IIFT 2015 LRDI | Previous Year IIFT Paper

Answer: Option C

Explanation :

Using all the statements, we can fill the following data directly,

From 5,

Chemical Engineer is offered job in India.

So, Brad is offered job in India

Hence, Carla is offered job at Germany.

From 4 and 6,

Evan is from Mechanical branch,

So, Frank is from IT branch;

Hence, Anthony is from Electrical branch and Carla is from Electronics branch.

So the final arrangement is,

From the table we get,

Carla is the one from electronics department.

Hence, option (c).

Answer: Option C

Explanation :

Using all the statements, we can fill the following data directly,

From 5,

Chemical Engineer is offered job in India.

So, Brad is offered job in India

Hence, Carla is offered job at Germany.

From 4 and 6,

Evan is from Mechanical branch,

So, Frank is from IT branch;

Hence, Anthony is from Electrical branch and Carla is from Electronics branch.

So the final arrangement is,

From the table we get,

The person in UAE has Metallurgy branch.

Hence, option (c).

Answer: Option B

Explanation :

Using all the statements, we can fill the following data directly,

From 5,

Chemical Engineer is offered job in India.

So, Brad is offered job in India

Hence, Carla is offered job at Germany.

From 4 and 6,

Evan is from Mechanical branch,

So, Frank is from IT branch;

Hence, Anthony is from Electrical branch and Carla is from Electronics branch.

So the final arrangement is,

From the table we get,

Brad -India -Chemical is the correct combination.

Hence, option (b).

Answer: Option A

Explanation :

Using all the statements, we can fill the following data directly,

From 5,

Chemical Engineer is offered job in India.

So, Brad is offered job in India

Hence, Carla is offered job at Germany.

From 4 and 6,

Evan is from Mechanical branch,

So, Frank is from IT branch;

Hence, Anthony is from Electrical branch and Carla is from Electronics branch.

So the final arrangement is,

From the table we get,

Frank joined IT Department in Australia.

Hence, option (a).

Answer: Option D

Explanation :

Using all the statements, we can fill the following data directly,

From 5,

Chemical Engineer is offered job in India.

So, Brad is offered job in India

Hence, Carla is offered job at Germany.

From 4 and 6,

Evan is from Mechanical branch,

So, Frank is from IT branch;

Hence, Anthony is from Electrical branch and Carla is from Electronics branch.

So the final arrangement is,

From the table we get,

Dinesh - UAE – Electronics is the correct combination.

Hence, option (d).

Answer: Option B

Explanation :

Let,

A = The breakfast doesn't have eggs

B = I will not go for a walk and will not have lunch.

From the logical deductions,

AB = ~B~A

Also, and is replaced by or and vice versa.

So,

~A = The breakfast has eggs.

~B = I will go for a walk or will have lunch.

So, we get,

~B~A = I will go for a walk or will have lunch, the breakfast has eggs.

Hence, option (b).

Answer: Option B

Explanation :

Let us rank their heights and their qualification on a scale from 1 to 6.
Rank 1 is the tallest and the most qualified and Rank 6 is the shortest and the least qualified.
Kamla is not the tallest. So Kamla's height can be any rank from 2 to 5.
Kripa is the least qualified amongst the daughters. So Kripa_Q = 6 (sub-script Q and H denote rank for qualification and height respectively).
From st iv, v and vi, we conclude that
Kamla_H > Kamlesh_H > Kriti_H > Kalpana_H > Kranti_H
But Kamla is not be the tallest. Therefore Kripa is the tallest and Kamla is the 2^nd tallest and the order follows as shown above.
From St ii, Kranti_Q = 1.
From St iii and from st vi, we conclude that
Kalpana_Q > Kamlesh_Q >Kriti_Q >Kamla_Q
Therefore on the basis of the qualification,
Kalpana's rank is 2, Kamlesh's rank is 3, Kriti's rank is 4 and Kamla's rank is 5.
The following table can be regarding their height and qualification:

​​​​​​​

Decreasing order of height are the ranks starting from 1 upto 6.

Third tallest amongst all is Kamlesh

Hence, option (b).

Answer: Option D

Explanation :

Most qualified is the person with rank 1.

The person with rank 1 is Kranti.

Hence, option (d).

Answer: Option B

Explanation :

The ranks in the increasing order of qualification are from 6 to 1.

Kriti's rank in decreasing order of qualification is 4.

