Analysis

Students must be a relieved lot today. CAT-14 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 “razor-thick 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 27-30 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 60-65 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 90-95% 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 17th 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.

 


Quantitative Ability & Data Interpretation
Question 1
Question:

What is the remainder when

is divided by 15
 

                (a)   3

                (b)   5

                (c)   6

                (d)   9


Solution:

Ans. 3

Remainiders will be 0 from 5! onwards. hence the required answer will be   

( 1 + 2 + 6 + 24  ……………..) / 15

                = 33/15 = 3


Question 2
Question:

let N=1 × 2 × 3 × 4 …………………… 20/6n

                        least no of n for which N is not an integer?

 

                 (a)   6

                (b)   9

                (c)   8

                (d)   2


Solution:

Highest power of 3 in 20!  is 8

Hence n = 9


Question 3
Question:

A man sell his goods at a discount of 48% & still marks a profit of 20%. What would be his profit percentage
if he offer a discount of 25% instead.


Solution:

40% discount mean

 

Marked

Selling Price

20% Profit Mean

5

3

Cost Price

Marked Price

Selling Price

 

5

3

5

 

6

5

10

6

 

If he offer 25% discount.

 

Cost Price

Marked Price

Selling Price

5

10

 

 

4

3

10

 

15 ® 5

50% Profit.

 


Question 4
Question:

Two container A & B contain oil. 60% of content of container A was transferred to container B. Then 50%

content of container B was transferred to A. the quantity of oil in contain A to that of container B is now 11 : 7.


Solution:

Let oil in containers .be A & B

After Ist operation

Container A = 0.4 A Container B = 0.6 A + B

After IInd  operation

Container A = 0.4A + 0.3A + 0.5B Container B = 0.3A + 0.5B

(0.7A + 0.5B )/ 11 = (0.3A + 0.5B )/ 7

1.6A = 2B = A / 5 = B / 4

= vol. of A : B = 5 : 4. 


Question 5
Question:

A can plough field in 3 hours, B can plough the some field in 4 hours. A, B alone with C can plough the same field in just 1 hours. In what time will ‘C’ along do it.

(a)  2 hours 24 min.

(b)  1 hours 20 min.

(c)  3 hours

(d)  1 hours 24 min.


Solution:

Assume work to be 12 units

 

A

B

A + B + C

C = (A + B + C) – (A + B)

Rate

4

3

12

5

Time

3

4

1

12/5

Work

12

12

12

12

 

Time needed = 12/5 = 2.4 hrs

0.4 hrs. = 0.4 x 60 = 24 min

Ans. 2 hrs. 24 min.

 


Question 6
Question:

A, B & C started travelling from the same point x in the same direction. A & B started at 10:00 am. Then C started at 12 : 00 pm.
C overtook A at 2 : 00 pm and then doubled his speed. C overtook B at 3 : 00 pm.
The ratio of speech of A & B is.


Solution:

Let speeds of A, B & C be A, B & C km/hr respectively.

Thus 4A = = 2C

= A / C = 1 / 2  ––––––– (1)

= 5B = Ac = B / C = 4 / 5 –––––– (2)

 

 

From (1) & (2)

A/C / B/C = 1/2 / 4/5

= A/B = 5 / 8

 


Question 7
Question:

Which of the following cannot represent the sides of a valid triangle.

 

(a)  2 – 1            2 – 3                     3 – 2

(b)  10                   38                         2

(c)  3                      4                              6

(d)  n2                    n2 + 1                     n2 + n


Solution:

10                           38                         2

 

Solution:             Answer is B. Simply check if sum of any two is less than the third will be false only for B.

                For 4th option simply check by talking n = 3.


Question 8
Question:

15th August 2014 (Independence day) was Friday. What day of the week was 15th August 2010?

  1. Monday
  2. Sunday
  3. Saturday
  4. Friday


Solution:

Since, we have to final the day of week for a date which is exactly 4 years earlier than the given day, Friday, also 2012 is or leap year. Thus, we have 5 odds in total to go back from Friday. Thus, the answer is Sunday.


Question 9
Question:

Two poles having heights of 6m & 10m are standing at a distance of 7 m. Then, through straight lines, the top of each pole is joined to the bottom of other pole. What is the height of the point of intersection of those lines?

Solution:

Directly, formula can be applied.

i.e.

a x b / a + b = 10 x 6 / 10 + 6 = 60 / 16 = 15 / 4 m.


Question 10
Question:

Three points A, B, C on a circle are such that angle ABC = 115°. If 0 is the center of the circle. What is the measure of angle OAC?

(a)  25°

(b)  20°

(c)  35°

(d)  55°


Solution:

 

angle ABC = 115

                Therefore  angle AXC = 180 – 115 = 65 (opposite angle of cyclic quadrilateral)

Thus angle AOC = 2 x 65 = 130°

(chord subtends an angle at the centre which is double of what it subtends on the circumference)

Triangle AOC is in isosceles triangle as AO = OC

Therefore angle OAC = angle OCA

Also angle OAC + angle OCA = 180 – angle AOC = 180 – 130 = 50

= 2 x angle OAC = 50   

Angle OAC = 25°


Question 11
Question:

Sandhya Nisha own land in the ratio 2 : 3. The total land was divided between Rice wheat in the ratio for contravention. Sandhya’s land was distributed between Rice & Wheat is 2 : 5. Find the ratio of rice and wheat on nisha’s land

Solution:

Total land owned by S & M is equal to that distributed between wheat & Rice cultivation this

Now S’5 land was distributed bet rice & wheat in the ratio 3 : 7


Question 12
Question:

The sides of an equilateral triangle are given as 3x – 7y + 1, x + 3y + 1, 2x – y. what is the

perimeter of the triangle

  1. 9
  2. 15
  3. 18
  4. 27


Solution:

Since, it is an equilateral triangle, all sides shall be equal.

i.e. 3x – 7y + 1 = x + 3y + 1 = 2x – y

By solving, x = 5 and y = 1.

