TCS 451 question answer with separate

Here you get all important question with separate doc file. it's very useful to revised all 451 questions....

//good

When 5% of the total wheat is lost in grinding, a country can export 9 million tons of wheat, but when6% of the total wheat is lost in grinding it needs to import 2 million tonnes of wheat. What is the totalproduction of wheat in the country? (in million tonnes)

(a) 1000 (b) 900 (c) 1100 (d)1150

total wheat=x million tons
wheat left after 5% lost=95x/100
now after giving out 9 million tons = (95x/100)-9 = required for the country
after 6% lost = 94x/100
now it need for country 2million more
required for country is =(94x/100)+2

(95x/100)-9=(94x/100)+2

x=1100

Arvind, a mason, decided to build a small house. For constructing the house he had to mix gravel, sandand cement in the ratio 7:5:3. The cement costs Rs.250 a bag which contains 10 kg of cement. He hasspent Rs.4500 to buy cement. How much will he have to spend for sand which costs Rs.70 for a 20 kgbag?(a) Rs.1050 (b) Rs.2050 (c) Rs.1075 (d) Rs.2000

cost of cement=Rs 4500
so wt. of cement required to construct house= (4500/250)*10= 180 Kg
now, total wt of material can be evaluate from following equation:
3x/15=180 kg
so,
x=900kg
hence total material req.=900kg
wt of sand used= 5/15 * 900= 300kg
cost of sand=70/20 * 300 =Rs 1050

Cadbury manufactures a chocolate box which contains “x” number of chocolates. There are three
houses A, B and C in the neighbourhood of Cadbury. Since it is a new variety of chocolate, themarketing manager of Cadbury decided to distribute free chocolates to the children in theneighbouring houses A, B and C. In House A there are 3 children, in B there are 5 children, and in Cthere are 7 children. After distributing the chocolates, he was left with one chocolate in each case. Howmany chocolates did he have in the beginning for distribution?(a) 204 (b) 211 (c) 214 (d) 217

211

LCM of 3,5,7 is 105.
Number of choclates can be
105*1+1=106
or
105*2+1=211
or
....
..

so from given options 211 is correct one.

Seven years ago, the combined ages of Amir and Akshay was twice that of Saif. If the sum of the ages of Akshay and Amir, fifteen years hence, is 98 years, what is the present age of Saif?(a) 27 yrs (b) 35 yrs (c) 34 yrs (d) 49 yr

if u consider aamir's present age as Am and Akshay's age as Ak ans Saif's age as s
then seven years before the equation should be
(Am-7)+(Ak-7)=2(S-7)......1
so,Am+Ak=2S
now after 15 years
(Am+15)+(Ak+15)=98
Am+Ak=68....2
so from equations 1&2
s=34

How many 3-digit numbers have even number of factors?(a) 22 (b) 879 (c) 21 (d) 878

10^2 =100

31^2 = 961

So 31-10 21+1=22

By using 1,2,3,4,5,how many 5 digit number can be formed which is divisible by 4 and repitition of the number is allowed?

5  5  5  1  1    (12,24,32,44,52)

125*5 == 625

Each letter in the sum stands for different digit. What number is represented by Baby?

P A Y
P A Y
P A Y
-----------------------------------
B A B Y

if P is assigned a value say 5, where ever P comes, it will have the value 5 only.. similarly for other digits.

ANS BABY=1710.
PAY
PAY
+PAY
------------
BABY
NOW: 1 case y+y+y=y means the sum of 3y = y it is possible when y=0. so y=0
and 2 case A+A+A=B and given p=5 now value of A may be 1,2,3,4,6,7,8,9 but only A=7 satisfy the question so that B = 1.

-------------------------

Y = 0
3A = B
A = 7
B = 1
P = 5

BABY = 1710

By selling 32 books for Rs. 100/- a man loses 40%. How many books shall he sell for Rs. 100/- to gain 20%?

Ans:::

Easy one

Ans is 16……

Jagan lies on Monday Tuesday Wednesday
Pradeep lies on Wednesday Thursday Saturday.
Jagan says I didn’t lie yesterday. Pradeep also says that I didn’t lie yesterday. What
day is today?

???????????

Tuesday.

On Tuesday,
Jagan tell lies : "I didn't lies Yesterday"
Pradeep tell truth : "I didn't lies Yesterday"

Two cars travel in the same direction at 40km/hr at a regular distance. A car comes in the opposite direction at 60km/hr. It meets each car in a gap of 8 seconds. What is the distance between the two cars?

x/100-y/100 = 8

x-y =800km *5/18 = 222.22mtrs

Find the least number which when divided by 7 gives the reminder 6, when divided by 6 gives reminder 5, when divided by 5 gives reminder 4 and so on ?

