TCS 2014

C.S.VEERARAGAVAN

LINES & PROGRAMMERS

• 2–15. Suppose 16 such programmers take 16 minutes to write 16 lines of code in total, how many lines of code can be written by 96 programmers in 96 minutes?

• (a) 16 (c) 432 (d) 96 (b) 576 • 2–34. Suppose 12 such programmers take 12 minutes to write 12 lines of

code in total. How many programmers will complete 96 lines in 96 minutes?

• (a) 12 (b) 96 (c) 1152 (d) 32 • 3–4. Suppose 24 such programmers take 24 minutes to write 24 lines of

code in total. How many programmers will be required in total to write 72 lines of code in 72 minutes?

• (a) 18 (b) 36 (c) 72 (d) 24 • 3–30. Suppose 10 such programmers take 10 minutes to write 10 lines of

code in total. How long will 90 programmers take to write 90 lines of code? (in minutes)

• (a) 180 (b) 900 (c) 10 (d) 90

Points on a Plane n1(P) n points:• For Maximum number of lines,

place the points on a circle. • Then number of lines = Number

of points = n Maximum n lines

• E.g.: For 5 points, lines are as shown in right-hand figure:

• Similarly for Minimum no. of lines, take points inside a triangle.

• If no condition 2 lines • With condition 3 lines

• 1. Given a collection of points P in the plane, a 1-set is a point in P that can be separated from the rest by a line, .i.e the point lies on one side of the line while the others lie on the other side. The number of 1-sets of P is denoted by n1(P). The minimum value of n1(P) over all configurations P of 5 points in the plane in general position (.i.e no three points in P lie on a line) is

• (a) 3 (b) 5 (c) 2 (d)7 • we need to draw a line with separation of the points . • As per condition , minimum = 3 lines needed.

•

• Given a collection of points P in the plane, a 1-set is a point in P that can be separated from the rest by a line, .i.e the point lies on one side of the line while the others lie on the other side. The number of 1-sets of P is denoted by n1(P). The maximum value of n1(P) over all configurations P of 5 points in the plane in general position (.i.e no three points in P lie on a line) is

• (a) 3 (b) 5 (c) 2 (d)7 • Solution:• we need to draw a line with separation of the points . • As per condition , maximum = 5 lines needed.

• Given a collection of 36 points in the plane P and a point “X” equidistant from all points in P, which of the following are necessarily true?

• A. The points in P lie on a circle.• B. The distance between any pair of points in P is larger than the

distance between X and a point in P

• (a) A only (b) B only (c) Both A and B (d) Neither A nor B• Solution :• We can able to arrange points with equidistant is possibly by

arranging those points in circle only.• So first condition is true .• But second condition is not possible .

HAND SHAKES

• 2–31. 20 people meet and shake hands. The maximum number of handshakes possible if there is to be no “cycle” of handshakes is (A cycle of handshakes is a sequence of k people a1, a2, ……, ak (k > 2) such that the pairs {a1, a2}, {a2, a3}, ……, {ak-1, ak}, {ak, a1} shake hands)

• (a) 19 (b) 18 (c) 20 (d) 21 • when no cycle, the MAXIMUM no. of

handshakes = (n-1) handshakes. =19 handshakes

• 3–24. 22 people meet and shake hands. The maximum number of handshakes possible if there is to be no ‘cycle’ of handshakes is (A cycle of handshakes is a sequence of people a1, a2…, ak such the pairs (a1, a2), (a2, a3), …, (a (k-1), a k) , (ak, a1) shake hands.

• (a) 22 (b) 7 (c) 21 (d) 11 • 21 hand shakes.

Dart Board

GHANA BOLIVIA – FOOT BALL MATCH – OCTOPUS

• Probability (Paul will correctly pick the winner) = Probability (Paul picks Ghana and Ghana wins) + Probability (Paul picks

Bolivia and Bolivia wins) = Probability (Paul picks Ghana)* Probability (Ghana wins/ Paul picks Ghana)

+ Probability (Paul picks Bolivia)* Probability (Bolivia wins /Paul picks Bolivia)

•

Red Ball Maximize out of 2 Boxes

Unpainted faces of a hollow cube• N cm cube• Answer = [(n3–(n–2)3)*6)–(No of faces painted * n2)• 1–20. A hollow cube of size 5 cm is taken with a

thickness of 1 cm. It is made of smaller cubes of size 1 cm. If the 4 faces of the outer surface of the cube are painted totally, how many faces of the smaller cubes remain unpainted?

• (a) 900 (b) 488 (c) 500 (d) 800 • [(53–(5–2)3)*6)–(4 * 52)• = 98*6 – 100= 588 – 100 = 488.

• 2–26. A hollow cube of size 5 cm is taken with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If only 2 faces of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?

• (a) 210 (b) 465 (c) 450 (d) 538 • [(53–(5–2)3)*6)–(2 * 52)• = 98*6 – 50= 588 – 50 =538.

Planet – Finger• 1–8.The citizens of planet

nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 8) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings?

