Practice Test for IIFT - 9 - CAT MCQ

Non-managers = total employees - managers 

Total employees from 2009 - 2011 = 2532 + 2237 + 2209 = 6978 Total managers from 2009 - 2011 = 444 + 424 + 525 = 1393. Number of non-managers from 2009-2011 = 6978 - 1393 = 5585 Required proportion = 1393/5585 = 0.2494 ~ 25% Hence, option 2.

4 x 342 = 1368 <1406 Hence, option 3 can be eliminated.
Hence, option 4.
Note: Though wekart need not be checked, it can be verified as under: Managers = 307 and Employees =1228.

4 x 307 = 1228

Non-Managers = Employees - Managers.
Hence, the number of non-managers for each company in each year is as shown below: 

The number of companies for which number of non-managers is higher than the previous year is given for each option as under: 2009: 8 companies (excluding Anaconda.com and Mentalguy.com) Similarly, 2011: 4 companies 2012: 6 companies 2014: 5 companies.
Hence, option 1.

By observation of the table of employees, the following companies have 2008 > 2013 and 2009 > 2014 koala cabs, houser.com, Bestfunda.com, Quicker i.e. 4 companies.
Hence, these companies will have more employees in the first two years than in the last two years, and can be directly eliminated.
Thus, the number of required companies cannot be greater than 6.
Hence, option 1 can be eliminated.
Similarly, the following companies have 2008 <2013 and 2009 <2014 Anaconda.com and Big Box.

Hence, these two companies definitely have less employees in the first two years than in the last two years.
Now, consider the remaining four companies: Blue bus: 2008-2009 = 202 + 221 = 423 and 2013-2014 = 140 + 343 = 483 Hence, Blue bus is valid. wekart: 2008-2009 = 152 + 114 = 266 and 2013-2014 = 144 + 123 = 267.

Hence, wekart is also valid.
Mentalguy.com: 2008-2009 = 226 + 234 = 460 and 2013-2014 = 225 + 293 = 518
Hence, Mentalguy.com is also valid.
Grabdeal: 2008-2009 = 212 + 322 = 534 and 2013-2014 = 125 + 339 = 464
Hence, Grabdeal is invalid.
Thus, there are five companies that meet the requirements.
Hence, option 3.

Start with option 2 as it can directly checked from the first table.
Option 2 is true, and can be hence, eliminated.
Now, consider options 1 and 4 as they can both be directly found from the table of non-managers found in one of the earlier questions.
Option 1 is true while option 4 is false as no company showed a continuous decrease in number of non-managers from 2009 to 2013.
Hence, option 4.

Considering the options, the percentages of marks obtained by the various project groups can be determined using the following ratios, 

Hence, option 2.

Bharti and Dushyant are working on a Chemistry project where Dushyant ends up scoring 290 out of 320. 290/320 > 60%
Hence, Dushyant gets 5 credits for this project, and so would have Bharti if she would have completed this project.
Since Bharti does not lose credits in the new project, she has to get either 5 or 6 credits.
Hence, any Mathematics or Biology project can be directly eliminated as the maximum credits in these are 4 each.
Since Charu-Ajay as well as Dushyant-Farhaan are working on Mathematics projects, Bharti would not have joined these.

Hence, options 2 and 4 are eliminated.

Now, Charu-Farhaan as well as Dushyant-Ajay are working on a project in the same research area i.e. Chemistry.
Hence, based on the two remaining options, the group that has scored more marks would have given Bharti more credits.
Charu-Farhaan have scored 181 while Dushyant-Ajay have scored 296.
Hence, she would have joined the Dushyant-Ajay group. Hence, option 3.

Note: It can be verified that the % score of Dushyant-Ajay = 296/320 = 92.5% (which gives 5 credits).

The numbers of credits obtained by Bharti, Charu and Dushyant have to be maximized. The number of credits scored by each of them can be summarized as follows, 

Since number of projects does not change for any one, each person still works on 5 projects. 2 projects provide 6 credits each - Charu-Dushyant and Charu-Esha. 2 projects provide 5 credits each - Bharti-Dushyant and Dushyant-Ajay All other projects give 4, 3 or 0 credits.
Hence, each person will try to definitely work on the four projects mentioned above, along with a fifth project giving 5 credits.
Maximum possible credits = 2(6) + 2(5) + 4 = 26 Now, see if this can be achieved by any of them.
Bharti: Will drop her projects with Ajay, Charu and Esha and join the following projects: Dushyant-Ajay, Dushyant-Charu and Charu-Esha She will continue the following projects Bharti-Dushyant and Bharti- Farhaan. Her total credits = 5 + 6 + 6 + 5 + 4 = 26 Since Bharti can get 26 credits, this is the maximum number of credits that any of them can get.
Hence, option 2

Note: Though you can show that Charu as well as Dushyant can also get 26 credits, you need not prove it.

