1) There are 15 poles in a straight line each equally placed from each other. The time taken to reach the 10th pole after a man starts walking from 1st pole is 10 seconds. What is the time required to reach 15th pole. 1A)Let any two poles be placed x units apart. Distance between 1st and 10th pole=(10-1)x=9x units. Distance between 1st and 15th pole=(15-1)x=14x units. If time taken to cover 9x units is 10 seconds, then time taken to cover 14x units is..... (10*14x)/9x = 15.55 seconds.

2)A prisoner tried to escape from the jail in five different ways in five consecutive days from monday to friday, but in vain.The five ways are.... (a)Tried to bribe the jailer of the prison. (b)Tried to escape in the giuse of a jailer. (c)Tried to dig a tunnel in the jailcell. (d)Escape in the guise of visitor to the jail. (e)Escape in the guise of work party who came to work in the jail. Condition: 1-Two days before he tried to dig the tunnel he tried to bribe the jailer. 2-It was not monday or friday when he tried to escape in the guise of a visior to the jail.

3-The day before he tried to escape in the guise of a jailer he tried tobribe the Jailer.

4-The day after he tried to escape in the guise of a visitor to the jail, he tried to escape in the guise of work party person.On each of the five days which method he used to escape.

2A)From condition 1 and 3 we can say that the rpisoner tried to escape in methods a,b,c in order on 3 consecutive days. from conditions 2 and 3 and the order for a,b,c we can say that the order for thursday and friday was e,d in order.So, the order from monday through friday will be a,b,c,e,d.

3)A cube of 3cm side is coloured red on all of its six faces. It is cut into small cubes of side 1cm each. (a) No. of small cubes with 3 faces coloured red : 8 (b) No. of small cubes with 2 faces coloured red : 12 (c) No. of small cubes with 1 face coloured red : 6 (d) No. of small cubes with no faces coloured red : 1

4)Two trains each 1/6 mile in length run in opposite dirctions to each other with equal speed of 60 miles/hour. What is the time taken to completely cross over the two trains.

4A)Speed at which two trains travel towards eachother = 60+60 = 120 miles/hour. Total length to be covered = (1/6)+(1/6) = 2/6 mile. Time to taken to cover 2/6 mile at speed of 120 miles/hour is..... (2/6)/120 = 1/360 hour = 10 seconds.

5)A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets atleast 2 balls of same colour. 5A) 6

6)A and B are two friends, together they go out for shopping. A buys a hat and a suit, all together A spends 15 rupees. B buys a hat which costs same as the suit bought by A. B even buys a dress which cost her one rupee less than the cost of hat bought by A. B says to A that "If at all I had bought my hat by spending one and half times the money u spent on buying ur hat, we both would have spent the same amount on buying our items(together, i.e., 15 rupees). (a) what is the cost of hat bought by A. (b) The money A and B apent together. 6A) Lets say A spent x and y rupees on hat and suit respectively. from the conditions we can say B spent y and x-1 rupees on hat and dress respectively. From what B told to A we can write that..... => x + y = 15 = (3/2)x + (x-1) => (3/2)x + (x-1) = 15 => (5/2)x - 1 = 15 => (5/2)x = 16 => x = 16 * (2/5) => x = 32/5 = Rupees 6.40 paise (a) answer is 6.40 Rupees. (b) Total A alone spent = x+y =15. B spent = y + (x-1) = (x+y)-1 = 15-1 = 14 Rupees.

So, together they spent = 15 + 14 = 29 Rupees. answer is 29 Rupees.

7)A man wearing a hat stands on a bridge across a river. At an instant his hat falls into the river and starts flowing along with the atream at the same speed of the stream. At the same instant the man jumps into the river and starts swimming in the direction opposite to the stream, like that he swims for 10 minute , then he turns back and starts swimming in the direction of the stream to retrieve the hat. He retrieves the hat under another bridge which is 1000 yards away down the stream from the first bridge. What is the speed of the river. 7A)Let the swimming speed of man be = a yds/min. speed of stream = b yds/min. Then the man swims at (a-b) yds/min. against the stream. and (a+b) yds/min in direction with the stream. Lets assume the man covered a distance of x in 10 min swim up the stream at speed (a-b). The time taken by him to cover the same distance down the stream at speed (a+b) is... => 10(a-b)/(a+b) The total time of travel(being same for both), hat and the man be 't'. Time taken by hat to travel from first bridge to the bridge down stream 1000 yards away at speed b is 't' minutes. Time taken by man to travel from first bridge to the bridge down stream 1000 yards away at speed (a+b) will be [t-{10+[(10(a-b)/(a+b))]}] minutes. There for we have for hat b*t=1000 yds. ------> (1) for man .... (a+b)[t-(10(a-b)/(a+b))] = 1000 yds. ------> (2) solving equations (1) and (2) we get => t=20 minutes. Answer : so, speed os the stream is 1000/20= 50 yds/min.

