## Aptitude-Puzzles for freshers

**Puzzles**

**Introduction: **

Puzzles are dealt in a detailed manner with certain solutions.Different puzzles are gathered from ShakuntalaDeviâ€™s puzzle books. Keeping in mind certain puzzles for Infosys some

reasoning problems are also dealt. Puzzle name at the top of each problem will give a brief idea regarding the mode of application.

SELECTING A CANDIDATE

**For an advertisement of six local posts,twelve persons applied for the job.Can you tell in how many different ways the selection can be made?**

Solution:

6^12

SET OF BAT AND BALL

**When I wanted to buy a bat and ball, the shopkeeper said they would together cost Rs.3.75.But I did not want to buy a ball.The shopkeeper said that bat would cost 75paise more than the ball.What was the cost of bat and the ball?**

Soluton:

Given that bat and ball together cost Rs.3.75 = 375paise

Let cost of the ball alone be x.

Given cost of the bat is 75p greater than cost of the ball.

So cost of the bat = x+75

x+x+75 = 375

2x = 375 â€“ 75

2x = 300

x = 150p

Hence cost of the ball = Rs.1.50

=>Cost of the bat = 1.50 + 75 = Rs.2.25

PLAYING CHILDREN

A group of boys and girls are playing.15 boys leave.There remain 2 girls for each boy.Then 45 girls leave.There remain 5 boys for each girl.How many boys were in the orginal group?

Solution:

Let B and G represent no.of boys and girls in the original group

respectively.

G ———> 2

B-15 ———-> 1

G/B-15 = 2/1

i.e., 2 girls are left for 15 boys who are alone.

G-45 ——————–>1

B-15 ————————->5

5 boys are left out when 15 girls are alone.

=>G/B-15=2/1 ——————————–(1)

=>G-45/B-15 =1/5 —————————-(2)

(1) & (2) =>

G = 2B-30

5G â€“ 225 = B – 15

5 ( 2B â€“ 30 ) = B â€“ 15 + 225

10B = B â€“ 15 + 225 + 150

9B = 360

B = 40

(1)=> G/40-15 = 2

G=50 girls.

**Problems**

1.Reshma appeared for a maths exam.She was given 100 problems to solve.She tried to solve all of them correctly but some went wrong.But she scored 85. Her score was calculated by subtracting

two times th no.of wrong answers from the no.of correct answers.How many problems did Reshma do correctly?

Soluton:

Assume W as wrong answers and R as correct answers

Given total no.of questions as 100

R+W=100 —————————(1)

Score is calculated by subtracting 2 times wrong answers(2W)

from right answers(R) and given as 85

R-2W=85 ——————————-(2)

(2)-(1)

R-2W=85

R+W=100

———————

W=5

Hence,100-5=95 is the no.of correct answers of Reshma.

**2.A RUNNNG RACE
Sneha,Shilpa,Sushma join a running race.The distance is 1500 metres.Sneha beats Shilpa by 30 metres and Sushma by 100 metres.By how much could Shilpa beat Sushma over the full distance if they both ran as before?**

Solution:

Total distance covered by Sneha=1500m

Shilpa=1500-30=1470

Sneha =1500-100=1400

Distance covered by Shilpa=1500*1400/1470=1428.6

Distance to be covered by Shilpa to beat Sushma over full distance

1500-1428.6=71.4m

3.FILLING A CISTERN

Pipe S1 can fill a cistern in 2 hours and pipe S2 in 3 hours.Pipe S3 can empty it in 5 hours.Supposing all the pipes are turned on when the cistern is competely empty,how long will it take to fill?

Solution:

S1 fills cistern in 1/2 hours

S2 fills cistern in 1/3 hours

S3 empties it in 1/5 hours

A the pipes S1,S2,S3 working i.e.,filling the cistern

1/2+1/3-1/5=15+10-6/30=25-6/30=19/30

No.of hours to fill=30/19=1 11\19hours.

**4.SEQUENCE PROBLEMS
What are the next two terms in the sequence?1,1,5,17,61,217………**

Solution:

The order in this cases is

Tn=3*Tn-1 +2*Tn-2

= 3(217)+2(61)

= 773

Tn+1=3(773)+2(271)

=2319+542

=2753

**5.SEQUENCES
What are the next two terms in the sequence?**

1,1,5,17,61,217……………..

Solution:

Tn=3Tn-1+2Tn-2

=3(217)+2(61)

=773

Tn+1=3(773)+2(217)=2753

**6.What are the next three terms to the series?
1+3+7+15+31+63………..**

Solution:

Actual term is 2exp n-1.

