these
Tuesday, November 17, 2009
LRDI 1 workshop
Questions 1-4:
A, B, C, D, E, & F are two sets of triplets, not necessarily in the order given. Each triplet contains 3 siblings.
1. None of the six is married to a sibling or to someone of the same sex.
2. Four of the six are male and two are female.
3. Neither set of triplets contains 3 siblings of the same sex.
4. A and D are married to each other.
5. B is E’s only brother.
1. Who cannot be sister and brother?
a) A and E b) C and F C) D and E d) D and F e) E and F
2) F must be a female if
a) A and E are siblings. b) C and F are siblings c) D and E are siblings. d) C is D’s sister-in-law. e) C is D’s brother-in-law.
3. Which of the following statements must be false?
a) C is D’s sister-in-law. b) E is D’s sister-in-law. c) B is D’s brother-in-law. d) C is D’s brother-in-law. e) E is D’s brother-in-law.
4. If E and F are married to each other, which of the following must be true?
a) C is a male. b) F is a male. c) A is a female. d) B is a female.
e) D is female.
Questions 5-8:
The 1100 students of Career launcher are encouraged to take part in a monthly CAT CARNIVAL, and many do so. Total participants were divided into teams of some members. Each team has one team representative. Each participant contributes 10 paisa a month, of which 5 percent is taken for expenses and 10 percent, is given to charity. The rest of the monthly total is equally divided among those whose team scores the highest total of marks; there is only one team who got the highest marks. This division ignores fractions of a paisa and any sum left over is added to the monthly charity contribution.
Eyebrows were raised when one of the team representatives won Rs. 3.10 in the first month (when there was no additional sum available for charity contribution). In the following month (with the same number of participants) there were murmurings when some other team representative won exactly Rs. 4.80.
5. How many students participated in the CAT carnival in the first month?
a) 558 b) 870 c) 620 d) 920 e) 930
6. The amount available for division in the second month is?
a) Rs. 52.70 b) Rs. 92.00 c) Rs. 93.00 d) Rs. 55.80 e) Rs. 62.00
7. How many winners were there in the second month?
a) 12 b) 9 c) 11 d) 21 e) 10
8. What was the sum given in charity in the second month?
a) Rs. 5.22 b) Rs. 10.90 c) Rs. 8.14 d) Rs. 4.70 e) Rs. 6.20
Questions 9-12:
From a normal pack of playing cards, Aditya, Nitin, Harsh, and Manoj were each given an ace, a king, a queen, a jack and a ten.
Harsh’s five cards were in 3 different suits and consisted of 3 red and 2 black cards. Manoj’s five cards were also in three different suits, his ace being in the same suit as his queen, and his king in the same suit as his jack. Nitin held more than one black card. Aditya’s five cards were all in the same suit. Harsh held the king of spades, and Manoj the ten of diamonds.
9. Among them, Nitin, Harsh and Manoj together held …..black cards.
a) 5 b) 7 c) 8 d) 10 e) cannot be determined
10. Who held the queen of hearts?
a) Manoj b) Harsh c) Nitin d)Aditya e) cannot be determined
11. Who held the ace of diamonds?
a) Manoj b) Harsh c) Nitin d)Aditya e) cannot be determined
12. Who held the king of diamonds?
a) Manoj b) Harsh c) Nitin d)Aditya e) cannot be determined
Questions 13-15:
When the four gamblers P, Q, R and S-sat down for their usual game of poker, each of them placed their stake money (a whole number of rupees). After a number of hands, P found that he had exactly doubled the money with which he had started. It was his turn to provide the evening’s refreshments, and he bought the first round of snacks. After a few-more hands, P exactly doubled the money he had left after buying the first round and he then bought the second round of snacks. Thereafter he repeated the process, i.e. he exactly doubled his remaining money and bought the third round of snacks.
During the fourth session, P took every rupee his three opponents still had on the table and found that he had exactly doubled the money he had left after buying the third round of snacks. He then bought a fourth round of snacks-which took the entire money he had.
Each of the four rounds of snacks cost precisely the same amount (a whole number).
R began the game with a sum (between 50 and 100) which was exactly half the total which Q and S had between them at the start.
13. What is the cost of one round of snacks? (At the minimum)
a) Rs. 96 b) Rs. 72 c) Rs. 78 d) Rs. 84 e) none of these
14. What was the money that P had with him at the beginning?
a) Rs. 70 b) Rs. 75 c) Rs. 80 d) Rs. 85 e) Rs.90
15. What was the money that R had with him at the beginning?
a) Rs. 82 b) Rs. 88 c) Rs. 98 d)Rs. 92 e) cannot be determined
A, B, C, D, E, & F are two sets of triplets, not necessarily in the order given. Each triplet contains 3 siblings.
1. None of the six is married to a sibling or to someone of the same sex.
2. Four of the six are male and two are female.
3. Neither set of triplets contains 3 siblings of the same sex.
4. A and D are married to each other.
5. B is E’s only brother.
1. Who cannot be sister and brother?
a) A and E b) C and F C) D and E d) D and F e) E and F
2) F must be a female if
a) A and E are siblings. b) C and F are siblings c) D and E are siblings. d) C is D’s sister-in-law. e) C is D’s brother-in-law.
3. Which of the following statements must be false?
a) C is D’s sister-in-law. b) E is D’s sister-in-law. c) B is D’s brother-in-law. d) C is D’s brother-in-law. e) E is D’s brother-in-law.
4. If E and F are married to each other, which of the following must be true?
a) C is a male. b) F is a male. c) A is a female. d) B is a female.
e) D is female.
Questions 5-8:
The 1100 students of Career launcher are encouraged to take part in a monthly CAT CARNIVAL, and many do so. Total participants were divided into teams of some members. Each team has one team representative. Each participant contributes 10 paisa a month, of which 5 percent is taken for expenses and 10 percent, is given to charity. The rest of the monthly total is equally divided among those whose team scores the highest total of marks; there is only one team who got the highest marks. This division ignores fractions of a paisa and any sum left over is added to the monthly charity contribution.
Eyebrows were raised when one of the team representatives won Rs. 3.10 in the first month (when there was no additional sum available for charity contribution). In the following month (with the same number of participants) there were murmurings when some other team representative won exactly Rs. 4.80.
5. How many students participated in the CAT carnival in the first month?
a) 558 b) 870 c) 620 d) 920 e) 930
6. The amount available for division in the second month is?
a) Rs. 52.70 b) Rs. 92.00 c) Rs. 93.00 d) Rs. 55.80 e) Rs. 62.00
7. How many winners were there in the second month?
a) 12 b) 9 c) 11 d) 21 e) 10
8. What was the sum given in charity in the second month?
a) Rs. 5.22 b) Rs. 10.90 c) Rs. 8.14 d) Rs. 4.70 e) Rs. 6.20
Questions 9-12:
From a normal pack of playing cards, Aditya, Nitin, Harsh, and Manoj were each given an ace, a king, a queen, a jack and a ten.