8)A train travels at a particular speed for a duration of one hour, after which one of its engine malfunctions reducing its speed to 3/5th of the actual speed before the occurance of fault in engine. It travels at this speed for 2 hours to reach at its destination. If the fault had occured 50 miles later on, the train would have reached its destination 45 minuts early. Find the distance traveled by the train. 8A)

Let the actual speed of train be X miles/hour. after fault its speed will be reduced to 3X/5 miles/hour. If we could find the speed of the train, we will get the solution!!! Short method: On observation one can find that time difference( 45 minutes) occur between two cases, in the 50 miles distance travelled by the train at two different speeds(X and 3X/5). => If the time taken to cover 50 miles at speed 3X/5 is 't' hours then time taken to travel 50 miles at speed X is t-(45/60) = (t-0.75) hours. We can write ...=> (3X/5)*t = X*[t-0.75] = 50 miles -->(i) solving we get... t=15/8 hours. then substituting value of 't' in first term of eq. (i) => (3X/5)*15/8=50 solving we get... X= 400/9 miles/hour. Answer : So, solving the problem with first case we get the distance travelled by the train as 97.77 miles. Long Method: From conditions we can write eq. => (X*1) + (3X/5)*2 = D miles. --->(1) where, D is distance travelled by the train. From second case, we can write, => (X*1) + 50 + [(3X/5)*(3-1-50/X-0.75)]=D miles.-->(2) from (1) we can write => 11X/5 = D => X=5d/11 from (2) we cab write => X+50+(3X/5)*(1.25-50/X)=D => 7X/4 + 20 = D -->(3), substituting value of X=5D/11 in (3) we get => D= 880/9 = 97.77 miles. So, answer is 97.77 miles.

9) In this problem five persons names are given, and also the ages of these persons and the names of the houses they live in , not in the order. Some conditions are given. we need to satisfy all the conditions and find the correct ages of all the persons and the houses they live in , in order.( 10 marks)

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Puzzle1) There are 15 poles in a straight line each equally placed from each other. The time taken to reach the 10th pole after a man starts walking from 1st pole is 10 seconds. What is the time required to

reach 15th pole.

1A)Let any two poles be placed x units apart.

Distance between 1st and 10th pole=(10-1)x=9x units.

Distance between 1st and 15th pole=(15-1)x=14x units.

If time taken to cover 9x units is 10 seconds,

then time taken to cover 14x units is.....

(10*14x)/9x = 15.55 seconds.

2)A prisoner tried to escape from the jail in five different ways in five consecutive days from monday to friday, but in vain.The five ways are....

(a)Tried to bribe the jailer of the prison.

(b)Tried to escape in the giuse of a jailer.

(c)Tried to dig a tunnel in the jailcell.

(d)Escape in the guise of visitor to the jail.

(e)Escape in the guise of work party who came to work in the jail.

Condition:

1-Two days before he tried to dig the tunnel he tried to bribe the jailer.

2-It was not monday or friday when he tried to escape in the guise of a

visior to the jail.

3-The day before he tried to escape in the guise of a jailer he tried tobribe the Jailer.

4-The day after he tried to escape in the guise of a visitor to the jail, he tried to escape in the guise of work party person.On each of the five days which method he used to escape.

2A)From condition 1 and 3 we can say that the rpisoner tried to escape in methods a,b,c in order on 3 consecutive days. from conditions 2 and 3 and the order for a,b,c we can say that the order for thursday and friday was

e,d in order.So, the order from monday through friday will be a,b,c,e,d.

3)A cube of 3cm side is coloured red on all of its six faces. It is cut into

small cubes of side 1cm each.

(a) No. of small cubes with 3 faces coloured red : 8

(b) No. of small cubes with 2 faces coloured red : 12

(c) No. of small cubes with 1 face coloured red : 6

(d) No. of small cubes with no faces coloured red : 1

4)Two trains each 1/6 mile in length run in opposite dirctions to each other with equal speed of 60 miles/hour. What is the time taken to completely cross over the two trains.

4A)Speed at which two trains travel towards eachother = 60+60 = 120 miles/hour.

Total length to be covered = (1/6)+(1/6) = 2/6 mile.

Time to taken to cover 2/6 mile at speed of 120 miles/hour is.....

(2/6)/120 = 1/360 hour = 10 seconds.

5)A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls,

and 1 red ball. How many minimum balls one should pick out so that to make sure

the he gets atleast 2 balls of same colour.