• (a) 192 (b) 64 (c) 54 (d) 102 • Here B = 8• 1–1000 ––––– 3B2 = 3*64=192

Planet –––– FingerB No of fingers1–100 ––––– 2B1–1000 ––––– 3B2

1–10000 -––––– 4B3

• 2–8.The citizens of planet nigiet are 6 fingered and have thus developed their decimal system in base 6. A certain street in nigiet contains 1000 (in base 6) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings?

• (a) 108 (b) 192 (c) 54 (d) 102 • 3*62 = 108.

• The Chennai corporation has a particular strategy for assigning numbers for buildings. The Corporation had strict rules that every building should have a number. In T Nagar, a place in Chennai there many buildings were under residential category. For buildings, the Corporation numbered from 1 to 100. For shops, corporation numbered between 150 and 200 only in prime numbers. How many times does the number “6” appear in the building numbers?

• (a) 20 (b) 19 (c) 10 (d) 13• Solution :• From the above formula , we can find out easily .• Here they used base 10 , to number from 1 to 100.• The specified number ( 6 ) used to numbering the buildings

from 1 to 100 with base 10 , is = 2 * 10 = 20

• 4. A research lab in Chennai requires 100 mice and 75 sterilized cages for a certain set of laboratory experiments. To identify the mice, the lab has prepared labels with numbers 1 to 100, by combining tags numbered 0 to 9. The SPCA requires that the tags be made of toxin-free material and that the temperature of the cages be maintained at 27 degree Celsius. Also, not more than 2 mice can be caged together and each cage must be at least 2 sq.ft in area. The 5 experiments to be conducted by lab are to be thoroughly documented and performed only after a round of approval by authorities. The approval procedure takes around 48 hours. How many times is, the tag numbered ’4 used by the lab in numbering these mice?′

• a)9 (b)19 (c) 20 (d) 21• Solution :• From the above formula , we can find out easily .• Here they used base 10 , to number from 1 to 100.• The specified number ( 4 ) used to numbering the buildings from 1 to

100 with base 10 , is = 2 * B = 2 * 10 = 20

Age of planet • Direct substitution of ‘ t ‘ and ‘ d ‘ value :• 1. On planet zorba, a solar blast has melted the ice caps on its equator. 8

years after the ice melts, tiny plantoids called echina start growing on the rocks. echina grows in the form of a circle and the relationship between the diameter of this circle and the age of echina is given by the formula d = 4 * √ (t – 8) for t ≥ 8 where d represents the diameter in mm and t the number of years since the solar blast. Jagan recorded the radius of some echina at a particular spot as 8mm. How many years back did the solar blast occur?

• (a) 8 (b) 12 (c) 16 (d) 24• Solution : • In this qn , we need to find t value i.e., years.• Given : radius = 8 mm• so Diameter , d = 2 * r = 16 mm• Substitute d value in given formula and get the value , t = 24 years.

• 2. A result of global warming is that the ice of some glaciers is melting. Twelve years after the ice disappears tiny plants, called lichens, start to grow on the rocks. Each lichens grows approximately in the shape of a circle. The relationship between the diameter of this circle and the age of the lichen can be approximated with the formula d = 13 * (t-11) for t > 11 where d represents the diameter of the lichen in millimeters, and t represents the number of years after the ice disappeared. Using the above formula, calculate the diameter of the lichen, 36 years after the ice has disappeared.

• (a) 468 mm (b) 457 mm (c) 325 mm (d) 11 mm• Solution : • In this qn , we need to find d value i.e., diameter,• Given : • years = 36• So Diameter , d = 13 * 25 = 325 mm• Substitute t value in given formula and get the value , d = 325 mm.

• 3. On planet Korba, a solar blast has melted the ice caps on its equator. 9 years after the ice melts, tiny planetoids called echina start growing on the rocks. Echina grows in the form of circle, and the relationship between the diameter of this circle and the age of echina is given by the formula, d = 4*√ (t-9) for t ≥ 9, where d represents the diameter in mm and t, the number of years since the solar blast. Anubhav recorded the radius of some echina at a particular spot as 7mm. How many years back did the solar blast occur?

• (a) 17 (b) 21.25 (c) 12.25 (d) 12.06• Solution : • In this qn , we need to find t value i.e., years.• Given : radius = 7 mm• so Diameter , d = 2 * r = 14 mm• Substitute d value in given formula and get the value , t = 12.06 years.

Triangle – Equidistant points

• If 3 lines in a plane are such that the points of intersection form a triangle of any kind, then the number of points equidistant from all the 3 lines is 4, consisting of 1 in-centre and 3 ex-centres.

• Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 20, 20 and 30, the number of points equidistant from all the 3 lines is

• (a) 1 (b) 0 (c) 4 (d) 2 • Answer = 4.•

Poison Cans – Mice • n cans • Convert (n-1) to binary • Answer is the number of digits in the binary number • 2–32. There are 45 cans out of which one is poisoned. If a person

tastes very little of this, he will die within 14 hours; so they decided to test it with mice. Given that a mouse dies in 24 hours and you have 24 hours in all to find out the poisoned can, how many mice are required to find the poisoned can?

• (a) 44 (b) 29 (c) 6 (d) 5 • there are 45 cans. 45 - 1 = 44; • Binary value of 44 is 101100 which has 6 digits. • Therefore, answer is 6.