Therefore Kriti's rank in the increasing order of qualification is 3.

Hence, option (b).

Answer: Option B

Explanation :

The ranks in the increasing order of height are from 6 to 1.

Kamla's rank in decreasing order of height is 2.

Therefore Kamla's rank in increasing order of height is 5.

Hence, option (b).

Answer: Option B

Explanation :

The Pattern here is,

18 × 2 + 1 = 37,

37 × 2 + 2 = 76,

76 × 2 + 3 = 155,

So,

155 × 2 + 4 = 314,

314 × 2 + 5 = 633,

633 × 2 + 6 = 1272,

Hence, option (b).

Answer: Option C

Explanation :

If no condition is imposed then we can select countries in 23 – 1 = 7 ways.

So, we have total 7 combinations.

Hence, option (c).

Note:- This question is ambiguous, as it is asking number of countries. In that case, we can select only 3 countries. But we don't have options. So here we are assuming number different combinations are asked.

Answer: Option B

Explanation :

If only condition 1 is imposed then we can select,

USA-UAE and USA-UAE-UK.

So, we have two combination.

Hence, option (b).
 
Note: This question is ambiguous, as it is asking number of countries. In that case, we can select 2 or 3 countries. But we don't have options. So here we are assuming number of different combinations are asked.

Answer: Option B

Explanation :

If only condition 2 and 3 are imposed then we can select,

UK-UAE, USA, USA-UAE.

So, we can select 1 or 2 countries considering all possible combinations.

Hence, option (b).

Answer: Option C

Explanation :

From the question W is the niece of X, So W is a female.

In option A, We see that all X , Y , Z and W are brothers and sisters.

In option B, W is father of Y. So, W is a male.

Consider option C.

It says X is Y's brother, Y is W's father and W is Z's sister.

Hence W is X's niece.

Hence, option (c).

Answer: Option C

Explanation :

​​​​​​​

From the figure, the final position is E and the starting point is D.

Distance DT = Distance DF + Distance FT

= Distance DF + Distance HP

= 1 + 2

= 3 miles.

Distance ET = DistanceEP – Distance TP

= Distance EP – Distance FH

= 9 – 5

= 4 miles

From the figure, the distance between starting point and the end point is

(DE)2 = 32 + 42

= 9 + 16

= 25

DE = 5 miles

Hence, option (c).

Answer: Option D

Explanation :

​​​​​​​

The statements i, ii and iii can be represented as shown above

From the figure, Set of iphones do not intersect with the set of ipads and the set of tablets, and the set of mobiles do not intersect with the set of tablets.

Consider the conclusions,

From the figure, first three conclusions are not true. In this set, the fourth conclusion cannot be concluded.

Hence, option (d).

Answer: Option C

Explanation :

​​​​​​​

From the figure, Excluding cities S and T cargo can go from city S to city T in 2 ways i.e S-U-Q-R-T and S-U-V-T. The shortest path from S to T is S-U-V-T. Therefore the minimum number of cities that the cargo has to cross in transit is 2.

Hence, option (c).

Answer: Option B

Explanation :

​​​​​​​

From the figure, Cargo can go from city P to city U in only one possible way i.e. P-Q-R-T-V-U. Going through this path, the cargo will have to pass through 4 cities.

Hence, option (b).

Answer: Option C

Explanation :

Consider option A. The path is P-Q-R-U. We have to cross 2 intermediate cities in it.

Consider option B. If we connect cities from P to S with a one way connection from cities S to P. The path formed will be P-Q-R-T-V-U. It becomes a longer path than option A.

Using option C, the path formed is the shortest path. The path formed is P-Q-U. This smaller than that in option A.

Consider option D. The path formed using option D is longer compared to the path formed using option C. Hence options A, B, and D are eliminated.

Hence, option (c).

Answer: Option B

Explanation :

Sum of 1½ ton Window ACs sold by Torrent Enterprises during April 2014 – March 2015

= 0.19(Sum of number of units sold in April-14, May-14, Sep-14, Feb-15, Mar-15 and Aug-14) + 0.33(Sum of number of units sold in Jun-14, Jul-14, Dec-14, Oct-14, Nov-14 and Jan-15)

= 0.19 × (1266 +1268 + 1296 + 1340 + 1350 + 1296) + 0.33 × (1272 + 1292 + 1300 + 1298 + 1300 + 1330)

= 0.19 × 7816 + 0.33 × 7772 = 1485.04 + 2571.36 ≈ 338

Hence, option (b).

