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2....TCS On line Written test questions and answer


1 Four friends namely Rahul, Ravi, Rajesh and Rohan contested for a dairy milk chocolate. To decide which friend will get the chocolate they decided to throw two dice. Every friend was asked to choose a number and if the sum of the numbers on two dice equals that number, the concerned person will get the chocolate. Rahul's choice was7, Ravi's choice was 9, Rajesh's choice was 10 and Rohan's choice was 11. Who has the maximum probability of winning the amount.

a) Rahul b) Ravi c) Rajesh d) Rohan
Rahul has chosen 7, similarly Ravi-9,Rajesh-10,Rohan-11
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6) Sum of the digits are= 2,3,4,5,6,7
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6) Sum of the digits are= 3,4,5,6,7,8
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)Sum of the digits are= 4,5,6,7,8,9
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)Sum of the digits are= 5,6,7,8,9,10
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)Sum of the digits are= 6,7,8,9,10,11
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)Sum of the digits are= 7,8,9,10,11,12

Among these 7 comes 6 times, 9 comes 4times, 10 comes 3times,11 comes only 2 times.
So the maximum probability goes to Rahul

2 Messrs. Siva Constructions, leading agents in Chennai prepared models of their lands in the shape of a rectangle and triangle. They made models having same area. The length and width of rectangle model are 24 inches and 8 inches respectively. The base of the triangle model is 16 inches. What is the altitude of triangle model from the base to the top?

a) 24 inches b) 8 inches c) 20 inches d) 32 inches
Area of a rectangle= Area of the Triangle
So, l*b=1/2(b*h)
24*8=1/2(16*8)
therefore 24inches will be the answer

3 There is a 12*6 grid..grid contains 1*1 squares..what is d total number of squares..(consider 1*1,2*2,3*3,4*4,5*5,6*6 size squares as well).Lady has 25 blues, 7 red,9 yellow gloves and hats.how many gloves of same colour.
(max + min + 2) = 25 + 9 +2=36
1!+2!+.....=50!=?
3.10350532 A- 10^64
Anita make cube with dimension 5*5*5 using 1*1*1 cubes.find no. Of cubes to make it hollow of same shape.
To make the 5*5*5 hollow, the inner cubic part must be removed. This would constitute one cube each from front,back,top,bottom,left and right ! Since the 5*5*5 is made up of 1*1*1 cubes, 3*3*3 cubes (i.e. 5-2*5-2*5-2) would have to be removed to make it hollow with the shape retained. 27 cubes would be removed, remaining 125-27=98.
1) if 3y + x > 2 and x + 2y
let 3y+x=2
y+2y+x=2
2y+x=2-y(ans)
3y > 2 -x
2y > 4/3 - 2x/3
x+2y > 4/3 + x/3
x + 2y > 1/3(4 + x)

4 Tim and Elan are 90 km from each other, they start to move each other simultaneously, Tim at speed 10 and Elan 5kmph, if every hour they double their speed what is the distance that Tim will pass until he meet Elan?
Tim and Elan will cover the distance in ratio of 2: 1 in same time.
so out of 90 kms, Tim will travel 60 kms and Elan will travel 30 mtrs.
he length and breadth of a field is 300x400ft,if there are 3 ants on average per square inch of field,find the approximate number of ants in field
as 300 ft=3600 inch and 400 ft=4800 inch.then area in sqr inch=17280000
1 sqr inch need 3 ant hence 17280000 sqr inch need 3*17280000 ants.
.diagonal of a square is double the side of an equilateral triangle find the ratio of area of triangle to area of square
let sides of equilateral triangle be 'a', therefore area= v3/4 * a^2

now, diagonal of square is 2a therefore sides of square will be v2*a
and area of square = 2 a^2

hence, ratio = (v3/4* a^2)/(2* a^2)
answer = v3/8

5 Given a collection of points P in the plane, a 1-set is a point in P that can be separated from the rest by a line, .i.e the point lies on one side of the line while the others lie on the other side.
The number of 1-sets of P is denoted by n1(P). The minimum value of n1(P) over all configurations P of 5 points in the plane in general position(.i.e no three points in P lie on a line) is
a)3 b)5 c) 2 d)1

The meaning of this question is little complicated. In easy language the question should be:

6 There are some points in the plane. We have to make a separation between these points by forming a line in such a way that one point could be separate from the others. No 3 points are in a line. Maximum no of methods for plotting the line and minimum no of methods for plotting the line will be (arrangement of points is your choice)

Now Answer Is:

For Maximize: The number of methods, Points should be drawn in the circumference of the CIRCLE So answer would be the number of total point in the plane.

