IIFT 2020

The below table gives exports as a percentage of production for the given products over the years.

Thus, the maximum percentage is shown by the product 108 in the year 2012.

Hence, the answer is option B.

Imports = Consumption + Exports - Production.

Using this, the below table is created showing the imports of each product over the years.

For 103, there were no imports in 2010.

For 107, there were no imports in 2012.

For 109, there were no imports in 109. 

Thus, there were no imports for 3 products in some years.

Hence, the answer is option D.

From the given table, we can observe that the number of 500ml packs sold in Lucknow is 2010, and the ratio of 500ml packs to 1000ml packs sold is 6.7 : 1.

Thus, the number of 1000ml packs sold in Lucknow = $$\frac{2010}{6.7}=300$$

Hence, the answer is option C.

Using the given information in the table, the complete table of samples sold becomes as shown below.

Thus, the total number of samples sold = $$7250+2130=9380$$

Thus, the number of samples distributed or wasted = $$13380-9380=4000$$

The number of wasted and used samples are in the ratio $$5:3$$.

Thus, the number of waster samples = $$\frac{4000}{8}\times 5=2500$$

The number of samples used = $$4000-2500=1500$$

Now, the number of packs wasted in Chennai = $$0.1\times 2500=250$$

The number of packs distributed in Chennai = $$0.12\times 1500=180$$

Thus, the percentage of units wasted to distributed in Chennai = $$\frac{250\times 100}{180}=138.88\approx139\%$$

Hence, the answer is option B.

Using the given information in the table, the complete table of samples sold becomes as shown below.

Thus, the total number of samples sold = $$7250+2130=9380$$

Thus, the number of samples distributed or wasted = $$13380−9380=4000$$

The number of wasted and used samples are in the ratio $$5:3.$$

Thus, the number of waster samples = $$\frac{4000}{8}\times 5=2500$$

The number of samples used = $$4000-2500=1500$$

Thus, the number of units of both packs distributed in Jaipur = $$0.14\times 1500=210$$

Hence, the answer is option B.

Using the given information in the table, the complete table of samples sold becomes as shown below.

Thus, the total number of samples sold = $$7250+2130=9380$$

Thus, the number of samples distributed or wasted = $$13380-9380=4000$$

The number of wasted and used samples are in the ratio $$5:3$$.

Thus, the number of waster samples = $$\frac{4000}{8}\times 5=2500$$

The number of samples used = $$4000-2500=1500$$

Since Delhi has the lowest ratio of wasted to sold, the answer is option A.

The below table shows the matching of families in each round as per the given conditions.

First, we will trace the matches of 57 to find who he will replace in the final round [1 or 9].

Round 1: 8, Round 2: 25, Round 3: 7, Round 4: 5, Round 5: 1, Round 6: 9.

Thus, 57 will take the place of 1 in round 6.

Thus, we will have to check which family among the options take the place of 9.

1. 35

Let’s trace the matches of 35 in all the rounds.

Round 1: 30, Round 2: 3, Round 3: 4, Round 4: 3, Round 5: 1, Round 6: 9.

Thus, 35 will also take the place of 1 round 6. Therefore, it cannot be the answer.

Similarly, we can find that 39 and 58 also take the place of 1 in round 6. Thus, they are also ruled out.

2. 47

Round 1: 18, Round 2: 16, Round 3: 15, Round 4: 13, Round 5: 9, Round 6: 1

Thus, 47 will take the place of 9 in the sixth round.

Hence, the answer is option B.

The below table shows the matching of families in each round as per the given conditions.

First, we will trace all the matches of 51 in various rounds.

Round 1 against 14, round 2 against 19, round 3 against 13, round 4 against 15, round 5 against 9, and round 6 against 1.

Thus, 51 takes the place of 9 in the final round.

The family that will also take the place of 9 in the finals cannot have a toss with 51.

1. 27

Round 1 against 38, round 2 against 6, round 3 against 5, round 4 against 7, round 5 against 1, and round 6 against 9. Thus, it will take the place of 1 in the final round. Hence, it can play against 51.

Similarly, 29 and 37 can also play against 51 in the final round. Thus, all three can be ruled out.

2. 24.

Round 1 against 41, round 2 against 9, round 3 against 10, round 4 against 11, round 5 against 13, round 6 against 1. 

Thus, 24 will also take the place of 9 in the final round.

Hence, 24 can never toss against 51.

Hence, the answer is option B.

The below table shows the matching of families in each round as per the given conditions.

Since 12 reaches the finals, we will trace its matches to see which family it replaces.

Round 1 against 53, round 2 against 21, round 3 against 11, round 4 against 9, round 5 against 13 and round 6 against 1. Thus, 12 replaces 9 in round 6.

Hence, the family facing it in the finals must replace 1 in the finals.

Since we have to calculate in descending order, let's first consider family 63.

63 plays round 1 against 2, round 2 against 31, round 3 against 1, round 4 against 3, round 5 against 5, and round 6 against 9.

Thus, 63 replaces 1 in the final round and can face 12.

Since we have to find the family in descending order and 63 satisfies the condition, the answer is option C.

The below table shows the matching of families in each round as per the given conditions.

From the above table, we can see that 9 - 13 is a toss in the original matchups of the tournament.

Thus, we will check the matches for the families given in the options.

A. 44 tosses against 21, 12, 11, 9, and 13. Thus, it does not toss against 25.

B. 32 tosses against 33, 1, 2, 5 and 1. Thus, it has a toss against 5 in round 4.

Hence, the answer is option B.