TCS 2012 papers

If p(x)=a x^4+b x^3+c x^2+d x+e.....has roots at x=1,2,3,4......p(0)=48...then p(5)=?????

Sol>P(x)=Q(x-1)(x-2)(x-3)(x-4)
as x=1,2,3,4 are roots of equation...
P(0)=48=Q(-1)(-2)(-3)(-4)
by solving Q=2
so, P(x)=2(x-1)(x-2)(x-3)(x-4)
put x=5
P(5)=2(5-1)(5-2)(5-3)(5-4)
P(5)=48 is the ans.

truck a = 10kg/min
truck b= 131/3kg/min
truck c = 5kg/min
if work simuntaneosuly
how min it take to load 2.4tons load

Sol>clearly 2.4 tons=24oo kg
A/q,truck is loaded simulaneously,
thetrefore,in 1 mins,we can load=10+5+131/3=58.6kg
so 2400kg can be loaded in=2400/58.6= 40.95 minutes

32. Difference between Bholu's and Molu's age is 2 years and the difference between Molu's and Kolu's age is 5 years. What is the maximum possible value of the sum of the difference in their ages, taken two at a time?

B-M=2
M-K=5
BY SOLVING WE GET

B-K=7

.............
SO MAX . POSSIBLE SUM IS (M-K)+(B-K)=5+7=12
.....

24. A 31" x 31" square metal plate needs to be fixed by a carpenter on to a wooden board. The carpenter uses nails all along the edges of the square such that there are 32 nails on each side of the square. Each nail is at the same distance from the neighbouring nails. How many nails does the carpenter use?

Sol> 124 nails.. since he will b placing nails aa over equally(32 *4= 128) but 4 nails at the corners will b common for 2 consecutive sides each.
this gives 128-4 =124

 

21. find the NEXT number in 0,4,5,11,....

0+4+1=5
4+5+2=11
5+11+3=19
therefore series is... 0 4 5 11 19

Silu and Meenu were walking on the road.Silu said, “I weight 51 Kgs. How much do you weight?†Meenu replied that she wouldn’t reveal her weight directly as she is overweight. But she said, “I weight 29 Kgs plus half of my weight†. How much does Meenu weight??

Sol>let meenu's age be x. therefore atq

1/2(x) + 29 = x
on solving this linear eqn in 1 variable we get x=58 kg

What is the reminder when 50! is divided by 16^15.

FOR N>=2 N! is always even.
last digit in 16^15 is 6 this is also a even no.
even/even is always have a remainder 0 or any even no only.
so go through options.

Mr.John has x children by his wife and Mrs.Sonia has (X+1) children by her husband. They got married and have been blessed with some children. Now , the whole family has 10childern. Assuming that two children of the same parents do not fight, find the maximum number of fights that can take place among children?
A.33 B.22 C.111 D.None of these

Sol>total children = (x+x+1)= 2x+1
now 2x+1!=10 because some new Children (more than 1)born.
thus 2x+1=7/11 but 11 is out of condition.
thus
3,3,4 are children of two different family.
however 2 children of same family cant fight.thus possibilities are:
3c1.3c1+3c1.4c1+4c1.3c1=33

Six plays, P, Q, R, S, T and U are to be held during the week i.e. from Sunday to Saturday. In one day, only one play can be shown, and the order of the plays is subject to the following:
A two day gap should exist between the showing of plays T and S.
The showing of U should be followed immediately by the showing of R.
P cannot be shown on Thursday.
Q should be shown on Tuesday and should not be followed by S.
There won’t be any play on one of the days. Friday or Sunday is not that day and just before that day, S has to be shown.
19. No play is shown on:
a) Sunday b) Saturday c) Monday d) Tuesday
20. On which day will the play R be shown?
a) Friday b) Saturday c) Thursday d) Monday
21. Which play is the last one to be shown?
a) S b) R c) P d) U
22. How many plays are shown between S and U?
a) One b) Two c) Three d) None of these

sun -s
mon - no play
tues- q
wed - t
thurs - r
fri - u
sat - p
now ans can b found easily and do let me know dat m i right or not
thurs - r
fri - u
sat – p

 

A sequence x1,x2,x3…. Is said to be in harmonic progression if the reciprocals 1/x1,1/x2,1/x3… are
in arithmetic progression. The 5th and 7th term for harmonic progression are 30 and 50 respectively what
is difference between 6th and 4th term ?