Harsh’s five cards were in 3 different suits and consisted of 3 red and 2 black cards. Manoj’s five cards were also in three different suits, his ace being in the same suit as his queen, and his king in the same suit as his jack. Nitin held more than one black card. Aditya’s five cards were all in the same suit. Harsh held the king of spades, and Manoj the ten of diamonds.
9. Among them, Nitin, Harsh and Manoj together held …..black cards.
a) 5 b) 7 c) 8 d) 10 e) cannot be determined
10. Who held the queen of hearts?
a) Manoj b) Harsh c) Nitin d)Aditya e) cannot be determined
11. Who held the ace of diamonds?
a) Manoj b) Harsh c) Nitin d)Aditya e) cannot be determined
12. Who held the king of diamonds?
a) Manoj b) Harsh c) Nitin d)Aditya e) cannot be determined
Questions 13-15:
When the four gamblers P, Q, R and S-sat down for their usual game of poker, each of them placed their stake money (a whole number of rupees). After a number of hands, P found that he had exactly doubled the money with which he had started. It was his turn to provide the evening’s refreshments, and he bought the first round of snacks. After a few-more hands, P exactly doubled the money he had left after buying the first round and he then bought the second round of snacks. Thereafter he repeated the process, i.e. he exactly doubled his remaining money and bought the third round of snacks.
During the fourth session, P took every rupee his three opponents still had on the table and found that he had exactly doubled the money he had left after buying the third round of snacks. He then bought a fourth round of snacks-which took the entire money he had.
Each of the four rounds of snacks cost precisely the same amount (a whole number).
R began the game with a sum (between 50 and 100) which was exactly half the total which Q and S had between them at the start.
13. What is the cost of one round of snacks? (At the minimum)
a) Rs. 96 b) Rs. 72 c) Rs. 78 d) Rs. 84 e) none of these
14. What was the money that P had with him at the beginning?
a) Rs. 70 b) Rs. 75 c) Rs. 80 d) Rs. 85 e) Rs.90
15. What was the money that R had with him at the beginning?
a) Rs. 82 b) Rs. 88 c) Rs. 98 d)Rs. 92 e) cannot be determined
Sunday, November 8, 2009
Number system
N is a natural number such that 300 < N < 750. P is the sum of the number N and the number formed by reversing the digits of N, even if upon reversing, it becomes a two-digit or a single-digit number. P is also equal to ‘K’ times the sum of the digits of the number N.
1. Find the minimum possible value of ‘K’.
1. 28.5 2. 32.75 3. 37.5 4. 40.25 5. 43.5
2. What is the product of all factors of the number N = 6^4 * 10^2 which are divisible by 5?
1. 2^210 * 3^102 * 5^140
2. 2^210 * 3^140 * 5^105
3. 2^140 * 3^210 * 5^102
4. 2^140 * 3^102 * 5^210
5. 2^102 * 3^210 * 5^140
3 .If the product of four positive integers is 10! Then what is the smallest possible value their sum can have?
1. 180 2. 181 3. 175 4. 176 5 . 174
4. 3^36 – 1 = 1A009463A296999120, where A is a single digit whole number, then the value of A is
1. 1 2. 2 3. 8 4. 5 5. 6
5. Out of the 200 even natural numbers. How many even numbers exit having even number of factors?
1. 90 2. 10 3. 190 4. None of these
6. Let d1,d2….dk be all the factors of a positive integer n in ascending order including 1 and n. Supposed d1+d2+d1……..+dk = 72
Then the value of 1/d1 + 1/d2 + 1/d3……….+1/dk is
1. N^2/ 72 2. N/72 3. 72/n 4.d36/ n
7. N = 2^15* 3^7* 5^10 . How many factors of N are multiples of 360 but not multiples of 10800?
1. 402 2. 240 3. 204 4. 420
8. A = 626!-625! How many consecutive zeros would be there at the end of A?
1. 156 2. 160 3. 1 4. None of these
9. A = 4^86 – 2^171. How many digits would be there in binary notation of A?
1. 171 2. 172 3. 87 4. 88
10. A girl starts counting on her fingers of one hand. She starts with her thumb, then her index finger, middle finger, ring finger and finally her little finger. She then continues to count on her ring finger, then her middle finger, then index finger and lastly her thumb. If she continues in this way, on which finger will her count8001.
1. Thumb 2. Little finger 3. Index finger 4. Ring finger
11. In a four digit number, the sum of the first two digits is equal to that of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of his second and fourth digit is twice the sum of the other digits. What is the third digit of the number?
1. 5 2. 8 3. 1 d. 4
12. What is the remainder when 6^83 + 8^83 is divided by 49?
13. S is a six digit number beginning with 1. If the digit 1 is moved from the leftmost Place to the rightmost place the number obtained is three times of S. Then he sum of The digits of S is
1. 24 2. 21 3. 27 4. 26
14. Read the information given below and answer the question that follows: Two natural numbers a and b are given in base 10. The a can be written as 212 in base b and 128 in base b+2.
The value of a+b in base 10 is
1. 114 2. 125 3. 107 4. 219
15. What is the remainder when (81) ^21 + (27) ^21 + (9) ^21 + (3) ^21 + 1 is divided by 3^20 + 1?
1. 61 2. 0 3. 1 4. 121
16. Five consecutive integers are chosen. Let S denote their sum and let P denote their product. If P/S is divisible by 100, what can be the smallest value of S?
1. 125 2. 615 3. 115 4. 625
17. How many zeroes are present at the end of 25! + 26! + 27! + 28! + 30!?
18. In a mathematical game, one hundred people are standing in a line and they are required to count off in a fives
as ‘one,two,three,four,five,one,two,three,four,five,’ and so on from the first person in the line. The person who says ‘five’is taken out of the line. Those remaining repeat this procedure until only four people remain in the line.
What was the original position in the line of the last person to leave?
1. 96 2. 98 3. 97 4. 93
19. The digits of a three digit number A are written in the reverse order to form another three digit number B. If B > A and B – A is perfectly divisible by 7, then which of the following is necessary true?
1. 100 < A < 299 2. 106 < A < 305 3. 112 < A < 311 4. 118 < A< 31 5. None of these
20. A test has 50 questions. A student scores 1 mark for a correct answer, –1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than
1. 6 2. 12 3. 3 4. 9
21. A number is formed by writing first 59 natural numbers in front of each other as 12345678910111213..... Find the remainder when this number is divided by 18.
