1. New Verbal Ability Section has been added, it will have 1 question for 10 minutes.
The question is to write an Essay / Business Letter of 70
-
100 words utilizing all
different words / phrases
of words given as part of question.
Sample: Using the following words write an email with minimum of 70 words to the Customer Mr John
Nixon about the delay in the project
Payment Processing System
-
Schedule
-
10th May (Friday)
-
will not recur
-
total delay
-
regret
How the Verbal ability section is assessed?
The test taker will be asked to frame a paragraph of 70
-
100 words using some words / phrases of words
given as the question. The Online Test engine of TouchStone will au
tomatically assess the Essay/
business letter / Paragraph written by the test taker in
-
terms of grammatical correctness and effective
utilization of words.
The result is Yes / No, so based on the correctness of Essay in terms of Grammar and utilization of
all
given words/phrases, the test taker will either be selected or rejected in the Test.
2. Change in Analytical ability section:
There will be two kinds of questions with two respective weightage of marks, ones with * mark and
others without * mark
Q
uestions with
*
mark
-
> All questions with
*
carry higher marks than the questions that do not have *
mark. And * marked questions are difficult questions.
There will be negative marking for all questions in Analytical Ability Section.
To be familiarized
with new pattern of TCS, please take practice (Open SeeSame) exam on Next Step
Portal of TCS. You may login to Nextstep portal with the logins created earlier.
Written Test Pattern:
Analytical Ability Test: 30 questions, 80 minutes
Verbal Ability Test:
10 Minutes
______________________________________________________________
1. Find X^y+y^x=46. Find X and Y values?
2. 3 white chips, 7 blue chips, 16 green chips, 2 chips drawn from the box in succession what is the probability that one is blue and other is white?
a) 7/50 b) 8/30 c) 7/25 d) 20/26
3. If a person has to work 8 continuous day & he gets a rest on the 9th day. If a person starts on Monday. What is the day of 12th rest day?
4. How many liters of a 90% of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 liter solution of 78% concentrated acid?
a) 8 b)9 c) 7 d)6
5. 3 cars A, B & C are in the race. A is twice as likely to win as B and B is thrice as likely to win as C. what is probability that B will win, if only one can win the race ?
a) ½ b) 2/5 c) 3/10 d) 1/10
6. A cow & Horse brought for Rs.2000. The cow is sold at a profit of 20% and the Horse is sold
at a loss of 10%. Of The overall gain is Rs.40. The cost price of the cow is :
a) 700 b) 800 c) 1200 d) 1300
7. The sum of 3 consecutive numbers of the four numbers A, B, C, D are 4613,4961,5010,5099 then what is the largest number among A,B,C,D ?
a) 1948 b) 1463 c) 1601 d) 1550
8. George printing press can print an edition of newspapers in 12 hours while Paul’s press can print the same edition in 18 hours. What is the total no. of hours the press working together but independent of one another to print the same edition?
a) 15 b) 17.4 c) 7 d) 7.2
9. 87th number in the series 2, 10, 26, 50…….
10. 70, 54, 45, 41……. What is the next number in the given series?
________________________________________________________________2. 3 white chips, 7 blue chips, 16 green chips, 2 chips drawn from the box in succession what is the probability that one is blue and other is white?
a) 7/50 b) 8/30 c) 7/25 d) 20/26
3. If a person has to work 8 continuous day & he gets a rest on the 9th day. If a person starts on Monday. What is the day of 12th rest day?
4. How many liters of a 90% of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 liter solution of 78% concentrated acid?
a) 8 b)9 c) 7 d)6
5. 3 cars A, B & C are in the race. A is twice as likely to win as B and B is thrice as likely to win as C. what is probability that B will win, if only one can win the race ?
a) ½ b) 2/5 c) 3/10 d) 1/10
6. A cow & Horse brought for Rs.2000. The cow is sold at a profit of 20% and the Horse is sold
at a loss of 10%. Of The overall gain is Rs.40. The cost price of the cow is :
a) 700 b) 800 c) 1200 d) 1300
7. The sum of 3 consecutive numbers of the four numbers A, B, C, D are 4613,4961,5010,5099 then what is the largest number among A,B,C,D ?
a) 1948 b) 1463 c) 1601 d) 1550
8. George printing press can print an edition of newspapers in 12 hours while Paul’s press can print the same edition in 18 hours. What is the total no. of hours the press working together but independent of one another to print the same edition?
a) 15 b) 17.4 c) 7 d) 7.2
9. 87th number in the series 2, 10, 26, 50…….