In HP 5th term is 1/30.
and in HP 7th term is 1/50.
so
a+4d=1/30-----(i)
a+6d=1/50-----(ii)
solve eq. (i) and (ii)..
we get
d= -1/150
a= 3/50
4th term is a+3d and 6th term a+5d...
put the value of a and d..
now 4th term is =1/25
6th term is =1/37.5
so the difference is 37.5-25=12.5

On increasing the SP of an article by Rs. 57, a shopkeeper earns a 5% more profit. calculate the cost price

Let CP of article be x
As given, On increasing the SP of an article by Rs. 57, a shopkeeper earns a 5% more profit
It means...
5%ofx = 57
That is, 5x/100 = 57
By solving, x=1140
So, CP= 1140

At college, 70% students studied Maths,75% students studied English,85% studied french and 80% studied german. What percentage at least must have studied all 4?

a)25% b)30% c)20% d)10% e)none of these

students not studied the subjects 30% math, 25% english, 15% french, 20% german.
30+25+15+20=90% it is the percentage that the students not common, the common are
100-90=10

answer will be = 10%

the sequence {An} is defined by A1=2 and An+1=An +2n what is the value of A(100)

9902..

a2= a1+2
a3= a2+4 == a1+2+4
a4= a1+2+4+6
....

a100= a1+2+4+.. +198
= a1+ 99/2 * (2+198)
=9902

In a telecom assembly factory..day r 250 men n 150 women..d avg productivity of all workers is 12 per day..d avg
productivity of men is 15 units per day..what is d avg productivity of women per day?

sum of men let m
and women n
so (m+n)/400 =12
m+n=400*12.....(1)
now given m/250=15 so m=250*15 ie 3750
ie n=(400*12)-3750
n= 1050
so avg productivity = 1050/150=7 ans

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

possible combinations can be either 2M+4W or 3M+3W

Total possible combinations = 8c2*5c4 + 8c3*5c3 = 700
Combinations in which both men are together = 5c4 + 6c1*5c2 = 65

Combination in which they are not together = 700-65 = 635

A man working in MNC, for going office he used daily local trains which was routed in the morning. There was two type of trains one was type A and other was type B . The frequency of both trains was 10 minutes if one trains (type A )start at 5:00 AM and second (Type B)start at 5:02 A.M and after that it continuously running in every 10 minutes .A Man went to the station at random time in the morning what is the number of chance he got type A train mostly rather than type B trains.

For getting train B, he has got only two minutes time i.e. only if he arrives in between 5:00 AM and 5:02 AM but if he comes in between 5:02 and 5:10 i.e 8 minutes time, he will get train A.
so chances of getting trains A and B are in ratio of 8:2 or 4:1.

If we express 41(3/17)% as a fraction.then it it equal to..
a) 17/7 b) 7/17
c)12/17 d) 3/17

This is a mixed fraction. dont get confused 41 whole 3/17. so mixed fraction gives 700/17 and when u take its % just divide by 100. so 7/17

if N is a natural no and N^3 has 16 factors then how many maximum factors can N^4 have??a)21 b)24 c)25 d)26

Sol>factors = (x+1)(y+1)

          =5*5=25

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+ 7sqrt(2)]:1
2) 1:[2+ 6sqrt(2)]
3) 1:[2+ 7sqrt(2)]
4) 1:(4+ 7sqrt(3)]

Let the radius of circle be r
let the side of square be a
then diagonal of square= a*sqrt(2)
This diagonal length = 12*r + 2r * sqrt(2)
(because the extreme circle's radius is perpendicular to side of square.)
Thus we get
12*r+2r*sqrt(2)=a*sqrt(2)
r(6*sqrt(2)+2)=a

r/a=1/(6*sqrt(2)+2)

Thus ratio: r:a = 1:(6*sqrt(2)+2)

Ans: 1:(6*sqrt(2)+2)

