Pattern (11-20)


PATTERN 11:

 Envelopes

 1. After the typist writes 12 letters and addresses 12 envelopes, she inserts the letters randomly into the envelopes (1letter per envelope). What is the probability that exactly 1 letter is inserted in an improper envelope?

(a) 11/12      (b) 0           (c) 1/12       (d) 1/6

 Reasoning:

     Placing exactly one letter in a wrong envelope and remaining letters in proper envelopes is not possible because if remaining letters are properly placed, then that one particular letter has to be placed into its own envelope. So given event is an impossible event.

 Probability (exactly 1 letter in improper envelope) = 0

 PATTERN 12:

 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) 512          (d) 513

PATTERN 13:

 1. 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

 Solution :

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.

(as in above  figure)

 PATTERN 14:

 Pace length

 1. 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 164cm. The formula applies to Bernard’s walking. Calculate Bernard’s walking speed kmph.

(a) 236.16      (b) 11.39        (c) 8.78         (d) 23.62

      n/ P (cm) = x

Walking speed = ( x * 60 * P2 ) / 10= 23.23814

PATTERN 15:

 1. 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

 Solution:

     Alice move should be a 1 move. Remain possibilities are leads to win of bob.

 2. Alok and Bhanu play the following coins in a circle game. 99 coins are arranged in a circle with each coin touching two other coins. Two of the coins are special and the rest are ordinary. Alok starts and the players take turns removing an ordinary coin of their choice from the circle and bringing the other coins closer until they again form a (smaller) circle. The goal is to bring the special coins adjacent to each other and the first player to do so wins the game. Initially the special coins are separated by two ordinary coins O1 and O2. Which of the following is true ?

 (a) In order to win, Alok should remove O1 on his first turn.

(b) In order to win, Alok should remove one of the coins different from O1 and O2 on his first turn

(c) In order to win, Alok should remove O2 on his first turn.

(d) Alok has no winning strategy.

 Solution:

There are two special coins separated by two ordinary coin. If Alok removes one out of these two ordinary coin say O1 or O2, Bhanu will remove the other ordinary coin and join both the special coins and win the game. Hence choice (a) is ruled out.

Total 99 coins. The remaining ordinary coins other than O1 and O2 and special coins S1 and S2, is 95 ordinary coins surrounded by special coins. If Alok starts removing one out of these 95ordinary coins and Bhanu will remove the second ordinary coin and this will continue and Alok will remove the 95 th ordinary coin and join the two special coin from out side and win the game.

 PATTERN 16:

 Ghana Bolivia – Football Match – Octopus

1. For the FIFA world cup, Paul the octopus has been predicting the winner of each match with amazing success. It is rumored that in a match between 2 teams A and B, Paul picks A with the same probability as A’s chances of winning. Let’s assume such rumors to be true and that in a match between Ghana and Bolivia, Ghana the stronger team has a probability of 2/3 of winning the game. What is the probability that Paul will correctly pick the winner of the Ghana- Bolivia game?

(a) 5/9          (b) 1/9                   (c) 2/3                   (d) 1/3

 Solution :

Condition:

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)

 Required Probability :

If Paul picks A with some probability as A’s chance of winning as 2/3 .

If they having chance of losing is 1/3, Paul’s predictions to be same.

From the Qn,

Ghana team is 2/3 strong.

Probability of predicting the winning team = {(2/3)*(2/3)} + {(1/3)*(1/3)} = 5/9

 2. Paul the octopus who has been forecasting the outcome of FIFA world cup matches with tremendous accuracy has now been invited to predict ICC world cup matches in2011. We will assume that the world cup contenders have been divided into 2 groups of 9 teams each. Each team in a group plays the other teams in the group. The top two teams from each group enter the semi finals (after which the winner is decided by knockout). However, Paul has a soft spot for India and when India plays any team, Paul always backs India. Alas, his predictions on matches involving India are right only 2 out of 3 times. In order to qualify for the semi finals, it is sufficient for India to win 7 of its group matches. What is the probability that India will win the ICC world cup?

(a) (2/3)^10                                        (b) (2/3)^9 + 8/3 * (2/3)^9

 (c) 8/3 * (2/3)^9                                  (d) (2/3)^10 + 8/3*(2/3)^9

 Condition : We have two chances to win the world cup

  1. India can win all 10matches + win the semi final and final matches.
  2. Win 7 out of 8 group matches + win the semi final and final matches.

Entry level probability :

Case 1:

To win all 10, probability is (2/3)^10.

Case 2:

Probability of winning 7mathces out of 8= (2/3)^7

and Probability of Loosing the 1 match= 1/3.

This process can occur in 8 ways.

Probability= 8*((2/3)^7 * 1/3)

Semi final and Final level probability :

To win semi final and final match probability is (2/3)^2

Total probability = 8*((2/3)^7 * 1/3) * (2/3)^2 = 8/3 * (2/3)^9.

So Total probability, Case1+Case 2 = (2/3)^10 + 8/3 * (2/3)^9

 PATTERN 17:

 Handshakes

 TYPE 1:

     Let n people {a1, a2 … an} meet and shake hands in a circular fashion. In other words,

There are totally n handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {an-1, an}, {an, a1}

For no cycle, we have to exclude the pair ( an-1, an).