420-1 = 419

A man has some socks in his drawer-18 identical blue,24 identical red, and 22 identical black.The lights are out and it is totally dark.How many socks must he take out to make sure he has a pair of each colour ?

answer is 48
in order to have pair of each color we hv to take 24 red then 22 identical black and only in the 2 blue socks are required.
so totally 24+22+2=48

Given 5 different green dyes, 4 different blue dyes and 3 different red dyes, the number of combination of dyes taking atleast one green dye and one blue dye, is

Answer: The least number of dyes that a combination can have is 2.
(one blue and one green).

Maximum number of dyes that a combination can have is 12
(5G, 4B, 6R).

Atleast one green dye can be selected out of 5 green dyes. The number of ways is



After selecting one or more green dyes, we can select atleast one blue dye out of 4 different blue dyes. The number of ways is


After selecting atleast one green dye and atleast one blue dye, at least one red dye or no red dye can be selected in

= 1 + 3 + 3 + 1 = 8 ways

The total number of combinations (31)(15)(8) = 3720.

////fresher world good /////

36 people {a1, a2, ..., a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, ..., {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is
" 11
" 18
" 12
" 13

Solution

Answer by SUMAN is correct.
because if we take person a1 in the set then it will cover a36 and a2 for shaking hand.similarly if we take person a4 in the set then it will cover a5 and a3 for shaking hand and so on, so minimum 12 persons are required.

20 men handshake with each other without repetition. What is the total number of handshakes made?
a) 190 b) 210 c) 150 d) 250

1+2+….+19 =190

The IT giant Tirnop has recently crossed a head count of 150000 and earnings of $7 billion. As one of the forerunners in the technology front, Tirnop continues to lead the way in products and services in India. At Tirnop, all programmers are equal in every respect. They receive identical salaries and also write code at the same rate. Suppose 12 such programmers take 12 minutes to write 12 lines of code in total. How many programmers required to write 72 lines of code in total in 72 minutes?

M1*R1/W1 =M2*R2/W2

12*12/12 = x*72/72

X = 12…..

// GOODDDD

A circular dartboard of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it
hits the dartboard at some point Q in the circle. What is the probability that Q is closer to the center of the circle than the periphery?
a) 0.75 b) 1 c) 0.5 d) 0.25

--à>>>> Here Distance is dosen’t metter……….

Inner circle radius ½ .. AREA = PI * (1/2)^2

Outer circle radius 1.. AREA = PI * (1)^2

So Probability = (PI/4) / PI

                       = ¼

On planet zorba, a solar blast has melted the ice caps on its
equator. 8 years after the ice melts, tiny plantoids called echina start
growing on the rocks. echina grows in the form of a circle and the
relationship between the diameter of this circle and the age of echina
is given by the formula
d = 4 * sqrt (t - 8)for t = 8
Where the represents the diameter in mm and t the number of years since
the solar blast.
Jagan recorded the time of some echina at a particular spot is 24 years
then what is diameter?

d = 4 * sqrt (t - 8)
d = 4 * sqrt (24 - 8)
d = 4 * sqrt (16)
d = 4 * 4
d = 16

14 ) On planet zorba, a solar blast has melted the ice caps on its equator. 8 years after the ice melts, tiny plantoids called echina start growing on the rocks. echina grows in the form of a circle and the relationship between the diameter of this circle and the age of echina is given by the formula
d = 4 * v (t - 8) for t = 8
where d represents the diameter in mm and t the number of years since the solar blast.
Jagan recorded the radius of some echina at a particular spot as 8mm. How many years back did the solar blast occur?
a) 24 b) 12 c) 8 d) 16

ANS:: here radius =8mm so d = 16mm

The equation should be d = 4*sqrt(t-8). Radius given as 8. So diameter = 16. Substitute d in equation.
16 = 4*sqrt(t-8) => 4 =sqrt(t-8) => t =24.

For the FIFA world cup, Paul the octopus has been predicting the
winner of each match with amazing success. It is rumored that in a match
between 2 teams A and B, Paul picks A with the same probability as A's
chances of winning. Let's assume such rumors to be true and that in a
match between Ghana and Bolivia, Ghana the stronger team has a
probability of 2/3 of winning the game. What is the probability that
Paul will correctly pick the winner of the Ghana-Bolivia game?