1. 6 2. 15 3. 9 4. 0
22. Given that N = (521) ^125 × (125) ^521, find the last two digits of N.
1. 75 2. 25 3. 45 4. None of these
23. If ab × cd = 1073 and ba × cd = 2117, find the value of (ab + cd) given that ab, ba and cd are all two digit positive integers.
1. 66 2. 65 3. 63 4. 95
24. A boy is having certain number of marbles, when he divides them in a group of two unequal numbers such that the sum of the squares of marble is equal to 150 times the difference of the two. What could be the total number of marbles?
1.47 2. 53 3. 33 4. 42 5. Cannot be determined
25. Find Remainder when N= 7^7^7^7 divided by 13.
26. The sum of 19 consecutive positive integers equals P^3 where p is prime no. The smallest of these 19 integers is
1.354 2. 353 3. 352 4. 351 5. None of these.
27. Find the number of zeros at the end of
25x30x35x40x45…………………1500.
a. 296 b. 293 c. 374 d. 370
28. A=1^1 x 2^2 x 3^3 x 4^4 x ……100^100. How many zeroes will be there at the end of A?
a. 1300 b. 1320 c. 1325 d. None of these
29. Find the remainder when ( 1×2 + 2×3 + 3×4 + …….. + 99×100 ) is divided by 202.
a. 0 b. 1 c. 99 d. 101 e. None of these
30. Find the highest power of 12 in 536 – 1.
a. 1 b. 2 c. 3 d. 4 e. None of These
1. Find the minimum possible value of ‘K’.
1. 28.5 2. 32.75 3. 37.5 4. 40.25 5. 43.5
2. What is the product of all factors of the number N = 6^4 * 10^2 which are divisible by 5?
1. 2^210 * 3^102 * 5^140
2. 2^210 * 3^140 * 5^105
3. 2^140 * 3^210 * 5^102
4. 2^140 * 3^102 * 5^210
5. 2^102 * 3^210 * 5^140
3 .If the product of four positive integers is 10! Then what is the smallest possible value their sum can have?
1. 180 2. 181 3. 175 4. 176 5 . 174
4. 3^36 – 1 = 1A009463A296999120, where A is a single digit whole number, then the value of A is
1. 1 2. 2 3. 8 4. 5 5. 6
5. Out of the 200 even natural numbers. How many even numbers exit having even number of factors?
1. 90 2. 10 3. 190 4. None of these
6. Let d1,d2….dk be all the factors of a positive integer n in ascending order including 1 and n. Supposed d1+d2+d1……..+dk = 72
Then the value of 1/d1 + 1/d2 + 1/d3……….+1/dk is
1. N^2/ 72 2. N/72 3. 72/n 4.d36/ n
7. N = 2^15* 3^7* 5^10 . How many factors of N are multiples of 360 but not multiples of 10800?
1. 402 2. 240 3. 204 4. 420
8. A = 626!-625! How many consecutive zeros would be there at the end of A?
1. 156 2. 160 3. 1 4. None of these
9. A = 4^86 – 2^171. How many digits would be there in binary notation of A?
1. 171 2. 172 3. 87 4. 88
10. A girl starts counting on her fingers of one hand. She starts with her thumb, then her index finger, middle finger, ring finger and finally her little finger. She then continues to count on her ring finger, then her middle finger, then index finger and lastly her thumb. If she continues in this way, on which finger will her count8001.
1. Thumb 2. Little finger 3. Index finger 4. Ring finger
11. In a four digit number, the sum of the first two digits is equal to that of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of his second and fourth digit is twice the sum of the other digits. What is the third digit of the number?
1. 5 2. 8 3. 1 d. 4
12. What is the remainder when 6^83 + 8^83 is divided by 49?
13. S is a six digit number beginning with 1. If the digit 1 is moved from the leftmost Place to the rightmost place the number obtained is three times of S. Then he sum of The digits of S is
1. 24 2. 21 3. 27 4. 26
14. Read the information given below and answer the question that follows: Two natural numbers a and b are given in base 10. The a can be written as 212 in base b and 128 in base b+2.
The value of a+b in base 10 is
1. 114 2. 125 3. 107 4. 219
15. What is the remainder when (81) ^21 + (27) ^21 + (9) ^21 + (3) ^21 + 1 is divided by 3^20 + 1?
1. 61 2. 0 3. 1 4. 121
16. Five consecutive integers are chosen. Let S denote their sum and let P denote their product. If P/S is divisible by 100, what can be the smallest value of S?
1. 125 2. 615 3. 115 4. 625
17. How many zeroes are present at the end of 25! + 26! + 27! + 28! + 30!?
18. In a mathematical game, one hundred people are standing in a line and they are required to count off in a fives
as ‘one,two,three,four,five,one,two,three,four,five,’ and so on from the first person in the line. The person who says ‘five’is taken out of the line. Those remaining repeat this procedure until only four people remain in the line.
What was the original position in the line of the last person to leave?
1. 96 2. 98 3. 97 4. 93
19. The digits of a three digit number A are written in the reverse order to form another three digit number B. If B > A and B – A is perfectly divisible by 7, then which of the following is necessary true?
1. 100 < A < 299 2. 106 < A < 305 3. 112 < A < 311 4. 118 < A< 31 5. None of these
20. A test has 50 questions. A student scores 1 mark for a correct answer, –1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than
1. 6 2. 12 3. 3 4. 9
21. A number is formed by writing first 59 natural numbers in front of each other as 12345678910111213..... Find the remainder when this number is divided by 18.
1. 6 2. 15 3. 9 4. 0
22. Given that N = (521) ^125 × (125) ^521, find the last two digits of N.
1. 75 2. 25 3. 45 4. None of these
23. If ab × cd = 1073 and ba × cd = 2117, find the value of (ab + cd) given that ab, ba and cd are all two digit positive integers.
1. 66 2. 65 3. 63 4. 95
24. A boy is having certain number of marbles, when he divides them in a group of two unequal numbers such that the sum of the squares of marble is equal to 150 times the difference of the two. What could be the total number of marbles?
1.47 2. 53 3. 33 4. 42 5. Cannot be determined
25. Find Remainder when N= 7^7^7^7 divided by 13.
26. The sum of 19 consecutive positive integers equals P^3 where p is prime no. The smallest of these 19 integers is
1.354 2. 353 3. 352 4. 351 5. None of these.
27. Find the number of zeros at the end of
25x30x35x40x45…………………1500.
a. 296 b. 293 c. 374 d. 370
28. A=1^1 x 2^2 x 3^3 x 4^4 x ……100^100. How many zeroes will be there at the end of A?
a. 1300 b. 1320 c. 1325 d. None of these
29. Find the remainder when ( 1×2 + 2×3 + 3×4 + …….. + 99×100 ) is divided by 202.
a. 0 b. 1 c. 99 d. 101 e. None of these
30. Find the highest power of 12 in 536 – 1.
a. 1 b. 2 c. 3 d. 4 e. None of These
How to prepare for CAT
I would like to recommend you to spend at least 5 hours a day for the ultimate preparation.