5A) 6

6)A and B are two friends, together they go out for shopping. A buys a hat and a suit,

all together A spends 15 rupees. B buys a hat which costs same as the suit bought by

A. B even buys a dress which cost her one rupee less than the cost of hat bought by A.

B says to A that "If at all I had bought my hat by spending one and half times the

money u spent on buying ur hat, we both would have spent the same amount on buying

our items(together, i.e., 15 rupees).

(a) what is the cost of hat bought by A.

(b) The money A and B apent together.

6A) Lets say A spent x and y rupees on hat and suit respectively.

from the conditions we can say B spent y and x-1 rupees on hat and dress respectively.

From what B told to A we can write that.....

=> x + y = 15 = (3/2)x + (x-1)

=> (3/2)x + (x-1) = 15

=> (5/2)x - 1 = 15

=> (5/2)x = 16

=> x = 16 * (2/5)

=> x = 32/5 = Rupees 6.40 paise

(a) answer is 6.40 Rupees.

(b) Total A alone spent = x+y =15.

B spent = y + (x-1) = (x+y)-1 = 15-1 = 14 Rupees.

So, together they spent = 15 + 14 = 29 Rupees.

answer is 29 Rupees.

7)A man wearing a hat stands on a bridge across a river. At an instant his hat falls

into the river and starts flowing along with the atream at the same speed of the stream.

At the same instant the man jumps into the river and starts swimming in the direction

opposite to the stream, like that he swims for 10 minute , then he turns back and

starts swimming in the direction of the stream to retrieve the hat. He retrieves the hat

under another bridge which is 1000 yards away down the stream from the first bridge.

What is the speed of the river.

7A)Let the swimming speed of man be = a yds/min.

speed of stream = b yds/min.

Then the man swims at (a-b) yds/min. against the stream.

and (a+b) yds/min in direction with the stream.

Lets assume the man covered a distance of x in 10 min swim up the stream at speed (a-b).

The time taken by him to cover the same distance down the stream at speed (a+b) is...

=> 10(a-b)/(a+b)

The total time of travel(being same for both), hat and the man be 't'.

Time taken by hat to travel from first bridge to the bridge down stream 1000 yards

away at speed b is 't' minutes.

Time taken by man to travel from first bridge to the bridge down stream 1000 yards

away at speed (a+b) will be [t-{10+[(10(a-b)/(a+b))]}] minutes.

There for we have for hat b*t=1000 yds. ------> (1)

for man .... (a+b)[t-(10(a-b)/(a+b))] = 1000 yds. ------> (2)

solving equations (1) and (2) we get

=> t=20 minutes.

Answer : so, speed os the stream is 1000/20= 50 yds/min.

8)A train travels at a particular speed for a duration of one hour, after which one

of its engine malfunctions reducing its speed to 3/5th of the actual speed before the

occurance of fault in engine. It travels at this speed for 2 hours to reach at its

destination. If the fault had occured 50 miles later on, the train would have reached

its destination 45 minuts early. Find the distance traveled by the train.

8A)

Let the actual speed of train be X miles/hour.

after fault its speed will be reduced to 3X/5 miles/hour.

If we could find the speed of the train, we will get the solution!!!

Short method:

On observation one can find that time difference( 45 minutes) occur between two cases,

in the 50 miles distance travelled by the train at two different speeds(X and 3X/5).

=> If the time taken to cover 50 miles at speed 3X/5 is 't' hours then time taken to

travel 50 miles at speed X is t-(45/60) = (t-0.75) hours.

We can write ...=> (3X/5)*t = X*[t-0.75] = 50 miles -->(i)

solving we get... t=15/8 hours.

then substituting value of 't' in first term of eq. (i) => (3X/5)*15/8=50

solving we get... X= 400/9 miles/hour.

Answer : So, solving the problem with first case we get the distance travelled

by the train as 97.77 miles.

Long Method:

From conditions we can write eq. => (X*1) + (3X/5)*2 = D miles. --->(1)

where, D is distance travelled by the train.

From second case, we can write, => (X*1) + 50 + [(3X/5)*(3-1-50/X-0.75)]=D miles.-->(2)

from (1) we can write => 11X/5 = D => X=5d/11

from (2) we cab write => X+50+(3X/5)*(1.25-50/X)=D

=> 7X/4 + 20 = D -->(3),

substituting value of X=5D/11 in (3) we get => D= 880/9 = 97.77 miles.

So, answer is 97.77 miles.

9) In this problem five persons names are given, and also the ages of these persons and

the names of the houses they live in , not in the order. Some conditions are given.

we need to satisfy all the conditions and find the correct ages of all the persons

and the houses they live in , in order.( 10 marks)