• 3–28. There are 100 cans out of them one is poisoned. If a person tastes very little of this, he will die within 14 hours; so they decided to test it with mice. Given that a mouse dies in 24 hours and you have 24 hours in all to find out the poisoned can, how many mice are required to find the poisoned can?

• (a) 10 (b) 99 (c) 7 (d) 6 • There are 100 cans. 100 – 1 = 99.• Binary equivalent = 110 0011 which has 7 digits.

Hence 7 mice required.

• 5–10. If there are 30 cans out of them one is poisoned. If a person tastes very little of this he will die within 14 hours so they decided to test it with mice. Given that a mouse dies in 24 hours and you have 24 hours in all to find out the poisoned can, how many mice are required to find the poisoned can?

• (a) 29 (b) 15 (c) 6 (d) 5• There are 30 cans. 30 – 1 = 29• Binary equivalent = 11101. hence no of mice = 5.

Clocks – Planet Oz

• One the Planet Oz, there are 8 days in a week – Sunday to Saturday and another day called Oz day. There are 36 hours in a day and each hour has 90 minutes while each minute has 60 seconds. As on earth, the hour hand covers the dial twice every day. Find the approximate angle between the hands of clock on Oz when the time is 12.40 am.

• (a) 89 (b) 251 (c) 111 (d) 79 • Angle = (20°) 12 ± (34°/9) 40 = 240° ± 151.11° =

89° (or) 391.11°

• One the Planet, Oz, there are 8 days in a week – Sunday to Saturday and another day called Oz day. There are 36 hours in a day and each hour has 90 minutes while each minute has 60 seconds. As on earth, hour hand covers the dial twice every day. Find the approximate angle between the hands of clock on Oz when the time is 11.40 a.m.

• (a) 83 (b) 74 (c) 129 (d) 65 • Angle = (20°) 11 ± (34°/9) 40 = 220° ± 151.11° =

69° (or) 371.11°

• One the Planet, Oz, there are 8 days in a week – Sunday to Saturday and another day called Oz day. There are 36 hours in a day and each hour has 90 minutes while each minute has 60 seconds. As on earth, hour hand covers the dial twice every day. Find the approximate angle between the hands of clock on Oz when the time is 9.40 a.m.

• (a) 83 (b) 74 (c) 129 (d) 65 • for 9: 40 am Angle = (20°) 9 ± (34°9) 40 = 180° ± 151.11° =

29° (or) 391.11°• most frequently asked are • 12: 40 am 89° • 11: 40 am 69° • 9: 40 am 29°

RED GREEN YELLOW SOCKS• Red Green Yellow socks /Gloves/Marbles• 1. For selecting a pair from each colour :• Red, Green, Yellow socks /Gloves/Marbles• (Maximum of Red, Green, Yellow) + (2nd

maximum of Red, Green, Yellow) + 2• 2. For selecting a thing from each colour :• Red, Green, Yellow socks /Gloves/Marbles• (Maximum of Red, Green, Yellow) + (2nd

maximum of Red, Green, Yellow) + 1• 3. For selecting a pair from any colour :• Total number of colors + 1

Red Green Yellow socks /Gloves/Marbles

• (Maximum of Red, Green, Yellow) + (2nd maximum of Red, Green, Yellow) + 2

• It is dark in my bedroom and I want to get two socks of the same colour from my drawer, which contains 26 red and 24 blue, 34 brown socks. How many socks do I have to take from the drawer to get at least two socks of the each colour?

• (a) 6 (b) 74 (c) 61 (d) 62 • 34 + 26 + 2 = 62.

• 25. It is dark in my bedroom and I want to get two socks of the same colour from my drawer, which contains 36 red and 24 blue, 14 brown socks. How many socks do I have to take from the drawer to get at least two socks of the each color?

• (a) 6 (b) 62 (c) 37 (d) 30 • 36 + 24 + 2 = 62.

• It is dark in my bedroom and I want to get two socks of the same color from my drawer, which contains 24 red and 24 blue socks. How many socks do I have to take from the drawer to get at least two socks of the same color?

• (a) 3 (b) 25 (c) 48 (d) 26• Solution:• Its under third category :• To get at least two socks of the same color

= Total number of colors + 1 = 2 +1 = 3

Number of matches – n rounds• Answer = 𝟐𝐧 − 𝟏• Middle-earth is a fictional land inhabited by hobbits, elves,

dwarves and men. The hobbits and elves are peaceful creatures who prefer slow, silent lives and appreciate nature and art. The dwarves and the men engage in physical games. The game is as follows. A tournament is one where out of the two teams that play a match, the one that loses get eliminated. The matches are played in different rounds, where in every round, half of the teams get eliminated from the tournament. If there are 9 rounds played in a knockout tournament, how many matches were played?

• (a) 511 (b) 512 (c) 256 (d) 255 • 29–1 = 511.

• Middle-earth is a fictional land inhabited by hobbits, elves, dwarves and men. The hobbits and elves are peaceful creatures who prefer slow, silent lives and appreciate nature and art. The dwarves and the men engage in physical games. The game is as follows. A tournament is one where out of the two teams that play a match, the one that loses get eliminated. The matches are played in different rounds, where in every round; half of the teams get eliminated from the tournament. If there are 6 rounds played in knock out tournament, how many matches were played?

