Puzzles
Introduction:
Puzzles are dealt in a detailed manner with certain solutions.
Different puzzles are gathered from ShakuntalaDevi’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 – 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 – 225 = B - 15
5 ( 2B – 30 ) = B – 15 + 225
10B = B – 15 + 225 + 150
9B = 360
B = 40
(1)=> G/40-15 = 2
G=50 girls.
Top
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 1119hours.
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.
Top
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
Top
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
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
Top
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
Top
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 – 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 – 1)/2] +2 = (x+6)/4 Rs/- -----(3)
Now he left with [(x-2)/2] – [(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] – [(x+14)/8] = (x-34)/8 Rs/-
Finally he had 1 Rupee in his purse.
i .e; Actual amount – 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,………………
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, 8.9.10,-------
(or) 504, 720,-----------------------
Top
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 – 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, “
how old are you? â€. She answered , “ 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 – 3)
Hence 3(x + 3) – 3(x – 3) = x
x = 3x + 9 – 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.
Image via CrunchBase
0 comments:
Post a Comment