ANS::::::

Here probability of Ghana is 2/3. The octopus also should have same probability in selecting Ghana as winner which is 2/3. So total probability of selecting the right team (winning team) by the octopus is (2/3)(2/3)=4/9. If Ghana has 2/3 chances, Bolivia will have 1-2/3=1/3 chances(Assuming that a tie is not possible). So octopus also has 1/3 chances of selecting Bolivia. So total probability for Bolivia would be (1/3)(1/3)=1/9. Any team can win the match. So total probability is 4/9+1/9=5/9

A sheet of paper has statements numbered from 1 to 40. For all values of n from 1 to 40, statement n says: 'Exactly n of the statements on this sheet are false.' Which statements are true and which are false?

The 39th statement is true and the rest are false

Explanation:

3rd and 4th option does not satisfies the given condition and all the statement cannot be false.

So, option  The 39th statement is true and the rest are false is answer

/////////// VERYYYYY GOOOODDDDDDD////

Alok and Bhanu play the following min-max game. Given the expression
N = 9 + X + Y - Z
Where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable.
Assuming both play to their optimal strategies, the value of N at the end of the game would be
27
0.0
20
18

// Always  remember

Note: For this type of questions:
x+y-z=11
x-y-z=2
x*(y-z)=18    //// x*(y+z)=18 wrong

///7 j levu pade bcs 3 varible 6e

First Read the question carefully, According to the given question all the numbers will be decided by alok and first 2 would be put by bhanu whereas alok will put the 3rd one.
So for getting maximum alok will tell 7 as the first Number
two cases arise
1) if bhanu will put this number as Z then 9, 9 will be the next 2 number. and 9+x+y-z = 20
2) if bhanu will put this number X, then alok will tell 4 or 5 as next number.
if bhanu will put 4 or 5 as Z then alok will put 9 as Y. So N will be
9+x+y-z= 9+7+9-4 = 21 OR 9+7+9-5 = 20

If bhanu will put 4 or 5 as Y then alok will put 0 as Z
9+x+y-z = 9+7+4-0 = 20 OR 9+7+5-0 = 21

So if both are playing with there best effort then 9+x+y-z will be 20

20 is the correct answer

Alok and Bhanu play the following min-max game. Given the expression
N = 15 + X*(Y - Z) Where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be?

Ans = 15 + 18 = 33

Alok and Bhanu play the following min-max game. Given the expression
N = X - Y - Z where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be
a)4
b)9
c)2
d)-18

Ans = 2

The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 buildings numbered 1 to 1000 in base 8. How many 3s are used in numbering these buildings?Express your answer in base 10
a) 54 b) 64 c) 265 d) 192

in between 1-100 there are 20 no of 3's.
as 03,13,23,30,31,32,33(here we use 2 3),34,35,36,37,38,39,43,53,63,73,83,93

similarly there are 20 3s in between each hundred no.
and there are 100 3's from 300-399.. so total n of 3's are.. 100+200=300

to convert it into base 10 system we use the same method as we use to convert from binary to decimel.

thus (300)base8=3*8^2+0+0=192 :)

The citizens of planet nigiet are 5 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 5) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings? Express result in terms of base 10.
a) 54 b) 64 c) 75 d) 100

300 3’s in 1 to 1000

Now (300)base5 = 3*5^2 = 75

A hare and a tortoise have a race along a circle of 100 yard diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race?
a)37.80
b)5
c)40
d)8

37.8

Hare starts after tortoise covers 1/5th distance .
When hare covers 1/8th of distance tortoise meets hare
So distance covered by tortoise = 1-(1/5 + 1/8)
= 27/40
Time taken by both is same
time = dist/speed
So (dist/ speed) of tortoise = (dist / speed) of hare
let speed of tortoise be t and of hare be h
27/40t = 1/8h
h = (40/(8*27) )* t = 5/27 * t
Now for the next part hare has to cover 7/8 th distance when tortoise covers 1/8 th distance
so we get 1/8t = 7/8h
h = 7t

so the factor by which h's speed increases = 7t/ (5/27 * t)
= 37.8

10 suspects are rounded by the police and questioned about a bank robbery. Only one of them is guilty. The suspects are made to stand in a line and each person declares that the person next to him on his right is guilty. The rightmost person is not questioned. Which of the following possibilities are true?
A. All suspects are lying or the leftmost suspect is innocent.
B. All suspects are lying and the leftmost suspect is innocent .
i> A only
ii> B only
iii> Both A and B
iv> Neither A nor B

Ans = Only A…….