• (a) 64 (b) 63 (c) 32 (d) 33 • 26 – 1 = 63.

• Middle-earth is a fictional land inhabited by hobbits, elves, dwarves and men. The hobbits and the elves are peaceful creatures who prefer slow, silent lives and appreciate nature and art. The dwarves and the men engage in physical games. The game is as follows. A tournament is one where out of two teams that play a match, the one that loses get eliminated. The matches are played in different rounds where in every round; half of the teams get eliminated from the tournament. If there are 10 rounds played in a knock-out tournament, how many matches were played?

• (a) 1024 (b) 1023 (c) 1025 (d) 1011• 210-1 = 1023.

Alok and Bhanu – Maximize Value for Equation

• If equation in the question is • N= (X+Y-Z) + K Answer = K+ 11 • N= (X-Y-Z) + K Answer = K+ 2 • N= (X*(Y-Z)) + K Answer = K+ 18

• Alok and Bhanu play the following min-max game. Given the expression N = 9 + X + Y – Z, where X, Y and Z are variables representing single digits (0 to 9). Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be

• (a) 20 (b) 18 (c) 27 (d) 0 • N = X + Y – Z + 9 • Answer 9 + 11 = 20.

• Alok and Bhanu play the following min-max game. Given the expression N = 25 + X + Y – Z, where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be

• (a) 43 (b) 16 (c) 36 (d) 34 • N = X + Y – Z + 25• ANSWER 25 + 11 = 36

• Alok and Bhanu play the following min-max game. Given the expression N = 32 + X* (Y – Z), where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be

• (a) 113 (b) 32 (c) -49 (d) 50 • N = X*(Y–Z) + 32• ANSWER 32 + 18 = 50.

• Alok and Bhanu play the following min-max game. Given the expression N = 25 + X – Y – Z, where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be

• (a) 36 (b) 27 (c) 14 (d) 43 • N = X – Y – Z + 25• ANSWER 25 + 2 = 27.

• Alok and Bhanu play the following min-max game. Given the expression N = 12 + X* (Y – Z), where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be

• (a) 30 (b) 12 (c) 20 (d) 23

Alice Bob – Coin Stack• Alice and Bob play the following coins-on-a-stack game. 50 coins are

stacked one above the other. One of them is a special (gold) coin and the rest are ordinary coins. The goal is to bring the gold coin to the top by repeatedly moving the topmost coin to another position in the stack. Alice starts and the players take turns. A turn consists of moving the coin on the top to a position i below the top coin (0 ≤ i ≤ 50). We will call this an i-move (thus a 0-move implies doing nothing). The proviso is that an i-move cannot be repeated; for example once a player makes a 2-move, on subsequent turns neither player can make a 2-move. If the gold coin happens to be on top when it's a player's turn then the player wins the game. Initially, the gold coin is the third coin from the top. Then

• (a) In order to win, Alice’s first move should be a 0-move. (b) In order to win, Alice’s first move should be a 1-move. (c) Alice has no winning strategy. (d) In order to win, Alice’s first move can be a 0-move or a 1-move.

• Answer Alice moves 1 move

• Alice and Bob play the following coins-on-a-stack game. 20 coins are stacked one above the other. One of them is a special (gold) coin and the rest are ordinary coins. The goal is to bring the gold coin to the top by repeatedly moving the topmost coin to another position in the stack. Alice starts and the players take turns. A turn consists of moving the coin on the top to a position i below the top coin (0 ≤ i ≤ 20). We will call this an i-move (thus a 0-move implies doing nothing). The proviso is that an i-move cannot be repeated; for example once a player makes a 2-move, on subsequent turns neither player can make a 2-move. If the gold coin happens to be on top when it's a player's turn, then the player wins the game. Initially, the gold coin is the third coin from the top.

• Then (a) In order to win, Alice’s first move should be a 0-move. (b) In order to win, Alice’s first move should be a 1-move. (c) Alice has no winning strategy. (d) In order to win, Alice’s first move can be a 0-move or a 1-move

• Answer Alice moves 1 move

• Alice and Bob play the following coins-on-a-stack game. 20 coins are stacked one above the other. One of them is a special (gold) coin and the rest are ordinary coins. The goal is to bring the gold coin to the top by repeatedly moving the topmost coin to another position in the stack. Alice starts and the players take turns. A turn consists of moving the coin on the top to a position i below the top coin (0 ≤ i ≤ 20). We will call this an i-move (thus a 0-move implies doing nothing). The proviso is that an i-move cannot be repeated; for example once a player makes a 2-move, on subsequent turns neither player can make a 2-move. If the gold coin happens to be on top when it's a player's turn then the player wins the game. Initially, the gold coin is the third coin from the top. Then

• • (a) In order to win, Alice’s first move should be a 0-move. (b) In order to

win, Alice’s first move should be a 1-move. (c) Alice has no winning strategy. (d) In order to win, Alice’s first move can be a 0-move or a 1-move

• Answer Alice moves 1 move

Hare Tortoise Race

• A hare and a tortoise race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/4 of the distance. By what factor should the hare increase its speed so as to tie the race?

• (a) 8 (b) 37 (c) 45 (d) 6.6 • Answer 6.6

• A hare and a tortoise race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race?