Concerned tasks for CAT preparation
Preparing Maths
Preparing Data Interpretation
Preparing Analytical reasoning
Reading comprehension
Preparing Verbal ability
Reading
There are two phases of preparation
First Phase:
In this phase you should get good grip on fundamentals and get familiar with all chapters and problems. Do not look for speed solving or feel any sort of pressure while preparing fundamentals. Learn with your own efficiency, without getting into fundamentals clearly speed-solving is not useful. Please manage to allocate at least one hour a day for reading newspapers, magazines. It is better to subscribe for any good magazines during the preparation. Get in touch with current affairs.
Second Phase (self analyzing and improving):
From second week of August you can take section tests along with revision of topics in alternate day , here if you think you have to allocate more time you should do. You must start taking comprehensive tests by the beginning of September.
Things you look to learn in comprehensive tests of 2-3 hour long, is how to manage paper so the attitude here should be the same like a batsman who wants to score as many runs as possible in the last 10 over’s in one dayers.
After you tested yourself spend time to analyze it. The key points to analyze your test:
Speed:
Concentration
Time Management
Accuracy
Finding traps
Self analysis of the test is very critical. Simple thumb rule says that you have to spend at least 3 times the amount of time that you are spending on test on analysis of test.
Concerned tasks for CAT preparation
Preparing Maths
Preparing Data Interpretation
Preparing Analytical reasoning
Reading comprehension
Preparing Verbal ability
Reading
There are two phases of preparation
First Phase:
In this phase you should get good grip on fundamentals and get familiar with all chapters and problems. Do not look for speed solving or feel any sort of pressure while preparing fundamentals. Learn with your own efficiency, without getting into fundamentals clearly speed-solving is not useful. Please manage to allocate at least one hour a day for reading newspapers, magazines. It is better to subscribe for any good magazines during the preparation. Get in touch with current affairs.
Second Phase (self analyzing and improving):
From second week of August you can take section tests along with revision of topics in alternate day , here if you think you have to allocate more time you should do. You must start taking comprehensive tests by the beginning of September.
Things you look to learn in comprehensive tests of 2-3 hour long, is how to manage paper so the attitude here should be the same like a batsman who wants to score as many runs as possible in the last 10 over’s in one dayers.
After you tested yourself spend time to analyze it. The key points to analyze your test:
Speed:
Concentration
Time Management
Accuracy
Finding traps
Self analysis of the test is very critical. Simple thumb rule says that you have to spend at least 3 times the amount of time that you are spending on test on analysis of test.
Sunday, October 11, 2009
DI
Drill 1 :Become Comfortable with calculations:
You should have a good command on tables and Squares (up to 30).
Now we will check how to do approximate calculations
For example Dividing 283 by 1983 and finding that value as a percentage.
283/1983: whenever denominator is greater than numerator try to approximate denominator. Hence you could take 1983 as 2000.
So now our question looks like 283/2000 which is noting but 0.1415 or 14.15%. Here we change our denominator by 17 which is approximately 1% 0f 1983(19.83).
So in our answer also approximately 1% error is there. 1% of 14.15% is nothing but 0.14%. So our final answer has to be 14.15% + 0.14% = 14.29%.
Now we will check what to do if numerator is greater than denominator, for example Dividing 1132 by 1069.
1132/1069: Here we can write 1132 as 1069 + 63 which will leads to 1 + 63/1069.
So now if we do approximation, it is on very small part. 63/1069≈0.06 and our final answer is 1.06 or 106%.
I hope now the concept of approximation is clear. Just try to solve few questions:
412/6950 a) 5.9% b) 4.7% c) 5.2% d) 6.2% e) 5.72%
511/1680 a) 31.5% b) 29.3% c) 30.1% d) 32.2% e) 30.4%
6.4/6.12 a) 1.045 b) 1.053 c) 1.035 d) 1.023 e) 1.031.
You should have a good command on tables and Squares (up to 30).
Now we will check how to do approximate calculations
For example Dividing 283 by 1983 and finding that value as a percentage.
283/1983: whenever denominator is greater than numerator try to approximate denominator. Hence you could take 1983 as 2000.
So now our question looks like 283/2000 which is noting but 0.1415 or 14.15%. Here we change our denominator by 17 which is approximately 1% 0f 1983(19.83).
So in our answer also approximately 1% error is there. 1% of 14.15% is nothing but 0.14%. So our final answer has to be 14.15% + 0.14% = 14.29%.
Now we will check what to do if numerator is greater than denominator, for example Dividing 1132 by 1069.
1132/1069: Here we can write 1132 as 1069 + 63 which will leads to 1 + 63/1069.
So now if we do approximation, it is on very small part. 63/1069≈0.06 and our final answer is 1.06 or 106%.
I hope now the concept of approximation is clear. Just try to solve few questions:
412/6950 a) 5.9% b) 4.7% c) 5.2% d) 6.2% e) 5.72%
511/1680 a) 31.5% b) 29.3% c) 30.1% d) 32.2% e) 30.4%
6.4/6.12 a) 1.045 b) 1.053 c) 1.035 d) 1.023 e) 1.031.
Basic DI tool
Here are some really basic tools to help you make calculations faster. I am not including the usual "vedic math" or "speed calc" techniques. I found that simpler methods help more!
I am sure you are all aware of that ubiquitous formula (a+b+ab/100). But I am equally certain that you (or even me, for that matter) have discovered even half of the uses of this formula. Well, for now, lets focus on its use in division.
As a really simple starter, lets take a division question:
If 1/3 = 33.33%, what is 1/2? Obviously, its 50%, silly! So, what does it have to do with that formula? Here's what:
Apart from its more popular uses like successive discounts etc, this formula holds true for ANY product of two or more quantities. In fact, this is the most generalized use of the formula. If P=AxB, then an change of a% in A and b% in B will result in a change of (a+b+ab/100)% in P. Conversely, if P has to remain constant, an a% increase/ decrease in A will necessitate a corresponding change of a/(100+a)% or a/(100-a)% in B respectively.
Now,
1/3=33.33% can also be written as 1=3 x 33.33% and
1/2=50% can also be written as 1=2 x 50%.
So, the product has remained constant at 1. 3 has decreased to 2. That's a decrease of 33.33%. According to the formula, the other factor will increase by 33.33/(100-33.33)= 50%. So, 33.33% will increase by 50% to 50%.