For minimize: The number of methods we should draw 3 points in a TRIANGLE and all other point inside this triangle in such a way that no 3 points could be in a line.

If it will be allowed to draw the points in a line then minimum possibilities will be 2 only, because if we take all points in a line then you can separate only the corner points from the others.

For example
If we have 10 points then maximum possibilities = 10
and minimum = 3

similarly for 5 points
Max = 5, Min = 3

for 19
max= 19 Min = 3
f(x)=f(f(x))=f(x^2)....how many such functions with degree >2 exist
does not exist....
In a sequence of integers, A(n)=A(n-1)-A(n-2),where A(n) is the nth term in the sequence, n is an integer and n>=3, A(1)=1,A(2)=1. Calculate S(1000), where S(1000) is the sum of first 1000 terms.
check it now
A(1)=1,A(2)=1.
A(3)=A(2)-A(1) acc to the given formula A(n)=A(n-1)-A(n-2)
A(3)=1-1=0.
similarliy,A(4)=A(3)-A(2)=0-1=-1.
A(5)=A(4)-A(3)=-1-0=-1.
A(6)=A(5)-A(4)=-1-(-1)=0.
after 6th term seq is repeating like...1,1,0,-1,-1,0...after every 6 terms
so the sum of these 6 terms are 1+1+0+(-1)+(-1)+0=0.
now we have to find sum of 1000 terms...
sum of first 600 terms is 0...(as consider 1,1,0,-1,-1,0 as one sequence,now consider 166 seq i.e 166*6=996 terms)
so the sum of first 996 terms is 0...after that 4 terms come i.e 1,1,0,-1
so the sum of last 4 terms = 1+1+0+(-1)=1.

7 planet four firesides in 4-dimensional space and thus the currency used by it's residentds are 3-dimensional objects. the rupee notes are cubical in shape while their conis are spherical. however the coni minting machinery lays out some stipulations on the size of the coins. the diameter of the conins should be at least 64 mm and not exceed 512 mm. given a conin the diameter of the next larger coin is at least 50% greater . the diameter of the coin must allways be an integer you are asked to disign a set of coins of different diameters with these requirements and your goal is to desing as many coins as possible. how many coins can you desing
1st coin=64cm
2nd "=64+32(50% of 64)=96cm
similarly 3rd=96+48=144
4th=144+72=216
5th=216+108=324
6th=324+112=436
7th not possible
ans is 6

8 There are 3 boys A, B, C and 2 Girls D, E. E always sit right to A. Girls never sit in extreme positions and in the middle position. C always sits in the extreme positions. Who is sitting immediate right to E?
Ans is either B or C
POSSIBLE WAYS TO SIT
1) B/C D A E C/B
2) A E B D C

9 What is reminder when 6^17+17^6 is devided by 7?
(7-1)^17 + (7*2+3)^6
binomial theorem
-1+3^6=728
728%7=0
(40*40* 40-31*31*31)/(40*40+40*31+31*31)=?a simle calcutation
(a^3-b^3)/(a^2+ab+b^2)
a-b
40-31=9

10 In base 7,a number is written only using the digits 0,1,....6.the number 135 in base 7 is 1x7^2+3x7+5=75 in base 10.what is the sum of the same of the base 7 numbers 1234 and 6543 in base 7?
11110
1234
6543

11110 which is 7^4+7^3+7^2+7= 2401+343 +49+7= 2800

11 There are 20 balls which are red,blue or green,If 7 balls are green and the sum of red balls and green balls is less than 13,atmost how many red balls are there
5, atmost there can be 5 red balls so that sum of green and red can be less than 13 (7+5=12)
3y+x>2 and x+2y-1
y
the ans is y>1.
explanation:
x+2y-1=0
so x+2y=1
now 3y+x>2 given
we can write in this way x+2y+y>2
then putting x+2y=1 we get 1+y>2 =>y>2-1=>y>1

12 Find the no of zeros in the product of 1^1*2^2*3^3*.....*49^49??
total zeros at the end of product 5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45 will be 5+10+15+20+50+30+35+40+45 = 250 zeros

13 n! has 13 zeros than wat is the higest and lowest value of n??
Here we have to consider only the 5,10,15,20.... these terms..
so form (5,10,15,20,30,35,40,45,55 ) we got 1 no of zero from each.
Now from 25(5*5) and 50(5*5*2) we got 2 no of zeroes form each.
so upto 55 factorial we get (9+4)= 13 zeroes.
so 55! is the smallest value and 59! is the largest value.