• (a)8 (b) 37.80 (c) 40 (d) 5 • Answer 37.80

Diameter of coins

• Planet fourfi resides in 4-dimensional space and thus the currency used by its residents are 3-dimensional objects. The rupee notes are cubical in shape while their coins are spherical. However, the coin minting machinery lays out some stipulations on the size of the coins.

• - The diameter of the coins should be at least 256 mm and not exceed 4096 mm.

• - Given a coin, the diameter of the next larger coin is at least 50% greater.

• - The diameter of a coin must always be an integer. • You are asked to design a set of coins of different diameters with

these requirements and your goal is to design as many coins as possible. How many coins can you design?

• (a) 7 (b) 8 (c) 9 (d) 6 • Answer 17

Circles – Anoop – x circles

• Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square.

• (a) 1:(2 + 7√2) (b) 1:(4 + 7 √ 3) • (c) (2 + 7 √2):1 (d) 1:(2 + 6 √ 2) • answer d.

• Anoop managed to draw 6 circles of equal radii with their centers on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the side of the square to the radius of the circles. Assume 2 is 1.4.

• (a) 9 : 1 (b) 6.2 : 1 (c) 10.4 : 1 (d) 7.6 : 1

• Anoop managed to draw 3 circles of equal radii with their centers on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the side of the square to the radius of the circles. Assume √2 is 1.4.

• (a) 10.8:1 (b) 6.2 : 1 (c) 4.2 : 1 (d) 4.8 : 1

AGES• 1. The difference between the ages of two of my three grandchildren is 3.

My eldest grandchild is three times older than the age of my youngest grandchild and my eldest grandchild’s age is two years more than the ages of my two youngest grandchildren added together. How old is my eldest grandchild?

• (a) 12 (b) 13 (c) 10 (d) 15• Solution :• Let X be the elder, Y be the younger, Z be the youngest.• From qn,• X=3Z, → 1• Y-Z=3 or X-Y=3 → 2• X=2+Y+Z. → 3• Comparing 1 and 2 ,we can say X-Y= 3 is not possible.• We need X value so get Y value and Z value from 1 and 2 respectively then

substitute in 3, we get X=15.

• 2. If a lady has 3 daughters and the difference in the ages of the middle one and the youngest is 3 and the eldest is thrice as old as the middle one and the eldest child’s age is two times of sum of the ages of the other two children. What is the age of eldest child?

• (a) 18 (b) 3 (c) 6 (d) 9• Solution:• Given:• X be elder, Y be middle and Z be younger• Y – Z = 3• X = 3Y• X = 2( Y + Z )• Solving above three equations, we get X = 18.

• 6 persons standing in the queue for ROBERT movie, are wearing different colour shirts. All of them belong to different age group, after two years their average age will be 43 and seventh person joined with them, hence the current average age has become 45. Find the age of seventh person?

• (a) 67 (b) 69 (c) 72 (d) 74• Solution :• Average property :• u If all values are increased or decreased by a certain quantity, the average also

increases or decreases by the same quantity. • So m = 41 u When a new value x is added to a set of n values with average m, if

the average increases by μ, then the added value x = m + (n + 1)μ. • Given :• ◦ m = 41 • ◦ n = 6• ◦ μ = 4• Age of seventh person = 41 + (6+1)4 = 69

• 2. X is 6 years younger to Y. X’s father is a businessman who invested 10000/- at 8% rate of interest and obtained his amount after 10 years. Y’s father is a job holder who invested around 20000 at 2% rate and obtained his amount after 20 years. Now Compounded both of them get around Rs. ABC .After 5 years the ratio of ages of X and Y is 1:2. Now X’s father is 20 years older to Y and Y’ father is 30 years more than X. After 20 years again X’s mother asks X’s father to purchase a LCD TV which costs around 45000/-. What is the age of X and Y together?

• (a) 12 (b) 8 (c) 18 (d) 6• Solution :• X + 6 = Y• ( X + 5 )/( Y + 5 ) = ( 1 : 2 )• X – Y = – 6• 2X + 10 = Y – 5 2X – Y = -5. • Solve above 2 eqn we get ans as x + y = 1+ 7 = 8

• 9years ago, Andromeda’s age was twice Achilles’ age .9 years hence, Andromeda’s age will be 4/3 times the age of Achilles’. Find Andromeda’s present age in binary numbers.

• (a)11011 (b) 11000 (c) 1001 (d) 1010• Solution :• Let Andromeda as A, Achilles as B.• A – 9 = 2( B – 9 ) A – 2B = – 9 ;• A + 9 = ( 4/3 ) ( B + 9 ) 3A – 4B = 9 ;• Solve the above 2 eqn, we get A=27.• Convert A value into binary, answer as 11011.

• The British mathematician Lewis Carol also loved to make up fantastic stories in which he embedded a number of clever puzzles and curious riddles. For example, his popular story, Alice in Wonderland, is about young girl called Alice who dreams of a strange world where she meets several unusual characters including the Red Queen and the March hare. In our story, after 2 years of time, Paul will be twice as old as Alice. Presently he is 6 times as old. How old is Paul now?

• (a) 2 (b) 4 (c) 3 (d) 6 • Solution:• Let present age of Paul be P and Alice be A.• From qn :• A=P/6;• P- 2A=2;• Sub A value, we get P=3.

