Students must be a relieved lot today. CAT14 is finally over and may be it feels anticlimactic – isn’t it? Its over??## Is this what we were preparing for all this time. We can not just understand but relate to the feeling. So what was it like. Here are some pointers of changes
1. VA was full of changes. Not a single question that was based on vocabulary against almost 33.33% of the VA questions that used to be word based over the last several years. The focus was squarely on reading skills. Even essence questions made an entry. Essence of a paragraph, Jumbled up paragraphs, Odd one out in a parajumble, inference to be drawn from a para and 16 questions from four RC sections. Everything screamed for better readers.
If you have been a regular reader of the newspapers and magazines and have been writing central ideas of the passages regularly as mandated at CHEM – remain assured you are expected to do well in such a section.
2. Overall, the VA section was much simpler in terms of toughness or complexity levels as compared to CATs over the last few years. If you have been preparing consistently, here was a section to score well competitively speaking. For a change the options weren’t “razorthick close”. One could eliminate options with fair ease.
3. Passages were taken from a variety of reading sources that comprise leading magazines globally to Wikipedia. Once again stressing the need to read regularly and varied.
4. QA was full of sitters. After a long time the QA section was such a simple one. Arithmetic was the flavor of the day. Percentages, Ratios, Time and Work, Distance and Time, Averages & Alligation and the Good old Number theory – everything found a place. The questions were as an aptitude test should be – simple but with that harmless twist in the language which careful readers will pick. Also questions required quick and good structuring from the test taker to be able to solve quickly. All in all those of you CHEMpions – if you practiced your ratios approach well, there should be no trouble in doing all of them quickly and accurately.
5. Geometry had its fair share with almost 7 questions. What the questions needed us to do was to quickly draw and visualize the situation. The questions were based on simple properties of Triangles and Circles primarily. The single mensuration question in the evening slot was volume questions almost similar to a Twister question on mensuration in the CHEM courseware. Once again even if you have done your “Thrust” material at CHEM – ALL of the geometry questions were sitters for you.
6. Algebra – kept itself restricted to Linear, quadratic equations and sequences and progressions – with a question on calendar, one on coordinate Geometry and another question on Trigonometry popping up here and there. Although the questions on these not so popular topics were very simple – even if you avoided them you can be excused because there were so many others equally simple and so mush practiced already that one had a field time scoring in the QA section.
7. Our Guess is, that as a CHEMpion, out of the 36 QA questions in the first section, you could have easily attempted 2730 questions in something like 60 min with near 100% accuracy. Now if that is a lofty expectation then too you should crack this section completely. J
8. Data Interpretation and Logical reasoning were Tough in the morning and not so tough in the evening slots. At least three cases in the morning slot – one in DI and two in LR were full of scattered data and multiple possibilities. Even the questions were a little twisted. Would have required a lot of patience to work themselves out. In comparison the cases in the evening slots were simpler. There was only once case based on vein diagrams that had clues given in a very scattered manner and could have posed a little problem in arranging all of it. However, the evening slot was far more manageable. LR had all the standard cases of Linear/circular sequencing, mathematical reasoning, conditional logical reasoning and the Analytical variety too. A notable exception was complete absence of Data Sufficiency. Verbal reasoning has not been appearing since many years now and so it continued.
9. If you have practiced the Material thoroughly there should be no trouble in negotiating most of these cases. We do think that the number of attempts in the morning slot would have to be lesser due to slightly tougher cases in DI/LR.
So overall, we think that a 55 to 60 attempts in the morning slot and 6065 attempts in the evening slot (with more or less even attempts in both the sections) would be a good number of attempts. If you are able to achieve that with approximately 9095% accuracy then you should be sitting pretty expecting a good score for top calls. You can adjust the numbers downward for lower percentiles.
A very detailed analysis of CAT 2014 follows. This is a hasty effort and hence has a lot of typo and formatting errors. They are being corrected and will be OK by evening of 17^{th} Nov 2014. Hope you appreciate the painstaking efforts of our faculty in bringing all of these together for you.
Do write to us at help@chgroup.in with your feedback and comments.
Wishing you a very Successful life ahead of you. Take Care.
Question:
An equilateral triangle has an incircle which is circumcircle to a square with Radius 1. What is the area of triangle?
 12
 6
Solution:
Given PQRS of side 1 unit Hence PR = 2 unit (diagonal) This PR will be the diameter of incircle
Therefore, radius = 1/2
Also if ‘a’ denotes the side of equilateral triangle then radius of incircle r = a/23
Therefore, a/23 = 1/2.
Therefore, a = 6
Hence area of equilateral triangle 3/4 a^{2} = 3/4 (6)^{2} = 33/2 sq.units.
Question:
The area of a regular hexagon ABCDEF is 60. What is the area of triangle ACE.
 30
 25
 40
 20
Solution:
Ans : 30
Given area of Hexagon ABCDEF = 60 sq cm
Let O is the center of the hexagon. Join O with A, C and E. then The figure is divided into six equal and congruent triangles. Hence
area of ACE = 30 sq cm
(Since it contains 3 triangles)
Question:
A month has 3 even dates on Monday. Which day is the 15^{th} of the month?
 Friday
 Sunday
 Saturday
 Wednesday
Solution:
Ans : Sunday
Solutions :
In a given month three Mondays comes on even dates only possible case is 2, 9,, 16, 23, 30
Therefore on 15^{th} of this month, these will be Sunday.
Question:
a^{x+2} b^{3x} = a^{1x} b^{5x}. What is the value of xlog_{10}(b/a).
 log a
 3/2 log a
 log 9/6
Solution:
Ans : log_{10 }(a^{1/2})
a^{x+2} b^{3x} = a^{1x} b^{5x}
a^{1 }= (b/a)^{2x}
= log a = 2x log b/a
= ½ log a = x log b/a
= log = xlog b/a
Question:
X started from a point A towards B and at the same time Y started from B towards X. X reached B in an hour and Y reached A in 84 minutes. In how much time after they started did they meet each other?
 35 min
 32 min
 24 min
 30 min
Solution:
35 minutes
Given x travels from city A to city B in 60 mins and y travels from city B to city A in 84 mins.
Let x be the distance between A and B
Then Speed of x = S_{x} = x/60
Speed of y = S_{y }= x/84
Relative Speed = Sx + Sy = = (x/60) +(x/84)
\ Time taken before they cross = Relative Distance/Relative speed = x / (sx + sy)
= 60×84/144 = 35 min
Question:
 How many five digit numbers formed using 1, 2, 3 using the numbers atleast once?
 150
 180
 240
 None of these
Solution:
There are 6 possible cases, we can make numbers using
I 1 (thrice), 2 (once), 3 (once) No. of ways = 5!/3! = 20
II 1 (once), 2 (thrice), 3 (once) No. of ways = 5!/3! = 20
III 1 (once), 2 (once), 3 (thrice) No. of ways = 5!/3! = 20
IV 1 (twice), 2 (twice), 3 (once) No. of ways = 5!/3!2! = 30
V 1 (thrice), 2(once), 3 (twice) No. of ways = 5!/2!2! = 30
VI 1 (once), 2 (twice) 3 (twice) No. of ways = 5!/2!/2! = 30
Total no. of ways = 20 + 20 + 20 + 30 + 30 + 30 = 150
Question:
No. of zeroes in 1 x 2 x 3 x ________x 1000
 249
 248
 250
 None
Solution:
Ans : a
Question:
In a can of 3 litres glycerin if one litre is taken and then the can is again filled with water. This process is again repeated with the mixture. What will be the ratio of glycerin to water in the can?
 4 : 5
 8 : 3
 5 : 4
 1 : 1
Solution:
Ans. (a)
3 
2 


