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8/10/2019 CTS QUESTIONS AND SOLUTIONS.pptx
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QUESTIONS
&
ANSWERS
COGNIZANT TECHNOLOGY SOLUTIONS
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You are having 31kg of rice. You are provided with a
1kg stone for weighing. In how many weights the 31kg
of rice can be weighed. ?
Answer: 5
Explanation: 1+2+4+8+16 = 31.
2
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A starts at 11:00AM and travels at a speed of 4km/hr. Bstarts at 1:00PM and travels at 1km/hr for the first 1hr and
2km/hr for the next hr and so on. At what time they will
meet each other ?
Ans: 8.00 pm
Explanation:
Time 11.00 12.00 1.00pm
2.00pm
3.00pm
4.00pm
5.00pm
6.00pm
7.00pm
8.00pm
8.30pm
8.45pm
A 0 4 8 12 16 20 24 28 32 36
B 1 3 6 10 15 21 28 363
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There are 81 coins, among them one coin weighs less
compared to other. You are given a physical balance to
weigh. In how many weighing the odd coin can be found. 4 times.
Explanation: divide into three groups of 27 each. Weigh two
groups and chose one Divide into three groups of 9 each and chose one out of two.
Divide into three groups of 3 each and chose one.
Choose any two and weigh it to find the least.
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LOGICAL DEDUCTION
Statements:
Some shirts are biscuits.
No Biscuit is book
Conclusions:
I Some shirts are books
II Some books are biscuits.
I + E = O
None of the given
conclusion follow
Choice is D.
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LOGICAL DEDUCTION
Statements:
No women can vote.
Some women are politicians.
Conclusions:
I Male politicians can vote.
II Some politicians can vote.
No Conclusion.
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LOGICAL DEDUCTION
Statements:
No man is a donkey.
Rahul is a man.
Conclusions:
I Rahul is not a donkey.
II All men are not Rahul.
I + E = O
Conclusion I follows.
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LOGICAL DEDUCTION
Statements:
All poles are guns.
Some boats are not poles.Conclusions:
I All guns are boats.
II Some boats are not guns.
No conclusion.
We cannot come to a
conclusion that Some
boats are not gunsdefinitely.
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LOGICAL DEDUCTION
Statements:
All rats are cows.
No cow is whiteConclusions:
I No white is rat.
II No rat is white.III Some whites are rats.
IV All cows are rats.
A + E = E
No rat is white.
Conclusion II follows.
Correct choice is E.
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LOGICAL DEDUCTION
Statements:
All apples are brinjals.
All brinjals are ladyfingers.
All ladyfingers are oranges.
Conclusions:
I Some oranges are brinjals.
II All brinjals are apples.
III Some apples are oranges.
IV All ladyfingers are apples.
A + A = A
All apples are
oranges.
Choice is E.
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LOGICAL DEDUCTION
Statements:
All poles are guns.
Some boats are not poles.Conclusions:
I All guns are boats.
II Some boats are not guns.
No conclusion.
We cannot come to a
conclusion that Some
boats are not gunsdefinitely.
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LOGICAL DEDUCTION
Statements:
All poles are guns.
Some boats are not poles.Conclusions:
I All guns are boats.
II Some boats are not guns.
No conclusion.
We cannot come to a
conclusion that Some
boats are not gunsdefinitely.
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LOGICAL DEDUCTION
Statements:
All poles are guns.
Some boats are not poles.Conclusions:
I All guns are boats.
II Some boats are not guns.
No conclusion.
We cannot come to a
conclusion that Some
boats are not gunsdefinitely.
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LOGICAL DEDUCTION
Statements:
All poles are guns.
Some boats are not poles.Conclusions:
I All guns are boats.
II Some boats are not guns.
No conclusion.
We cannot come to a
conclusion that Some
boats are not gunsdefinitely.
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LOGICAL DEDUCTION
Statements:
All poles are guns.
Some boats are not poles.Conclusions:
I All guns are boats.
II Some boats are not guns.
No conclusion.
We cannot come to a
conclusion that Some
boats are not gunsdefinitely.