• In the year 2002, Britain was reported to have had 4.3m closed-circuit televisions (CCTV) cameras–one for every 14 people in the country. This scrutiny is supposed to deter and detect crime. In one criminal case, the police interrogates two suspects. The ratio between the ages of the suspects is 6:5 and the sum of their ages is 66 years. After how many years will the ratio be 8:7?

• (a) 11 years (b) 12 years (c) 6 years (d) 7 years• Solution:• A/6 = B/5 = x;• A = 6x; B = 5x;• A + B = 66 ; == x = 6;• A = 36, B = 30;• After X years, (( A + X ) / ( B + X )) = ( 8/7 )• Solve the above eqn, we get X=12.

• A greengrocer was selling apple at a penny each, chickoos at 2 for a penny and peanuts at 3 for a penny. A father spent 7 pennies and got the same amount of each type of fruit for each of his three children. What did each child get?

• (a) 1 apple, 2 chickoos, 2 peanuts• (c) 1 apple, 3 chickoos, 2 peanuts• (b) 1 apple, 2 chickoos, 1 peanut• (d) 1 apple, 1 chickoos, 1 peanut• Solution:• Get the answer from options.• No of items * no of children*cost of item=total cost for the item.• 1apple*3 =3*1 =3,• 2chickoos*3 =6*(1/2) =3,• 1peanuts*3 =3*(1/3) =1.• The sum of cost is 7. • It matches with father cost. So this is correct answer.

• There is a pie to be divided among 20 people. A man eats3 pizza, a women eats two pieces and a child eats half a piece of pie. Find the number of men, women and children so that they are 20 people in total and everyone gets some pie. There are 20 pieces of pie in all.

• (a) 7 women, 1 men and 12 children • (b) 5 women, 1 men and 14 children• (c) 6women, 2 men and 12 children• (d) 4women, 2 men and 14 children • Solution:• a)• 7w*2=14• 1m*3=3• 12*1/2 =6,• Total=23people.• b)• 5w*2=10• 1m*3=3• 14*1/2=7,• Total=20 people.

• Mr. Alex is the father of children Jane, Joe, and Jill. He goes to a nearby park twice a week. He loves his children very much. On a certain day, on his way to the park he finds fruit vendors selling different fruits. Watermelon is one penny each, dates at 2 for a penny and plums at 3 for a penny. Mr. Alex spent 7 pennies and got the same amount of each type of fruit for each of his three children. What did each child get?

• (a) 1 Watermelon, 2 Dates, 1 plum• (b) 1 Watermelon, 1 Date, 1 plum• (c) 1 Watermelon, 3 Dates, 2 plums • (d) 1 Watermelon, 2 Dates, 2 plums• Solution:• Get the answer from options.• No of items * no of children*cost of item=total cost for the item.• A)• 1watermelon*3 =3*1 =3,• 2dates*3 =6*(1/2) =3,• 1plum*3 =3*(1/3) =1.• The sum of cost is 7.• It’s matches with Mr. Alex’s cost. So this is correct answer.

logical • 1. One day Rapunzel meets Dwarf and Byte in the

Forest of forgetfulness. She knows that Dwarf lies on Mondays, Tuesdays and Wednesdays, and tells the truth on the other days of the week. Byte, on the other hand, lies on Thursdays, Fridays and Saturdays, but tells the truth on the other days of the week. Now they make the following statements to Rapunzel – Dwarf: Yesterday was one of those days when I lie. Byte: Yesterday was one of those days when I lie too. What day is it?

• (a) Monday (b) Sunday (c) Thursday (d) Saturday

Mon Tue Wed Thur Fri Sat SunDwarf L L L T T T TByte T T T L L L T

Condition: Both of them says yesterday is one of the day I lied.* Monday is not possibleBecause both the persons speaks truth.* Tue and wed is not possibleBecause Byte speak true on Mon and Tue.In Thursday,* Dwarf speaks True.So the statement he said is True.In the case of Byte,he speaks Lies on Thursday.So the statement he said is false and satisfies the condition.So Answer is Thursday.

• There were three persons A,B,C. A always says the truth. B lies on Mondays, Tuesday and Wednesday but C lies on Thursday, Friday & Saturday .One day A said that “ B & C came and spoke me, B said -Yesterday way one of the days when I lie and C said that – yesterday way one of the days when I lie too”. Then which day was that?

• (a) Tuesday (b) Thursday (c) Saturday (d) Sunday

• Solution :• This problem is similar to previous . Variables alone

changed .

• TYPE 1: Question states “At most n are true” . All true / 1st half false• Question states “At most n are false” . All false / 1st half true• TYPE 2: Question states “At least n are true” . 1st half true, 2nd half

false• Question states “At least n are false” . 1st half false, 2nd half true• TYPE 3: Question states “Exactly n true” . One statement true /

Second last statement true• Question states “Exactly n false” . One statement false / Second last

statement false

• 1. A sheet of paper has statements numbered from 1 to 40. For each value of n from 1 to 40, statement n says “At least n of the statements on this sheet are true.” Which statements are true and which are false?