Answer: Option A

Explanation :

Units sold in Apr-14, May-14 and Sep-14 = 3830

Units sold in Jun-14, Jul-14 and Dec-14 = 3864

Units sold in Aug-14, Feb-15 and Mar-15 = 3986

Units sold in Oct-14, Nov-14 and Jan-15 = 3928

​​​​​​​​​​​​​

Thus, the absolute difference between average annual sales (in units) of 1 ton and ½ ton is maximum.

Hence, option (a).

Answer: Option C

Explanation :

Sales for various WAC types from Oct-14 to Mar-15:

Two ton: 0.25 × 1300 + 0.31(1340 + 1350) + 0.25(3928) = 2141

One & Half ton: 0.33 × 1300 + 0.19(1340 + 1350) + 0.33(3928) = 2236

One ton: 0.12 × 1300 + 0.34(1340 + 1350) + 0.26(3928) = 2092

One & Half ton: 0.3 × 1300 + 0.16(1340 + 1350) + 0.16(3928) = 1449

Referring to previous solution, the sales for various WAC types from Apr-14 to Sep-14:

Two ton: 3950 – 2141 = 1809

One & Half ton: 4056 – 2236 = 1820

One ton: 3376 – 2092 = 1284

One & Half ton: 4226 – 1449 = 2777

Half yearly sales performance for various types is:

$2 ton=\frac{\left({\displaystyle \frac{2141}{6}}\right)-\left({\displaystyle \frac{1809}{6}}\right)}{\left({\displaystyle \frac{1809}{6}}\right)}$

$=\frac{2141-1809}{1809}\approx 0.18$

$1 \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. ton=\frac{2236-1820}{1820}\approx 0.23$

1 ton $=\frac{2092-1284}{1284}\approx 0.63$

$\frac{1}{2} ton=\frac{1449-2777}{2777}\approx -0.48$

Thus, second best performance in Half Yearly Sales Performance is of 1½ ton WACs.

Hence, option (c).

Answer: Option D

Explanation :

Monthly sales performance:
i.    May 2014 = 2/1266
ii.    June 2014 = 4/1268
iii.    October 2014 = 2/1296
iv.    February 2015 = 10/1330 
10/1330 is the highest fraction.

Hence, option (d).

Answer: Option D

Explanation :

Sales volatility of 2 ton WAC

$=\frac{419-253}{329}\approx 0.5$

Sales volatility of 1¹/₂ ton WAC

$=\frac{439-241}{338}\approx 0.59$

Sales volatility of 1 ton WAC

$=\frac{459-153}{281}\approx 1.09$

Sales volatility of ¹/₂​​​​​​​ ton WAC

$=\frac{609-207}{352}\approx 1.14$

Thus, 2 ton of WAC has the least Sales Volatility.

Hence, option (d).

Answer: Option C

Explanation :

​​​​​​​The first table gives the difference between the average sales of various products over seven years.
e.g. A – B = 214.29 crores
Hence, total difference between A and B over the seven year period is:
A – B = 214.29 × 7 = 1500 crores
Similarly,
C – A = 900 cr, A – D = 1000 cr, A – E = 300 cr, C – B = 2400 cr, D – B = 500 cr, E – B
= 1200 cr, C – D = 1900 cr, C – E = 1200 cr,
E – D = 700 cr
Let the sales of A in 2013 be a(‘000s), C in 2010 be c(‘000s) and D in 2012 be d(‘000s).
The total sales, by volume, of each product are:
A = (6800 + a) × 10³
B = 5800 × 10³
C = (7400 + c) × 10³
D = (5700 + d) × 10³
E = 7000 × 10³
Hence, difference between number of units sold of B and E = 1200 × 10³
Assuming that each product is sold at the same cost in each year,
Cost per product = (1200 cr)/(1200 thousand) = (1200 × 10⁷)/ (1200 × 10³)
= 10000
∴ Total sales of B over the given period = 5800 × 10³ × 10000 = 5800 × 10⁷ = 5800 crores
Similarly, total sales of E over the given period = 7000 × 10³ × 10000 = 7000 × 10⁷ = 7000 crores
Using these values in the equations above, A = 7300 crores, C = 8200 crores, D
= 6300 crores
Also, total sales of A = (6800 + a) × 10³ × 10000 = (6800 + a) × 10⁷
= (6800 + a) crores
Comparing this value with 7300 crores (found as total sales of A earlier),
a = 7300 – 6800 = 500 crores
Similarly, c = 800 crores and d
= 600 crores
Hence, option (c).