Because all boats
could be guns.All poles are guns does not imply that all things which are not
poles are not guns.
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LOGICAL DEDUCTION
Statements:
Most clocks are fans;
Some fans are walls
Conclusions :
I. Some walls are fans
II. Some clocks are walls
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I + I
NO CONCLUSION
GIVEN CONCLUSION 1 ISREPLICA OF SECOND
STATEMENT.
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LOGICAL DEDUCTION
Statements:
All birds are dogs;
Some dogs are cats
Conclusions :
I. Some cats are not dogs
II. All dogs are not birds
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A + I
NO CONCLUSION
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LOGICAL DEDUCTION
Statements:
Some fools are intelligent;
Some intelligent are great
Conclusions :I. Some fools are great
II. All great are intelligent.
NO CONCLUSION.
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LOGICAL DEDUCTION
Statements:
All Men are married;
Some men are educated
Conclusions :I. Some married are educated
II. Some educated are married.
BOTH CONCLUSIONS FOLLOWS..
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LOGICAL DEDUCTION
Statements: All bags are chalks; All chalks are bottles.Conclusions :
I Some bottles are bags.
II. All bags are bottles
III. All bottles are bags
IV. Some chalks are not bagsA) Only I, II and IV follow B) Only I, III and IV follow.C) Only II, III and IV follow. D) All Follow E) none of these
E) None of these. Only II follows.
Because I says some bottles are not bags (False)
By III all bottles need not be bags. (False)
Clearly IV is negative and contradicts the given statement 2. (False)
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LOGICAL DEDUCTION
Statements: Some trees are buses; All buses are hatsConclusions :
I Some trees are hats
II. Some hats are trees
III. All hats are buses
IV. Some buses are hatsA) None follows B) Only I, II and IV follow.C) Only II, III and IV follow. D) All Follow E) none of these
Conclusion is Some trees are hats.
Hence both I & II only follows.
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LOGICAL DEDUCTION
Statements :
All cars are tables;
Some children are tables
Conclusions :
I. Some cars are children
II. Some children are cars
No conclusion.
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LOGICAL DEDUCTION
Statements : Some dogs bark; All dogs bite
Conclusions :
I. Those dogs who do not bark, also bite.II. Those dogs who do not bark, not necessary
bite.
Conclusion I follows.
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LOGICAL DEDUCTION
Statements :
No magazine is cap;All caps are cameras
Conclusions :
I. No camera is magazine
II. Some caps are magazines
The conclusion is some magazines are not cameras
Hence none of the Given conclusion follows.
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LOGICAL DEDUCTION
Statements :
Some coolers are watches; no watch is bed
Conclusions :
I. No watch is cooler
II. No cooler is watch
III. Some watches are bedsIV. Some coolers are beds
A. None follows B. Only I and IV follow C. Only either II or III follows
C. Only either III or IV follows E. Only either II or IV follows
I + E = OHence Some coolers are not beds implies some coolers are beds. Hence onlyIV follows.
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LOGICAL DEDUCTION
Statements :
Some frogs are bricks; all bricks are cakes
Conclusions :
I. Some cakes are not frogs.
II. Some cakes are frogs.
III. No cake is frogIV. All frogs are cakes
A. None follows B. Only I and II follow C. Only either I, II and IIIfollow D. Only II, III or IV follows E. Only III and IV follow
I + A = I
Some frogs are cakes implies some cakes are frogs and some cakes arenot frogs.
Hence Only I and II follows. CHOICE B.
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LOGICAL DEDUCTION
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LOGICAL DEDUCTION
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ANALYTICAL ALGEBRA
48
A group of friends goes for dinner and gets bill of Rs 2400 . Two of them says
that they have forgotten their purse so remaining make an extra contribution of
Rs 100 to pay up the bill. Tell the no. of person in that group
Let N = number of people and S = Each personsshare.
NS = 2400S =2400
[1]
Now N 2 friends will pay S + 100 each.(N 2)(S + 100) = 2400NS + 100N 2S = 2600 [2]
Using [1] in [2], 2400 + 100N 2 2400
=2600
2400N + 100N24800 = 2600 N100N2200N 4800 = 0.
