## Puzzles for freshers in aptitude exam

**21.A JUMPING FROG
A frog starts climbing a 30 ft wall. Each hour it climbs 3 ft and slips back 2. How many days does it take him to reach the top and get out?**

Solution:

Hours Feets

1 3ft – 2ft = 1ft

2 3ft – 2ft = 1ft

3 3ft – 2ft = 1ft

: :

: :

27 3ft – 2ft = 1 ft

At the end of 27th hour the frog climbs 27fts and on 28th hour it

climbs the remaining 3fts and comes out.

**22.QUESTION OF PROBABILITY
My friend Parveen teaches at a school. One day she conducted a test for 3 of her students and when they handed back the test papers, they had forgotten to write their names. Parveen returned
the papers to the students at random.What is the probability that none of the 3 students will get the right paper?**

Solution:

The possible combinations for the given condition are:

1 2 3 condition met

1 2 3 No

1 3 2 Yes

2 1 3 No

2 3 1 Yes

3 1 2 Yes

3 2 1 No

The required probability = 2/6 = 1/3

**23.MATHEMATICAL ODDITY
In the 20th century there are only seven years whose numbers are a mathematical oddity because their numbers signify a prime number.The first one of its kind was the year 1951.Can you name the other sin?**

Solution:

1973,1979,1987,1993,1997,1999.

24.DOWN THE ESCALATOR

Recently,while in London,I decided to walk down the escalator of a tube station.I did some quick calculation in my mind.I found that if I wa down 26 steps, I require 30 seconds to reach the bottom.

However ,if I am able to step down 34 stairs I would only require 18 secionds to get to the bottom. If the time is measured from the moment the top step begins to descend to the time I step off the

last step at the bottom , can you yel the height of the stairway in steps?

Solution:

Given that after walking 26 steps time needed to reach the bottom is

30 seconds —(1)

Similarly after walking 34 steps, it needs

18 seconds more to reach the bottom —(2)

from (1) & (2) it is clear that

for (34-26) steps it took (30-18) seconds.

i.e; 12 seconds for 8 steps

in 1 second ——–?

(1*8)/12=2/3 steps/sec.

i.e; 2 steps in 3 seconds

for 30 seconds ———–how many steps?

=> (30*2)/3=20 steps.

Finally Total number of steps = 26 + (steps covered in 30 seconds)

=26+20=46 steps.

**25.A COMPUTING PROBLEM
Compute (100-1)(100-2)(100-3)……………..(100+1)(100+2)(100+3)=?**

Solution:

=>(100-1)(100-2)(100-3)………(100-100)(100+1)(100+2)(100+3)

=>(100-1)(100-2)(100-3)………(0)(100+1)(100+2)(100+3)

=0

**26.A CIRCLE AND A TRIANGLE
What do you call a circle which passes through the vertices of of a triangle?**

Solution:

Circumscribed.The meaning to circumscribe is to describe a figure

round another so as to touch it at points without cutting.This is

exactly what takes place with the circumscribed circle.

To find the center of such a circle,we have to bisect the sides of

a triangle and errect perpendiculars which are concurrent at the

circumcentre. The radius r of the circumscribed circle of the

triangle ABC is given by R=a/2SinA=b/2SinB=c/2Sinc

27.MISSING TERMS

48,60,58,72,68,104…….

Here is a sequenc.Can you find the two missing terms?

Solution:

The odd terms are in the decimal system and differ by 10.And each

even term is the preceeding odd term expressed in the octonamy

system.78-8=70,remainder 6:9:8=1,remainder 1.Therefore the next

two terms are: 78,116

28.PACKETS OF CANDY

If 6 men can pack 6 packets of candy in 6 minutes.How many are required to pack 60 packets in 60 minutes?

Solution:

Given that for 6 men to pack 6packets of candy it takes 6 miutes

i.e., for 1 man to pack 1 packet it takes 1 miute.

Hence,for 60 packets to be packed in 60 miutes we need 60 men.

**29.A PROBLEM OF WEIGHT
In my neighbourhood lives a man who weighs 200 pounds.He has two sons.They both weigh 100 pounds each.On a festival day they decide to go across the river on a boat to vissit some relations.
But the boat could carry a maximum load at only 200 pounds.Yes they managed to come across the river by boat.How did they?**

Solution:

Let us assume that c1 —————–>first son

c2 —————->second son

f ——————>father

First the two sons c1,c2rowed across the river and c1 stayed

behind while c2 returned in the boat to his father.The son remained

behind while the father crossed the river.Then the other son

brought back and the two brothers c1,c2 rowed over together.