This may sound too obvious and easy, but take a look at what you can do with this. How would you calculate, say, 113/256? Now, we know that 113/226 = 1/2=0.50. Using this fact, we can calculate the deviation from 0.5. Since 256 is 30/226 or roughly 1/7.5 or 2/15 or 13.33% more than 226, it means that the final result will be 13.33/(100+13.33)% less than 0.5. That is, approx. 12% less than 0.5, or 0.44. That is fairly close to the correct answer that is 0.4414!
Even if you had taken 13% (instead of dividing it by 113.33), you would have been quite close at 0.45! The key is to round off the denominator to a "convenient" number. And if your denominator deviation is not more than 15%, you can simply take this same deviation as correction in your answer. To summarize, follow these steps:
1. Round off the denominator to either a multiple/ factor of the numerator, or to the nearest hundred. This way, the calculation will be easier for the new fraction.
2. Calculate the percentage deviation from the actual denominator. You may use approximations depending on the need of the hour.
3. Provided that the percentage deviation is less than 15%, adjust your answer by the same percentage in the "other direction".
Obviously, this involves a fairly high amount of calculation, but the good news is that it will get easier with a better knowledge about percentages.
Why don't you go ahead and try out the following:
135/283=? (Hint: Take 283 as 270 or 300)
178/1327=? (Hint: Take 1327 as 1300)
I am sure you are all aware of that ubiquitous formula (a+b+ab/100). But I am equally certain that you (or even me, for that matter) have discovered even half of the uses of this formula. Well, for now, lets focus on its use in division.
As a really simple starter, lets take a division question:
If 1/3 = 33.33%, what is 1/2? Obviously, its 50%, silly! So, what does it have to do with that formula? Here's what:
Apart from its more popular uses like successive discounts etc, this formula holds true for ANY product of two or more quantities. In fact, this is the most generalized use of the formula. If P=AxB, then an change of a% in A and b% in B will result in a change of (a+b+ab/100)% in P. Conversely, if P has to remain constant, an a% increase/ decrease in A will necessitate a corresponding change of a/(100+a)% or a/(100-a)% in B respectively.
Now,
1/3=33.33% can also be written as 1=3 x 33.33% and
1/2=50% can also be written as 1=2 x 50%.
So, the product has remained constant at 1. 3 has decreased to 2. That's a decrease of 33.33%. According to the formula, the other factor will increase by 33.33/(100-33.33)= 50%. So, 33.33% will increase by 50% to 50%.
This may sound too obvious and easy, but take a look at what you can do with this. How would you calculate, say, 113/256? Now, we know that 113/226 = 1/2=0.50. Using this fact, we can calculate the deviation from 0.5. Since 256 is 30/226 or roughly 1/7.5 or 2/15 or 13.33% more than 226, it means that the final result will be 13.33/(100+13.33)% less than 0.5. That is, approx. 12% less than 0.5, or 0.44. That is fairly close to the correct answer that is 0.4414!
Even if you had taken 13% (instead of dividing it by 113.33), you would have been quite close at 0.45! The key is to round off the denominator to a "convenient" number. And if your denominator deviation is not more than 15%, you can simply take this same deviation as correction in your answer. To summarize, follow these steps:
1. Round off the denominator to either a multiple/ factor of the numerator, or to the nearest hundred. This way, the calculation will be easier for the new fraction.
2. Calculate the percentage deviation from the actual denominator. You may use approximations depending on the need of the hour.
3. Provided that the percentage deviation is less than 15%, adjust your answer by the same percentage in the "other direction".
Obviously, this involves a fairly high amount of calculation, but the good news is that it will get easier with a better knowledge about percentages.
Why don't you go ahead and try out the following:
135/283=? (Hint: Take 283 as 270 or 300)
178/1327=? (Hint: Take 1327 as 1300)
Perfect Number
In mathematics, a perfect number is defined as a positive integer which is the sum of its proper positive divisors, that is, the sum of the positive divisors not including the number itself. Equivalently, a perfect number is a number that is half the sum of all of its positive divisors, or σ(n) = 2 n.
Example:
The factors of 6 are 1, 2, 3 and 6.1 + 2 + 3 = 6 = 2^ (2-1)*(2^2 – 1)
The factors of 28 are 1, 2, 4, 7, 14 and 28.1 + 2 + 4 + 7 + 14 = 28.
28 = 2^ (3-1)*(2^3 – 1)
The factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496.1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496.
496= 2^ (5-1)*(2^5 – 1)
The factors of 8128 are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and 8128. I'll let you add them up.
8128= 2^ (7-1)*(2^7 – 1)
Formula to calculate Perfect Number:
Perfect Number = 2^ (n-1) * (2^n - 1)
Where n and 2^n – 1 has to be prime numbers.
Note: It is not necessary that when n is prime 2^n – 1 is always prime.
Put n = 11, 2^11 – 1 = 2047= 23*89 .
Example:
The factors of 6 are 1, 2, 3 and 6.1 + 2 + 3 = 6 = 2^ (2-1)*(2^2 – 1)
The factors of 28 are 1, 2, 4, 7, 14 and 28.1 + 2 + 4 + 7 + 14 = 28.
28 = 2^ (3-1)*(2^3 – 1)
The factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496.1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496.
496= 2^ (5-1)*(2^5 – 1)
The factors of 8128 are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and 8128. I'll let you add them up.
8128= 2^ (7-1)*(2^7 – 1)
Formula to calculate Perfect Number:
Perfect Number = 2^ (n-1) * (2^n - 1)
Where n and 2^n – 1 has to be prime numbers.
Note: It is not necessary that when n is prime 2^n – 1 is always prime.
Put n = 11, 2^11 – 1 = 2047= 23*89 .
How to calculate Last two digit
Dear student, Agenda is how to calculate last two digits of the number ending on 1.
Last digit or unit digit will be always 1.
Just observe tenth place digit in following example
11^2 = 121 (1 * 2 = 2)
11^3= 1331 (1 * 3 = 3)
21^2 = 441 (2 * 2 = 4)
21^3= 9261 (2 * 3 = 6)
Just apply this technique to the following examples
Example: 31^8
If number is ending on 1 last digit is always 1.Important is how to calculate Tenth place digit. It is also very easy. Tenth place digit will be (3 * 8 =24) so tenth place digit is 4
So last two digits of 31^8 is 41.
Find the last two digits of 41^2789
In no time at all you can calculate the answer to be 61 (4 × 9 = 36. Therefore, 6 will be the tens digit and one will be the units digit)
Find the last two digits of 71^56747
Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit)
Now try to get the answer to this question within 10 s:
Find the last two digits of 51^456 × 61^567
The last two digits of 51^456 will be 01 and the last two digits of 61^567 will be 21. Therefore, the last two digits of 514^56 × 61^567 will be the last two digits of 01 × 21 = 21
Last two digits of numbers ending in 3, 7 or 9
Find the last two digits of 19266.