Solving, N = 6 or 8. Hence there are 8 friends in the group.
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ALGEBRA
11/25/2014C.S.VEERARAGAVAN 98948 34264
Given the following functions
(1) f(n a b c ) = ac if n=1
(2) f(n a b c) = f( n-1 a c b) + f( 1 a b c) + f( n-1 b a c ) if n > 1.
Then what is the value f( 2 a b c ) = ?f( 2 a b c) = f(1 a c b) + f(1 a b c) + f(1 b a c)
= ab + ac + bc
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ALGEBRA
11/25/2014C.S.VEERARAGAVAN 98948 34264
There are 600 tennis players. 4% wear wrist band on one wrist Of
the remaining, 25% wear wrist bands on both hands.
How many players don't wear a wrist band?
There are 600 tennis players
4% wear wrist band on one wrist = 600 * 04 = 24.
Remaining players = 576
Of the remaining, 25% wear wrist bands on both hands
and Of the remaining, 75% do not wear wrist bands .
so players not wearing wrist bands =3
4*576 =144*3 = 432
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NUMBER SYSTEM
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NUMBER SYSTEM
11/25/2014C.S.VEERARAGAVAN 98948 34264
If all the 6 are replaced by 9, then the algebraic sum of all the numbers from 1 to 100(bothinclusive) varies by
Sum of all numbers from 1 to n is given by n(n+1)/2so sum of all numbers from 1 to 100 is 100*(100+1)/2 = 50506 comes 20times in the numbers from 1 to 100
6 comes in 'units' place 10 times in the numbers 1 to 100 i.e. 6,16,26,...., 96
6 comes in 'tens' place 10 times in the numbers 1 to 100 i.e. 60,61,62,....., 69
sum of these 6's is 10(6*10)+10(6*1)=600+60=660
if all 6's are replaced by 9
then sum of those 9's will be 10(9*10)+10(9*1)=990
now algebraic sum will be 5050-660+990= 5380
SHORTCUT:from 1 to 100, 6 is used 10 time at unit place and 10 timesat tenth place so on replaces it by 9 ,its value increased 30(10*3=30)due
to unit place and 300 (10*30 = 300) due to tenth place so total increase
will be 330
PROBABILITY
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PROBABILITY
Out of forty students, there are 14 who are taking Physics and29 who are taking Calculus. What is the probability that arandomly chosen student from this group is taking only theCalculus class?
n(Physics) = 14 & n(Calculus) = 29.
n(total) = 14 + 29 n (both) = 40.
n(both) = 3.
n(Calculus only) = 29 3 = 26.
P(Calculus only) =26
40
=13
20
=65%.
53
VENN DIAGRAM
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VENN DIAGRAMIn town of 500 people, 285 read Hindu and 122 read Indian express and
127read Times of India 20 read Hindu and times of India and 29 read Hindu and
Indian express and 35 read Times of India and Indian express. 50 read no newspaper. Then how many read only one paper?
N(Total) = 500 N(Hindu) = 285
N(Indian) = 212 N(Times) = 127
N(H T) = 20 N(H I) = 29
N(T I ) = 35
N(H T I) = 450(285 + 122 + 127) + (20 + 29 + 35) = 450 (534) + (84)
N(H) = 285 49 = 136. N (I) = 12264 = 58 N(T) = 127 55 = 72
Number of persons who read only one paper = 266
H T
I50
X
20 - X
29 - X 35 - X
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VENN DIAGRAM
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VENN DIAGRAMIn a group of persons traveling in a bus, 6 persons can speak Tamil, 15 can
speak Hindi and 6 can speak Gujarati. In that group, none can speak any other
language. If 2 persons in the group can speak two languages and one personcan speak all the three languages, then how many persons are there in the
group ?
A) 21 B) 23 C) 22 D)24
X + Y + Z = 2
H = 15 14(X + Y)
T = 6 5 (X + Z)
G = 6 5 (Y + Z)
14 (X + Y) + 5 (X + Z) + 5 (Y + Z) + 3 = 242(X + Y + Z) = 27 4 = 23.