The sequence {A(n)} is defined by A(1)=2 and A(n+1)=A(n)+2n. What is the value of A(100).
The sequence {A(n)} is defined by A(1)=2 and A(n+1)=A(n)+2n.
The value of A(100)=A(99+1)=A(99)+2*99
=A(98)+2*98+2*99
=A(97)+2*97+2*98+2*99

A(100)=A(1)+2*1+2*2+2*3+......+2*98+2*99
= 2+(2*1+2*2+2*3+......+2*98+2*99)
= 2+2(1+2+3+.....+98+99)
= 2+2(sum of the first 99 natural numbers)
= 2+2(99*100/2) [sum of the first n natural numbers={n*(n+1)/2}]
= 2+9900
= 9902
14 A,B COMMON TO ALL THREE. THEN ATLEAST 1 ELEMNT COMMON TO 2 IE X1,X2,X2,X3,X3,X1. FINDMINIMUM ELEMENTS IN EACH GROUP.
ANSWER IS 4.

15 Directions for Q. 1 to Q. 5: Refer the data:
J, K, L, M and N collected stamps. They collected a total of 100 stamps. None of them collected less than 10.
No two among them collected the same number.
(i) 3 collected the same number as K and M together.
(ii) L collected 3 more than the cube of an integer
(iii) The no. collected by J was the square of an integer.
(iv) Total no. collected by K was either the square or cube of an integer.

1. The no. collected by J was:
(1) 27 (2) 49 (3) 36 (4) 64

2. The no. collected by K was:
(1) 16 (2) 27 (3) 25 (4) 36

3. The difference of numbers collected by L & M was:
(1) 3 (2) 2 (3) 5 (4) 9

in the question the number collected by J should be cube of the integer and l should collect 3 more than square of integer.
J+L+N=K+M
so the sum of J+L+N=K+M=50
l=3^2+3=12
so J=3^3=27
n=50-27-12=11
now k is either square or cube so possibilty are 36 or 16.and m=14 or 34
so j=27
now value for k can come after solving question 3
m-l=14-12=2 or 34-12=22
so there is 2 in option so value of m is 14 and k=36.

16 Bhanu spends 30% of his income on petrol on scooter. ? of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol then what is the expenditure on house rent?

let the income x,so in petrol he spends 3x/10 rs,remaining x-3x/10=7x/10 on house rent and balance on.now 3x/10=300,so x=1000,on house rent he spends rs 700-balane..
Let exp(m,n) = m to the power n. If exp(10, m) = n exp(2, 2) where to and n are integers then n = exp(m,n) = m to the power n= m^n

exp(10, m) = n exp(2, 2) = n*2^2 = 4*n
10^m =4*n
when m=2,
n=25
so min integer values of m and n are 2 and 25 respectively.
A Grocer bought 24 kg coffee beans at price X per kg. After a while one third of stock got spoiled so he sold the rest for $200 per kg and made a total profit of twice the cost. What must be the price of X?
A. $33 1/3 B. 66 2/3 C.44 4/9 D.50 1/3
ns: C. 44 4/9

cost price= 24*X $,
spoiled= 24*1/3=8 kg, now total amount= (24-8)=16 kg
selling price= 16*200$ =3200$
profit= (3200-24*X)
profit=2* cost , (3200-24*X)= 2*24*X, x=44 4/9

17 The age of two people is in the ratio 6:8. the sum of their ages is 77. after 2 years the ratio of their ages becomes 5:7. wat is their present age?
The ratio of two people be 6x and 8x
sum of their ages is 6x+8y=77
after 2 yrs,ratio is 5:7,so,
(6x+2)/(8x+2)=5/7
on cross multiplying this,
x=2
hence age of a person 6(2)=12

hence another person ages is 8(2)=16

answer is 12,16

MOTHER +DAUGHTER+INFANT AGE IS 74. MOTHER AGE IS 46 MORE THEN DAUGHTER AND INFANT.
AND INFANT AGE IS 0.4 OF DAUGHTER. FIND DAUGHTERS AGE.
M+D+I = 74 ..,.,.Equ..1