     * when no cycle, the MAXIMUM no. of handshakes = (n-1) handshakes.

      * If cyclic handshake, no of handshakes = no of people.

TYPE 2:

     For minimum people in a set we can consider handshakes as {a1, a2, a3}, {a4, a5, a6}, {a7, a8, a9},…….{ an-2,an-1,an}

For minimum people, we can consider the set {a2, a5, a8… an-1}.

So, MINIMUM people in a set =  n / 3

 1. 36 people {a1, a2, …, a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is

(a) 18          (b) 13                    (c) 34                   (d) None

 Solution :

     From the  type 2 :

MINIMUM people in a set =  n/3 = 12 

 2. 21 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 that the pairs (a1, a2), (a2, a3), … , (a (k-1), a k) , (ak, a1) shake hands.

(a) 17             (b) 18              (c) 19                      (d) 20

 Solution :

 If non cyclic handshake, no of handshakes = no of people – 1 = 20

3. 66 people {a1, a2,…, a66} meet and shake hands in a circular fashion. In other words, there are totally  36 handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {a65, a66}, {a66, a1}. The size of the smallest  set of people such that the rest have shaken hands with at least one person in the set is

(a) 22     (b) 33          (c) 65          (d) 11

Solution :

From the  type 2 :

MINIMUM people in a set =  n/3 = 22 

4. The thanksgiving banquet at No.2, Richter Street, had 49 guests which consisted of 6 statesmen, 26 relatives and their families. At the end of a banquet 19 people shake hands with other , some of which were between the statesmen alone, some between relatives alone and some between the statesmen and relatives. How many handshakes will there be in total?

(a) 342                (b) 171              (c) 180                        (d) 162 

 Solution :

nCr

n=19, r=2    ==>171

 PATTERN 18:

Clock

Speed of the hour hand = 360 / (18×90) minutes = (2 / 9) ° per minute.

       Speed of the minute hand = 360 / 90 minutes = 4° per minute.

       Relative speed of the two hands of clock = (4 – 2 / 9 )° per minute = 34/9 ° per minute.

Suppose we have to find angle at H hr and M minutes.

Then

Angle = { ( 360 / half of the number of hours in the day of a planet ) * H +  ( Relative speed of the two hands of clock ) * M  }

 or

{ (  360 / half of the number of hours in the day of a planet ) * H -  ( Relative speed of the two hands of clock ) * M }

 1. 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 hours has 90 min while each minute has 60 sec. As on earth, hour hand covers the dial twice every day. Find the approximate angle between the hands of clock on Oz when time is 12.40 am

(a) 89        (b) 251                  (c) 111                 (d) 79

 Solution :

     Angle = [(360/ 18 ) * 12 + ( 34/9 ) *40 ] or [ (360/ 18 ) * 12 - ( 34/9 ) *40 ]

            = [ 240 + 151.11 ] or [ 240 – 151.11 ]

            = 89o or 151.11o

PATTERN 19:

 Unpainted Faces of a Hollow Cube

Unpainted face = (( n3 − (n − 2)3 ) * 6) – (No of faces painted *n2)

 1. 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 4 faces of the outer surface of the cube are pained totally how many faces of the smaller cubes remain unpainted?

(a) 900      (b) 488                   (c) 500            (d) 800

 Solution :

  5 cm cube with 4 faces painted

 Unpainted Faces of a Hollow Cube = (( 53 − (5 − 2)3 ) * 6) – (4 * 52 )

                                  = 98 * 6 – 100

                                 = 488 .

Another method :

     Formula for hollow cube=n3 –(n-2)3

Each 1*1 cube has 6 faces
Total 1*1 cubes =53- (5-2) 3 = 98
Total faces of small cubes = 98*6 = 588
Now, four sides of the 5*5 cube is painted
So, total small cube faces painted = 4*5*5 = 100
Unpainted faces  = 588 – 100 = 488

 

2. Susan made a block with small cubes of 8 cubic cm volume. To make the block she used 3 small cubes long, 9 small cubes wide and 5 small cubes deep. She realizes that she has used more small cubes than she really needed. She realized that she could have glued a fewer number of cubes together to lock like a block with same dimensions, if it were made hollow. What is the minimum number of cubes that she needs to make the block?

 

(a) 114         (b) 135    (c) 21          (d) 71

Solution :

     Formula for Hollow cube i.e., Variable sides = x*y*z – (x – 2)(y – 2)(z – 2)

     Minimum number of cubes    = ( 3 *5 * 9 ) – (3-2)(5-2)(9-2)

                                       = 135 – 21

                                 = 114

PATTERN 20:

 

1. Planet fourfi resides in 4 – dimensional space and thus the currency used by its residents are 3 – dimensional objects. The rupees 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 64mm and not exceed 512mm

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

The diameter of the 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) 5                (b) 8                       (c) 9            (d) 6

 Solution :

Condition: Next coin should be 50 % more than the previous

Start with

1st coin :   64 mm

2nd coin:    64+32       =   96 mm

3rd coin:    96+48       =  144mm

4th coin:    144+72      =  216 mm

5th coin:    216+108     =  324 mm

6th coin:    324+162     =  486 mm

7th coin:  486+243    =  729 mm is greater than 512 so we stopped here.

Ans is 6 coins.

 

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