Answer: Option B

Explanation :

Because the cost per product is assumed to be the same for each product for each year, the average sales (or total sales) for any year across products can be found on the basis of units sold itself.

i.e., the year with highest units sold will also have highest average annuals sales.

2010: 500 + 600 + c + 1200 + 1600

= 3900 + c = 3900 + 800 = 4700

2011: 800 + 600 + 1400 + 1100 + 300 = 4200
 
2012: 1200 + 1000 + 1500 + d + 200

= 3900 + d = 3900 + 600 = 4500

2013: a + 1000 + 800 + 100 + 2000 = a + 3900 = 4400
 
Thus, the minimum sales are in 2011.

Hence, option (b).

Answer: Option B

Explanation :

​​​​​​​The first table gives the difference between the average sales of various products over seven years.
e.g. A – B = 214.29 crores
Hence, total difference between A and B over the seven year period is:
A – B = 214.29 × 7 = 1500 crores
Similarly,
C – A = 900 cr, A – D = 1000 cr, A – E = 300 cr, C – B = 2400 cr, D – B = 500 cr, E – B
= 1200 cr, C – D = 1900 cr, C – E = 1200 cr,
E – D = 700 cr
Let the sales of A in 2013 be a(‘000s), C in 2010 be c(‘000s) and D in 2012 be d(‘000s).
The total sales, by volume, of each product are:
A = (6800 + a) × 10³
B = 5800 × 10³
C = (7400 + c) × 10³
D = (5700 + d) × 10³
E = 7000 × 10³
Hence, difference between number of units sold of B and E = 1200 × 10³
Assuming that each product is sold at the same cost in each year,
Cost per product = (1200 cr)/(1200 thousand) = (1200 × 10⁷)/ (1200 × 10³)
= 10000
∴ Total sales of B over the given period = 5800 × 10³ × 10000 = 5800 × 10⁷ = 5800 crores
Similarly, total sales of E over the given period = 7000 × 10³ × 10000 = 7000 × 10⁷ = 7000 crores
Using these values in the equations above, A = 7300 crores, C = 8200 crores, D
= 6300 crores
Also, total sales of A = (6800 + a) × 10³ × 10000 = (6800 + a) × 10⁷
= (6800 + a) crores

The product having the least total sales over the given period will also have the least average.

Hence, this product is B, with total sales of Rs. 5800 crores.

Hence, option (b).

Answer: Option D

Explanation :

This question can be directly answered from the table given in the question.

The least difference between average sales (among the given options) is between products D and E.

Hence, option (d).

Answer: Option C

Explanation :

​​​​​​​The first table gives the difference between the average sales of various products over seven years.
e.g. A – B = 214.29 crores
Hence, total difference between A and B over the seven year period is:
A – B = 214.29 × 7 = 1500 crores
Similarly,
C – A = 900 cr, A – D = 1000 cr, A – E = 300 cr, C – B = 2400 cr, D – B = 500 cr, E – B
= 1200 cr, C – D = 1900 cr, C – E = 1200 cr,
E – D = 700 cr
Let the sales of A in 2013 be a(‘000s), C in 2010 be c(‘000s) and D in 2012 be d(‘000s).
The total sales, by volume, of each product are:
A = (6800 + a) × 10³
B = 5800 × 10³
C = (7400 + c) × 10³
D = (5700 + d) × 10³
E = 7000 × 10³
Hence, difference between number of units sold of B and E = 1200 × 10³
Assuming that each product is sold at the same cost in each year,
Cost per product = (1200 cr)/(1200 thousand) = (1200 × 10⁷)/ (1200 × 10³)
= 10000
∴ Total sales of B over the given period = 5800 × 10³ × 10000 = 5800 × 10⁷ = 5800 crores
Similarly, total sales of E over the given period = 7000 × 10³ × 10000 = 7000 × 10⁷ = 7000 crores
Using these values in the equations above, A = 7300 crores, C = 8200 crores, D
= 6300 crores
Also, total sales of A = (6800 + a) × 10³ × 10000 = (6800 + a) × 10⁷
= (6800 + a) crores

The same concept can be used for this question.

The values for 2010, 2011, 2012 and 2013 were already known.

