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<body bgcolor="aqua"><center><h2>Puzzles</h2></center>
<pre>
<font size="2">
<strong>Introduction:</strong>
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.
<b>SELECTING A CANDIDATE</b>
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
<b>SET OF BAT AND BALL</b>
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
<b>PLAYING CHILDREN</b>
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.
<font size=3> <a href="puzzles.html"><b>Top</b></a></font>
<b>Problems</b>
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.
<font size=3> <a href="puzzles.html"><b>Top</b></a></font>
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
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