H T
G
1
X
Y Z
55
VENN DIAGRAM
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VENN DIAGRAM
Out of a total of 120 musicians in a club , 5% can play all the three instruments-
Guitar, Violin and Flute. It so happens that the number of musicians who can
play any two and only two of the above instruments is 30.
The number of musicians who can play the guitar alone is 40.
What is the total number of those who can play violin alone or flute alone ?
A) 30 B) 38 C) 44 D) 45
5% of 120 = 6 can play all 3.
X + Y + Z = 30
V + F = 120 (40 + 30 + 6)
= 44
G V
F
6
X
Y Z
40
56
VENN DIAGRAM
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VENN DIAGRAM
57
In a town 65% people watched the news on television, 40% read a
newspaper and 25% read a newspaper and watched the news on
television also.
What percent of the people neither watched the news on television
nor read a news paper ?
A) 5 B) 10 C) 15 D) 2065 + 40 25 = 80
80% watched tv or read newspaper or both.
Hence 20% neither watched tv nor read newspaper.
POWERS
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POWERS
58
A secret can be told only 2 persons in 5 minutes .the same person
tells to 2 more persons and so on. How long will take to tell it to
768 persons ?
a)47.5 min b)50 min c) 500 min d)49 min
29= 512 & 210= 1024.
In 9 x 5 = 45 minutes secret spread to 512 persons.
768 = 512 + of 512.
Hence 45 + of 5 = 47.5 min.
AGE
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AGE
59
When I was married 10 years ago my wife is the 6th member of the
family.
Today my father died and a baby born to me.
The average age of my family during my marriage is same as today.
What is the age of Father when he died?
Let the age of Father ten years ago be x. Remaining 6 persons total be y.
Average 10 years ago =x+y6
After 10 years Average =
y+5(10)
6 =
50+y
6
Equating both we have x = 50
Present Age of Father = 60
BOAT & STREAMS
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BOAT & STREAMS
60
A son and father goes for boating in river upstream .
After rowing for 1 mile son notices the hat of his father falling in the river.
After 5 min. he tells his father that his hat has fallen.
So they turn round and are able to pick the hat at the point from where theybegan boating after 5min.
Tell the speed of river?
The hat fallen after travelling one mile and after that they travelled for 5MIN.
so that hat travelled down stream for one mile and they collected the hatafter 5mins of travel down stream.
So the total time they spent after losing hat is 10 minutes.
In this 10 minutes the hat exactly travelled 1 mile(i.e with river speed).
the river speed is 6miles/hour.
HCF & LCM
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HCF & LCM
61
There are three departments having students 64,58,24 .
In an exam they have to be seated in rooms such that each roomhas equal number of students and each room has students of one
type only (No mixing of departments).
Find the minimum number rooms required ?
The HCF is 2.
Hence 32 + 29 + 12= 73.
PERMUTATION & COMBINATION
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PERMUTATION & COMBINATION
62
Argentina had football team of 22 player of which captain is from
Brazilian team and goalki from European team.
For remaining player they have picked 6 from Argentinean and 14
from European.
Now for a team of 11 they must have goalki and captain so out of 9
now they plan to select 3 from Argentinean and 6 from European.
Find out number of methods available for it.
6C3* 14C6= 160600
DATA SUFFICIENCY
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DATA SUFFICIENCY
63
(a) if the data in statement I alone are sufficient to answer the question;
(b) if the data in statement II alone are sufficient to answer the question;
(c) if the data in either in I or II alone are sufficient to answer the question;
(d) if the data in both the statements together are not sufficient to answer the question;
(e) if the data in both the statements together are needed;
How many visitors saw the exhibition yesterday?