3 
2 
9 

4 
Ratio of G : W = 4 : 5.
Solutions:
Total 

Glycerine after 1^{st} exchange water 

Glycerine after 2^{nd} exchange water 

3 
: 
2 





3 
: 
2 

9 



4 
5 
\ Glycerine : Water = 4 : 5
Question:
What will be the remainder of 2014^{2014} when divided by 5.
 1
 0
 2
 3
Solution:
1.
Reminder
Question:
For which points (a + b) x + (a – b) y = 2a + 3b satisfies
 (2, 1)
 (5/2, 1/2)
 (3 1/2)
 None of these
Solution:
Ans (5/2 ,1/2)
(a + b) x + (a – b) = 2a + 3b
(a b)y = a[2 – x] + b[3 – x]
y = [a(2x)+b(3x)] /(ab)
Put the given options (x, y) = (5/2 , 1/2)
Question:
For which condition log_{10}(1 + n) – log_{10}(n) < log_{10}(1.01)
 n > 100
 n < 100
 n > 10
 n < 10
Solution:
Ans. n > 100
Given log_{10}(n + 1) – log_{10}n < log_{10}(1.01)
Question:
Ages of Three brothers are in the ratio of 4 : 5 : 8. If age of Elder brother is two less than nine times the difference of other two brothers. Then what is the sum of ages of all of them?
 17
 51
 34
 68
Solution:
Ans. –34
Solutions
Given ages of three brothers, say A, B, C be 4 : 5 : 8
Then given 9[5x – 4x] = 8x + 2
= x = 2
Question:
In a class having two section the average weight is 55.8kg if average of section A is 56.4kg and of section B 54.2 kg. then what will be the ratio of students of section A and B.
 3 : 8
 8 : 3
 5 : 7
 1 : 4
Solution:
Ans. 8 : 3.
Ratio of students = 8 : 3.
Question:
Subodh sold two watches. One of Neha at profit of 25% and one to Seema at loss of 25%. If Neha sold it to Gautam at a loss of 20%. Then which of the following statements is true.
 Gautam & Subodh bought the watch at same price.
 Gautam spent 40% less than Seema
 Neha bought at 20% less than Gautam
 All of these
Solution:
Ans. (a)
Subodh 
Neha 
Gautam 
4 
5 