I=0.4D
M=D+0.4D+46

M=1.4D+46
put in equ 1
1.4D+46+D+0.4D=74 as I=0.4D
2.8D=74-46
D=28/2.8
D=10.,.,.
,
16 Mr and mrs smith had invited 9 of their friend and their spouses for party at wiki beachresort.the stand for group photograph if mr smith never stand next to mrs smith then how many way group arrange in row(A)20!(B)19!+18!(C)18*19!(D)2*19!
simplest... ans is 18*19!
method ..: mr and mrs smith never b together so out of 20(with both them) 19 can b any place so total = 19!
also 1 of them can stand with other so 18 places /..
so total is 18*19!..
19!*18bcoz along with mr and mrs smith can be arranged in 19!*2 (2 bcoz of mr and mrs smith arrangement) all together can be arranged in 20! ways so no mr and mrs smith can be arranged in 20!-19!*2 =19!*18
Is this solution Helpfull? Yes (8) | No (2) viresh (4 Day ago)
20!-(19!*2)
total possibility minus they will be together
A father purchases dress for his three daughter. The dresses are of same color but of different size .the dress is kept in dark room .What is the probability that all the three will not choose their own dress...
a dosesnot choose his dress 2/3
b dosesnot choose his dress 1/2
c dosesnot choose his dress 1 total 1/3

17 Leena cuts small cubes of 3 cubic cm each.She joined it to make a cuboid of length 10 cm,width 3cm, and depth 3 cm.How many more cubes does she need to make a perfect cube?
a)910 b)250 c)750 d)650
ns is a)910. To make cube we need 10*10*10=1000. now we have only 10*3*3=90.so the more cube we need are 1000-90=910.

18 Q:- A boy wants to make cuboids of dimension 5m,6m & 7m from small cubes of .03m^3.Later he realized that he can make some cuboids by making it hollow.Then it takes some cubes less.What is the number of cubes to be removed
a)2000 b)5000 c)3000 d)7000
Volume of the cuboid= 5*6*7
No of cubes needed = (5*6*7)/0.03 = 7000
Volume of hollow cube= (5-2)(6-2)(7-2)
No of cubes needed = (5-2)(6-2)(7-2)/0.03 = (3*4*5)/0.03= 2000
So, number of cubes to be removed = 7000 - 2000 = 5000

19 An empty tankbe filled with aninlet pipe 'A' in 42 minutes. after 12 minutes an outlet pipe 'B' is opened which can empty the tank in 30 minutes. After 6 minutes another inlet pipe 'C' opened into the same tank, which can fill the tank in 35 minutes and the tank is filled find the time taken to fill the tank?
The milk and water in two vessels A and B are in the ratio 4 : 3 and 2: 3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C containing half milk and half water?
let c.p. of 1liter milk is 1RS.
now milk in 1st mixer is 4/7 and in 2nd mixure is 2/5.
milk in finle mixer is 1/2
now from mixure formula..
4/7 2/5
1/2
1/10 1/14
so ratio is 14/10 =7/5
20 How many kgs. of wheat costing Rs. 8 per kg must be mixed with 86 kg of rice costing Rs. 6.40 per kg so that 20% gain may be obtained by Belling the mixture at Rs. 7.20 per kg ?

Sol. S.P. of 1 kg mixture = Rs. 7.20,Gain = 20%.
? C.P. of 1 kg mixture = Rs.[(100/120)*7.20]=Rs. 6.
By the rule of alligation, we have:
C_P. of 1 kg wheat of 1st kind C.P. of 1 kg wheat of 2nd kind
(800p) . (540 p)
Mean price
( 600 p)
60 200
Wheat of 1st kind: Wheat of 2nd kind = 60 : 200 = 3 : 10.
Let x kg of wheat of 1st kind be mixed with 36 kg of wheat of 2nd kind.
Then, 3 : 10 = x : 36 or lOx = 3 * 36 or x = 10.8 kg.

21 In what ratio must water be mixed with milk to gain 20 % by selling the mixture at cost price?
let the sp op 1 l mix is =re 1
cp =(100/120)*1=5/6
cp of one let. water =0
so ratio=1-(5/6)/(5/6)-0=1/5
so required ratio =1:5

 






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