Considering the options, the percentage of marks obtained by Ajay can be given as follows, 

Verify each option:

Option 1: Dropping her project with Ajay and not joining any other group will clearly not help Bharti in increasing her number of credits.
Hence, option 1 is eliminated. 

Option 2: Bharti and Esha originally get (180/320) = 56.25% i.e. they get 3 credits.
If they score 190, their percentage = 190/320 = 59.375% i.e. they get 3 credits.
Hence, she cannot increase her number of credits.
Hence, option 2 is eliminated.
Option 3: As seen earlier, Bharti and Farhaan receive 4 credits while Bharti and Charu also receive 4 credits. Since there is no increase in the number of credits, option 3 is eliminated.
Hence, option 4.

Note: To verify whether option 4 is correct, consider the number of credits Charu and Dushyant get. Charu and Dushyant get 6 credits. Hence, if Bharti drops her project with Charu and works with Charu and Dushyant, she loses 4 credits and gets 6 credits instead. Hence, there is an increase in the number of credits.

Observe the first chart. The total number of students placed in IT & ITeS, FMCG, BFSI, Consulting, Telecom and Others in the year 2010-2011 are 93, 33, 69, 54, 36 and 15 respectively.
From the second chart, observe that the percentage increase in each of these sectors in the same order in 2011-2012 is 15%, 10%, 5%, 10%, 15% and 5%.

After the increase, the total numbers of students (rounded to the nearest integer) placed in these sectors in the same order for the year 2011-2012 are 107, 36, 72, 59, 41 and 16.
From the third graph, sector-wise international offers as a percentage of total offers for 2011-2012 are 35%, 8%, 20%, 10%, 25% and 0%. Hence, the total number of international offers (rounded to the nearest integer) for the given sectors in the same order are 37, 3, 14, 6, 10 and 0.
Hence, the total number of international offers in 2011-2012 is 70.
Hence, option 4.

In 2010-2011, 69 people were placed in BFSI. In 2011-2012, the number increased to 72.
There was an increase of 5% in 2012-2013. Hence, the number increased to 76.
Finally, in 2013-2014, after an increase of 20%, the number increased to 91.
Since 91/72 ~ 1.26, the percentage increase is approximately 25%.
Hence, option 1.

In 2011-2012, the total number of offers in IT/ITeS was 107.
In 2012-2013, after the total increase of 15%, the total number of offers was 123.
Out of these, since 55% were international offers, the remaining 45% were domestic offers.
Hence, number of offers = 45% of 123 = 55 Hence, option 3.

As seen earlier, the total number of students placed in IT & ITeS, FMCG, BFSI, Consulting, Telecom and Others in the year 2011 - 2012 are 107, 36, 72, 59,41 and 16 respectively.
Similarly, the total number of students placed in these sectors in the same order in 2012 - 2013 can be given as 123, 38, 76, 65,45 and 19.
Similarly, the total number of students placed in these sectors in the same order in 2013 - 2014 can be given as 129, 42, 91, 68,47 and 22. In FMCG, 8% of the offers are international.
Hence, 92% of the offers are domestice i.e. 39 offers are domestic.
The total number of offers in 2013 - 2014, (129 + 42 + 91 + 68 + 47 + 22) = 399
Now, (39/399) ~ 10% 

Hence, option 2.

As seen earlier, Total number of offers made in Consulting (54 + 59 + 65 + 68) = 246 Total number of offers made in Telecom (36 + 41 + 45 + 47) = 169.  Total number of offers = 246 + 169 = 415 Hence, option 3.

Hence, option 4.

Solution: Let the terms in A.P be x₁, x₂, ... , x₁₀₀. The sum of 100 terms = 100 * (x₁ + x₁₀₀) / 2 = 1. Therefore  x₁+ x₁₀₀ = 1/50 i.e. x₈ + x₉₃ = 1/50 .. (i)

The sum of first 15 terms = 1 5x₈ 

The sum of last 15 terms = 15x₉₃ Since, the sum of last 15 terms is twice the sum of the first 15 terms, we get 2x₈ = X₉₃ ...(ii)

Substituting (ii) in (i), we get 3x₈ = 1/50. Therefore x₈ = 1/150 = 2/300 Hence, option 1.
Alternatively, Let the terms of A.P be: (a - 99d), (a -97d), ..., (a - 3d), (a - d), (a + d), (a + 3d), ...,(a + 97d), (a + 99d)

Sum = 100a = 1, a = 1/100

Sum of first 15 terms = (a - 99d), (a - 97d), ... (a - 71d) = 15a - = 15(a-85d)

Sum of last 15 terms = (a + 99d), (a + 97 d), ... (a + 71d) = 15(a + 85d)

Sum of last 15 terms is twice the sum of first 15 terms i.e. 30(a - 85d) = 15(a + 85d) d = a / (85 x 3)

8th term = a - 85 d = a(1 - 8 5/8 5 x 3 ) = 2a/3 = 2/300 Hence, option 1.

Hence, option 4.