19^266 = (19^2)133. Now, 19^2 ends in 61 (19^2 = 361) therefore, we need to find the last two digits of (61)133.
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tens digit will be 8 and last digit will be 1)
Find the last two digits of 33^288.
33^288 = (33^4)72. Now 33^4 ends in 21 (33^4 = 33^2 × 33^2 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21^72. By the previous method, the last two digits of 21^72 = 41 (tens digit = 2 × 2 = 4, unit digit = 1)
So here’s the rule for finding the last two digits of numbers ending in 3, 7 and 9:
Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method.
Now try the method with a number ending in 7:
Find the last two digits of 87^474.
87^474 = 87^472 × 87^2 = (87^4)^118 × 87^2 = (69 × 69)^118 × 69 (The last two digits of 87^2 are 69) = 61^118 × 69 = 81 × 69 = 89
If you understood the method then try your hands on these questions:
Find the last two digits of:
1. 27^496
2. 79^93
3. 583^512
Last digit or unit digit will be always 1.
Just observe tenth place digit in following example
11^2 = 121 (1 * 2 = 2)
11^3= 1331 (1 * 3 = 3)
21^2 = 441 (2 * 2 = 4)
21^3= 9261 (2 * 3 = 6)
Just apply this technique to the following examples
Example: 31^8
If number is ending on 1 last digit is always 1.Important is how to calculate Tenth place digit. It is also very easy. Tenth place digit will be (3 * 8 =24) so tenth place digit is 4
So last two digits of 31^8 is 41.
Find the last two digits of 41^2789
In no time at all you can calculate the answer to be 61 (4 × 9 = 36. Therefore, 6 will be the tens digit and one will be the units digit)
Find the last two digits of 71^56747
Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit)
Now try to get the answer to this question within 10 s:
Find the last two digits of 51^456 × 61^567
The last two digits of 51^456 will be 01 and the last two digits of 61^567 will be 21. Therefore, the last two digits of 514^56 × 61^567 will be the last two digits of 01 × 21 = 21
Last two digits of numbers ending in 3, 7 or 9
Find the last two digits of 19266.
19^266 = (19^2)133. Now, 19^2 ends in 61 (19^2 = 361) therefore, we need to find the last two digits of (61)133.
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tens digit will be 8 and last digit will be 1)
Find the last two digits of 33^288.
33^288 = (33^4)72. Now 33^4 ends in 21 (33^4 = 33^2 × 33^2 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21^72. By the previous method, the last two digits of 21^72 = 41 (tens digit = 2 × 2 = 4, unit digit = 1)
So here’s the rule for finding the last two digits of numbers ending in 3, 7 and 9:
Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method.
Now try the method with a number ending in 7:
Find the last two digits of 87^474.
87^474 = 87^472 × 87^2 = (87^4)^118 × 87^2 = (69 × 69)^118 × 69 (The last two digits of 87^2 are 69) = 61^118 × 69 = 81 × 69 = 89
If you understood the method then try your hands on these questions:
Find the last two digits of:
1. 27^496
2. 79^93
3. 583^512
Interesting Maths Fact
1. If you join together the feet of the altitudes of a triangle, they make another triangle called the pedal triangle, and the orthocentre is the incentre of this pedal triangle.
2. The orthocentre, circumcentre and centroid are collinear (Euler’s line) and the centroid is two thirds the distance from the orthocenter to the circumcenter. The incenter is not on the Euler line except when the triangle is isosceles. When the triangle is isosceles, the Euler Line also passes through one of the vertices!
3. Take a triangle with vertices at A, B and C, and let H be its orthocenter. The orthocenter for any of the triangles formed from three of these four points is the fourth point. (4 points which satisfy this condition are called an orthocentric set.)
4. The distance from the orthocenter to a vertex is twice the distance from the circumcenter to the opposite side of the vertex.
One great math site that I love to visit frequently is www.mathforum.org and it is rich with all the knowledge that you will ever need!
2. The orthocentre, circumcentre and centroid are collinear (Euler’s line) and the centroid is two thirds the distance from the orthocenter to the circumcenter. The incenter is not on the Euler line except when the triangle is isosceles. When the triangle is isosceles, the Euler Line also passes through one of the vertices!
3. Take a triangle with vertices at A, B and C, and let H be its orthocenter. The orthocenter for any of the triangles formed from three of these four points is the fourth point. (4 points which satisfy this condition are called an orthocentric set.)
4. The distance from the orthocenter to a vertex is twice the distance from the circumcenter to the opposite side of the vertex.
One great math site that I love to visit frequently is www.mathforum.org and it is rich with all the knowledge that you will ever need!
Points to remember
Quantitative Ability – POINTS TO REMEMBER
1.If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x^3+3x^2+2x+6=0 has no positive roots
2.For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.
3.For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)
4.Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be
2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
5.For a cubic equation ax^3+bx^2+cx+d=o
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = -d/a
6.For a bi-quadratic equation ax^4+bx^3+cx^2+dx+e = 0
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Sum of the product of the roots taken three at a time = -d/a
Product of the roots = e/a
7.If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)
8.Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
Then,
If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
If a1/a2 <> b1/b2, then we have a unique solution.
9.a + b = a + b if a*b>=0
else, a + b >= a + b
10.The equation ax^2+bx+c=0 will have max. value when a<0>0. The max. or min. value is given by (4ac-b^2)/4a and will occur at x = -b/2a
11.If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.
12. If for two numbers x*y=k (a constant), then their SUM is MINIMUM if x=y (=root(k)). The minimum sum is then 2*root (k).
13.Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.
14. For any 2 numbers a, b where a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
(GM)^2 = AM * HM
15.For three positive numbers a, b, c
(a + b + c) * (1/a + 1/b + 1/c)>=9
16.For any positive integer n
2<= (1 + 1/n)^n <=3
17. a^2 + b^2 + c^2 >= ab + bc + ca
If a=b=c, then the case of equality holds good.
18.a^4 + b^4 + c^4 + d^4 >= 4abcd (Equality arises when a=b=c=d=1)
19.(n!)^2 > n^n
20.If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s
21.x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)
22.e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 < e < 3
23.log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]
24.(m + n)! is divisible by m! * n!
25.When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.
26.Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)
The sum of first n natural numbers = n(n+1)/2
The sum of squares of first n natural numbers is n(n+1)(2n+1)/6
The sum of cubes of first n natural numbers is (n(n+1)/2)2/4
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n^2
1.If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x^3+3x^2+2x+6=0 has no positive roots
2.For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.