10. 70, 54, 45, 41……. What is the next number in the given series?
1.
If a month is having four Sundays then what is
the first day of the next month?
If month has 28 days i.e. 4 complete weeks:
The month could start on any day, there will be exactly 4 Sundays. Thus, the
month could end on any weekday. And the first day of the next month could be
any week-day.
If the month has 29 days i.e. 4 complete
weeks and 1 odd day: The month could start on ANY day EXCEPT Sunday (else it
will have 5 Sundays). Thus, the next month could star on ANY day EXCEPT Monday.
If the month has 30 days i.e. 4 complete
weeks and 2 odd days: The month could start on ANY day EXCEPT Saturday or
Sunday (else it will have 5 Sundays). Thus, the next month could start on ANY
day EXCEPT Monday or Tuesday.
If the month has 31 days i.e. 4 complete
weeks and 3 odd days: The month could start on ANY day EXCEPT Friday, Saturday
or Sunday (else it will have 5 Sundays). Thus, the next month could start on
ANY day EXCEPT Monday or Tuesday or Wednesday.
2.
If January 10 2007 is Friday then what is the
day of February 9 2010?
Number of days from 10 January 2007 to 10
January 2010 = 365+366+365=1096 i.e. 4 odd days (more than complete week).
Number of days from 10 January 2010 to 9
Feb 2010 = 21+9=30 i.e. 2 odd days.
Total number of odd days = 6 i.e. 9 Feb
2010 will be 6 week-days ahead of Friday i.e. the day will be Thursday.
3.
If March 11 2003 is Tuesday, then what is March
11 2004?
Number of days from 11 March 2003 to 11
March 2004 = 366 i.e. 2 odd days. Thus 11 March 2004 will be Thursday.
4.
What is the no. of Sundays in a leap year
(non-leap year)?
Incomplete question.
A leap year has 366 days i.e. 52 weeks + 2
days. Thus, there can be 52 Sundays or 53 Sundays.
Anon- leap year has 365 days i.e. 52 weeks
+ 1 day. Thus, there can be 52 Sundays or 53 Sundays, in a non-leap year also.
5.
In a month there are 4 Thursdays and 4 Sundays.
S o what is the first day of that month?
A month has 28 or 29 or 30 or 31 days.
Thus, there are 4 complete weeks and 0 or 1 or 2 or 3 extra days.
If month has 28 days: The month could start
on any day, it will always have 4 Thursday and 4 Sundays.
If month has 29 days: The month CANNOT
start on a Thursday, else it will have 5 Thursday. Also the month CANNOT start
on a Sunday, else it will have 5 Sundays. The month can start on nay other day
than Thursday or Sunday.
If month has 30 days: The month CANNOT
start on Wednesday or Thursday (else it will have 5 Thursdays). The month
CANNOT start on Saturday or Sunday (else it will have 5 Sundays). Thus, month
can start on Monday, Tuesday, Friday.
If month has 31 days: The month cannot
start on Tuesday, Wednesday or Thursday (else it will have 5 Thursdays). The
month cannot start on Friday, Saturday or Sunday (else it will have 5 Sundays).
Thus month can start ONLY on Monday.
6.
If 0>a>b>c>d, then which is greatest among the following:
a)
(c+d)/(d+a) b) (d+b)/(b+c) c) (c+d)/(b+c)
Among option a and c, the numerator is the
same. We cannot determine which denominator, (d+a) OR (b+c) is going to be
smaller. So we cannot decide which will be greater of the two.
Among option b and c, the denominator is
the same. But the numerator, c+d, will be greater than (d+b). Thus option b is
greater than option c.
Among option a and b, we cannot decide
which is greater.
E.g: If a=1, b=2, c=3, d=10, then option a
is 13/11 and option b is 12/5. Obviously option b is greater. BUT if a=1, b=2,
c=11, d=12, then option a is 23/13 and option b is 14/13. Option a is greater
in this case.