Each side is 9. Thus, perimeter is 27.


Question 13
Question:

In a graduation party, every student exchanged his photograph with every other student. A total

                of 1560 photographs were exchanged. How many students were there in the party

  1. 30
  2. 40
  3. 50
  4. 60


Solution:

since every pair of student has exchanged 2 photographs. Thus, total number of pairs (combinations) should be 1560 / 2 = 780.

Thus nC2 = 780

n(n – 1) / 2 = 780

Thus, n = 40 students.


Question 14
Question:

The surface of water of a swimming pool is a rectangle of area 26 x 10 m.

The depth of the swimming varies uniformly with the shallow end at 1.60 m. and the deep end 4.4 m.

What is the volume of water that the swimming pool will hold?


Solution:

The vertical cross section along the length is a trapezium. Thus volume will be equal to Area of trapezium x width

Area of trapezium = 1/2 (0.6 + 4.4) x 26 = 6 x 13 = 78 sq.m.

Volume = Area of trapezium x 10

= 78 x 10

= 780 m3


Question 15
Question:

Sum of ages of A, B & C is 21.

Five year hence Average of A & C’s age B is 1.5 years more than B’s age

What is age of B 10 years hence?


Solution:

A + B + C = 21 ––––––––– (1)

(A + 5) + (C + 5) / 2 = (B + 5) + 1.5

= A + C + 10 = 2 B + 13

A – 2 B + C = 3 –––––––––(2)

Subtracting (2) from (1)

3B = 18

B = 6

Therefore 10 years hence B’s age = 6 + 10 = 16


Question 16
Question:

Sum of first n terms of GP is 2n.

Sum of first 2n terms of the same GP is n

What will be the sum of first 3n terms of the G.P.


Solution:

If sum of first n terms of GP is 2n

& sum of first 2n terms of GP is n

Then, sum of second n terms shall be (n – 2n) = – n.

This shall be true for n = 1 also

Let the 3 terms of GP be a, ar and ar2

Then, a = 2n

ar = – n

Thus, r = – 1/2

So, ar2 = n / 2

The sum of 3n terms or a + ar + ar2 shall be    

2n + –n + n / 2 = 3n / 2 Ans.


Question 17
Question:

Five different books are to be distributed among A, B & C such that A & B each gets at least one book. In how many ways can this be done.


Solution:

Following cases can be made for the given distribution.

 

 

A

B

C

Number of ways

1

1

3

5C1 x 4C1 = 20

1

2

2

5C1 x 4C2 = 30

1

3

1

5C1 x 4C3­ = 20

1

4

0

5C1 x 4C4 = 5

2

1

2

5C2 x 3C1 = 30

2

2

1

5C2 x 3C2 = 30

2

3

0

5C2 x 3C3 = 10

3

1

1

5C3 x 2C1 = 20

3

2

0

5C3 x 2C2 = 10

4

1

0

5C4 x 1C1 = 5

 

 

 

Total ways = 180

 


Question 18
Question:

Given that xi + xi+2 = 2.xi+1

and x1 = 1 and x2 = 3

Find x20


Solution:

x1 = 1 x2 = 3

xi­ + xi + 2 = 2. xi + 1

therefore x1 + x3 = 2. x2

= 1 + x3 = 2.3

= x3 = 5

Similarly x2­ + x4 = 2.x3

= 3 + x4 = 2.5

= x4 = 7

Thus x1, x2, x3, x4, x5……………… from the series of odd term 1, 3, 5, 7, 9………………..

Therefore x20 = 39.


Question 19
Question:

let A, B & C be angles of A triangle. if angle ACB is acute. and

3 Sin A + 4 Cos B = 6 & 4 Cos A  + 3 Sin B = 1

Find angle C.


Solution:

Square both equations

9 sin2 A + 16 Cos2 B + 24. Sin + cos B = 36

16 sin2 A + 9 cos2 B + 24 cos A sin B = 1

Adding both the equation

25 (Sin2 A + Cos2 B) + 2 triangle (Sin A. Cos B + Cos A – Sin B) = 37

25 + 24. Sin (A + B) = 37

Sin (A + B) = 12/24 = 1/2

A + B = 30° or 150

Because C is acute A + B = 150 & hence angle C = 30°.


Question 20
Question:

These are six numbers in A.P. There sum is 3.

The first terms is four times the third term. Then the value of fifth term is.


Solution:

Let the term be

a – 5d a – 3d a – d a + d a + 3d a + 5d

Sum 6a = 3 Þ ½

(a – 5d) = 4 (a – d)

– d = 3a

d = – 3a = -3 / 2

\ fifth term = a + 3d

= 1/2 + 3 (–3/2) = 1/2 – 9/2 = –8/2 = – 4.


Question 21
Question:

If Xi + Xi+1 = k and X10 = 1

Find X91


Solution:

We know X10 = 1

Therefore X10 + X11 = K

Þ 1 + X11 = k

X11 = K – 1

Now X11 + X12 = K

(k – 1) + X12 = K

(K – 1) + X12 = K

Therefore X12 = 1

and X12 + X13 = K

X13 = K –1

We understand that the term will alternate between 1 & k – 1

Even term will be 1 & odd will be k – 1

Therefore  X91 = K – 1


Question 22
Question:

Two number are in the ratio 7 : 8. If there sum is 300, then find their lowest common multiple (LCM)

Solution:

The number are 140 & 160.