One day Rapunzel meets Dwarf and Byte in the Forest of forgetfulness. She knows that Dwarf lies on Mondays, Tuesdays and Wednesdays, and tells the truth on the other days of the week. Byte, on the other hand, lies on Thursdays, Fridays and Saturdays, but tells the truth on the other days of the week. Now they make the following statements to Rapunzel — Dwarf: Yesterday was one of those days when I lie. Byte: Yesterday was one of those days when I lie too. What day is it?

THINK ABOUT IT

ANs is Tuesday….

21 people meet and shake hands. The maximum number of handshakes possible if there is no cycle of handshakes is
(A cycle of handshakes is a sequence of people a1, a2,....., ak such that the pairs (a1, a2), (a2, a3),.....(a(k-1), ak), (ak, a1) shake hands.
a) 17 b) 18 c) 19 d) 20

///GOOODDDDDDD

If handshakes are taken in random manner then max. handshakes = 21*20/2*1 = 210
If all the persons are there in a straight line then number of handshakes = 21-1 = 20

There are two boxes, one containing 10 red balls and the other containing 10 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is

ANS:::::

1/2 ( 1 + 9/19) == 14/19

A sheet of paper has statements numbered from 1 to 40. For each value of n from 1 to 40, statement n says "At least n of the statements on this sheet are true." Which statements are true and which are false?

	The odd numbered statements are true and the even numbered are false
	The even numbered statements are true and the odd numbered are false.
	The first 13 statements are true and the rest are false.
	The first 26 statements are false and the rest are true.

Two bowls are taken, one contains water and another contains tea. One spoon of water is added to second bowl and mixed well, and a spoon of mixture is taken from second bowl and added to the first bowl. Which statement will hold good for the above?

	Second liquid in first bowl is less than the first mixture in second bowl
	Second liquid in first bowl is more than the first mixture in second bowl
	Second liquid in first bowl is equal to the first mixture in second bowl
	none of these

A die is rolled and a coin is tossed. Find the probability that the die shows an odd number and the coin shows a head.

ANS = 3/6  *  1/2  = 1/4…..

 

????????????????

Alice and Bob play the following coins-on-a-stack game. 50 coins are stacked one above the other. One of them is a special (gold) coin and the rest are ordinary coins. The goal is to bring the gold coin to the top by repeatedly moving the topmost coin to another position in the stack.Alice starts and the players take turns. A turn consists of moving the coin on the top to a position i below the top coin (0 ≤ i ≤ 20). We will call this an i-move (thus a 0-move implies doing nothing). The proviso is that an i-move cannot be repeated; for example once a player makes a 2-move, on subsequent turns neither player can make a 2-move. If the gold coin happens to be on top when it's a player's turn then the player wins the game. Initially, the gold coinis the third coin from the top. Then
a) In order to win, Alice’s first move should be a 0-move.
b) In order to win, Alice’s first move should be a 1-move.
c) Alice has no winning strategy.

d) In order to win, Alice’s first move can be a 0-move or a 1-move

/GOOODDDDDDD

(b)In order to win, Alice's first move should be a 1-move....
reason:there are two possibility after 1- move by Alice:-
1. when bob takes 0-move....then the coin config is same ..... but now alice can neither take 1-move nor 0-move.... so she has to take 2 or greater move(let it be 2 move)............
now we have only one coin above the gold coin......now bob can neither take 1-move nor 0-move nor 2-move.... so she has to take 3 or greater move.......
but the move greater then 2-move brings gold coin on top....makes alice win...

2.now bob takes 2-move after alice's first move....... now thre is only one coin above the gold coin..... so Alice takes 0-move........now we have only one coin above the gold coin......now bob can neither take 1-move nor 0-move nor 2-move.... so she has to take 3 or greater move.......
but the move greater then 2-move brings gold coin on top....makes alice win...
This answer is given by ROHIT KUMAR and is the correct answer

Rearrange and categorize the word 'RAPETEKA'?

	Answer: bird Explanation: parakeet -bird

Qn.24

A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If 4 faces of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?

	800
	500
	900
	488
	Answer: 488 Explanation: The outside of the large cube = 5cm The inside space = 3cm Total number of small cubes = (5*5*5) - (3*3*3) = 125 - 27 = 98 Total number of faces = 98 * 6 = 588 now surface area of 4 faces = 4 * 5*5 = 100 cm^2 so subtract 100 cubes as they have surface area 1cm^2 each so unpainted faces of cubes = 588 - 100 = 488

The pacelength P is the distance between the rear of two consecutive footprints. For men, the formula, n/P = 144 gives an approximate relationship between n and P where, n = number of steps per minute and P = pacelength in meters. Bernard knows his pacelength is 164cm. The formula applies to Bernard's walking.

 Calculate Bernard's walking speed in kmph.