Glenn and Jason each have a collection of cricket balls. Glenn said that if Jason would give him 2 of his balls they would have an equal number; but, if Glenn would give Jason 2 of his balls, Jason would have 2 times as many balls as Glenn. How many balls does Jason have?

ans is 14.jason have 14 and glenn has 10.
let jason has x and glenn has y.
according to first condition..
y-2=x+2
y-x=4.....(1)
second condition .
2(x-2)=y+2......(2)
solve 1st and second we will get x=10 and y=14

how many sero's r dere in 100!

counting number of zeros at the end of n!
value will be
n/5 + n/5^2 + n/5^3 + n/5^4 + ..........
the integral value of this sum will be the total number of zeros.
in this question n=100
100/5 + 100/5^2 + .......
the integral value is 20+4=24.

if a ladder is 100m long, and distance b/w bottom of ladder and wall is 60. top side of bottom and wall is joint.
what is the maximum size of cube that place b/t them.

The answer is 35. it is a simple pythagorus thm problem.let the edge of cube be x. now make a square between a right angled triangle of base 60 hypotnuse 100 and height is calculated to be 80. now due to square two right triangles are formed. so the hypo of 1st tri. + hypo. of 2nd tri. should be equal to 100.
therefore (60-x)^2 +x^2 + x^2 + (80-x)^2 = 100.

A man buy some bunch of banana, but of conversing he got 2 more.
if the banana is 12 rs dorzen and after founding the 2 more banana his overall price in dorzen is less 1 rs.

As its Rs 12/dozen. So 1 banana=Rupee 1.
Suppose he puchases x bananas.
so price of x bananas= rs x
but he purchased x+2 bananas for rs x
so price per dozen= 12x/(x+2)
ATQ:
12x/(x+2) = 11
=> x=22

A man hav 5 shirt 3 tie and 3 pair of shoes,
how many type he can wear the garment.

multiply all three 5*3*3= 45

each subject is allowed atleast one period. there are 5 subjects & 6 periods. . in how many ways we can arrange the periods ?? the qs. has been repeated so many times here. . still m nt clear with this.
the ans i carried out is( 6!/2!)*5
however evryone sayin its 6!*5
which one is correct ??

there is 6 period. let we consider subjects are A B C D E
now if A is used 2 times then we have to arrange A A B C D E in 6 periods which will have 6!/2! ways
similarly for B C D E, there will be 6!/2! ways
there are 5 subjects thus total ways=(6!/2!)*5=1800

what is the unit place of 2^(3^3458) ??

Lets take the power of 3 first...
3458%4=2 (2 is remainder that is unit place of power of 3)
so unit digit of 3^3458=3^2 =9
So whole eqaution can we written as...
2^9
Now,For 2^(9) .... 9%4=1(1 is remainder that is unit place of power of 2)
so unit digit is=2^1 = 2

What is the last non zero digit of 96! ?

given :1*2*3*....96
first take 1st 10 digits
1*2*3*4*5*6*7*8*9*10
now ignore those which give a product of 10
thus take only 1*2*3*4*6*8*9
note that we havnt taken 2 and 5 as it gives a product of 10.similarly we ignored 10
now the product of 1 2 3 4 6 8 9 gives 8 in units place
every 10 digits set is taken and each set gives 8 digit i units place in the product of those 10 digits
now from 1 to 90 we have 8*8*....nine times
now for the above product we have 8 as units digit
Now find out the units digit for 91*92*93*94*95*96....again here we xclude 10
and we take non zero digit i.e 2
now 8*2 gives 6 in the units place
thus ans is 6

In a 100m race , A beats B by 15m and B beats C by 10m . If A beats C by 5 sec then find the speed of C.

A-----15 m----B--10 m-------C
[----------25 m-------------]
[--------time= 5 sec--------]
speed = D/T
=25/5
=5 m/s.
Hope this is correct.