Their L.C.M. is 1120.


Question 23
Question:

There are 10 people standing in a queue. Except the first & last, every other person’s age is between the ages of persons standing in front of him and standing behind him. The third person is younger than the eighth person. Who will be the youngest person in the queue?

Solution:

As per the data given, the people are standing in increasing order of their age. Thus, the youngest person will be the first person is the queue.


Question 24
Question:

Given that x2 + bx – c = 0. If u and v are the roots of the given equation and absolute value of u

                exceeds that of v. then, which of the following is true?

  1. u > 0 and u < 0
  2. u > 0 and u > 0
  3. u < 0 and v < 0
  4. u < 0 and v > 0


Solution:

Eg. x3 + 3x  – 4 = 0

(x + 4) (x – 1) = 0

X = – 4 or x = 1.

hence U<0 and v>0


Question 25
Question:
Solution:


Question 26
Question:
Solution:


Question 27
Question:
Solution:


Question 28
Question:

In a hospital, temporary staff of 500 people was recruited in which 300 were employed in nursing department while 200 were employed in cleaning department. The average salary of each temporary employee is Rs. 300. If each employee in nursing department earns Rs. 50 more than that in cleaning department. Then, what is the salary of each employee in nursing department?

Solution:

300x + 200(x – 50) / 500 = 300

3x + 2 (x – 50) / 5 = 300

x = 320


Question 29
Question:

Diagonals of a rhombus are in the ratio 4 : 5 & its area is 12 sq. cm

If the side of the square is equal to the longer of the diagonals. What is the area of such a square.


Solution:

Let diagonals be 4x & 5x

Area of Rhombus = 1/2 x 4x x 5x = 10x2 = 12

x2 = 6/5 x = 6 / 5

\ Longer diagonal = 5 x 6 / 5

Area of square where side 5 x 6/5 = (5 x 6/5)2

= 25 x 6/5 = 30 sq.cm.


Question 30
Question:

ABC is a triangle. D is a point on AB and E is point on CD. If area of triangles ADE, DBE & AEC are respectively 4, 5 & 6

Area of BEC will be


Solution:

ADE & ADC have the same base AD &

\ Their heights will be in the ratio 4 : 10 or 2 : 5 As ABDE & BDC also have same base & their heights are also in the ratio 2 : 5 thus their areas will also be in the ratio 2 : 5

Thus area of ABC = 5/2 x 5 = 12.5

BEC = 12.5 –5 = 7.5.


Question 31
Question:

what is the number of real roots for the equation x6 + x3 – 2 = 0


Solution:

x6 + x3 – 2 = 0 can be written as

(x3 + 2) (x3 – 1) = 0

Thus x3 = – 2 (are real root)

or x3 = 1 (are real root)

so, total number of real roots are 2.


Question 32
Question:

(x + y + z) (xy + yz + zx) – xyz

 

  1. (z – x) (x – y) (y – z)
  2. (x + y) (y + z) (z + x)
  3. (x + y + z)2
  4. (x + y +z)3


Solution:

By putting any values of x, y and z in the options, we can get the right answer as (x + y) (y + z) (z + x)


Question 33
Question:

what is the perpendicular distance between two parallel lines given as x +y = 4 and x + y = – 2

Solution:

d = |c1 – c2 / a2 + b2|

= 4 – (–2) / 12 + 12

6 / 2  = 32


Question 34
Question:

How many distinct real values of x satisfy 2|x| ≤ 1.97x?

  1. 0
  2. 1
  3. 2
  4. Infinte


Solution:

Ans(b)

CASE 1 – If  x > 0, |x|  = x

Hence, the given equation can be written as 2x ≤ 1.97x or .03x ≤ 0 or x ≤ 0. But, this is not true as  x > 0.

CASE 2 – If  x < 0, |x|  = -x

Hence, the given equation can be written as -2x ≤ 1.97x or -3.97x ≤ 0 or x ≥ 0. But, this is not true as  x<0 .

CASE 3 – If x = 0, both the sides of the given inequality are zero. Hence, x = 0 is only one solution.


Question 35

CASE

The following line graph shows the production(in thousand units ) of two types of goods P1 and P2 by four companies W,X,Y, and Z in two cities Rajmahal  and Kanpur.

The larger circle/square represents the units sold in Rajmahal while the smaller circle/square represents the units sold in Kanpur.

Selling price of P1 = Rs. 40/unit & that of P2 = Rs. 30/ unit.

Assume that all the units produced by the companies is sold.

Question:

For which company, was the total production of P2, the highest?

                (a)  w

                (b)  x

                (c)  y

                (d)  z


Solution:

From the given graph, we can make the following table;

               

 

 

x

y

z

w

Rajmahal

P1

6

3.5

8

3

P2

6

4

5

2

Kanpur

P1

7

2.5

1

3

P2

2

7

3

1

Total

P1

13

6

9

6

Total

P2

8

11

8

3

 

From the table, we can see that the total production of P2 was the highest for company y.  Hence answer is (c).

              


Question 36

CASE

The following line graph shows the production(in thousand units ) of two types of goods P1 and P2 by four companies W,X,Y, and Z in two cities Rajmahal  and Kanpur.

The larger circle/square represents the units sold in Rajmahal while the smaller circle/square represents the units sold in Kanpur.

Selling price of P1 = Rs. 40/unit & that of P2 = Rs. 30/ unit.