7.
* question:
If f(1)=0, f(m+n)=f(m)+f(n)+4*9mn-1,
M,n>0
Then what is f(17)=?
Put m=n=1, we get f(2)=0 + 0 + 36 – 1 = 35
Put m=n=2, we get f(4)=35 + 35 + 36*2*2 – 1
= 213
Put m=n=4, we get f(8)=213 + 213 + 36*4*4 –
1 = 1001
Put m=n=8, we get f(16)=1001 + 1001 + 36*8*8
– 1 = 4305
Putting m=16, n=1, we get f(17)=4305 + 0 +
36*16*1 – 1 = 4880.
8.
8n /4444
=?(Remainder)
4444 is a large number. Thus,
for n = 1, 2, 3, the remainders will be 8, 64, 512. And so the answer will not
be unique.
9.
I f a no. is divided by 375 then the remainder
is 5 , then what remainder would be obtained if the same no. is divided by 17?
The number is of the form 375a+5. When
divided by 17, the remainder will be a+5 i.e. again the answer will not be
unique and depend on a.
10.
If
Suraj’s sum of the salaries of 2003,
2004, 2005 is $36400 and the salary gets incremented 20% every year then find
the salary drawn by Suraj in the year 2005?
If salary in 2003 is 100x, then salary in
2004 is 120x and in 2005 is 144x. Thus, the sum = 364x = 36400 i.e. x=100.
Salary in 2005 is 14400.
11.
Ax4 – bx2 +x+5, f(-3)=2,
f(3)=?
Putting x=-3 , we get 81A-9b-3+5=2 i.e.
81A-9b=0
Puttinx x=3, we get 81A-9b+3+5 i.e. 0+3+5=8.
12.
If ROADIES word is permutated in all the ways in
the alphabetical order, then find the 44th word?
Words starting with A: 6! = 720. Thus, the
44th word will start with A. Finding which factorial is less than
44, we realize than 4!=24. Thus the first 24 words will be only when 4 letters
have to be arranged i.e. the first 3 letters will be ADE_ _ _ _ and in the
blanks we will arrange I, O, R, S.
The series of words will be ADI _ _ _ _ and
there will be 24 such words. The word we are searching will be in this group.
Words of the type ADIE _ _ _ will be 3! = 6
words. Cumulative words so far = 24+6=30.
Words of the type ADIO _ _ _ will be 3! = 6
words. Cumulative words so far = 30+6=36.
Words of the type ADIR _ _ _ will be 3! = 6
words. Cumulative words so far = 36+6=42.
The 43rd word will be ADISEOR
and the 44th word will be ADISERO.
13.
A child is asked by his mom to find out the
mistake in the series and replace it with the right one, (odd man out)
60,48,38,28,24,20,18.
The differences are 12, 10, 10, 4, 4, 2.
If we correct the term 28 to 30, the
differences will be 12, 10, 8, 6, 4, 2.
14.
F(1)=4, f(x+y)=f(x)+f)y)+7xy+2, then f(2)+f(5)=?
Putting x=y=1, we get f(2) = 4 + 4 + 7*1*1
+ 2 = 17
Putting x=y=2, we get f(4) = 17 + 17 + 7*2*2
+ 2 = 64.
Putting x=4, y = 1, we get f(5)=64 + 4 +
7*4*1 + 2 = 98
We want f(2) + f(5) = 17 + 98 = 115.
15.
An equilateral triangle and a regular hexagon
have equal parameters then what is the ration of their areas?
A)
6:1 b) 1:6 c) 3:2 d) 2:3
Let perimeter of both the equilateral
triangle and of the hexagon be 6x.
Side of equilateral triangle = 2x. Thus its
area = root(3) * 4x2 / 4.
Side of hexagon = x. Thus, its area = 6 *
root(3) * x2 / 4.
Thus, ratio is 4 : 6 i.e. 2 : 3.
16.
How many of no.s X( X being integer) with
10<=X <=99 are 18 more than the sum of their digits
a)
9 B)12 c) 10 d) 18
If the number is xy, then we have
10x + y = x + y + 18 i.e. x = 2 and y could be any value from 0 to 9. Thus,
there are 10 such numbers, 20, 21, 22, …, 29
17.