I. Each entry pass holder can take up to three persons with him / her.
II. In all, 243 passes were sold yesterday.
Ans: D
DATA SUFFICIENCY
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DATA SUFFICIENCY
64
(a) if the data in statement I alone are sufficient to answer the question;
(b) if the data in statement II alone are sufficient to answer the question;
(c) if the data in either in I or II alone are sufficient to answer the question;
(d) if the data in both the statements together are not sufficient to answer the question;
(e) if the data in both the statements together are needed;
How much was the total sale of the company?I. The company sold 8000 units of product A each costing Rs. 25.
II. The company has no other product line
Ans: E
DATA SUFFICIENCY
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DATA SUFFICIENCY
65
(a) if the data in statement I alone are sufficient to answer the question;
(b) if the data in statement II alone are sufficient to answer the question;
(c) if the data in either in I or II alone are sufficient to answer the question;
(d) if the data in both the statements together are not sufficient to answer the question;
(e) if the data in both the statements together are needed;
In what proportion would Raj, Karan and Altaf distribute profit among them
I. Raj gets two-fifth of the profit.
II. Karan and Althaf have made 75% of the total investment.
Ans: D
DATA SUFFICIENCY
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DATA SUFFICIENCY
66
(a) if the data in statement I alone are sufficient to answer the question;
(b) if the data in statement II alone are sufficient to answer the question;
(c) if the data in either in I or II alone are sufficient to answer the question;
(d) if the data in both the statements together are not sufficient to answer the question;
(e) if the data in both the statements together are needed;
What time did the train leave today.
I. The train normally leaves on time
II. The scheduled departure is at 14.30.
Ans: D
DATA SUFFICIENCY
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DATA SUFFICIENCY
67
(a) if the data in statement I alone are sufficient to answer the question;
(b) if the data in statement II alone are sufficient to answer the question;
(c) if the data in either in I or II alone are sufficient to answer the question;
(d) if the data in both the statements together are not sufficient to answer the question;
(e) if the data in both the statements together are needed;
On which day in January, Subhas left for Germany?
I. Subhas has so far spent 10 years in Germany.
II. Subhas' friend Anil left for Germany on 15th February and joined
Subhas 20 days after Subhas' arrival.
Ans: D.
CONVERT THE GIVEN BINARY NUMBERS
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CONVERT THE GIVEN BINARY NUMBERS.
68
(11100111)2= ( )16
1110 0111
14 7
(E7)16
CONVERT THE GIVEN BINARY NUMBERS
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CONVERT THE GIVEN BINARY NUMBERS.
69
(01011010)2=( )8
01 011 010
1 3 28
CONVERT THE GIVEN BINARY NUMBERS
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CONVERT THE GIVEN BINARY NUMBERS.
70
(11110000)2= ( )10
1 1 1 1 0 0 0 0
7 6 5 4 3 2 1 0
27 26 25 24 23 22 21 20
128 64 32 16 0 0 0 0
(240)10
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CONVERT THE GIVEN BINARY NUMBERS
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CONVERT THE GIVEN BINARY NUMBERS.
72
(01001110)2= ( )8
01 001 110
1 1 6
1168
2 sides coloured = 12 * (N-2)CUBES & CUBOIDS
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N = 3 sides coloured = 8
( )
1 side coloured = 6 * (N-2)2
0 sides coloured = (N-2)3
CUBES & CUBOIDS
a cube is coloured orange on one face, pink on the opposite face,
brown on one face and silver on a face adjacent to the brown face.
The other two faces are left uncoloured.
It is then cut into 125 smaller cubes of equal size.
Now, answer the following questions based on the above statements:
73
How many cubes have at least one face coloured pink ?
A) 1 B)9 C) 16 D) 25
Ans: D
2 sides coloured = 12 * (N-2)CUBES & CUBOIDS
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N = 3 sides coloured = 8
1 side coloured = 6 * (N-2)2
0 sides coloured = (N-2)3
CUBES & CUBOIDS
a cube is coloured orange on one face, pink on the opposite face,
brown on one face and silver on a face adjacent to the brown face.
The other two faces are left uncoloured.
It is then cut into 125 smaller cubes of equal size.
Now, answer the following questions based on the above statements:
74
How many cubes have all the faces uncoloured ?
