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)
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