A&B starts from their home at 10kmph, then
travels from their home on MG road at 20 kmph and 40 kmph there is a ‘T’
junction on their path.’ A’ turn left at ‘T’ junction at 12 noon.
‘B’ reaches the ‘T’ junction earlier and
turns right . Both of them continued travelling till
2:00 pm. What is the distance between
A&B at 12:00PM
a)120km b)160km c) 150km d) 140km
The question is not clear. After A reaches
T junction at 12 noon, then till 12 pm, he will be travelling for 12 hours i.e.
240 kms. B would have reached T junction earlier and would have travelled more
than 480 kms. Thus, distance between them has to be more than 720 km. Also it
is given that they continue travelling till 2:00 pm, which does not fit with
the given data. We also need to know at what time they left home (I think the
10 kmph at start will be 10 AM). Overall, in the current format, the question
is ambiguous.
18. There are 6 tasks and 6 persons
Task 1 cannot be assigned either to person 1 or person 2
Task 2 must be assigned to either person3 or person 4.
Every person is to be assigned one task.
In how many ways can be assignment done?
a)192 b) 360 c) 144 d) 180
Task 2 can be assigned in 2 ways. Post this task 1 can be
assigned in 3 ways. The remaining 4 tasks can be assigned to remaining 4
persons, one task to each person, in 4! = 24 ways. Thus, total number of ways =
2*3*24 = 144.
19. In the sample
subtraction problem below, single digits are replaced by letters, find the
values of
3*A + &*B + 4*C
*D=?
A5C1 a)
80 b) 99 c) 89 d)96
3U79
---------------
397D
---------------
There will be a 1 borrowed from ten’s column to unit’s
column. Thus, D = 2.
If there is no borrowing from hundred’s column to ten’s
column, then (C – 1) – 7 = 7 i.e. C = 15, which is not possible. Thus, there
necessary has to be a borrowing from hundred’s column to ten’s column as well.
And we have 10 + (C – 1) – 7 = 7 i.e. C = 5.
If there is no borrowing from thousand’s column to hundred’s
column, then (5 – 1) – U = 9 i.e. U = -5, which is not possible. Thus, there
necessary has to be a borrowing from thousand’s column to hundred’s column as
well. And we have 10 + (5 – 1) – U = 9 i.e. U = 5.
And finally we have (A – 1) – 3 = 3 i.e. A = 7.
Thus, the subtraction is 7551 – 3579 = 3972.
The required expression is wrongly written, there is no B in
the digits. Also the symbol & is used which does not mean anything. But the
subtraction is explained.
20. A series of books were published at seven years
intervals when the seventh book was issued the sum of the publications years
was 13524, when the first book was published?
a)1911 b) 1800 c) 1900 d) 1811
a + ( a + 7) + (a + 14) + …. Sum of 7 terms = 13524.
Thus, 
i.e. 2a + 42 =
3864 i.e. a = 1911
i.e. 2a + 42 =
3864 i.e. a = 1911
21. What are the next 3no.s for the series
11,23,47,83,131
Checking the differences, 12, 24, 36, 48. Thus, the next
term will be 131 + 60 = 191.
22. A circular swimming pool is surrounded by a concrete
wall 4ft wide. If the area of the wall is 11/25 of the area of the pool then
radius of the pool in feet is
a) 20 b) 8 C) 16 d) 30
If area of the pool is 25k, then area of the wall = 11k.
Thus, area of the outer circle, pool plus wall, will be 25k + 11k = 36k. Thus,
ratio of the areas of the inner circle and outer circle is 25 : 36. Thus ratio
of radius will be 5 : 6. And it is given that the difference in these radii is
4 ft. Thus, radius of pool is 20 ft.
23. A pair of dice are rolled together till a sum of either
5 0r 7 is obtained what is the probability that 5 comes before 7
A) 0.45 b) 0.6 c) 0.5 d) 0.4
Probability of getting a sum of 5 when a pair of dice is
rolled = 4/36 = 1/9.
Probability of getting a sum of 7 when a pair of dice is
rolled = 6/36 = 1/6.