A) 24 B)36 C) 48 D) 64Ans: C
2 sides coloured = 12 * (N-2)CUBES & CUBOIDS
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N = 3 sides coloured = 8
1 side coloured = 6 * (N-2)2
0 sides coloured = (N-2)3
CUBES & CUBOIDS
a cube is coloured orange on one face, pink on the opposite face,
brown on one face and silver on a face adjacent to the brown face.
The other two faces are left uncoloured.
It is then cut into 125 smaller cubes of equal size.
Now, answer the following questions based on the above statements:
75
How many cubes have at least two faces coloured ?
A) 19 B)20 C) 21 D) 23Ans: C
2 sides coloured = 12 * (N-2)CUBES & CUBOIDS
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N = 3 sides coloured = 8
1 side coloured = 6 * (N-2)2
0 sides coloured = (N-2)3
CUBES & CUBOIDS
a cube is coloured orange on one face, pink on the opposite face,
brown on one face and silver on a face adjacent to the brown face.
The other two faces are left uncoloured.
It is then cut into 125 smaller cubes of equal size.
Now, answer the following questions based on the above statements:
76
How many cubes are coloured orange on one face and have the
remaining faces uncoloured?A) 8 B) 12 C) 14 D) 16
Ans: D
2 sides coloured = 12 * (N-2)CUBES & CUBOIDS
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N = 3 sides coloured = 8
1 side coloured = 6 * (N-2)2
0 sides coloured = (N-2)3
CUBES & CUBOIDS
a cube is coloured orange on one face, pink on the opposite face,
brown on one face and silver on a face adjacent to the brown face.
The other two faces are left uncoloured.
It is then cut into 125 smaller cubes of equal size.
Now, answer the following questions based on the above statements:
77
How many cubes one coloured silver on one face, orange or pink on
another face and have four uncoloured faces?
A) 8 B) 10 C) 12 D) 16
Ans: A
REASONING LINEAR ARRANGEMENT S P
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REASONING LINEAR ARRANGEMENT
78
Five children are sitting in a row.
S is sitting next to P but not T.
K is sitting next to R who is sitting on the extreme left and
T is not sitting next to K.
Who are sitting adjacent to S?
A) K and P B) R and P C) Only P D) P and T E) Insufficient Information.Ans: D
S P
P S
R K
K T
T S
S TR K S P T
REASONING LINEAR ARRANGEMENT
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REASONING LINEAR ARRANGEMENT
79
In the Olympic Games, the flags of six nations were flown on the masts in the
following way.
The flag of America was to the left of Indian tricolour and to the right of the flag of
France.
The flag of Australia was on the right of the Indian flag but was to the left of the flag
of Japan, which was to the left of the flag of China.
Find the two flags which are in the centre.
A) India and Australia B) America and India C) Japan and Australia D) America
and Australia
Ans: A
REASONING LINEAR ARRANGEMENT
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REASONING LINEAR ARRANGEMENT
80
There are five friends Sachin, Kunal, Mohit, Anuj and Rohan.
Sachin is shorter than Kunal but taller than Rohan.
Mohit is tallest.
Anuj is a little shorter than Kunal and little taller than Sachin.
Who is the shortest?
(a) Rohan(b) Sachin
(c) Anuj
(d) Kunal
(e) None of these
Ans: A
Rohan Sachin Anuj Kunal Mohit
REASONING LINEAR ARRANGEMENT
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REASONING LINEAR ARRANGEMENT
81
There are five friends Sachin, Kunal, Mohit, Anuj and Rohan.
Sachin is shorter than Kunal but taller than Rohan.
Mohit is tallest.
Anuj is a little shorter than Kunal and little taller than Sachin.
If they stand in the order of their heights, who will be in the middle?
(a) Kunal
(b) Rohan
(c) Sachin
(d) Anuj
(e) None of these
Ans: D
Rohan Sachin Anuj Kunal Mohit
REASONING LINEAR ARRANGEMENT
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REASONING LINEAR ARRANGEMENT
82
There are five friends Sachin, Kunal, Mohit, Anuj and Rohan.