Assume that all the units produced by the companies is sold.

Question:

Which company has the highest revenue from the sale of P2 in Rajmahal?

                (a)  w

                (b)  x

                (c)  y

                (d)  z


Solution:

We can construct the following table from the given data for Rajmahal

 

x

y

z

W

Units of P2

6

4

5

2

Revenue fom P2

180

120

150

60

                Clearly, x has the highest revenue. Hence answer is (b).


Question 37

CASE

The following line graph shows the production(in thousand units ) of two types of goods P1 and P2 by four companies W,X,Y, and Z in two cities Rajmahal  and Kanpur.

The larger circle/square represents the units sold in Rajmahal while the smaller circle/square represents the units sold in Kanpur.

Selling price of P1 = Rs. 40/unit & that of P2 = Rs. 30/ unit.

Assume that all the units produced by the companies is sold.

Question:

Which company has the least income by sale of both P1 & P2 in Kanpur?

                (a)  w

                (b)  x

                (c)  y

                (d)  z


Solution:

For the sale in Kanpur, we can construct the following table;

               

 

X

Y

Z

y

Units of P1

7

2.5

1

3

Revenue from P1

280

100

40

120

Units of P2

2

7

3

1

Revenue from P2

60

210

90

30

Total Revenue

340

310

130

150

                So, the least income is for company z.  Answer (d)


Question 38

CASE

The following line graph shows the production(in thousand units ) of two types of goods P1 and P2 by four companies W,X,Y, and Z in two cities Rajmahal  and Kanpur.

The larger circle/square represents the units sold in Rajmahal while the smaller circle/square represents the units sold in Kanpur.

Selling price of P1 = Rs. 40/unit & that of P2 = Rs. 30/ unit.

Assume that all the units produced by the companies is sold.

Question:

If later on, company w merges in company x and company y merges in company z, then which of the following statements is/are true?

                Statement 1       :               Total units of P1 for wx is greater then that for yz

                Statement 2       :               Total units of P2 for yz is greater than that for wx.

 

                (a)  only 1

                (b)  only 2

                (c)  both 1 and 2

                (d)  neither 1 nor 2


Solution:

               

 

 

wx

yz

Rajmahal

P1

9

11.5

P2

8

9

Kanpur

P1

10

3.5

P2

3

10

Total  P1

19

15

Total  P2

11

19

Hence, from the table, both statements 1 & 2 are true. Hence answer is (c).


Question 39

CASE

In a group, there are 18 men and 18 women who like at least one and atmost two of the three like at least one and atmost two of the three activities among reading books, watching movines and playing sports.

 

  • There are total 12 persons in the group who like exactly two of the three activates.
  • There are twice as many women as men who like both Reading books and watching moves.
  • There is no one who like reading books and playing sports.
  • The ratio of men to women of people who are reading books and watching movies is same as the ratio of mean to women who are watching movies and playing sports.
  • The number of people who are reading books and watching movies is same as the number people who are watching moves and playing sports.
  • The ratio of men and women who are only reading books is 2 : 1 and the number of men and women who are only playing
  • Total number of people who are only reading books is same as the number of  people who are only reading books is same as the number of people who are only watching movies.
Question:

What is the ratio of number of men and women like exactly one of the three activities?

                                (a)  7 : 5

                                (b)  6 : 5

                                (c)  8 : 7

                                (d)  11 : 10


Solution:

From the given information, we can draw the following diagrams

Number of men liking exactly one activity = 6 + 3 + 5 = 14

Number of women liking exactly are activity = 3 + 6 + 1 = 10

Therefore required ratio = 14 : 10 = 7 : 5 Hence answer is (a).


Question 40

CASE

In a group, there are 18 men and 18 women who like at least one and atmost two of the three like at least one and atmost two of the three activities among reading books, watching movines and playing sports.

 

  • There are total 12 persons in the group who like exactly two of the three activates.
  • There are twice as many women as men who like both Reading books and watching moves.
  • There is no one who like reading books and playing sports.
  • The ratio of men to women of people who are reading books and watching movies is same as the ratio of mean to women who are watching movies and playing sports.
  • The number of people who are reading books and watching movies is same as the number people who are watching moves and playing sports.
  • The ratio of men and women who are only reading books is 2 : 1 and the number of men and women who are only playing
  • Total number of people who are only reading books is same as the number of  people who are only reading books is same as the number of people who are only watching movies.
Question:

what is the ratio of number of women reading books to the number of men watching movies?

  1. 2 : 1
  2. 1 : 2
  3. 1 : 1
  4. 3 : 2


Solution:

Number of women reading books = 3 + 4 = 7

Number of men watching movies = 3 + 2 + 2 = 7

The required ratio = 1 : 1. Ans.(3)


Question 41

CASE

In a group, there are 18 men and 18 women who like at least one and atmost two of the three like at least one and atmost two of the three activities among reading books, watching movines and playing sports.

 

  • There are total 12 persons in the group who like exactly two of the three activates.
  • There are twice as many women as men who like both Reading books and watching moves.
  • There is no one who like reading books and playing sports.
  • The ratio of men to women of people who are reading books and watching movies is same as the ratio of mean to women who are watching movies and playing sports.
  • The number of people who are reading books and watching movies is same as the number people who are watching moves and playing sports.
  • The ratio of men and women who are only reading books is 2 : 1 and the number of men and women who are only playing
  • Total number of people who are only reading books is same as the number of  people who are only reading books is same as the number of people who are only watching movies.
Question:

what proportion of women only watch movies?               