Probability of not getting either a 5 or a 7 = 1 – 1/9 – 1/6
= 26/36 = 13/18.
Required proability is that on the last roll, we get a 5 and
on all the previous rolls we get neither a 5 nor a 7. Thus, required
probability is 
, which is a infinite GP with sum = 
= 0.6.
, which is a infinite GP with sum = = 0.6.
24. One day Esta started 30 min late from home and reached
the office 50 min late while driving 25% slower than her usual speed. How much
time in minutes does Esta take to reach her office from home?
a) 40 b) 60 c) 80 d) 20
Of the 50 minute being late, 30 minutes is due to starting
late and remaining 20 minutes is due to travelling at reduced speed. Ratio of
reduced speed to normal speed is 3 : 4. Ratio of time at reduced speed to time
at normal speed is 4 : 3. And difference in these times is 20 minutes. Thus,
time at normal speed is 60 minutes.
25. After an election, the newly constituted geocity office
is exploring the use of hollow cylinders for water towers and has built a model
in their office model.
The circumference of the top of the cylinder is 78cm, d-27
cm, h= 53 cm, then what is the distance (in cm) crawled by bug before reaching
the honey?
a)
119 b) 80 c) 100
d) 89
Circumference = πd = 78 which does not match with the
given d-27 cm. So some ambiguity in the question. Also it is not clear if bug
is crawling in a spiral manner or a vertical straight line. Anyways if bug is
crawling in a spiral manner, the distance it covers in reaching the top when it
make one complete circle in the process is 
, where c is the circumference of the cylinder and h
is the height. If I ignore the d-27 cm data in the question, and assume that
bug crawls in a spiral fashion and completes one circle in reaching the top the
answer will be 
= 94.3
, where c is the circumference of the cylinder and h
is the height. If I ignore the d-27 cm data in the question, and assume that
bug crawls in a spiral fashion and completes one circle in reaching the top the
answer will be = 94.3
26. An organization has 3 committees, only
2 persons are members of all 3 committees, but every pairs of committees has 3
members in common. What is the least possible no. of members on any one committee?
a) 5 b) 6 c) 7 d) 4
Committee 1: a, b, c, d
Committee 2: a, b, c, e
Committee 3: a, b, d, e
Thus, least possible members in each
committee = 4.
27. If A ^B means A raised to the power of
B, in which of the following choices must
P be greater than Q
a) 0.9^P=0.9^Q b) 0.9^P=0.92^Q c) 0.9^P>0.9^q
Option a: P=Q
Option b: Since the two given values are
equal, but the base are unequal, the one with the lower base will have a higher
index. Thus, P > Q. But this is NOT
the only case. The equality will hold even if P = Q = 0. And the questions asks
P “must” be greater than Q.
Thus, the answer is option c.
28.2 gears one with 12 teeth and other one
with 14 teeth are engaged with each other. One teeth in smaller and one tooth
in bigger are marked and initially those 2 marked teeth are in contact with
each other. After how many rotations of the smaller gear with the marked teeth
in the other gear will again come into contact for the first time?
a)
7 b) 12 c) Data insufficient) 84
The number of rotations of the two
wheels will be inversely proportional to the number of teeth the wheel has.
Since the ratio of the number of teeth is 6 : 7, the ratio of the number of
rotations of the wheels is 7 : 6. The two marked teeth will be together only
after both the wheels have done an integral number of rounds. Thus, smaller
wheel will have done 7 rounds.
29. A& B
completed a work together in 5 days. A worked at twice the speed and B at half
the speed it would have taken then 4 days to complete the job. How much time
would it take for A alone to do the job?
a) 25 b) 10 c) 20 d) 15
Let the job be 20 units. And let A and B do a and b units
per day at normal speed. Thus, 5(a + b) = 20 i.e. a + b = 4. Also 4(2a + b/2) =
20 i.e. 4a + b = 10. Subtracting the two equations, 3a = 6 i.e. a = 2. Thus A
will take 20 units/2 units per day = 10 days.
30. find the missing in the series:
70,54,45,41,____.
Checking the differences, 16, 9, 4. These are all perfect
squares. Thus the next difference will be 1 and the required number will be 40.
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