Sachin is shorter than Kunal but taller than Rohan.
Mohit is tallest.
Anuj is a little shorter than Kunal and little taller than Sachin.
If they stand in the order of increasing heights, who will be the second?
(a) Anuj
(b) Sachin
(c) Rohan
(d) Kunal
(e) None of these
Ans: B
Rohan Sachin Anuj Kunal Mohit
REASONING LINEAR ARRANGEMENT
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SO G G
83
There are five friends Sachin, Kunal, Mohit, Anuj and Rohan.
Sachin is shorter than Kunal but taller than Rohan.
Mohit is tallest.
Anuj is a little shorter than Kunal and little taller than Sachin.
Who is the second tallest?
(a) Sachin (b) Kunal (c) Anuj (d) Rohan (e) None of these
Ans: B
Rohan Sachin Anuj Kunal Mohit
REASONING LINEAR ARRANGEMENT
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84
There are five friends Sachin, Kunal, Mohit, Anuj and Rohan.
Sachin is shorter than Kunal but taller than Rohan.
Mohit is tallest.
Anuj is a little shorter than Kunal and little taller than Sachin.
Who is taller than Anuj but shorter than Mohit?
(a) Kunal
(b) Rohan
(c) Sachin
(d) Date Inadequate (
(e) None
Ans: A
Rohan Sachin Anuj Kunal Mohit
REASONING LINEAR ARRANGEMENT
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85
In a March Past, seven persons are standing in a row.
Q is standing left to R but right to P.
O is standing right to N and left to P.
Similarly, S is standing right to R and left to T.
Find out who is standing in the middle.
A) P B) Q C) R D) O
REASONING CIRCULAR ARRANGEMENT
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86
Mr. A, Miss B, Mr. C and Miss D are sitting around
a table and discussing their trades.
1. Mr. A sits opposite to cook
2. Miss B sits right to the barber.
3. The washer man is on the left of the tailor
4. Miss D sits opposite Mr. CWhat are the trades of A and B?
A. Tailor and Barber
B. Tailor and cook
C. Barber and cook
D. washer man and cook
A
TABLE
COOK
TABLEBARBER
B
TAILOR
WASHER
MAN
D
/C
C
/D
REASONING LINEAR ARRANGEMENT
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87
In a pile of 10 books, there are 3 of History, 3 of Hindi, 2 of mathematics and 2 of
English. Taking from above,
there is an English book between a history and mathematics book,
a history book between a mathematics and an English book,
a Hindi book between an English and a mathematics book,
a mathematics book between two Hindi books and
two Hindi books between a Mathematic and a History book.
Book of which subject is at the sixth position from top ?
A. English B. Hindi C. Mathematics D. History
ENGLISH
HISTORY
MATHS
HISTORY
REASONING
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88
One boy can eat 100 chocolates in half a minute, and anothercan eat half as many in twice the length of time. How many
chocolates can both boys eat in 15 seconds?1st boy eats 100 chocolates / 30 sec so he can eat 50 in 15sec
2nd boy can eat one-half of 100 in twice of 30 sec
so he can eat 50 in 60 sec or that is 12.5 in 15 sec
so, together they can eat 50 + 12.5 or 62.5 chocolates in 15seconds.
REASONING PERCENTAGE
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89
Potatoes are made up of 99% water and 1% "potato matter."Jack bought 100 pounds of potatoes and left them outside in the sun for a while.
When he returned, he discovered that the potatoes had dehydrated and were now only madeup of 98% water.
How much did the potatoes now weigh?
initially water weigh 99 pounds & patato matter weigh 1 pound.
after dehydration water = 98%
=> potato matter = 2% = 1 pound
100% = (1/2) * 100 = 50 pounds
So, potato now weigh 50 pounds.
REASONING
8/10/2019 CTS QUESTIONS AND SOLUTIONS.pptx
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You own a pet store.
If you put in one canary per cage, you have one canary too many.
If you put in two canaries per cage, you have one cage too many.
How many canaries and cages do you have?
Ans: four canaries and three cages.