4 
5 
Hence subodh and Gautam bought water at same price
Question:
What will be the remainder of x^{2014} + x – 1 / x – 1
 1
 0
 2014
 2
Solution:
Ans. (a).
Reminder of x^{2014} + x – 1 / x^{–1} ® x^{2014} /x^{–1} + x–1/ x – 1
(x^{2014} – 1^{2} / x – 1) + 0 [Remainder Zero]
1
[x^{a}  b^{a} will be completely divisible by x – b if a is even]
Question:
A contractor hires 150 workers for a project to be completed in 180 days. If in 60 days only onefourth work is done. Then how many extra workers needed to complete the project in time?
 75
 60
 225
 50
Solution:
Ans (a)
150/1 = 2x/3
= x = 225
= Extra workers required = 225 – 150 = 75
Question:
In a triangle if one of the angle is twice the sum of other two angles. Then what can be the maximum possible value of the smallest angle?
 pi/6
 pi/3
 pi/4
 pi/2
Solution:
In a triangle ABC
Let angle B = 2[angle A + angle C]
= angle B = 120°
= angle A + angle C = 60°
Hence maximum possible values of smallest angle would be 30° = pi/6
Question:
If Arithsmetic means of ten numbers is 25. and none of them is less than 20. Then the value of one of them can be atmost
 70
 75
 100
 80
Solution:
Ans. 70
Given Arithmetic mean of ten numbers is 25 Hence total is 250
Now to get the maximum value of the set fwe have to take the minimum value of other numbers i.e 20.
Hence x + 20 x 9 = 250
x = 70
Question:
If Rimi travels from city A to city B and then do city A. friend Mini travels from city B to city A on the same road and Then back again They meet for the first time at a distance of 60 m from A and second time at a distance of 30 m from B. what is the total distance between A and B^{2}.
 150 m
 120 m
 180 m
 200 m
Solution:
From both, we can say that
60/(d + 30) = (d + 60)/(d + 120)
Solving, we get d = 60
Therefore, total distance = 150 m.
Question:
AB is side of square with centre O and then on the circle with centre O, on the minor arc AB there exist a point C. then what is the angle angle OCA?
 135°
 45°
 75°
 120°
Solution:
Ans. 135°.
Given
ABCD is a square
angle ADB = 45°
= angle ACB = 135° [Cycle in quadrilateral opp. angles are supplementary]
Question:
In a triangle in which AB = 9 cm, AC = 12 cm and BC = 15 cm and a point 0 lies on BC which is circumcentre of triangle ABC what will be the area of triangle 0AB
 27
 90
 54
 18
Solution:
Ans 27.
Given O is the circumcenter, Now triangle ABC is a right angled triangle. Since it satisfies pythagorous theorem [9^{2 }+ 12^{2} = 15^{2}]
= arc BAC forms a semicircle [because angle A = 90°]
= BC is diameter = 15 cm
= OB = 15/2 cm
= OA = 15/2 cm
Area of triangle ABC = 1/2 x 9 x 12 = 54 therefore area of AOB = 54/2 = 27.
Question:
In a rhombus the diagonals are in the ratio of 1 : 3.Then what is the smallest angle of rhombus
 60°
 45°
 30°
 15°
Solution:
Ans. (a)Solutions :
1 x 2 x 3 x ………….. x 1000 = 1000
No. of zeroes in 1000 = 1000/5 + 1000/5^{2} + 1000/5^{3} + 1000/5^{4} +…………..
= 200 + 40 + 8 + 1 + 0 + …………..
= 249
Let AC & BD be diagonals and interest at 0.
Given AC : BD = Ö3 : 1 Þ Let AC = 2Ö3 BD = 2
Then AO : OD = Ö3 : 1
= AD^{2} = (Ö3)^{2} + (1)^{2} = 2
Now AB = AD = 2cm Also BD = 2cm
therefore, ADB is equilateral triangle
Hence smallest angle in rhombus is 60°
Question:
If x^{3}, y^{3}, z^{3} are in G.P. then which of the following holds
a
 log x, log y, log z are in A.P.
 (log x)^{3}, (log y)^{3}, (log z)^{3} are in A.P.
 log x, log y, log z are in G.P.\
 log x, log y, log z are in H.P.
Solution:
If x^{3}, y^{3}, z^{3} are in G.P.
y^{6} = x^{3} z^{3}
Taking log, 6 log y = 3[log x + log z]
2log y = log x + log z
[If 2b = a + c then a, b, c are in A.P.]
Therefore, log x, log y, log z are in A.P.
Question:
For natural no. X, Given x^{2} + 31 is a perfect square. How many possible values of x exist
 1
 2
 Infinite
 Not possible
Solution:
1
x^{2} + 31 = perfect square = y^{2}
Þ y^{2} – x^{2} = 31
(y – x) (y + x) = 31 x 1
\ y – x = 1
And y + x = 31
Only one such value is possible
[x = 15, y = 16]
Question:
For positive value of a and b given
f(x) = a – x^{2} + 1 and g(x) = x^{2} + b + 3 and f(2) g(1) < 0
 a < 3
 b > 3
 a > 2
 b < 2
Solution:
Ans. a < 3
f(x) = x – x^{2} + 1
g(x) = x^{2} + b + 3
f(2) g(1) < 0
(a – 4 + 1) (1 + 6 + 3) < 0
(a – 3) (b + 4) < 0
= a < 3
Question:
What cannot be the value of 4sin x cosx (1 + sin x cosx)
 –1
 0
 3
 4
Solution:
4 sin x cos x [1 + sin x cos x]
The max. value exist at x = 45°
Therefore, 4 x 1/2 x 1/2 [1 + (1/2)^{2}]
= 4/2 [1 + 1/2] = 3
Hence 4 cannot be the value of 4 sin x cos x [1 + sin x cos x]
Question:
Given a list of consecutive odd integers, the first number denotes the total number of integers in the list. If six times the first number exceeds the average of all the numbers by 2013, then what is the total number of integers in the list.
 503
 501
 505
 507
Solution:
Here all the values depend on the total no. of Integers. Hence we will check by picking up each option individually. Take option (a) 503
If 503 is the total no. of integers in the list. It will also be the first no. of square. The last no. could be find out
Last no. = 503 + (503 – 1)2
= 503 + 1004 = 1507
The series / list will be [503, 505, 507,………..1507]
Avg. will be (a + l)/2 = (503 + 1507)/2 = 1005
Given condition 6(503) – 1005 = 3018 – 1005 = 2013 satisfies.
Question:
If m ◇ n = m^{2n} then what is the value of
(3 ◇ 4) ◇ 2 =
 3^{32}
 4^{12}
 3^{16}
 4^{6}
Solution:
3^{32}
(3 ◇ 4) ◇ 2 = (3^{8}) ◇ 2 = 3^{32}
Question:
An exam consists of twenty multiple choice questions having three options each. In how many ways a student answer the questions correctly but alternatively?
 2^{11}
 2^{10}
 2^{10}x^{ 20}P_{10}
 None
Solution:
There are two ways in which he can answer the question as required
I 