30.A PROBLEM OF CANDY BARS

Recently I attended a birthday party.All the children in the party were given candy bars.All the children got 3 candy bars each except the child sitting in the end.She got only 2 candy bars.If

only child had been given 2 candy bars there would have been 8 candy bars remaining.How many candy bars were there altogether to begin with?

Solution:

Suppose that there were x children at the party.

If we distribute the candies in the above mentioned ways,then

the resulting expressions

3(x-1)+2———————(1)

2x+8————————-(2)

3(x-1)+2=2x+8

3x-3+2=2x+8

3x-1=2x+8

x=9

Therefore the no.of candies for distribution 2x+8=2*9+8=18+8=26

**31.FIND OUT THE SUM
What is the sum of all numbers between 100 and 1000 which are divisible by 14?**

Solution:

Let us assume that the sum is S

s=112+126+……..+994

s=14(8+9+10+…………………+71)

s=14(8+71)(71-8+1)/2=7(79)(64)=35392

**32.WALKING ALL THE WAY
One day I decided to walk all the way from Banglore to Tumkur.I started exactly at noon and some one I know in Tumkur decided to walk all the way from Tumkur to Banglore and she started
exactly at 2 PM on the same day. We met on the Banglore – Tumkur road at 5 past four and we both reached our destination at exactly the same time.At what time did we both arrive?**

Solution:

There fore Total time = 2:00 PM + 3:55 PM + 1:55 PM = 7:50PM

**33.THE TRAINS AND THE FALCON
Two trains start from two opposite directions towards each other.The stations from which they start are 50 miles apart.Both the trains start at the same time towards the other train. As soon as it
reaches the second one, it fies back to the first train and so on and so forth. It continues to do so, flying bacwards and forwards from one train to other until the trains meet. Both the trains travel
at a speed of 25 miles per hour,and the bird flies at 100 miles per hour. How many miles will the falcon have flown before the trains meet?**

Solution:

The trains travel at 25 miles per hour.

Hence they will meet after travelling for one hour and the falcon also

must have been flyingfor one hour. Since it travels at 100 miles per

hour the bird must have flown 100 miles

**34.VALUE OF ‘S’
S434S0, what number must be substituted with to make it divisible by 36?**

Solution:

To be divisible by 36, the number has to be divided by 4 and 9

To be divisible by 4 , the number ‘S’ must be an even number and to be

divisible by 9, the sum of all the digits of the number must be either

equal to 9 or a multiple of 9

i.e; S + 4 + 3 + 4 + S + 0 = 9n

The only digit that meets these two condition is 8

35.HEIGHT OF A ROOM

Given the floor area of a room as 24 feet by 48 feet,and the space diagonal of the room as 56 feet,Can you find the height of the room?

Solution:

We know that,

Volume of a cube =l pow(2) + b pow(2) + h pow(2).

Here the values of l & b are given.

We also know that

(diagonal)pow(2) = (length)pow(2) + (breadth)pow(2).

=> (x)pow(2) = 24 pow(2) + 48 pow(2)

=> x = 24squareroot(5);

Therefore,volume= h pow(2) + x pow(2) =56 pow(2)

=> h=16

Therefore,height of the room=16 ft.

**36.A QUESTION OF DISTANCE
It was a beautiful sunny morning. The air was fresh and a mild wind was blowing against my wind screen I was driving from Banglore to Brindavan Gardens. It took me 1 hour and 30 miutes to complete the journey. After lunch I returned to Banglore. I drove for 90 minutes.How do you explain it?**

Solution:

90 minutes = 1 hour 30 minutes.

Hence,the driving time there and back is absolutely the same because

90 minutes and 1 hour and 30 minutes are one and the samething.

**37.ARRANGE THE DIGITS:
Arrange the digits 1,2,3,4,5,6,7,8,9 in order from left to right and use only + or _ signs so as to produce a result of 100?**

Sol:

123-45-67+89

**38.DIVISION OF 45:
Can you divide the number 45 into four parts such that when 2 is added to the first part, 2 is subtracted from the second part, 2 is multiplied by the third part, and the fourth part is divided by 2.
All the four results to be the same number.**

Sol:

Let us take A,B,C,D are the four equal parts and their sum is

equal to 45.