If a container contains 100% alcohol initially, then the percentage of alcohol remaining after removing x% of alcohol 'n' number of times with replacement by other liquid is given by:

= 0.78%
Hence, option 1.

Since the children’s swimming area is created by joining the centre of the hexagon to the non-common vertices of two adjacent sides, the children’s area is as shown below (shaded).

This area comprises two of the six equilateral triangles that form the hexagon. Let each of these triangles have an area of k sq.units.

Let Sunil have invested for x months out of the entire year.
Since profits are divided in ratio of investments, ratio of profits is (26000 x 12): (16000 x 9): (25000 * x) i.e. 312000: 144000 : 25000x 

Thus, Sunil joined the business for the last 6 months i.e. he joined 3 months after Mukesh.
Hence, option 2.

Total words in the set S = 5!/2 = 60

There are two vowels in the word - A and E.
Thus, the original word now has three parts - P, L and the group comprising APE.
These three parts can be arranged among themselves in 3! ways i.e. 6 ways.
However, A and E can also be interchanged to give an arrangement EPA.

Total ways in which P is between two vowels = 6 x 2 = 12

Required Probability = 12/60 = 0.2

Hence, option 4.

Let the three vessels contain 1 litre, 3 litres and 5 litres of water and milk mixture.
In the 1st vessel, 

Because the trains travel at a uniform speed, start together and then meet somewhere, they travel for the same time.
When the time is constant, the distance is proportional to the speed of the trains.
Let the speed of A and B be a and b respectively, and let the distance between the two cities be x km.
When they meet for the first time, train A has covered 500 km and train B has covered (x - 500) km respectively.

They meet for the second time at a point that is 400 km away from B.
Hence, train A has first travelled the (x - 500) km to B and then come back another 400 km.
Hence, distance covered by train A between the two meetings = x - 500 + 400 = (x - 100) km. Similarly, distance covered by train B between the two meetings = 500 + x - 400 = (x + 100) km.

500x + 50000 = x² - 600x + 50000.  x² - 1100x = 0. Therefore x = 0 or 1100 Since x is the distance between the two cities, x = 1100 cm. Hence, option 1.

The Taylor series for sinx, cosx and e^x are as given below: 

By checking each option, the given series can only be obtained for (e²- e¹)/2 Hence, option 4.

Hence, option 2

Let the second and the third side of the triangle be a and b. a : b = 7 : 6 Let P be the perimeter of the triangular garden.
P = 261 + 7k + 6k = 261 + 13k.

P - 261 = 13k, k includes an even power of 3.
P - 261 = 13 x 3²ⁿ x m.

Now, consider each value of P from the options. 

1476 - 261 = 1215 (which is not divisible by 13). Hence, option 1 is eliminated.

1249 - 261 = 988 = 13 x 76. Since 76 does not contain an even power of 3, option 2 is eliminated.

1314 - 261 = 1053 = 13 x 81. Since 81 = 3⁴ i.e. 81 contains an even power of 3, this is a possible value of the perimeter. 1353 - 261 = 1092 = 13 x 84. Since 84 does not contain an even power of 3, option 4 is eliminated.
Hence, option 3.

Since the PIN number is odd, and the only odd digit given is 1, it is of the form _ _1.
Now, there are two cases possible: Case 1: 0 is one of the digits of the number.
Hence, 0 has to be the ten’s digit.
The hundreds digit can be chosen from 2, 6 and 8 in three ways.
Thus, there are 3 such numbers possible.

Case 2: 0 is not one of the digits of the number.
Here, the two numbers for the tens and hundreds place can be chosen from 2, 6 and 8 in ³C₂ = 3 ways.
These 2 numbers can then be arranged among themselves in 2! = 2 ways.
Thus, there are 3 x 2 = 6 such numbers possible.
Total possible PIN numbers = 3 + 6 = 9 Hence, option 2.

Number of people who speak French and German but not Spanish = 29 - 24 = 5.

Let the number of persons who speak only Spanish be x.
Since 124 people were surveyed and 5 people do not speak any of the three languages, the remaining 119 speak atleast one language. Therefore 43 + 5 + 17+ 9 + 24 + 7 + x = 119 x = 14
Number of people who speak Spanish = 9 +24+ 7 + 14 = 54

Hence, option 2.

Hence, amount due to the Rs. 10 lakhs is, = 10 x (1.04)¹⁵ = Rs. 18.01 Lakhs.

Now, if A is the additional sum that is continually added at compound interest r, then the first deposit of A will accrue interest for 14 years, the second deposit of A for 13 years and so on. The last deposit of Rs. 2 Lakhs will not accrue any interest.

So, the amount due to the additional deposits will be AR^{(n - 1)} + AR^{(n - 2)} +AR^{(n - 3)} + . . . + A .

This is a Geometric Progression with (n - 1) terms and common ratio equal to R.

Hence, for A = Rs. 2 lakhs and r = 4%, this amount,

Adding the two amounts due to these two separate investments, we get, 40.05 + 18.01 = Rs. 58.06 Lakhs Hence, option 1.