R 

W 

R 

W 
………… 

(R ® Right) 
II 

W 

R 

W 

R 
………… 

W ® Wrong) 
Now in case I; the 1^{st} question can be answered correctly in one way only and for the 2^{nd} question it can be answered wrong in two ways & so on………..
1 x 2 x 1 x 2 x……… = 1^{10 }x 2^{10} = 2^{10} ways
Same number of ways lies for Case II also.
Therefore total no. of ways = 2^{10} + 2^{10} = 2^{11}.
Question:
The graph of the function f(x) = ax^{2} + bx + c is symmetric around x = p then the value of p is
 P= (b/a)
 P= b/a
 P= b/2a
 P= b/2a
Solution:
Given f(x) = ax^{2} + bx + c
= a [x^{2} + b/a x + c/a]
= a [x^{2} + 2 x b/2a x x + (b/2a)^{2} – (b/2a)^{2} + c/a]
= a [[x + b/2a]^{2} + (4ac – b^{2})/4a^{2}]
This graph is symmetric around x = –b/2a
Therefore, P = –b/2a
Question:
Given a, b are roots of quadratic equation x^{2} + ax + b then the roots of the quadratic equation bx^{2} + ax + 1 would be
 1/α,1/β
 1/ α^{2},1/β^{2}
 a + b, a– b
 None
Solution:
Given quadratic equation x^{2} + ax + b = 0
Then product of roots αβ = b
Sum of roots α+β =q
Next quadratic equation bx^{2} + ax + 1 = 0
Then product of roots αβ = 1/b =1/ αβ
Hence clearly by visualising options
New roots are α = 1/ α and β=1/ β
Question:
Given (3p + 6q + r + 2s) (3p – 6q – r + 2s) = (3p + 6q – r – 2s) (3p – 6q + r – 2s)
then which of the following is true?
 p : r = q : s
 p : q = s : r
 p : r = s : q
 None of these
Solution:
(3p + 6q + r + 2s)/(3p + 6q – r – 2s) = (3p – 6q + r – 2s)/(3p – 6q – r + 2s)
Reversing Componendo & Dividendo.
= (3p + 6q)/(r + 2s) = (3p – 6q)/(r2s)
= (p + 2q)/(p2q) = (r + 2s)/(r – 2s)
Again Reversing Componendo & Dividendo
p/2q = r/2s
= p/r = q/5 = p : r = q : s
Question:
The sum of A.P. of n numbers is n(n + 8) and n > 1 then common difference would be
d = 3
d = 2
d = 1
d = 4
Solution:
Ans. (b)
Given S_{n} = n (n + 8)
Put n = 2 S_{2} = 2 (2 + 8) = 20 = T_{1 }+ T_{2}
Put n = 3 S_{3} = 3 (3 + 8) = 33 = T_{1} + T_{2} + T_{3}
n = 4 S_{4} = 4 (12) = 48 = T_{1} + T_{2} + T_{3} + T_{4}
Therefore, T_{3} = 33 – 20 = 13
T_{4} = 48 – 33 = 15
Hence common difference is 2.
Question:
Solution:
CASE
500 Students participated in an exam for 3 colleges X, Y and Z. 106 students got a call from at least one of the colleges. Students getting a call from only X, students getting a call from only Y and students getting a call from only Z are in the ratio 1:3:4. The number of students getting a call from only X and Z, the number of students getting a call from only X and Y and the number of students getting a call from only from Y and Z are in the ratio 1:2:3. The number of students getting a call from only Y is twice of the number of students getting a call from all the three colleges. Half of the students getting a call from Z got call from at least one more college.
Question:
How many got a call from all the three colleges?
Solution:
12
Let the number of people with offers from all the three colleges by x. Then according to the condition, the number of people with offer from Y only is 2x. Also the ratio of number of people with offers by X only, Y only and Z only is 1:3:4. So let us consider total number of people with offers from all the colleges be 3x and Y only as 6x to make them integers. So the number of people with calls from Z only will be 8x. Let us also take X and Z only, X and Y only and Y and Z only be y, 2y and 3y.
Now according to the given condition
8x = y + 3y + 3x
or 5x = 4y
Now adding all the values of the Venn diagram we get
2x + 6x + 8x + 3x + y + 2y + 3y = 106
19x + 6y = 106
Substituting the value of y we get x = 4
CASE
500 Students participated in an exam for 3 colleges X, Y and Z. 106 students got a call from at least one of the colleges. Students getting a call from only X, students getting a call from only Y and students getting a call from only Z are in the ratio 1:3:4. The number of students getting a call from only X and Z, the number of students getting a call from only X and Y and the number of students getting a call from only from Y and Z are in the ratio 1:2:3. The number of students getting a call from only Y is twice of the number of students getting a call from all the three colleges. Half of the students getting a call from Z got call from at least one more college.
Question:
How many offers were made by the colleges?
Solution:
160
Let the number of people with offers from all the three colleges by x. Then according to the condition, the number of people with offer from Y only is 2x. Also the ratio of number of people with offers by X only, Y only and Z only is 1:3:4. So let us consider total number of people with offers from all the colleges be 3x and Y only as 6x to make them integers. So the number of people with calls from Z only will be 8x. Let us also take X and Z only, X and Y only and Y and Z only be y, 2y and 3y.
Now according to the given condition
8x = y + 3y + 3x
or 5x = 4y
Now adding all the values of the Venn diagram we get
2x + 6x + 8x + 3x + y + 2y + 3y = 106
19x + 6y = 106
Substituting the value of y we get x = 4
CASE
500 Students participated in an exam for 3 colleges X, Y and Z. 106 students got a call from at least one of the colleges. Students getting a call from only X, students getting a call from only Y and students getting a call from only Z are in the ratio 1:3:4. The number of students getting a call from only X and Z, the number of students getting a call from only X and Y and the number of students getting a call from only from Y and Z are in the ratio 1:2:3. The number of students getting a call from only Y is twice of the number of students getting a call from all the three colleges. Half of the students getting a call from Z got call from at least one more college.
Question:
Based on above case
Solution:
Let the number of people with offers from all the three colleges by x. Then according to the condition, the number of people with offer from Y only is 2x. Also the ratio of number of people with offers by X only, Y only and Z only is 1:3:4. So let us consider total number of people with offers from all the colleges be 3x and Y only as 6x to make them integers. So the number of people with calls from Z only will be 8x. Let us also take X and Z only, X and Y only and Y and Z only be y, 2y and 3y.
Now according to the given condition
8x = y + 3y + 3x
or 5x = 4y
Now adding all the values of the Venn diagram we get
2x + 6x + 8x + 3x + y + 2y + 3y = 106
19x + 6y = 106
Substituting the value of y we get x = 4
CASE
500 Students participated in an exam for 3 colleges X, Y and Z. 106 students got a call from at least one of the colleges. Students getting a call from only X, students getting a call from only Y and students getting a call from only Z are in the ratio 1:3:4. The number of students getting a call from only X and Z, the number of students getting a call from only X and Y and the number of students getting a call from only from Y and Z are in the ratio 1:2:3. The number of students getting a call from only Y is twice of the number of students getting a call from all the three colleges. Half of the students getting a call from Z got call from at least one more college.
Question:
Based on above case
Solution:
Let the number of people with offers from all the three colleges by x. Then according to the condition, the number of people with offer from Y only is 2x. Also the ratio of number of people with offers by X only, Y only and Z only is 1:3:4. So let us consider total number of people with offers from all the colleges be 3x and Y only as 6x to make them integers. So the number of people with calls from Z only will be 8x. Let us also take X and Z only, X and Y only and Y and Z only be y, 2y and 3y.
Now according to the given condition
8x = y + 3y + 3x
or 5x = 4y
Now adding all the values of the Venn diagram we get
2x + 6x + 8x + 3x + y + 2y + 3y = 106
19x + 6y = 106
Substituting the value of y we get x = 4
CASE
Eight teams A, B, C, D, E, F, G and H participate in a tournament. The teams are divided into two groups of four each. Every team of the group plays one match each with all the other teams of that group. Each day two games are played and thus the tournament gets over in 6 days. The result of a match is either a win or draw. If a team wins the match it gets 3 points. In case of a draw 1 point is awarded to both the teams. If a team loses a match, it gets 0 points. Following is the table of points at the end of each day.