• a) All statements are false.• b) The odd numbered statements are true and the even

numbered statements are false.• c) All statements are true.• d) The even numbered statements are true and the odd

numbered statements are false.• TYPE 2: Question states “At least n are true” . 1st half

true, 2nd half false• Hence d is correct.

• 2. A sheet of paper has statements numbered from 1 to 40. For each value of n from 1 to 40, statement n says “At least n of the statements on this sheet are true.” Which statements are true and which are false?

• (a)First half of the statements are true and the rest are false.• (b) The odd numbered statements are true and the even numbered

statements are false.• (c) First half of the statements are false and the rest are true.• (d) The even numbered statements are true and the odd numbered

statements are false.• TYPE 2: Question states “At least n are true” . 1st half true, 2nd

half false• Correct answer a.

• 3. A sheet of paper has statements numbered from 1 to 40. For each value of n from 1 to 40, statement n says “At most n of the statements on this sheet are true.” Which statements are true and which are false?

• (a) All statements are true.• (b) The odd numbered statements are true and the even numbered

statements are false.• (c) The first half of the statements are true and the remaining

statements are false.• (d) The even numbered statements are true and the odd numbered

statements are false.• TYPE 1: Question states “At most n are true” . All true / 1st half

false• Correct answer is a.

• 4. A sheet of paper has statements numbered from 1 to 40. For each value of n from 1 to 40, statement n says “Exactly n of the statements on this sheet are true.” Which statements are true and which are false?

• a) All statements are false.• (b) The odd numbered statements are true and the even numbered

statements are false.• (c) Second last statement is true and the remaining statements are

false.• (d) The even numbered statements are true and the odd numbered

statements are false.• TYPE 3: Question states “Exactly n true” . One statement true /

Second last statement true• Correct answer is c.

Suspects

• TYPE 1:• A –> All lying• B –> Left most culprit/ guilty• Answer –> A and B• TYPE 2:• A –> All lying• B –> Left most innocent• Answer –> A or B

• 10 suspects are rounded by the police and questioned about a bank robbery. Only one of them is guilty. The suspects are made to stand in a line and each person declares that the person next to him on his right is guilty. The rightmost person is not questioned. Which of the following possibilities are true?

• (A) All suspects are lying or the leftmost suspect is innocent.

• (B) All suspects are lying and the leftmost suspect is innocent.

• (a) A only (b) B only (c) Both A and B • (d) Neither A nor B• Answer : a and b.

• 2. 15 suspects are rounded by the police and questioned about a bank robbery. Only one of them is guilty. The suspects are made to stand in a line and each person declares that the person next to him on his right is guilty. The rightmost person is not questioned. Which of the following possibilities are true?A) All suspects are lying.

• B) The leftmost suspect is guilty.• C) Rightmost suspect is guilty.• (a) A only (b) A and B (c) B only (d) B and C

Permutations and combinations• Divisible by 4, digits 1, 2, 3, 4, 5 form n digit number when repetition of digits is

allowed• Number of values = 5 (n−1)

• 1. Alok is attending a workshop “How to do more with less” and today’s theme is working with fewer digits. The speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind) had only worked with fewer digits. The problem posed at the end of the workshop is How many 5 digit numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4?Can you help Alok find the answer?

• (a) 375 (b) 625 (c) 500 (d) 3125• Solution:• Condition for divisibility by 4 is last 2 digits must be divisible by 4.• Among the five numbers (1, 2, 3, 4, 5) have to find the combination which are

divisible by 4.• The combinations are 12, 24, 32, 44, 52.• First three digit is 5*5*5 and remain two digit is 5 so answer is 5*5*5*5

• 2. How many 13 digit numbers are possible by using the digits 1, 2, 3, 4, 5 which are divisible by 4 if repetition of digits is allowed?

• (a) 4*511 (b) 512 (c) 4*512 (d) 513

• Condition for divisibility by 4 is last 2 digits must be divisible by 4.

• Among the five numbers (1, 2, 3, 4, 5) have to find the combination which are divisible by 4.

• The combinations are 12, 24, 32, 44, 52.• First 11 digit is 511 and remain two digit is 5 so answer

is 512

• The pace length P is the distance between the rear of two consecutive footprints. For men, the formula, n/P = 150 gives an approximate relationship between n and P where, n = number of steps per minute and P = pace length in meters. Bernard knows his pace length is 152 cm. The formula applies to Bernard’s walking. Calculate Bernard’s walking speed in kmph.

• (a) 207.936 (b) 7.72 (c) 228 (d) 20.794• n/ P (cm) = x• Walking speed = ( x * 60 * P2 ) / 107 = 20.7936

• The pace length P is the distance between the rear of two consecutive footprints. For men, the formula n/P = 180 gives an approximate relationship between n and P where, n = number of steps per minute and P = pace length in meters. Bernard knows his pace length is 120 cm. The formula applies to Bernard’s walking. Calculate Bernard’s walking speed in kmph.