	Answer: 23.24 Explanation: n/P = 144 gives an approximate relationship between n and P where, n = number of steps per minute and P = pacelength in meters. Bernard knows his pacelength is 164cm. Number of steps in one minute = 144*1.64 Distance travelled in 1 minute = 144*1.64*1.64 mtrs Distance travelled in one hr = 144*1.64*1.64*60/1000 km = 23.24 km

Qn.27

A sheet of paper has statements numbered from 1 to 30. For all values of n from 1 to 30, statement n says "At most n of the statements on this sheet are false". Which statements are true and which are false?

	All statements are true.
	The odd numbered statements are true and the even numbered are false.
	All statements are false.
	The even numbered statements are true and the odd numbered are false.

////??????

On the planet Oz, there are 8 days in a week- Sunday to Saturday and another day called Oz day. There are 36 hours in a day and each hour has 90 min while each minute has 60 sec. As on earth, the hour hand covers the dial twice every day. Find the approximate angle between the hands of a clock on Oz when the time is 12:40 am.

Since total hours in a day = 36 and
as on earth, the hour hand covers the dial twice every day.
so there will be 18 numbers from 1 to 18 in the clock.
when the time will be 12:40(hands will be at 12 and 8 marks in the clock), two hands of the clock will make an angle of 20*4 = 80 degree, here 20 is the angle between two numbers in the clock.
now for 80 min, advancement of the hour hand will be (20/90)*40 = 80/9
so total angle = 80 + (80/9) = 800/9 = 88.889 degree
if we consider the obtuse angle, then it will be 360-88.889 = 271.111 degree

Qn.33

Planet fourfi resides in 4-dimensional space and thus the currency used by its residents are 3-dimensional objects. The rupee notes are cubical in shape while their coins are spherical. However the coin minting machinery lays out some stipulations on the size of the coins.

The diameter of the coins should be at least 64mm and not exceed 512mm. Given a coin, the diameter of the next larger coin is at least 50% greater.

The diameter of the coin must always be an integer. You are asked to design a set of coins of different diameters with these requirements and your goal is to design as many coins as possible. How many coins can you design?

	8
	9
	5
	6
	Answer: 6 Explanation: 1st coin is of 64mm diameter 2nd coin is of 64+32 = 96mm 3rd coin is of 96 + 48 = 144mm 4th coin is of 144+ 72 = 216mm 5th coin is of 216+108 = 324mm 6th coin is of 324+ 162= 486 7th coin no wards diameter > 512mm So, ans is 6

Qn.34

Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square.

	1:[2+ 6sqrt(2)]
	Answer: 1:[2+ 6sqrt(2)] Explanation: The extreme circles will have radius perpendicular to sides of square so the part of diagonal till the centre of circle will be sqrt(2) * r and remaining portion is r , 5 more circles will contribute 10r and last circle will contribute r + sqrt(2) * r . Length of diagonal = 12r + 2sqrt (2) r  = sqrt (2) * s  where s -> side of square. So, ratio of r:s = 1/( 2 + 6 * sqrt(2)) = 1: [ 2+ 6 sqrt(2)]
	1:(4+ 7sqrt(3)]
	[2+ 7sqrt(2)]:1
	1:(2+ 7sqrt(2)]

Qn.35

Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 20, 20 and 30, the number of points equidistant from all the 3 lines is
 

	1
	Answer: 1 Explanation: Only incenter point is equidistant from all sides
	4
	3
	0

Alok and Bhanu play the following coins in a circle game. 99 coins are arranged in a circle with each coin touching two other coin. Two of the coins are special and the rest are ordinary. Alok starts and the players take turns removing an ordinary coin of their choice from the circle and bringing the other coins closer until they again form a (smaller) circle. The goal is to bring the special coins adjacent to each other and the first player to do so wins the game. Initially the special coins are separated by two ordinary coins O1 and O2. Which of the following is true?

a) In order to win, Alok should remove O1 on his first turn.
b) In order to win, Alok should remove one of the coins different from O1 and O2 on his first turn.
c) In order to win, Alok should remove O2 on his first turn.
d) Alok has no winning strategy.

Ans is B
alok should remove coin other than o1,o2..
there are 95 other coins.
if for first turn if alok remove other than o1 or o2 we are left wit 94 coins
again bhanu has 2 choices either(o1 or o2) or coins within 94
if he removes a coin from 94..
alok and bhanu continues...
when there are only 5 ordinary coins its alok's turn and if he removes the 5 th coin(o3 or o4 or o5)not the o1 or o2
then bhanu will have no choice and Alok will win.
This answer is given by ROHIT NEGI and is correct answer