The next three terms are:

2exp7-1=127

2exp8-1=255

2exp9-1=511

**7.A PROBLEM OF SHOPPING
Samsrita went out for shopping by taking with her Rs.15/- in one rupee notes and 20p coins.On return she had as many one rupee notes as she originally had and as many 20p coins as she had one rupee
notes.She came back with 1/3rd with what she had.How much did Samsrita spend and how much did she take?**

Solution:

Let x be no.of rupee notes y be no.of 20pcoins.

So,when going for shopping 100x+20y paise were there with Samsrita.

On return she had 100y+20x paise.

Also it is given that she had 1/3 rd of the orginal amount.

1/3(10x+20y)=100y+20x

=>4x=280y

=>x=7y

y=1 => x=7 total =7.20 <15
y=2 =>x=14 total=14.40~=15

y=3 =>x=21 total=21.60 >>15

Hence the suitable value nearer to the amunt is 14.40 and so is the

amount Samsrita carried with her.

1/3(1440)=480paise.

Rs.4.80/- is amount spent by Samsrita.

**8.A PUZZLE OF CULTURAL GROUPS
Literary,Dramatic,Musical,Dancing and Painting are the 5 groups of a club.Literary group meets every other day,dramatic every third day,musical every fourth day,dancing every fifth day,painting every
sixth day.Five groups meet on NewYears day of 1975 and starting from that day they met regularly on schedule. How many times did all the 5 groups meet on same day in first quarter excluding
Jan1,1975.How many days did none of them met?**

Solution:

LCM of 2,3,4,5,6 is 60.

Hence excluding Jan1,1975 they met on every 61st day.

60/2=30 60/3=20 60/4=15 60/5=12 60/6=10

Literary meet for 30 2 day intervals.

Dramatic meet for 20 3 day intervals.

Musical meet for 15 4 day intervals.

Dancing meet for 12 5 day intervals.

Paintng meet for 10 6 day intervals.

First quartr implies 3 months with 90 days.

so inorder to a nswer that how many days do they don/t meet

atleast once in first quarter is got by rounding all other categories.

By counting all the intervals for other groups no.of days in

Jan 8,Feb 7,Mar 9.

Total is 24.

9.STOLEN MANGOES

Three naughty boys stole some mangoes from a garden.Among them one counted and ate one.From the remainder he took precise third and went back to sleep. After sometime second boy woke up,counted the mangoes,ate one,took an exact third of the remaining and went back to sleep. After sometime third boy also did the same.In the morning they found one which was rotten and hence threw it away from the remainder,they made an exact division.How many mangoes did they steal?

Solution:

Let the noof mangoes be x

After the first boy had eaten noof mangoes =x-1

After taking 1/3 rd of remaining it is 2x-2/3

Second boy ate one and tok 1/3 then it is 2(2x-2/3 -1)=4x-10/9

Third boy ate and tok 1/3 as 2(4x-19)/27=8x-38/27

Deducting the rotten one from remaining noogf mangoes left

=8x-38/27=8x-765/27

This is divided among the three equally 8x-65/27=3n

8x=81n+65

Let n be equal to odd number 2b+1

8x=81(2b+1)+65

4x=81b+73

Let b=2c+1

4x=81(2c+1)+73

2x=81c+77

Let c=2d+1

x=81d+79

Least value of x for d=0 is 79

for d=1 is 160

for d=3 is 241

On verfication,79-1=78/3=26

Hence 79 is the correct answer.

**10.AN ELECTION PROBLEM
My club had a problem recently.They had to appoint a Ssecretary from among the men and a joint secretary from among the women. We have a membership of 12 men and 10 women.In how many ways can the selection be made?**

Solution:

As per the permutations and combinatins concept of mathematics,

out of 12 men one selected as secretary can be done in 12c1 ways

out of 10 women one selected as joint secretary can be done in 10c1 ways

Hence one secrtary and one joint secretary is 12*10=120

11.SNAPPING A PLANE

A plane has a span of 12 metres.It was photographed as it was flying directly overhead with a camera with a depth of 12cm.In the photo the span of the plane was same.Can you tell how higher was the plane when it was snapped?