  1. 1 : 4
  2. 1 : 5
  3. 2 : 3
  4. 1 : 3


Solution:

Number of women only watch movies = 6 out of total 18

therefore, the required ratio is 6 : 18 = 1 : 3 Ans (4).


Question 42

CASE

In a group, there are 18 men and 18 women who like at least one and atmost two of the three like at least one and atmost two of the three activities among reading books, watching movines and playing sports.

 

  • There are total 12 persons in the group who like exactly two of the three activates.
  • There are twice as many women as men who like both Reading books and watching moves.
  • There is no one who like reading books and playing sports.
  • The ratio of men to women of people who are reading books and watching movies is same as the ratio of mean to women who are watching movies and playing sports.
  • The number of people who are reading books and watching movies is same as the number people who are watching moves and playing sports.
  • The ratio of men and women who are only reading books is 2 : 1 and the number of men and women who are only playing
  • Total number of people who are only reading books is same as the number of  people who are only reading books is same as the number of people who are only watching movies.
Question:

women who are only reading books persons what proportion of the persons watching movies?               

  1. 1 : 7
  2. 2 : 7
  3. 3 : 4
  4. 5 : 4


Solution:

The required ratio = 3 : 21 = 1 : 7


Question 43

CASE

The following tables give the data about the placement records of the students by two colleges. e.g. in college A, in mechanical; 30 students got placement offer of 3 lakh.

 

Both the colleges provide 100% placements.

 

College A    

 

3

4

5

6

7

Mechanical

30

65

55

23

7

Civil

18

38

30

24

10

Electronics

36

60

40

32

12

Electrical

35

68

52

25

0

Computer Science

25

59

63

20

13

 

College B

 

3

4

5

6

7

Mechanical

20

68

52

30

10

Civil

24

35

43

13

5

Electronics

23

49

62

36

10

Electrical

18

64

65

26

7

Computer Science

16

62

56

30

16

 

Question:

If all the students who were offered a package at 3 lakh, do not accept the offer, which of the given statements is true regarding the change in the average package?

 

Statement 1          Average package of college A decreases by more than Rs. 30000. for mechanical branch.

Statement 2          Average package of college B for mechanical branch decreases by mechanical branch decreases by less than Rs. 30000.

 

  1. Only 1
  2. Only 2
  3. Both 1 & 2
  4. Neither 1 nor 2


Solution:

For college A, average package = 4.5 lakh earlier and after removing the package of 3L, it will be 4 lakh (approx.)

Therefore, reduction will be approx 50000.

Therefore, statement 1 is true.

 

For college B, package before and after is respectively 4,67 & 4.34 lakh.

Therefore, reduction is approx 33,000.

Therefore, statement 2 is not true.

 

Ans (a)


Question 44

CASE

The following tables give the data about the placement records of the students by two colleges. e.g. in college A, in mechanical; 30 students got placement offer of 3 lakh.

 

Both the colleges provide 100% placements.

 

College A    

 

3

4

5

6

7

Mechanical

30

65

55

23

7

Civil

18

38

30

24

10

Electronics

36

60

40

32

12

Electrical

35

68

52

25

0

Computer Science

25

59

63

20

13

 

College B

 

3

4

5

6

7

Mechanical

20

68

52

30

10

Civil

24

35

43

13

5

Electronics

23

49

62

36

10

Electrical

18

64

65

26

7

Computer Science

16

62

56

30

16

 

Question:

For college A, for which branch, the proportion of number of students who are offered a package not exceeding 4 lakh, is the highest?

  1. Mechanical
  2. Civil
  3. Electronics
  4. Electrical


Solution:

 

Mechanical

95/180

Civil

56/120

Electronics

96/180

Electrical

103/180

 

Clearly, the ratio is the highest for electrical.

Ans (d)

 


Question 45

CASE

The following tables give the data about the placement records of the students by two colleges. e.g. in college A, in mechanical; 30 students got placement offer of 3 lakh.

 

Both the colleges provide 100% placements.

 

College A    

 

3

4

5

6

7

Mechanical

30

65

55

23

7

Civil

18

38

30

24

10

Electronics

36

60

40

32

12

Electrical

35

68

52

25

0

Computer Science

25

59

63

20

13

 

College B

 

3

4

5

6

7

Mechanical

20

68

52

30

10

Civil

24

35

43

13

5

Electronics

23

49

62

36

10

Electrical

18

64

65

26

7

Computer Science

16

62

56

30

16

 

Question:

If the companies offered 6 lakh, reduced the package to 5 lakh, then which of the following statements is true?

Statement 1          Average package for civil students of college A is reduced by 20000.

Statement 2          Average package for civil students of college B is reduced by more than 10000.

  1. Only 1
  2. Only 2
  3. Both 1 & 2
  4. Neither 1 nor 2

 


Solution:

Average package of civil for college A before & offer reduction are respectively 570/120 & 546/120. Reduction = 24/120 = 1/5 lakh i.e. 20, 000.

Therefore, statement I is true.

for college B, reduction is 13/120 » 10800.

Therefore, statement 2 is also true.  

Ans (c)


Question 46

CASE

The following tables give the data about the placement records of the students by two colleges. e.g. in college A, in mechanical; 30 students got placement offer of 3 lakh.

 

Both the colleges provide 100% placements.