Based on the options, find the values for 2009, 2014 and 2015.

2009: 1200 + 300 + 1500 + 600 + 1300

= 4900

2014: 1800 + 700 + 1200 + 900 + 1300

= 5900

2015: 1300 + 1600 + 1000 + 1800 + 300

= 6000

Since the sales increase from 2014 to 2015, option 4 can be directly eliminates as we need to find % decline.

For the other three values, it can be observed that the maximum % decline is in 2011.

Hence, option (c).

Answer: Option D

Explanation :

Factor performance

= 0.3 × Factor score in 2014 + 0.35 × Factor score of 2012 + 0.35 × Factor score in 2010

Among the given options, Factor score of Macroeconomic Environment is the highest for 2010 and 2014.

For the year 2012, Factor Score of Business Sophistication is the highest and that of Macroeconomic Environment is second highest.

Thus, by comparing the values of Business Sophistication and Macroeconomic Environment, it can be concluded that Macroeconomic Environment has the best Factor Performance.

Hence, option (d).

Answer: Option C

Explanation :

From the graph, it is clear that among the given options, the difference between the Factor Score of 2014 and Factor Score of 2012 is maximum and the Factor Score of 2010 is the least for Infrastructure.

Therefore, Infrastructure has the best Factor Performance.

Hence, option (c).

Answer: Option B

Explanation :

By observation,

i.    Sum of Factor Scores of Infrastructure > Sum of Factor Scores of Technological Readiness
ii.    Sum of Factor Scores of Institutions > Sum of Factor Scores of Market Efficiency
 
So, it is sufficient to compare values of Infrastructure and Institution.

The total score for Infrastructure

= 4 + 4 + 5.25 = 13.25

The total score for Institution

= 4.25 + 4.75 + 4.5 = 13.5

Therefore, Institutions has the highest average across indices of 2010, 2012 and 2014.

Hence, option (b).

Answer: Option B

Explanation :

Among the given options, the difference between Factor Score in 2014 and that in 2010 is minimum for Institutions.

Hence, option (b).

Answer: Option C

Explanation :

www.jay.com spent $(27% of 557000) for 240 impressions.

So, cost per advertisement on Website A = 0.27 × 557000/240 ≈ $ 626.63

Similarly,

Cost per advertisement on Website B = 0.22 × 557000/370 ≈ $ 331.19

Cost per advertisement on Website D = 0.13 × 557000/300 ≈ $ 241.37

Cost per advertisement on Website E = 0.20 × 557000/150 ≈ $ 742.67

Thus, Website D provide facility of least cost per advertisement.

Hence, option (c).

Answer: Option D

Explanation :

Quality traffic provided by:

$\mathrm{Website} \mathrm{A}=\frac{2800}{120}=23.33$

$\mathrm{Website} \mathrm{B}=\frac{2500}{60}=41.67$

$\mathrm{Website} \mathrm{D}=\frac{3000}{80}=37.5$

$\mathrm{Website} \mathrm{E}=\frac{3500}{40}=87.5$

Thus, Website E has provided maximum quality traffic.

Hence, option (d).

Answer: Option C

Explanation :

Least the value of (Complete buying/Start buying) for a website, maximum is the leakage in online buying.

Values of (Complete buying/Start buying) are (1200/2500 =) 0.18, (900/2000 =) 0.45, (1300/3000 =) 0.43 and (1600/3500 =) 0.46 for Website B, Website C, Website D and Website E respectively.

So, Website Dsent traffice to www.jay.com with maximum leakage.

Hence, option (c).

Answer: Option A

Explanation :

Efficiency of online display advertising expenditure on

$\mathrm{Website} \mathrm{A}=\frac{2700}{27\% \mathrm{of} 557000}=\frac{10}{557}$

Efficiency of online display advertising expenditure on

$\mathrm{Website} \mathrm{B}=\frac{1200}{22\% \mathrm{of} 557000}=\frac{5.45}{557}$

Efficiency of online display advertising expenditure on

$\mathrm{Website} \mathrm{C}=\frac{900}{18\% \mathrm{of} 557000}=\frac{5}{557}$

Efficiency of online display advertising expenditure on$$

$\mathrm{Website} \mathrm{E}=\frac{1600}{20\% \mathrm{of} 557000}=\frac{8}{557}$

Advertising budget was spent most efficiently on Website A.

Hence, option (a).