A 
B 
C 
D 
E 
F 
G 
H 
Day 1 
3 
1 
0 
0 
0 
0 
0 
1 
Day 2 
3 
4 
1 
0 
0 
1 
0 
0 
Day 3 








Day 4 








Day 5 








Day 6 








Question:
Which 2 teams teams are in the same group?
(a) A and B
(b) B and C
(c) D and E
(d) E and F
Solution:
CASE
Eight teams A, B, C, D, E, F, G and H participate in a tournament. The teams are divided into two groups of four each. Every team of the group plays one match each with all the other teams of that group. Each day two games are played and thus the tournament gets over in 6 days. The result of a match is either a win or draw. If a team wins the match it gets 3 points. In case of a draw 1 point is awarded to both the teams. If a team loses a match, it gets 0 points. Following is the table of points at the end of each day.

A 
B 
C 
D 
E 
F 
G 
H 
Day 1 
3 
1 
0 
0 
0 
0 
0 
1 
Day 2 
3 
4 
1 
0 
0 
1 
0 
0 
Day 3 








Day 4 








Day 5 








Day 6 








Question:
Which of the following statements are true?
(a) C played on Day 3
(b) G played on day 5
(c) …….
(d) ……
Solution:
CASE
Eight teams A, B, C, D, E, F, G and H participate in a tournament. The teams are divided into two groups of four each. Every team of the group plays one match each with all the other teams of that group. Each day two games are played and thus the tournament gets over in 6 days. The result of a match is either a win or draw. If a team wins the match it gets 3 points. In case of a draw 1 point is awarded to both the teams. If a team loses a match, it gets 0 points. Following is the table of points at the end of each day.

A 
B 
C 
D 
E 
F 
G 
H 
Day 1 
3 
1 
0 
0 
0 
0 
0 
1 
Day 2 
3 
4 
1 
0 
0 
1 
0 
0 
Day 3 








Day 4 








Day 5 








Day 6 








Question:
Based on above case
Solution:
CASE
Eight teams A, B, C, D, E, F, G and H participate in a tournament. The teams are divided into two groups of four each. Every team of the group plays one match each with all the other teams of that group. Each day two games are played and thus the tournament gets over in 6 days. The result of a match is either a win or draw. If a team wins the match it gets 3 points. In case of a draw 1 point is awarded to both the teams. If a team loses a match, it gets 0 points. Following is the table of points at the end of each day.