 

College A    

 

3

4

5

6

7

Mechanical

30

65

55

23

7

Civil

18

38

30

24

10

Electronics

36

60

40

32

12

Electrical

35

68

52

25

0

Computer Science

25

59

63

20

13

 

College B

 

3

4

5

6

7

Mechanical

20

68

52

30

10

Civil

24

35

43

13

5

Electronics

23

49

62

36

10

Electrical

18

64

65

26

7

Computer Science

16

62

56

30

16

 

Question:

From the set above

Solution:


Question 47

The following table gives the data of the grades given to four students A, B, C & D in 5 subjects S1, S2, S3, S4 & S5. The grades are given a numerical value from 0, 1, 2, 3 and 5. The table also gives Grade average (GA) which is the average of marks in five. Subjects & SGA, which is the average marks in a particular subject.

It is known that the maximum grade of A is 3. 

 

S1

S2

S3

S4

S5

GA

A

0

 

 

2

 

2

B

1

 

 

3

 

1.4

C

 

 

3

2

4

 

D

 

3

 

 

 

1.8

Maximum

3

 

 

3

 

Minimum

 

 

2

 

0

SGA

1.5

1.75

2.75

1.75

2.25

 

Question:

what is the grade of D is S4?

(a) 0

(b)1

(c) 3

(d) 4


Solution:

 

 

S1

S2

S3

S4

S5

GA

A

0

 

 

2

 

10

B

1

 

 

3

 

7

C

 

 

3

2

4

 

D

 

3

 

 

 

9

Maximum

3

 

 

3

 

Total = 40

Minimum

 

 

2

 

0

SGA

6

7

11

7

9

Total = 40

 

From the table above, grade of D in S4 is 0 and GA of C = 40 – (10 + 7 + 9) / 5 = 2.8


Question 48

The following table gives the data of the grades given to four students A, B, C & D in 5 subjects S1, S2, S3, S4 & S5. The grades are given a numerical value from 0, 1, 2, 3 and 5. The table also gives Grade average (GA) which is the average of marks in five. Subjects & SGA, which is the average marks in a particular subject.

It is known that the maximum grade of A is 3. 

 

S1

S2

S3

S4

S5

GA

A

0

 

 

2

 

2

B

1

 

 

3

 

1.4

C

 

 

3

2

4

 

D

 

3

 

 

 

1.8

Maximum

3

 

 

3

 

Minimum

 

 

2

 

0

SGA

1.5

1.75

2.75

1.75

2.25

 

Question:

what is GA of C?

(a) 1.6

(b) 2.8

(c) 3.2

(d) 4.8


Solution:

 

 

S1

S2

S3

S4

S5

GA

A

0

 

 

2

 

10

B

1

 

 

3

 

7

C

 

 

3

2

4

 

D

 

3

 

 

 

9

Maximum

3

 

 

3

 

Total = 40

Minimum

 

 

2

 

0

SGA

6

7

11

7

9

Total = 40

 

From the table above, grade of D in S4 is 0 and GA of C = 40 – (10 + 7 + 9) / 5 = 2.8


Question 49

The following table gives the data of the grades given to four students A, B, C & D in 5 subjects S1, S2, S3, S4 & S5. The grades are given a numerical value from 0, 1, 2, 3 and 5. The table also gives Grade average (GA) which is the average of marks in five. Subjects & SGA, which is the average marks in a particular subject.

It is known that the maximum grade of A is 3. 

 

S1

S2

S3

S4

S5

GA

A

0

 

 

2

 

2

B

1

 

 

3

 

1.4

C

 

 

3

2

4

 

D

 

3

 

 

 

1.8

Maximum

3

 

 

3

 

Minimum

 

 

2

 

0

SGA

1.5

1.75

2.75

1.75

2.25

 

Question:

what can be said about the grade of B in S2?

(a) 1

(b) 2

(c) at most 1

(d) at most 2


Solution:

 

 

S1

S2

S3

S4

S5

GA

A

0

 

 

2

 

10

B

1

 

 

3

 

7

C

 

 

3

2

4

 

D

 

3

 

 

 

9

Maximum

3

 

 

3

 

Total = 40

Minimum

 

 

2

 

0

SGA

6

7

11

7

9

Total = 40

 

The best grade of A is 3.

therefore, For A to have total 10, the only possibility is 3, 3, 2, 2. Hence, A’s grade in S2 can be either 2 or 3 \B’s grade (to get sum of S2 as 7) can be 0, 1 or 2

therefore, the grade can be at most 2.

Ans (4).


Question 50

The following table gives the data of the grades given to four students A, B, C & D in 5 subjects S1, S2, S3, S4 & S5. The grades are given a numerical value from 0, 1, 2, 3 and 5. The table also gives Grade average (GA) which is the average of marks in five. Subjects & SGA, which is the average marks in a particular subject.

It is known that the maximum grade of A is 3. 

 

S1

S2

S3

S4

S5

GA

A

0

 

 

2

 

2

B

1

 

 

3

 

1.4

C

 

 

3

2

4

 

D

 

3

 

 

 

1.8

Maximum

3

 

 

3

 

Minimum

 

 

2

 

0

SGA

1.5

1.75

2.75

1.75

2.25

 

Question:

What BEST can be said about C’s grade in S1? 

  1. 1
  2. 2
  3. 3
  4. 4


Solution:

Total of S1 is 6 & maximum score is 3 \ the grade of C in S1 can be 2 or 3. The best grade between the two is 3.

Ans (3).