Solution:

Actual span of the plane was 12m

Span of the plane n photograph was 800m

Depth of the plane is 12000m=12cm

Hence,height of the plane when photographed be x

12000:800 = x:12

x=180m

12.A THRST PROBLEM

Pramatha and Pranathi went camping.They took their own water in bg plastic bottles. Pramatha got thirsty and drank half the water in her bottle.A little later on she drank 1/3 f what was left.Sometime

afterwards she drank 1/4 of what remained and so on Pranathi also had a bottle of the same size.She drank half the bottle at the first instance ,half of what remained when she drank next and so on.

Aftr each took 10 drins ,the water Pramatha left was how many times greatr than the water Pranathi had left?

Soltuion:

Pramatha for the first drink 1/2

for the second drink 1/3

for the third drink 1/4

She drank 10 times and hence by the end of the 10th drink 1/11

of water she had in the bottle was over.

Pranathi for the first drink 1/2

for the second drin 1/4

for the third drink 1/8

So Pranathi as per the given information has drunk 1/1024 of water

she had in the bottle.

Water left for Pramatha/Water left for Pranathi=1/11 / 1/1024 =1024/11

**13.NAME OF THE EXCHANGE
In GreatBritain some years ago the first threeletters of a telephone number usd to indicate the name of the exchangeHow many such arrangements of 3 letters is it possible to devise from the 26
letters of the alphabet?**

Solution:

For permutations the no.of ways to select is npr=n!/(n-r)!

Hence out of 26 letters the possible outcomes are 26p3=26!/23!=15600

14.VALUE OF THE SERIES

Take a good look at the following series.

1-1/3+1/5-1/7+1/9-1/11+1/13………………..

Find the value of the series and multiply the answer by 4.You will notce that a well-known vale approximates this product.Even more interestng is that as you add more terms the approximation becomes closer.

Solution:

Ths is an Arithmetic progresson with value .76 when two terms

added becomes .77 and multiplidd by 4 it becomes 3.04 and 3.08

repeatition it is 3.14 which pi value adjusted to 2 decimals.

15.PLANTING TREES

If you wished to plant some trees so that each was equidistant from every pther tree,what is the largest number you would plant?

Solution:

From the above informatin,as per equidstant formula of triangle,it is

an equilateral triangle.

Planting at all the three corners only 3 can be planted.

The centroid is the middle point placed exactly equidistant from

all the corners.

Hence 4 plants can be planted at euqidistant.

**16.LENGTH OF A TRAIN
A train is travelling at the speed of 96 kmph.It takes 3 seconds to enter a tunnel and 30 seconds more to pass thorugh it completely.What is the length of the train and the tunnel?**

Solution:

Speed of the train=95*5/18 m/sec

Time taken=3ssec

Length of the train=96*5/18*3=80m

Length of the tunnel=96*5/18*30=800m

17.A GAME OF BILLIARDS

Rajv,Sanjiv,Vinay were playing a game of Billiards.Rajiv can give Sanjiv 10 points in 50 and Sanjiv can give 10 points in 50.How many points in 50 must Rajiv gve Sanjiv to make an even game?

Solution:

Rajiv 50 Sanjiv 40

Sanjiv 50 Vinay 40

Sanjiv 40 Vinay 40*40/50=32

Rajiv 50 Sanjiv 40 Vinay 32

Rajiv gains 18 points than Vinay (50-32=18)

18.WOMEN AT CLUB SOCIALS

Women outnumbered men by 16 at a club social.Seventimes the no.of women exceeds nine times the no.of men by 32.What was the number of men and women at club?

Solution:

Let W and M be the no.of women and men respectively.

Given W=M+16……………..(1)

7W=9M+32…………….(2)

7*(1)=>7W=7M+112……..(3)

(3)-(2) =>M=40

W=56

19.FILLING WINE IN BARRELS

A friend of mine in London has a very nice cellar.He has two large barrels in the cellar.The larger barrel is mostly empty.But the smaller barrel is only 5/6 th full f wine while it can hold 536 litres.Supposing he empties the smaller barrel and fills the bigger barrel to find that the wine fills only 4/9 of it.How much

wine would the larger barrel hold when full?

Solution:

5/6——————-536

4/9——————-?

5/6*536=4/9*x

=>x=1005 litrs

**20.WEIGHT OF A BRICK
We have a brick of regular size.It weighs 4 kilograms.How much do you think asmaller brick four times small, but made of the same material weigh?**

Solution:

The weight of the given brick = 4 Kilograms = 4000 grams

It is given that the smaller brick’s volume is 4 times smaller

than the given one.

The volume of smaller brick = 4 * 4 * 4 = 64 times smaller

The smaller brick’s weight = 4000/64 = 62.5 grams

## If you have questions, please ask below