3.For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)
4.Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be
2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
5.For a cubic equation ax^3+bx^2+cx+d=o
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = -d/a
6.For a bi-quadratic equation ax^4+bx^3+cx^2+dx+e = 0
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Sum of the product of the roots taken three at a time = -d/a
Product of the roots = e/a
7.If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)
8.Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
Then,
If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
If a1/a2 <> b1/b2, then we have a unique solution.
9.a + b = a + b if a*b>=0
else, a + b >= a + b
10.The equation ax^2+bx+c=0 will have max. value when a<0>0. The max. or min. value is given by (4ac-b^2)/4a and will occur at x = -b/2a
11.If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.
12. If for two numbers x*y=k (a constant), then their SUM is MINIMUM if x=y (=root(k)). The minimum sum is then 2*root (k).
13.Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.
14. For any 2 numbers a, b where a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
(GM)^2 = AM * HM
15.For three positive numbers a, b, c
(a + b + c) * (1/a + 1/b + 1/c)>=9
16.For any positive integer n
2<= (1 + 1/n)^n <=3
17. a^2 + b^2 + c^2 >= ab + bc + ca
If a=b=c, then the case of equality holds good.
18.a^4 + b^4 + c^4 + d^4 >= 4abcd (Equality arises when a=b=c=d=1)
19.(n!)^2 > n^n
20.If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s
21.x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)
22.e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 < e < 3
23.log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]
24.(m + n)! is divisible by m! * n!
25.When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.
26.Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)
The sum of first n natural numbers = n(n+1)/2
The sum of squares of first n natural numbers is n(n+1)(2n+1)/6
The sum of cubes of first n natural numbers is (n(n+1)/2)2/4
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n^2
Prime Nos.
A prime numbers is a number greater than one whose factors are only one and itself. In other words, 4 is not prime, because its factors are 1, 2 and 4 (1x4, 2x2). But 5 is prime, because the only way you can get a product of 5 is by multiplying 1 and 5 (1x5). Composite numbers are those who has more than two factors. The number 1 is not prime OR composite because it has only one factor.
Here are all the prime numbers up to 500:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, ...
How to find big Prime Numbers:
Following is a somewhat celebrated result of Euler:
The function f(x) = x^2 + x + 41 is always prime for x = 0, 1 ,..., 39.
This quadratic was the record holder for centuries as a consecutive,
Distinct quadratic prime-producer for an initial range of input
Values. It is not, however, the new function is there that can give more range of prime numbers than Euler’s:
The Function:
f(x) = 36x^2 - 810x + 2753
Which is always prime for x = 0, 1, ... , 44.
Here are all the prime numbers up to 500:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, ...
How to find big Prime Numbers:
Following is a somewhat celebrated result of Euler:
The function f(x) = x^2 + x + 41 is always prime for x = 0, 1 ,..., 39.
This quadratic was the record holder for centuries as a consecutive,
Distinct quadratic prime-producer for an initial range of input
Values. It is not, however, the new function is there that can give more range of prime numbers than Euler’s:
The Function:
f(x) = 36x^2 - 810x + 2753
Which is always prime for x = 0, 1, ... , 44.
Tuesday, September 1, 2009
Saturday, August 29, 2009
CALENDAR.
Dear student
Go through these the 10 bullet points to solve questions on calender.
1. 1st January 0001 was a Monday.
2. Calendar repeats after every 400 years.
3. Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
4. Odd days- remainder obtained when no. of days is divided by 7. Normal year has 1 odd day and leap year has 2 odd days.
5. Calendar moves ahead by number of odd days.
6. While checking leap year just analyze whether February falls in that period or not.
7. Century has 5 odd days and leap century has 6 odd days.
8. Take out net odd days.( add all the odd days and again divide by 7)
9. In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
10. 1st January 1901 was Tuesday.
We calculate odd days on the basis of the previous month.
Example – what day is it on 29th August 1982?
Method- As we know 1/1/1901 was a Tuesday now we take 1982 and 1901 - 1982-1901=81 years. - 81/4=20…. ( disregard decimal) - 81+20=101 - 101/7- remainder is 3, so 3 days from Tuesday is Friday. -Now check whether it is a leap year or not. In this case it is not a leap year. Therefore 2nd July will be Friday - Now we have to go month wise . 2nd august = 3odd days= so from Friday 3 odd days will be Monday so 29th August will be a Sunday.
Dear student
Go through these the 10 bullet points to solve questions on calender.
1. 1st January 0001 was a Monday.
2. Calendar repeats after every 400 years.
3. Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
4. Odd days- remainder obtained when no. of days is divided by 7. Normal year has 1 odd day and leap year has 2 odd days.
5. Calendar moves ahead by number of odd days.
6. While checking leap year just analyze whether February falls in that period or not.
7. Century has 5 odd days and leap century has 6 odd days.
8. Take out net odd days.( add all the odd days and again divide by 7)
9. In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
10. 1st January 1901 was Tuesday.
We calculate odd days on the basis of the previous month.
Example – what day is it on 29th August 1982?
Method- As we know 1/1/1901 was a Tuesday now we take 1982 and 1901 - 1982-1901=81 years. - 81/4=20…. ( disregard decimal) - 81+20=101 - 101/7- remainder is 3, so 3 days from Tuesday is Friday. -Now check whether it is a leap year or not. In this case it is not a leap year. Therefore 2nd July will be Friday - Now we have to go month wise . 2nd august = 3odd days= so from Friday 3 odd days will be Monday so 29th August will be a Sunday.
Seven new IIMs in India by 2011
B K Upmanyu
Fri, Aug 28, 2009 17:21:14 IST
THE COUNTRY’S three youngest states - Jharkhand, Chhattisgarh and Uttarakhand, along with four others, Tamil Nadu, Haryana, Jammu & Kashmir and Rajasthan, will get their branches of the prestigious Indian Institutes of Management (IIM) during the period 2009-2011.
The Cabinet has approved a proposal for setting up 7 new IIMs, taking the total number of the premier management school to 14. India already has seven functional IIMs in Ahmedabad, Bangalore, Kolkata, Lucknow, Indore, Kozhikode and Shillong.
In yet another development that would benefit five lakh students belonging to economically weaker sections with an annual income limit of Rs 4.5 lakh, the Cabinet on Economic Affairs (CCEA) approved a scheme to provide full interest subsidy during the period of moratorium on loans taken by them for pursuing approved technical and professional courses.
While the scheme will be applicable from the current academic year 2009-2010 itself, the Government will bear the interest payable by the student during his/her entire course period, plus one year or six months after he/she gets a job (moratorium period), whichever is earlier. Once the moratorium period is over, the student will have to pay the interest on the outstanding loan amount. The number of loans, as on March 31, 2009, was 16 lakhs and the total outstanding amount Rs 24,000 crore.