Verbal Ability & LR
Question 1
CASE

Let a < b < c < d < e be five positive distinct integers A list of 10 items has been created as follows.  

a + b, a + c, a + d, a + e, b + c, b + d, b + e, c + d, c + e, d + e 

Following additional information has been given

1)  The list contains 18, 29, 75, 84

2)  The other items in the list are at least 30 and at most 74

Question:

Find d – c

                (a)  9

                (b)  3

                (c)  6

                (d)  2


Solution:

Max sum will be when two largest numbers will be added

                Therefore, d + e = 84

                e + c = 75

                Subtracting both,

                d – c = 9


Question 2
CASE

Let a < b < c < d < e be five positive distinct integers A list of 10 items has been created as follows.  

a + b, a + c, a + d, a + e, b + c, b + d, b + e, c + d, c + e, d + e 

Following additional information has been given

1)  The list contains 18, 29, 75, 84

2)  The other items in the list are at least 30 and at most 74

Question:

Find c + e

                (a)  77

                (b)  70

                (c)  75

                (d)  17


Solution:

From Solution e + c = 75


Question 3
CASE

Let a < b < c < d < e be five positive distinct integers A list of 10 items has been created as follows.  

a + b, a + c, a + d, a + e, b + c, b + d, b + e, c + d, c + e, d + e 

Following additional information has been given

1)  The list contains 18, 29, 75, 84

2)  The other items in the list are at least 30 and at most 74

Question:

If b + d = 42, what is b + c?

Solution:


Question 4
CASE

Let a < b < c < d < e be five positive distinct integers A list of 10 items has been created as follows.  

a + b, a + c, a + d, a + e, b + c, b + d, b + e, c + d, c + e, d + e 

Following additional information has been given

1)  The list contains 18, 29, 75, 84

2)  The other items in the list are at least 30 and at most 74

Question:

………..

Solution:


Question 5

CASE

5 persons line in a building of three floors. Only one person lines on the ground floor while two each line on the first and the second floor. They own five animals – cow, dog, parrot, rabbit, horse  

Following additional details are provided as follows

1)  D lives on the top floor and his neighbor does not runs dog.

2)  The person who lives on the ground floor owns a cow.

3)  B owns a parrot and his neighbor does not own either a horse or rabbit

4)  C does not own either a cow or a dog and his neighbor does not own horse

5)  A does not live on the ground floor and he does not own a dog.

Question:

Who lines on the ground floor?

Solution:

Following arrangement can be made from the clues:

floors

person 1

owns

person 2

owns

2

D

Dog

B

Parrot

1

A

Horse

C

Rabbit

Ground

E

cow

 

 

 

 

 


Question 6

CASE

5 persons line in a building of three floors. Only one person lines on the ground floor while two each line on the first and the second floor. They own five animals – cow, dog, parrot, rabbit, horse  

Following additional details are provided as follows

1)  D lives on the top floor and his neighbor does not runs dog.

2)  The person who lives on the ground floor owns a cow.

3)  B owns a parrot and his neighbor does not own either a horse or rabbit

4)  C does not own either a cow or a dog and his neighbor does not own horse

5)  A does not live on the ground floor and he does not own a dog.

Question:

Who lines on the ground floor?

Solution:

Following arrangement can be made from the clues:

floors

person 1

owns

person 2

owns

2

D

Dog

B

Parrot

1

A

Horse

C

Rabbit

Ground

E

cow

 

 

 

 

 


Question 7

CASE

5 persons line in a building of three floors. Only one person lines on the ground floor while two each line on the first and the second floor. They own five animals – cow, dog, parrot, rabbit, horse  

Following additional details are provided as follows

1)  D lives on the top floor and his neighbor does not runs dog.

2)  The person who lives on the ground floor owns a cow.

3)  B owns a parrot and his neighbor does not own either a horse or rabbit

4)  C does not own either a cow or a dog and his neighbor does not own horse

5)  A does not live on the ground floor and he does not own a dog.

Question:

Who owns a dog?

Solution:

Following arrangement can be made from the clues:

floors

person 1

owns

person 2

owns

2

D

Dog

B

Parrot

1

A

Horse

C

Rabbit

Ground

E

cow

 

 

 

 

 


Question 8

CASE

5 persons line in a building of three floors. Only one person lines on the ground floor while two each line on the first and the second floor. They own five animals – cow, dog, parrot, rabbit, horse  

Following additional details are provided as follows

1)  D lives on the top floor and his neighbor does not runs dog.

2)  The person who lives on the ground floor owns a cow.

3)  B owns a parrot and his neighbor does not own either a horse or rabbit

4)  C does not own either a cow or a dog and his neighbor does not own horse

5)  A does not live on the ground floor and he does not own a dog.

Question:

Who is D’s neighbor?

Solution:

Following arrangement can be made from the clues:

floors

person 1

owns

person 2

owns

2

D

Dog

B

Parrot

1

A

Horse

C

Rabbit

Ground

E

cow

 

 

 

 

 


Question 9

Case

Six persons A, B, C, D, E, F sitting around a round table. Two of them are guitarists, two of them were singers, one of them was flautist and one was percussionist. The two guitarists are sitting adjacent to each other and every other person is taller than either of them.

1)   C, the singer was sitting left of B.

2)  The percussionist is taller than both the guitarists but shorter than the flautist

3)  The percussionist was sitting opposite to one of the guitarist.

4)  A is the shortest and sitting to the left of him is the singer

5)  The tallest one was sitting opposite to B, who is taller than the guitarists.

6)  E is not sitting opposite to either C or a guitarist and sitting to the right of F is a singer

Question:

Who is the tallest?

                (a) C

                (b) D

                (c) E

                (d) F


Solution:

Ans : E

 

Following arrangement can be made


Question 10

Case

Six persons A, B, C, D, E, F sitting around a round table. Two of them are guitarists, two of them were singers, one of them was flautist and one was percussionist. The two guitarists are sitting adjacent to each other and every other person is taller than either of them.