• (a) 236.16 (b) 8.78 (c) 15.56 (d) 23.62

• The pace length P is the distance between the rear of two consecutive footprints. For men, the formula, n/P = 144 gives an approximate relationship between n and P where, n = number of steps per minute and P = pace length in meters. Bernard knows his pace length is 164 cm. The formula applies to Bernard’s walking. Calculate Bernard’s walking speed in kmph. (a) 236.16 (b) 11.39 (c) 8.78 (d) 23.24

•

Ferrari

• 2. Ferrari S. p. A. is an Italian sports car manufacturer based in Maranello. Italy. Founded by Enzo Ferrari in 1928 as Scuderia Ferrari, the company sponsored drivers and manufactured race cars before moving into production of street-legal vehicles in 1947 as Ferrari S.p.A. Throughout its history, the company has been noted for its continued participations in racing, especially in Formula One, where it has enjoyed great success Rohit once bought a Ferrari. It could go 4 times as fast Mohit’s old Mercedes. If the speed of Mohit’s Mercedes is 35 km/hr and the distance traveled by the Ferrari is 968km, find the total time taken for Rohit to drive that distance.

• (a) 242 hours (b) 27 hours (c) 7 hours (d) 6.91 hours • Solution :• Time taken by Ferrari = Distance / ( n times * speed)• = 968 /( 4 * 35)• = 6.91hrs

• Formula to find the intercept between x and z • ax+by+cz = d• Distance = Sqrt of ((d/a)2+(d/c)2) • 1. Determine the distance between x-intercept and z-

intercept of the plane whose equation is 6x+8y-3z=72.• (a) 31.92 (b)26.83 (c) 32.66 (d) 25.63• Solution :• d=72, a=6, b=8, c=-3;• (d/a)^2=144 (d/c)^2=576 • Sqrt(144+576)=26.83

• 2.Determine the distance between the x-intercept and the z-intercept of the plane whose equation is 2x+9y-3z=18

• (a) 6.32 (b) 10.82 (c)3.00 (d)5.00 • Given :• d=18, a=2, b=9, c=-3;• Sqrt(81 + 36)=10.816

Series completion• 1. Lucy finds around 25 groups of stars that appear to her as constellations. She

draws 7 patterns of the constellations in her notebook and notes down the number of stars in each of them. She counts 5 stars in first constellation and 15 on next. She counts a number the third time and forgets to note it down. The next four constellations she counts 51.53,159,161. Next day her father looks at the notebook and wants to know the number of stars in the third constellation. Lucy only remembers that number of starts counted in each of the constellation followed a pattern 5,15, x, 51, 53, 159, 161.

• (a) 19 (b) 17 (c) 47 (d) 31• Given :• 15 51 159 ; 5 x 53 161• 15*3+6 = 51• 51*3+6 = 159;• 5*3+2 =17• 17*3+2 =53• 53*3+2 =161;

Probability • The great musician Rahman has organized a live concert.

The concert is organized in a big auditorium. Rahman plays both English and Tamil songs on his Yamaha Casio. The audience in the Eastern part of the auditorium love listening to Tamil Songs and those in the western part of the auditorium. He plays songs in random. The probability that he plays English songs for 6 consecutive times is 1 in

• (a) 32 (b) 16 (c) 64 (d) 128• Solution:• Possible events is English and Tamil = 2• Favorable is English =1• So probability for playing one English song is 1/2.• For consecutive 6 times is 1/ (26) =1/64.

• There are 7 children. You are told that the youngest child is a boy. The probability that all of them are boys is 1 in

• (a) 64 (b) 21 (c) 2 (d) 128 • d.

Speed, Distance, Time• A person was fined for exceeding the speed limit by 10mph.

Another person was also fined for exceeding the same speed limit by twice the same. If the second person was traveling at a speed of 35 mph, find the speed limit.

• (a) 35mph (b) 15 mph (c) 20 mph (d) 30 mph

• Solution :• I st person exceeds the limit is 10mph.• 2nd person exceeds twice the limit i.e. 20mph• 2nd persons speed is 35 mph.• So the speed limit is (35-20) =15mph.

• With four fifths of then tank full, a vehicle travels 12 miles. How much distance will the vehicle travel with one third tank full?

• (a) 8.05 km (b) 6.05 km (c) 12km (d) 5 km• Solution:• 4/5 of full tank travels = 12 miles.• Full tank travels = 15 miles.• One-third full tank = 15/3 =5 miles.• 1 miles = 1.609km.• So 5miles =5*1.609=8.05km.

• It was the semester exam day, Vidhya caught the college bus. She enjoyed traveling by bus. Moving at 6 mph, the bus took Vidhya to college at the right time. She finished her exam and had a chit chat with her friends and suddenly she realized that it was 6 pm and she had missed the college bus. She decided to walk back home at 4 mph. What is her average speed for the day?

• (a) 4 mph (b) 5 mph (c) 2.4 mph (d) 4.8 mph• Solution :• Its need to find the average speed = (2*6*4)/(6+4)• = 4.8mph

• Spores of a fungus, called late blight, grow and spread infection rapidly. These pathogens were responsible for the Irish potato famine of the mid-19th century. These seem to have attacked the tomato crops in England this year. The tomato crops have reduced and the price of the crop has risen up. The price has already gone up to $45 a box from $27 a box a month ago. How much more would a vegetable vendor need to pay to buy 27 boxes this month over what he would have paid last month?

• (a) $27 (b) $ 18 (c) $45 (d) $ 486• Solution :• Vendor paid extra = amount paid this month – amount paid last

month.• = 27*45 -27*27 = $486