A+B+C+D = 45

Given that,

A+2 = B-2 = C*2 = D/2 —————-(1)

=>A = B-4 ; C = (B-2)/2 ; D = 2(B-2)

=>B-4 + [(B-2)/2] + 2(B-2) = 45

=>B=12

:. A=8 , C=5 , D=20

And condition (1) is satisfied.

i .e; 8+2 = 12-2 = 5*2 = 20/2

**39.SPECIAL NUMBER:
What is the special about the number 1729?**

Sol:

This is popularly known as Ramanujan’s number. This is the known

number that is a sum of two cubes in two different ways.

i .e; (10*10*10) + (9*9*9) = 1729

( 12*12*12) + (1*1*1) = 1729

**40.PRICE OF A BOTTLE:
A bottle and its cork together cost Rs 1.10, and the bottle costs Rs 1.00 more than its cork. What is the price of the bottle?**

Sol:

Let us assume that,

B = Price of the bottle

C = Price of the cork

It is given that,

B + C = Rs 1.10 ——————-(1)

and B â€“ C = Rs 1.00 ——————-(2)

From the equations (1) and (2) it is clear that

B = Rs 1.05

C = Rs 0.05

**41.A QUESTION OF DISTANCE:
It was a beautiful sunny morning. The air was fresh and a mild wind was blowing against my wind screen. I was driving from Banglore to Brindavan Gardens. It took me one hour and 30 minutes to complete the journey. After lunch I returned to banglore. I drove for 90 minutes. How do you explain it?**

Sol:

90 minutes = 1 hour 30 minutes.

Hence, the driving time there and back is absolutely same because

90 minutes and 1 hour 30 minutes are one and the same.

**42.FOR THE CHARITIES:
One day when I was walking on the road in New Delhi, a group of boys approached me for donation for their poor boys’ fund. I gave them a Rupee more than half the money I had in my purse. I must have walked a few more yards when a group of women approached me for donation, for an orphanage. I gave them 2 Rupees more than half the money I had in my purse. Then after a few yards I was approached by a religious group for a donation to the temple they were building. I gave them 3 Rupees more than half of what I had in my purse.At last I returned to my hotel room, I found that I had only one Rupee remaining in my purse. How much money did I have in my purse
when I started?**

Sol:

Suppose that the money in his purse when he started = x ——–(1)

For poor boys fund he gave x/2 + 1 Rs/- —————–(2)

i .e; 1 Rupee more than half the amount he had.

Now he left with [x – (x/2 + 1)] =(x-2)/2 Rs/-

For Orphanage he gave [(x/2 â€“ 1)/2] +2 = (x+6)/4 Rs/- —–(3)

Now he left with [(x-2)/2] â€“ [(x+6)/4] = (x-10)/4 Rs/-

For temple building he gave [(x-10)/4]/2 + 3 = (x+14)/8 Rs/- —(4)

Now he left with [(x-10)/4] â€“ [(x+14)/8] = (x-34)/8 Rs/-

Finally he had 1 Rupee in his purse.

i .e; Actual amount â€“ Expended amount = 1

:. from (1),(2),(3) and (4) we have

x-{ [ (x+2)/2 ] + [ (x+6)/4 ] + [ (x+14)/8 ] } = 1

=>x-34=8

=> x=42

:. The original amount in his purse at the beginning = Rs 42/-

43.A PAIR OF PALLINDROMES

Multiply 21978 by 4.Comment about the result?

Solution:

21978*4=87912.

If we clearly observe the two numbers 21978 and 87912, the resultant

number ie; 87912 is the reverse number of the number 219780.

There fore these two numbers are a pair of paindromes.

44.A COMPUTING PROBLEM

Compute: [5-2/(4-5)]pow(2).

Solution:

[5-2/(4-5)]pow(2)

=[5-2/(-1)]pow(2)

=[5+2]pow(2)=49.

**45.CONTINUE THE SERIES
1,3,6,10.Name the next three numbers in the series.**

Solution:

The series is +2,+3,+4,—————–

There fore next three numbers are:

10+5,10+5+6,10+5+6+7 = 15,21,28.