HRD Minister Kapil Sibal said both the establishment of seven IIMs and the interest subsidy scheme were part of his 100-day agenda and that he kept the promises he made.
In the first phase, four IIMs at Tiruchirappalli, Ranchi, Raipur and Rohtak will be set up in 2009-10, to become functional from the 2010-11 academic session. The remaining three in J&K, Uttarakhand and Rajasthan, where the locations are yet to be decided, will be set up in the second phase in 2010-11. Though postgraduate programme in management will be the flagship programme, several executive programmes focusing on civic and municipal services will be started in the first year.
In all, an amount of Rs 1,057 crore would be required for the establishment of the seven new IIMs during the XI Plan and wherein admissions will be through the Common Admission Test (CAT). Phase-I will have an intake of 140 students in the PGP course and which will reach 560 students by the end of phase-II.
B K Upmanyu
Fri, Aug 28, 2009 17:21:14 IST
THE COUNTRY’S three youngest states - Jharkhand, Chhattisgarh and Uttarakhand, along with four others, Tamil Nadu, Haryana, Jammu & Kashmir and Rajasthan, will get their branches of the prestigious Indian Institutes of Management (IIM) during the period 2009-2011.
The Cabinet has approved a proposal for setting up 7 new IIMs, taking the total number of the premier management school to 14. India already has seven functional IIMs in Ahmedabad, Bangalore, Kolkata, Lucknow, Indore, Kozhikode and Shillong.
In yet another development that would benefit five lakh students belonging to economically weaker sections with an annual income limit of Rs 4.5 lakh, the Cabinet on Economic Affairs (CCEA) approved a scheme to provide full interest subsidy during the period of moratorium on loans taken by them for pursuing approved technical and professional courses.
While the scheme will be applicable from the current academic year 2009-2010 itself, the Government will bear the interest payable by the student during his/her entire course period, plus one year or six months after he/she gets a job (moratorium period), whichever is earlier. Once the moratorium period is over, the student will have to pay the interest on the outstanding loan amount. The number of loans, as on March 31, 2009, was 16 lakhs and the total outstanding amount Rs 24,000 crore.
HRD Minister Kapil Sibal said both the establishment of seven IIMs and the interest subsidy scheme were part of his 100-day agenda and that he kept the promises he made.
In the first phase, four IIMs at Tiruchirappalli, Ranchi, Raipur and Rohtak will be set up in 2009-10, to become functional from the 2010-11 academic session. The remaining three in J&K, Uttarakhand and Rajasthan, where the locations are yet to be decided, will be set up in the second phase in 2010-11. Though postgraduate programme in management will be the flagship programme, several executive programmes focusing on civic and municipal services will be started in the first year.
In all, an amount of Rs 1,057 crore would be required for the establishment of the seven new IIMs during the XI Plan and wherein admissions will be through the Common Admission Test (CAT). Phase-I will have an intake of 140 students in the PGP course and which will reach 560 students by the end of phase-II.
Friday, August 21, 2009
A Tool to understand graphs
I found these mathopenref tools extremely helpful in understanding graphs:
1. Linear Equations: http://www.mathopenref.com/linearexplorer.html
2. Quadratic Equations: http://www.mathopenref.com/quadraticexplorer.html
3. Cubic Equations: http://www.mathopenref.com/cubicexplorer.html
Things to do: Change the variable values with the help of the sliders and see the change in the shape of the curve.
For instance, in a quadratic equation y=ax^2+bx+c:
changing c moves it up and down
changing b changes the slope
changing a alters the curvature of the parabolic element
Enjoy!
1. Linear Equations: http://www.mathopenref.com/linearexplorer.html
2. Quadratic Equations: http://www.mathopenref.com/quadraticexplorer.html
3. Cubic Equations: http://www.mathopenref.com/cubicexplorer.html
Things to do: Change the variable values with the help of the sliders and see the change in the shape of the curve.
For instance, in a quadratic equation y=ax^2+bx+c:
changing c moves it up and down
changing b changes the slope
changing a alters the curvature of the parabolic element
Enjoy!
Concept of Factor
N = P^a * Q^b * R^c (Where P,Q and R are prime numbers and a, b and c are natural numbers)
Then
1. Total no of factors of N = (a+1)*(b+1)*(c+1)
2. Total no of perfect square factors of N= (a+1)/2 * (b/2+1) *(c+1)/2 [If a and c is an odd no, b is an even no]
3. Total no of odd factors of N= (1 * (b+1)*(c+1)) [If P=2]
4. Total no of odd factors of N= (a+1)*(b+1)*(c+1) [If 2 is not the factor of N]
5. Total no of even factors of N= (a * (b+1)*(c+1)) [If P=2]
6. Total no of even factors of N= 0 [If 2 is not the factor of N]
7. Total no of non ordered pairs of (x,y) such that (x * y = N) = [(a+1)*(b+1)*(c+1)]/2 note : If total no of factors is an even no.
8. Total no of non ordered pairs of (x,y) such that (x * y = N) = [(a+1)*(b+1)*(c+1)+ 1]/2 note : If total no of factors is an odd no.
9. Total no of co-prime pairs using all the factors of N = a+b+c+2 (ab+bc+ca) + 4abc
10. Sum of all the factors of N = [(P^(a+1) – 1)/(p – 1)] * [(Q^(b+1) – 1)/(Q – 1)] * [(R^(c+1) – 1)/(R – 1)]
Then
1. Total no of factors of N = (a+1)*(b+1)*(c+1)
2. Total no of perfect square factors of N= (a+1)/2 * (b/2+1) *(c+1)/2 [If a and c is an odd no, b is an even no]
3. Total no of odd factors of N= (1 * (b+1)*(c+1)) [If P=2]
4. Total no of odd factors of N= (a+1)*(b+1)*(c+1) [If 2 is not the factor of N]
5. Total no of even factors of N= (a * (b+1)*(c+1)) [If P=2]
6. Total no of even factors of N= 0 [If 2 is not the factor of N]
7. Total no of non ordered pairs of (x,y) such that (x * y = N) = [(a+1)*(b+1)*(c+1)]/2 note : If total no of factors is an even no.
8. Total no of non ordered pairs of (x,y) such that (x * y = N) = [(a+1)*(b+1)*(c+1)+ 1]/2 note : If total no of factors is an odd no.
9. Total no of co-prime pairs using all the factors of N = a+b+c+2 (ab+bc+ca) + 4abc
10. Sum of all the factors of N = [(P^(a+1) – 1)/(p – 1)] * [(Q^(b+1) – 1)/(Q – 1)] * [(R^(c+1) – 1)/(R – 1)]
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