A 
B 
C 
D 
E 
F 
G 
H 
Day 1 
3 
1 
0 
0 
0 
0 
0 
1 
Day 2 
3 
4 
1 
0 
0 
1 
0 
0 
Day 3 








Day 4 








Day 5 








Day 6 








Question:
Based on above case
Solution:
CASE
Following is the table containing Students who appeared and passed in exam of Maths and Physics from three schools A, B and C. The data is written in the form of X/Y where X means number of students who passed the exam and Y represents total number of students who appeared in the exam. There are 1800 students in School A, 1500 students in school B and 1200 students in school C.

Maths 
Physics 
Maths and Physics 


Males 
Females 
Males 
Females 
Males 
Females 
A 
650/1000 
450/800 
700/1000 
600/800 
500/1000 
400/800 
B 






C 






Question:
If proportion of males clearing Maths paper and proportion of males clearing physics paper is X and Y respectively and proportion of females clearing Maths paper and proportion of females clearing physics paper is M and N respectively then which of the following is true
 X > Y ; M > N
 X > Y; M < N
 X < Y; M > N
 X < Y; M < N
Solution:
CASE
Following is the table containing Students who appeared and passed in exam of Maths and Physics from three schools A, B and C. The data is written in the form of X/Y where X means number of students who passed the exam and Y represents total number of students who appeared in the exam. There are 1800 students in School A, 1500 students in school B and 1200 students in school C.

Maths 
Physics 
Maths and Physics 


Males 
Females 
Males 
Females 
Males 
Females 
A 
650/1000 
450/800 
700/1000 
600/800 
500/1000 
400/800 
B 






C 






Question:
Based on above case
Solution:
CASE
Following is the table containing Students who appeared and passed in exam of Maths and Physics from three schools A, B and C. The data is written in the form of X/Y where X means number of students who passed the exam and Y represents total number of students who appeared in the exam. There are 1800 students in School A, 1500 students in school B and 1200 students in school C.

Maths 
Physics 
Maths and Physics 


Males 
Females 
Males 
Females 
Males 
Females 
A 
650/1000 
450/800 
700/1000 
600/800 
500/1000 
400/800 
B 






C 






Question:
Based on above case
Solution:
CASE
Following is the table containing Students who appeared and passed in exam of Maths and Physics from three schools A, B and C. The data is written in the form of X/Y where X means number of students who passed the exam and Y represents total number of students who appeared in the exam. There are 1800 students in School A, 1500 students in school B and 1200 students in school C.

Maths 
Physics 
Maths and Physics 


Males 
Females 
Males 
Females 
Males 
Females 
A 
650/1000 
450/800 
700/1000 
600/800 
500/1000 
400/800 
B 