46.NAME FIVE TERMS OF ANOTHER SERIES

These are the numbers that are the first five terms of a series that add upto 150.Can you name five terms of another series without fractons that add upto 153?

[ex: 10,20,30,40,50. sum=150.]

Solution:

Each term in this series is a factorial, in other words,the product

of a the numbers from 1 to that particular term considered.

The first five terms of the series are,there fore 1,2,6,24,120.

Their sum is 153 and are factorials of 1,2,3,4,5 respectively.

**47.FIND OUT THE TIME
What does 1408 hours mean?**

Solution:

1408 hours is actually 8 minutes past 2 PM.

This is the system of twenty-four-hour cock.Writing the hours and

minutes this way is a sensible means of avoiding confusion

between AM and PM.

**48.FIND OUT TTHE PATTERN
What do you think the pattern is?6,24,60,120,210,336,â€¦â€¦â€¦â€¦â€¦â€¦**

Solution:

The series is 1.2.3, 2.3.4, 3.4.5, 4.5.6, 5.6.7, 6.7.8,———-

The next numbers would be 7.8.9,

**49.THE TRAIN AND THE CYCLIST
A railway track runs parallel to a road until a bend brings the road to a level crossing. A cyclist rides along to work along the road every day at a constant speed of 12 miles per hour.He normally meets a train that travels in the same direction at the crossing One day he was late by 25 minutes and met the train 6 miles ahead of the level crossing. Can you figure out the speed of the train?**

Solution:

Suppose that the train and the cyclist meet everyday at the crossing

at 8:00A.M. i. e; starts at 7:00A.M

Since the cyclist is late by 25 minutes, he starts at 7:25A.M

As his speed is 12 miles per hour, he reaches the crossing at

7:25A.M + 1 Hour = 8:25A.M

By 8:30A.M the train is 6 miles ahead of the cyclist

The difference between their timings = 8:30A.M â€“ 8:25A.M = 5 Minutes

The difference between their distances = 6 Miles

Therefore,the train travels 6 miles in 5 minutes

In 1 minute it travels ———————?

= [(1 * 6) /5] * 60

= 72 Miles/hour

**50.HEIGHT OF THE PALM TREE
A palm tree was 90 cm high, when it was planted. It grows by an equal number of cm each year, and at the end of the seventh year it was one ninth taller than at the end of the sixth year. Can you tell how tall was the tree at the end of the twelfth year? **

Solution:

Suppose that the tree grows x cm each year

Height of the tree at the end of the sixth year = (90 + 6x) cm

Growth in seventh year is,

X = 1/9(90 + 6x) cm

x = 10 + 2x/3

x = 30

Therefore the height of the tree at the end of the twelfth

year=(90+12*30)=450cm

**51.PROBLEM OF AGE
Recently I attended a cocktail party. There was a beautiful young lady,who seemed very vitty and intelligent. She was posed a question, â€œ how old are you? â€. She answered , â€œ my age 3 years hence munltiply by 3 and from that subtracted 3 times my age 3 years ago will give you my exact age? How od is the lady?**

Solution:

Let the age be x

Age after 3 years wi be (x + 3)

Age before 3 years = (x â€“ 3)

Hence 3(x + 3) â€“ 3(x â€“ 3) = x

x = 3x + 9 â€“ 3x + 9

x = 18

Therefore the age of the lady = 18 years

52.CONSECUTIVE NATURASL NUMBERS

There are two consecutive natural numbers whose product is equal to the product of three consecutive natural numbers,for example x(x+1) = y(y+1)(y+2).What are the two numbers?

Solution:

14 * 15 = 5 * 6 * 7

53.SOME GLUTTON

A man sitting beside me at a hotel ate idlis one after the other by ordering plate by plate. He said to me after drinking some water the last one I ate was my hundredth idli in last five days. Each day

I ate 6 more than the previous day. Can you tell me how many he ate yesterday?

Soluton:

First day the number of idlis he ate be x

Second day the count is (x+6)

Third day ————————- (x+12)

Fourth day ————————(x+18)

Fifth day —————————(x+24)

Total is 5x + 6(1+2+3+4) =100

5x + 60 =100

x = 8

Day Idlis

1 8

2 14

3 20

4 26

5 32

So, on fourth day the number of idlis the man ate were 26.

## If you have questions, please ask below