1)   C, the singer was sitting left of B.

2)  The percussionist is taller than both the guitarists but shorter than the flautist

3)  The percussionist was sitting opposite to one of the guitarist.

4)  A is the shortest and sitting to the left of him is the singer

5)  The tallest one was sitting opposite to B, who is taller than the guitarists.

6)  E is not sitting opposite to either C or a guitarist and sitting to the right of F is a singer

Question:

Who is the percussionist?

                (a) B

                (b) C

                (c) D

                (d) F


Solution:

Ans D

 

Following arrangement can be made


Question 11

Case

Six persons A, B, C, D, E, F sitting around a round table. Two of them are guitarists, two of them were singers, one of them was flautist and one was percussionist. The two guitarists are sitting adjacent to each other and every other person is taller than either of them.

1)   C, the singer was sitting left of B.

2)  The percussionist is taller than both the guitarists but shorter than the flautist

3)  The percussionist was sitting opposite to one of the guitarist.

4)  A is the shortest and sitting to the left of him is the singer

5)  The tallest one was sitting opposite to B, who is taller than the guitarists.

6)  E is not sitting opposite to either C or a guitarist and sitting to the right of F is a singer

Question:

Who is the guitarist?

                (a) B

                (b) C

                (c) E

                (d) F


Solution:

Ans F

 

Following arrangement can be made


Question 12

Case

Six persons A, B, C, D, E, F sitting around a round table. Two of them are guitarists, two of them were singers, one of them was flautist and one was percussionist. The two guitarists are sitting adjacent to each other and every other person is taller than either of them.

1)   C, the singer was sitting left of B.

2)  The percussionist is taller than both the guitarists but shorter than the flautist

3)  The percussionist was sitting opposite to one of the guitarist.

4)  A is the shortest and sitting to the left of him is the singer

5)  The tallest one was sitting opposite to B, who is taller than the guitarists.

6)  E is not sitting opposite to either C or a guitarist and sitting to the right of F is a singer

Question:

Who is sitting to the right of flourish?

                (a) A

                (b) B

                (c) C

                (d) D


Solution:


Question 13

CASE

A company has to send four persons Kunal, Lynda, Mona & Sunny to cities – Sydney, Los Angeles, Lisbon & Banglore over a course of two years, according to the following conditions:

  1. Kunal & Lynda are a couple and they will go to any location together. Except them, no person will be assigned with any other person.
  2. Each of the cities must be assigned at least two persons over a period of two years.
  3. Lynda dislikes Sydney & sunny dislikes Lisbon & hence, they will not be assigned there.
  4. None will be assigned to the city with which it shares the starting letter, for the 1st year.
  5. Lynda is not assigned to Los Angles for the second year.
Question:

Which of the following is necessarily true?

  1. Lynda is assigned to Los Angles for the second year
  2. Sunny is assigned to Los Angles for the first year
  3. Mona is assigned to Sydney for the first year
  4. No one is assigned to Lisbon for the first year


Solution:

First Year
 

XL

XS XL

XL

 

Los Angles

Lisbon

Sydney

Banglore

M / S

-

S / M

L + K

 

Second Year
 

XL

XS

XL

 

Los Angles

Lisbon

Sydney

Banglore

M / S

L+K

S / M

-

 


Question 14

CASE

A company has to send four persons Kunal, Lynda, Mona & Sunny to cities – Sydney, Los Angeles, Lisbon & Banglore over a course of two years, according to the following conditions:

  1. Kunal & Lynda are a couple and they will go to any location together. Except them, no person will be assigned with any other person.
  2. Each of the cities must be assigned at least two persons over a period of two years.
  3. Lynda dislikes Sydney & sunny dislikes Lisbon & hence, they will not be assigned there.
  4. None will be assigned to the city with which it shares the starting letter, for the 1st year.
  5. Lynda is not assigned to Los Angles for the second year.
Question:

Lynda is assigned to which city for the first year?

  1. Los Angles
  2. Lisbon
  3. Sydney
  4. Banglore


Solution:

First Year
 

XL

XS XL

XL

 

Los Angles

Lisbon

Sydney

Banglore

M / S

-

S / M

L + K

 

Second Year
 

XL

XS

XL

 

Los Angles

Lisbon

Sydney

Banglore

M / S

L+K

S / M

-

 


Question 15

CASE

A company has to send four persons Kunal, Lynda, Mona & Sunny to cities – Sydney, Los Angeles, Lisbon & Banglore over a course of two years, according to the following conditions:

  1. Kunal & Lynda are a couple and they will go to any location together. Except them, no person will be assigned with any other person.
  2. Each of the cities must be assigned at least two persons over a period of two years.
  3. Lynda dislikes Sydney & sunny dislikes Lisbon & hence, they will not be assigned there.
  4. None will be assigned to the city with which it shares the starting letter, for the 1st year.
  5. Lynda is not assigned to Los Angles for the second year.
Question:

: If Mona is assigned to Sydney for the second year, then Sunny is assigned to?

  1. Los Angles
  2. Lisbon
  3. Sydney
  4. Banglore


Solution:

First Year
 

XL

XS XL

XL

 

Los Angles

Lisbon

Sydney

Banglore

M / S

-

S / M

L + K

 

Second Year
 

XL

XS

XL

 

Los Angles

Lisbon

Sydney

Banglore

M / S

L+K

S / M

-

 


Question 16
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Question 17
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Question 18
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Question 19
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Question 20
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Note:
These questions are similar in nature to CAT-2014 and are sourced from CHEM question bank.
We do not claim them to be actual CAT questions.