C 






Question:
Based on above case
Solution:
CASE
There are four judges J1, J2, J3, J4 who heard 200, 100, 150 and 250 cases. Out of the cases they heard the verdict were as follows:
P1 
The percentage of people who were sentences Guilty 
P2 
The percentage of People out of Guilty, where the dependent challenged the decision in SC 
P3 
The percentage of People out of Not Guilty, where the Prosecution challenged the decision in SC 
P4 
The percentage of cases out of challenged by defendant and overturned by SC 
P5 
The percentage of cases out of challenged by Prosecution and overturned by SC 
Question:
Out of the judgments consider by J4, how many were finally convicted guilty?
Solution:
CASE
There are four judges J1, J2, J3, J4 who heard 200, 100, 150 and 250 cases. Out of the cases they heard the verdict were as follows:
P1 
The percentage of people who were sentences Guilty 
P2 
The percentage of People out of Guilty, where the dependent challenged the decision in SC 
P3 
The percentage of People out of Not Guilty, where the Prosecution challenged the decision in SC 
P4 
The percentage of cases out of challenged by defendant and overturned by SC 
P5 
The percentage of cases out of challenged by Prosecution and overturned by SC 
Question:
Among all the cases how many cases were overturned by the Supreme Court?
Solution:
CASE
There are four judges J1, J2, J3, J4 who heard 200, 100, 150 and 250 cases. Out of the cases they heard the verdict were as follows:
P1 
The percentage of people who were sentences Guilty 
P2 
The percentage of People out of Guilty, where the dependent challenged the decision in SC 
P3 
The percentage of People out of Not Guilty, where the Prosecution challenged the decision in SC 
P4 
The percentage of cases out of challenged by defendant and overturned by SC 
P5 
The percentage of cases out of challenged by Prosecution and overturned by SC 
Question:
Based on above case
Solution:
CASE
There are four judges J1, J2, J3, J4 who heard 200, 100, 150 and 250 cases. Out of the cases they heard the verdict were as follows:
P1 
The percentage of people who were sentences Guilty 
P2 
The percentage of People out of Guilty, where the dependent challenged the decision in SC 
P3 
The percentage of People out of Not Guilty, where the Prosecution challenged the decision in SC 
P4 
The percentage of cases out of challenged by defendant and overturned by SC 
P5 
The percentage of cases out of challenged by Prosecution and overturned by SC 
Question:
Based on above case
Solution:
CASE
Passage1
Frugal Engineering (Business Today)
To get a handle on what frugal engineering is, it helps to understand what it is not. Frugal engineering is not simply lowcost engineering. It is not a scheme to boost profi t margins by squeezing the marrow out of suppliers' bones. It is not simply the latest take on the decadeslong focus on cost cutting.
Instead, frugal engineering is an overarching philosophy that enables a true "clean sheet" approach to product development. Cost discipline is an intrinsic part of the process, but rather than simply cutting existing costs, frugal engineering seeks to avoid needless costs in the fi rst place. It recognises that merely removing features from existing products to sell them cheaper in emerging markets is a losing game. That's because emerging market customers have unique needs that usually aren't addressed by maturemarket products, and because the cost base of developed world products, even when stripped down, remains too high.
Maturemarket customers continue to accept price premiums for new features, leading companies to overengineer their product lines – at least from the point of view of emergingmarket customers. The virtual extinction of manual car windows in the United States is just one example. Frugal engineering, by contrast, addresses the billions of consumers at the bottom of the pyramid who are quickly moving out of poverty in China, India, Brazil, and other emerging nations. They are enjoying their fi rst taste of modern prosperity, and are shopping for the basics, not for fancy features. According to the late C.K. Prahalad, author of The Fortune at the Bottom of the Pyramid, these potential customers, "unserved or underserved by the large organised private sector, including multinational fi rms," total 4 to 5 billion of the 6.7 billion people on Earth. Although the purchasing power of any of these new consumers as an individual is only a fraction of a consumer's purchasing power in mature markets, in aggregate they represent a market nearly as large as that of the developed world.
However, many may not fully grasp the challenges that competition in emerging markets entails. The prospect of highvolume profi t streams may be enticing, but those profi ts must be earned in the face of lower prices, lower perunit profi ts, and stringent cost targets.
In addition, too few companies realise how demanding emerging market customers can be. They don't spend easily, because they don't have much to spend. They require a different set of product features and functions than their developedworld counterparts, but still insist on high quality. Global companies, therefore, must change the way they think about product design and engineering. Simply selling the cheapest products on hand or reusing technologies from higher priced products will not cut costs enough and is unlikely to result in the kind of products these new customers will buy.
Question:
Direct question based on above passage
Solution:
Question:
Direct question based on above passage
Solution:
Question:
Inference based question based on above passage
Solution:
Question:
Inference based question based on above passage
Solution:
CASE
Star Party
A star party is a gathering of amateur astronomers to observe the night sky. Such parties may last for a night or extend over a week. The larger star parties can attract hundreds or even thousands of participants.
Regional star parties are now held annually and form a part of amateur astronomy. It is important to choose a proper location for a star party. Ideally, the location should offer a dark sky unpolluted with city lights.
The sheer thrill of meeting likeminded enthusiasts with varieties of telescopes and binoculars offering spectacular views of constellations, stars, nebulae, planets, comets and deep sky objects makes a star party worth attending. It is common to organize popular lectures, contests, sightseeing tours, raffles and exhibitions of homebuilt telescopes at some of the larger star parties. Such events also attract commercial vendors who use the opportunity to meet potential customers. The ambience is that of fun, camaraderie and learning, all under the magnificent canopy of the enchanting night sky.
Question:
Inference
Solution:
CASE
Obesity
Environmental causes of obesity cross community, national, and international lines. Key drivers of the epidemic include technology advances that have cut the cost of processed foods and drinks, in turn, leading to larger portion sizes; relentless marketing of highcalorie, lownutrient foods and drinks, especially to children; and a dramatic rise in the prices of more nutritious foods, such as fruits, vegetables, lean meats, and lowfat dairy products. Other factors include less time for gym classes and recess at school; limited access to supermarkets in lowincome areas; and a built environment that encourages driving rather than walking or biking. Each of these factors is a target for policy change, and those changes can take many forms and work at many levels.
Question:
Inference
Solution:
CASE
Effect of fish diet on brain structure (Medical Journal)
Eating fish has long been linked with heart health. Now new research adds to growing evidence that fish is good for the brain as well. Researchers at the University of Pittsburgh found that people who ate baked or broiled fish just once a week had a lower risk of developing Alzheimer’s disease. They were also at lower risk of mild cognitive impairment, a type of memory loss that sometimes leads to Alzheimer’s.
The fish eaters had more brain gray matter, as measured by M.R.I., or magnetic resonance imaging, brain scans, than those who didn’t regularly eat fish. Greater brain volume may indicate intact memory and thinking functions, whereas brain shrinkage has been linked to Alzheimer’s disease and other forms of dementia. “If you eat fish just once a week, your hippocampus—the big memory and learning center—is 14% larger than in people who don’t eat fish that frequently. 14%. That has implications for reducing Alzheimer’s risk,” Raji said. “If you have a stronger hippocampus, your risk of Alzheimer’s is going to go down.”