Father is 5 times faster than son. Father completes work in 40days before son.If both of them work together when will work get complete?

sol is if son takes 5x days, father wil take x days.thus 4x is the difference in days which is equal to 40.therefore,x=10
and 5x=50
thus the work is done together in 50/6 days

in how many ways a team 11 must be selected from 5men and 11 wonmen such that team must comprise of not more than 3men

2256 WAYS
Men Women
3 8
2 9
1 10
0 11

Q>the milk and water in vessels a and b are in ratio 4:3 and 2:3.in what ratio the liquids in both vessels are mixed to obtain a new mixture in vessel c contain half milk and half water ?

let the total liquid in container = x

quantity of A in container 1 = 4x/7

quantity of B in container 1 = 3x/7

quantity of A in container 2 = 2x/5

quantity of B in container 2 = 3x/5

new ratio of A and B = 1/2:1/2

let quantity container 1 mix = n

let quantity container 2 mix = m

=>now Quantity of A = 4xn/7+2xm/5 = x(4n/7+2m/5)

=>now Quantity of B = 3xn/7 + 3xm/5 = x(3n/7 + 3m/5)

As new quantity of A and B are equal

=> 4n/7+2m/5 = 3n/7 + 3m/5

=> n/7= m/5

=> n/m =7/5

numbers are formed by using 1,2,3,4,5(repetetion not allowed)sum of all nos. formed by

these nos.

ans is 3999960I think this formula useful for this type problems..

(sum of digits)*(n-1)!(11111......n times)

here n means no. of numbers in the problem

for example in given problem n=5;

Ram leaves to his office 30 min late as compared to other days.He travel with speed 25% less than other days speed and finally he reach office 50 min late.what is the time taken by ram in normal days.

Ans-60 min

if a,b,c,d,e are integer in increasing order n a+b+c+d+e=FG,FG can be written as 10F+G then what is the max. value of G.

a)4

b)5

c)3

ans is b.) 5

if we take five numbers as...18,19,20,21,22...their sum is 100...bt FG is two digit...so the no. are 17,18,19, 20, 21...which gives sum as 95..

95= 10*9 + 5..so max values of G is 5

how many 6 digit n0. can b formed frm digits 1 2 3 4 5 6 7 so that d digit shouldn't repeat and the second last digit is even

a) 6480

b) 320

c) 2160

d) 720

The second last digit can be filled by only three numbers i.e 2,4and6 in 3p1 ways.Remaining 6 nos can be placed in 5 places in 6p5 ways,hence the ans =6p5 * 3p1=2160

a girl sees mirror reflection of analog clock while going for school at home n reaches school in 20 min .when she reaches school she finds tht she ws 2hrs 20 min late. wat as the time when she rchd school.?

She reached school at 7:20am.

Left home at 7:00am but she saw time in mirror reflection as 5:00am.

The school time was 5:00 am.

 more questions on time work regarding profit distribution accrdng to work done.

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

possible combinations can be either 2M+4W or 3M+3W

Total possible combinations = 8c2*5c4 + 8c3*5c3 = 700

Combinations in which both men are together = 5c4 + 6c1*5c2 = 65

Combination in which they are not together = 700-65 = 635

There are 3 buckets of 8,5 and 3 liter...out of whicg only 8 liter bucket is fullyfill.... u have to exact 4-4 lit liquid in 8 and 5 liter bucket by using only these buckets in minimum no of steps ?the correct answer is seven steps

explain:-

8(lit bucket) 5(lit bucket) 3(lit bucket)

initialy-8 0 0

1) 3 5 0

2) 3 2 3

3) 6 2 0

4) 6 0 2

5) 1 5 2

6) 1 4 3

7) 4 4 0

If a number 272758 and 273437 is divided by N then they give remainder as 13 and 17 respectively. find the sum of digits of number N.

since 13 n 17 are remainders

thus subtracting gives 272758-13=272745 n 273437-17=273420

then GCF of 272745,273420 gives 45

ANS:4+5=9 (easy to find gcf,N may be less than 45 also,GO ACCORDING TO THE OPTIONS)

Q>Rank of the word GOOGLE in the dictionary of words formed by letters of the word GOOGLE taking all letters at a time?

Given word = GOOGLE

Arrange in Alphabetical order:-

=> EGGLOO

We will follow the sequence of given word characters…like

First take G then…O…then….O……G..…L…..E…

So, For G:

Coefficient is 1 bcz number of characters before G in ‘EGGLOO’ is only 1 (that is E) and,

And remaining letters can be arranged in 5!/(2!*2!) ways

For O:

Coefficient is 3 bcz number of characters before O in EGGLOO is 4 but we will not count the above already traversed character (ie E) so Coeffiecint of O =3 and,

And remaining letters can be arranged in 4!/2! ways

Now we can calculate so on.... for remaining characters also....

So,

G => 1*5!/(2!*2!) = 30

O => 3*4!/(2!) = 36

O => 3*3! = 18

G => 1*2! = 2

L => 1*1! = 1

E => 0*0! = 0

So total words before GOOGLE are:-

30+36+18+2+1 = 87

Hence, GOOGLE IS 88th Word.That is, Rank = 88

Q>A merchant buys 20 kg wheat at Rs.30/kg and 40 kg wheat at Rs.25/kg and mixed them. He sold one-third of mixture at Rs.26, what should be the cost/kg of the remaining mixture, so as to make a profit of 25% on the whole?

Total investment CP= 20*30 + 40*25= 1600

Total Mixture= 20+40=60

Sold 1/3 of 60= 20kg at Rs26/kg

Remaining mixture= 60-20=40

Earned 20*26= Rs520

To get a profit of 25% on whole total SP= 1.25*1600= 2000

remaining amount= 2000-520=1480

Hence, cost/kg of remaining mixture 1480/40=37

i think it shld be 37rs/kg

Q>1-2+3-4+5......-98+99 =?

sol>ans is 50.

as 1-2+3-4+5-6+7-....-98+99

can be written as

=(1+3+5+7+....+99)-(2+4+6+8+...98)

=(1+3+5+7+....+99)-2*(1+2+3+..+49)

=(1+3+5+7+....+99)-2*49*50/2

(1+3+5+7+....+99)-2450

=2500-2450 (as it is in AP as 1 1st term 99 last term)

50 ans

Q>Prime factorisaton of int N is A*A*B*C. a,b,c all are distinct prime integers.

how many factors does N have ??

sol>A^p*B^q*C^r=(p+1)*(q+1)*(r+1)

hence the ans is 12

Three friends Mohit,Manohar and Prashant decided to share a lot of bananas. Each of them had half of the total plus half a banana in order. After each of them took their share twice, no bananas were left.

1.How many bananas were there?

2.How many bananas did Mohit get?

3.How many bananas did Manohar get?

(2^n)-1 is the formula

here n is roatation

according to question its twice means 2*3=6

(2^6)-1=63

If the sum of n terms of two series of A.P are in the ratio 5n+4:9n+6 .find the ratio of their 13th terms

Ratio of sum of n terms 2 A.P = 5n+4 : 9n+6

So, Ratio Sum of 13 terms = 5*13+4 : 9*13+4 (put n=13)= 69 : 123

Now, Ratio Sum of 12 terms = 5*12+4 : 9*12+4 (put n=12)= 64 : 114

So to claculate 13th term we will simple subtract sum of first 12 terms from sum of first 13 terms i.e, Sum(13) - Sum(12) = = 69 : 123 - 64 : 114

(The important thing is tht we will not subtract it by taking LCM as these are ratios, We will subtract Numerator to Numerator & Denomenator to Denomenator) so, Ratio of 13th term of 2 APs will be = 5 : 9

A grand father has 3 grand children. Age difference of two children among them is 3. Eldest child age is 3 times the youngest child’s age and the eldest child age is two year more than the sum of age of other two children. What is the age of the eldest child?

take ages x,y,z;

y-x=z-y=3;=>y=x+3;

z=3x;=>z=x+y+2 =>z=2x+5;

.ie 3x=2x+5=>x=5 then z=15

                A boy buys 18 sharpeners (brown or white) for Rs. 100. For every white sharpener, he pays one rupee more than the brown sharpener. What is the cost of white sharpener & how much did he buy?

If he bought x white sharpeners @ Rs (y+1) and (18-x) brown sharpeneres @ Rs y per sharpener, then

x*(y+1)+x*(18-x)=100

x= 100-18y

Only integral value of x less than 18 will be 10.

then x=10, y=5 so he bought 10 white sharpeners @ Rs 6 per sharpeners and 8 brown sharpeners @ Rs 5 per sharpener.

when m+n is divided by 12 the reminder is 8 and m-n is divided by 12 the reminder is 6.what is the reminder when mn is divided by 6??

Remainder is always 1

This answer can be proved with multiple values... 2 of them are :-

1. m=7 n=1

2. m=19 n=1

For m=7,n=1

m+n = 8 divide by 12 gives remainder 8

m-n = 6 divide by 12 gives remainder 6

so, m*n = 7 divide by 6 gives remainder 1

For m=19,n=1

m+n = 20 divide by 12 gives remainder 8

m-n = 18 divide by 12 gives remainder 6

so, m*n = 19 divide by 6 gives remainder 1

If p(x)=a x^4+b x^3+c x^2+d x+e.....has roots at x=1,2,3,4......p(0)=48...then p(5)=?????

P(x)=Q(x-1)(x-2)(x-3)(x-4)

as x=1,2,3,4 are roots of equation...

P(0)=48=Q(-1)(-2)(-3)(-4)

by solving Q=2

so, P(x)=2(x-1)(x-2)(x-3)(x-4)

put x=5

A man asks 5 people to make a guess about the amount of money in his pocket which is less than 50.

A guess that the amt is a multiple of 10..

B guess that the amt is a multiple of 12..

C guess that the amt is a multiple of 15..

D guess that the amt is a multiple of 18...

E guess that the amt is a multiple of 30

Which of the following guesses are correct?

a.AE

b. AB

c. BC

d. DE

P(5)=2(5-1)(5-2)(5-3)(5-4)

P(5)=48 is the ans.

AE:30 is multiple of both 10 and 30 which is less than 50. ... possible answer.

AB: there is no multiple of 10 and 12 less than 50. ..... not possible answer.

BC: Same logic as that of AB. ..... not possible answer.

DE: Same logic as that of AB. ..... not possible answer.

Hence answer is AE.

Addition of 641+852+913=2456 is incorrect. Whats the largest digit that can be changed to make the decision corret?

4 in 641 must be replaced with 9.

so 4 is the ans.

1) If X^Y denotes X raised to the power of Y, find out last two

digits of (2957^3661)+(3081^3643)

Options

o 42

o 38

o 98

o 22

sol. 98 is the ans

7^1 is 7

and

5*1=5

so last to digit of (2957^3661) is 57

if a sum of 9 is lent to be paid back in 10 equal monthly installments of rs1 each the rate of interest is?

A three digit number was divided successively in order by 4, 5 and 6 leaving out the remainders. The remainders were respectively 2, 3 and 4. How many such three digit

numbers are possible? ans 7

3

5

7

9

Raju can do a piece of work in 10 days..vicky 12days,tinku 15 days..day all start the work together,but raju leaves aftr 2 days,vicky leaves 3 days before the work is completed..how many days work is completed?

7,5,9,6

raju+vicky+tinku one day work = (1/10)+(1/12)+(1/15)= 1/4

now for 2 days together work done= 2*(1/4)= 1/2

......now work done by tinku in last 3 days alone=(3/15)=(1/5)

remaining work= 1-(1/2 + 1/5)= 3/10

....now vicky+tinku one day work=(1/12)+(1/15)=9/60

therefore 3/10 work will be done by both of them in= (60/9)*(3/10)= 2 days

now answer= 2days(as got from above; it is days required to do rem. work)+ 3days(when vicky leaves)+2days(raju leaves)=7 days ans.

In how many rearrangements of word AMAZED is the letter 'E' positioned in between the 2 'A's(not necessarily flanked)?

1)24

2)72

3)120

4)240

sorry its not necessary to be flanked

(AEA)MZD=4!=24

(AEMA)ZD=3!*2!=12

(AEMDA)Z=2!*3!=12

(AEMDZA)=4!=24

72 ANS

How many keystrokes are needed to type numbers from 1 to 1000?

1 to 9.............9 = 9

10 to 99...........90*2 = 180

100 to 999.........900*3=2700

1000...............4 = 4

+

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

2893

Girish tore away one of the pages from Vaidhy’s book. Later Girish figured that the average of the remaining pages of the book is 67 10/19. Find the page numbers that Girish had torn away assuming that the pages are starting from 1.

1 2 2 3 3 3 4 4 4 4 1 1 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 4 ..................what is the 2852 term?

same for

1^3 is 1

8*3=24 unit digit of 24 is 4

last 2 digit of(3081^3643) is 41

41+57=98....

A clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock it looses uniformly and was observed to be 10 min slow on the subsequent Tuesday at 9pm . when did the clock show the correct time

A clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock it looses uniformly and was observed to be 10 min slow on the subsequent Tuesday at 9pm .

In 54 hrs , it lost 15 mins.

so 5 mins are lost in 5*54/15 = 18 hrs

so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM Monday

there are 5 sweets - jamun, kulfi, peda. laddu and jilebi that i wis t eat on 5 consecutive days, monday throufh friday, one sweet a day, based on following self imposed constraints:

1) laddu is not eaten on monday

2) if jamun is eaten on monady , then laddu must be eaten on friday

3) if jamun is eaten on tuesday, kulfi should be eaten on monday

4) peda is eaten the day following the day eating jelabi

based on the above , peda can be eaten on any day expect??

monday bcoz the day before jalebi is required to be taken bt its sunday and no sweet is eaten on sunday

Question 1:

  George can do a work in 60 days. George and Paul can do the same work in 20 days.How much time will Paul take to complete the same work if works alone?

Answer: 30days

Question 2:

 Two people A and B starts running at 10:00 AM  towards north at a speed of 40Kmph and 60Kmph respectively. After 2 hours(12:00 noon ), A turned left and B turned right. They stopped at 2:00 PM.Find the distance between A and B at 2:00 PM.

Question 3:

 You are given 87 matchsticks. They are arranged in rectangular manner.Find the number of triangles formed?

Question 4:

 Three unbiased dice are thrown. Find the probability of getting sum 10.

Answer: 1/8

 Question 5:

 If ab3133ab is divisible by 12.What is the minimum value of (a+b)?

a)     4    b)6   c)7   d)2

Answer: c) 7

Question 6:

Five people Pradeep,Qutub,Raj,Sandy and Tej are in a row. Raj sits to the immediate right of Tej. If Pradeep and Qutub have only two persons sitting between them, which of following must be true?

    Pradeep is at one extreme end of row.

    Qutub is sitting beside Pradeep.

    Tej or Raj must be at one extreme end of the row.

    Sandy must be at one of ends of the row.

Answer:

     4)Sandy must be at one of ends of the row.

Question 7:

Five rivers  Ya-Fen,Volga,Yamuna,Hudson and Siene flow through the five cities v-z. New York,Paris,Moscow,Tokyo and Delhi,but not necessarily in same order.No two rivers flow in the same city and no two cities have same river flowing through them.Vogla flows through neither Moscow nor Delhi.Ya-Fen is neither in New York nor in Paris. Yamuna is neither in Delhi nor in Moscow.Hudson is in Moscow while new York has Yamuna flowing through it. Which of following is definitely FALSE?

     If Volga flows through Tokyo,then Ya-Fen flows through Delhi.

     If Tokyo has Volga,then Siene must flow through Delhi.

     If Ya-Fen flows through Tokyo, then Siene must flow through Delhi.

    Ya-Fen may flow through Tokyo or Delhi.

Silu and Meenu were walking on the road.Silu said, “I weight 51 Kgs. How much do you weight?†Meenu replied that she wouldn’t reveal her weight directly as she is overweight. But she said, “I weight 29 Kgs plus half of my weight†. How much does Meenu weight??

let meenu's age be x. therefore atq

1/2(x) + 29 = x or simply 1/2(x)=29

on solving this linear eqn in 1 variable we get x=58 kg

22) A boy buys 18 sharpeners (brown or white) for Rs. 100. For every white sharpener, he pays one rupee more than the brown sharpener. What is the cost of white sharpener & how much did he buy?

If he bought x white sharpeners @ Rs (y+1) and (18-x) brown sharpeneres @ Rs y per sharpener, then

x*(y+1)+x*(18-x)=100

x= 100-18y

Only integral value of x less than 18 will be 10.

then x=10, y=5

so he bought 10 white sharpeners @ Rs 6 per sharpeners and 8 brown sharpeners @ Rs 5 per sharpener.