Question:
Para jumble
Solution:
CASE
Germ War
During World War II, Britain prepared anthrax animal food cakes to infect German cattle. They also stockpiled vaccines should the Germans strike first. The Scottish research isle of Gruinard was rendered unsafe as a result of escaped anthrax spores until 1987.
Question:
Para jumble
Solution:
CASE
Mental Health
Those who worry about their mental health remain disturbed as they find themselves different from other but as they can feel the fear of worry they have normal mental state. It is only the consciousness of mental state that worries them
Question:
Para jumble (odd one out)
Solution:
CASE
Steps of Scuba Diving
Question:
Para jumble
Solution:
CASE
Negotiation is an Art
Many people believe that negotiations are "all or nothing", and that there has to be one winner and one loser. Nothing could be farther from the truth. While the goal of negotiation is most certainly getting what you want, the fact is that the best deals (the ones that stick) incorporate terms and ideas from both parties.
Question:
Para jumble (odd one out)
Solution:
CASE
What is Theory
The word “theory” implies that something is unproven or speculative. As used in science, however, a theory is an explanation or model based on observation, experimentation, and reasoning, especially one that has been tested and confirmed as a general principle helping to explain and predict natural phenomena.
Question:
Para jumble (odd one out)
Solution:
CASE
plainchant (also called "Gregorian" chant), the vocal religious practice of the Roman Catholic Church. Plainchant was transmitted by memory until the early 9th century, when the Holy Roman Emperor Charlemagne arranged for it to be notated, and for standardized plainchant books to be distributed to churches and monasteries across Europe. Limited in pitch range and monophonic (i.e., composed of a single melody with no accompaniment), plainchant was sung largely by monks, nuns, and clerics rather than by professional singers. Plainchant was sung in the Divine Offices, eight daily prayer services using Old Testament texts, and in the Mass, a midmorning celebration of the life and death of Jesus Christ. The Alleluia reproduced here was a chant of jubilation ("Alleluia" = "Hallelujah"), sung as part of the Mass.
Passage 2 Evolution of Technology.
A lengthy passage with detailed example of Guntenberg Printing Press
Question:
Inferential based question based on above passage
Solution:
Question:
Inferential based question based on above passage
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Question:
Inferential based question based on above passage
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Question:
Inferential based question based on above passage
Solution:
Case
Passage 3Colonial rule in India has undermined the intellectual selfconfidence among the people of India.
The passage discusses the impact of colonial rule on the minds and behavior of Indians while gives examples of Aryabhatta and other ancient mathematicians.
Question:
Inferential based question based on above passage
Solution:
Question:
Inferential based question based on above passage
Solution:
Question:
Inferential based question based on above passage
Solution:
Question:
Strengthen the given idea from the passage
Solution:
Case
Passage 4 Post independence Kurdistan and its distinction from Iraq in terms of culture, policy etc.
Question:
Direct question based on above passage
Solution:
Question:
Inferential based question based on above passage
Solution:
Question:
Inferential based question based on above passage
Solution:
Question:
Application based question based on above passage
Solution:
Case
Effective Branding
Successful branding is not only about the product but also the way it associates to the customer. To become a successful brand one needs to study the trend and follow the role models that customers follow and project the product used by their role models. The customer gets associated to a brand when he receives a sense of satisfaction that the purchase is made was the right choice.
Question:
What can be summarized from the above paragraph? (independent.co.uk)
Solution:
Case
Some great authors have published their worst works from beyond the grave. A few though, keep getting better when they’re dead, such as the Chilean novelist and short story writer, Roberto Bolaño. His seminal fivepart novel, 2666, came out posthumously, won the National Book Critics Circle Award and convinced the world he was not just a master of the short form but could put out his life’s best work at nearly 900 pages, even after death.
Question:
Which of the following can best summarize the above?
Solution:
Question:
Inference A simple paragraph on change in global climate.
Solution:
Question:
Summary A moderate paragraph on Guayaki Indians and studies done by Vellard.
Solution:
Question:
ErrorIdentification
Solution:
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ErrorIdentification
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ErrorIdentification
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Solution:
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Solution:
CASE
138 fruits were to be distributed among four people A, B, C and D. The fruit distributed are Apple, Oranges, Plums and Guava. There were 12 apples to be distributed.
There are four students out of which there are 2 females.
Every student gets at least one of each variety of fruit and no students can have more than 4 of any variety. D got least and it was 5 less than the highest.
A got 4 each of Apple and Guava but he didn’t get the highest. B and C got the same number of fruits which was also one of the females. B got same number of apple, Oranges and Plum while C ha dsame number of Oranges and Guava.
Question:
Which of the following is definitely true?
Solution:
Question:
What could be the total number of Oranges?
Solution:
Question:
A question based on above case
Solution:
Question:
A question based on above case
Solution:
CASE
A mobile application has been created which has four consecutive buttons which can have either of the two status on or off. By pressing a button its status changes from on to off and vice versa.
Question:
If the initial arrangement is (on,off,on,off) then how many different combinations are possible by pressing 2 consecutive buttons?
(a) 2
(b) 4
(c) 8
(d) 16
Solution:
Question:
Which of the following will give all the possible combinations of the switches?
Solution:
Question:
If the application is modified such that you can press any two buttons (not necessarily consecutive buttons) then how many total combinations are possible?
(a) 2
(b) 4
(c) 8
(d) 16
Solution:
Question:
There was one more similar question……….
Solution:
Case
A, B, C, D, and E are five people who take exactly one of the services from 5 service Provider Sata Sky, Sairtel, Tish, Audicone, Citi.
In the first year
Exactly one of the service provider had none of the subscription.
Tish had 2 of this people as subscriber.
E is the only subscriber of his service provider.
In the second year
Audiocone retain its only subscriber and got one more subscriber from other subscriber.
A changed from its earlier subscriber to Sairtel.
Citi increased its subscriber by 1 from 1^{st} year.
Sata sky retails its only subscriber.
Question:
In the first year which of the following had no subscriber?
Solution:
Question:
In the second year which of the following had subscription of Sata sky?
Solution:
Question:
A question based on above case
Solution:
Question:
A question based on above case
Solution:
Case
In a school different students take part in activities like Academics, Sports and Culture. They were given the codes 1 or 0. For example 101 will give 1 in academics, 0 in sports and 1 in culture. By the strange coincidence, the number of boys which is abc is same as number of girls bcd. Here abc and bcd are binary numbers.
Boys 


Sports 

Culture 
0 
1 
1 

60 



Girls 


Culture 

Acad 
0 
1 
0 


1 
50 
70 
Question:
How many boys are there in Academics?
Solution:
Question:
A question based on above case
Solution:
Question:
A question based on above case
Solution: