CAT 2008 LRDI | Previous Year CAT Paper

Answer: Option C

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Explanation :

Since Aditya didn’t get a call from any of the colleges, so for each college, he either didn’t clear one of the sectional cut-offs or he didn’t clear the aggregate cut-off or both.

If he didn’t clear one of the sectional cut-offs, then for that section he scored less marks than the least cut-off among the given cut-offs of all the colleges.

For example, for section A, it is given that the cut-offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. The least cut-off among them is 42.

So, in order to not clear the sectional cut-off of section A for colleges 1, 4 and 5, he should have scored less than 42.

Similarly,

For colleges 1, 2 and 6, Aditya’s Section B marks < 41

For colleges 1, 2, 3 and 5, Aditya’s Section C marks < 42

For colleges 4 and 6, Aditya’s Section D marks < 44

If he scores less in Section C and D, he would not get calls for any colleges. Also in order to maximise the score we would assume that he got just one less than the cut-off in section C and D and he scored maximum marks (50) in other sections.

∴ Maximum marks obtained by Aditya such that he doesn’t get any calls = 41 + 43 + 50 + 50 = 184

Hence, option (c).

Answer: Option B

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Explanation :

Since Bhama got calls from all colleges, she must have cleared each of the 4 sections. This means that for a particular section she scored more marks than the greatest cut-off for that section across the six colleges.

For example, for section A, it is given that the cut-offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. The greatest cut-off among them is 45.

So, in order to clear the sectional cut-off of section A for all the colleges, she should have scored at least 45.

Since we wish to minimise her marks, we should take her score in section A as 45.

Similarly, in sections B, C and D, she scored 45, 46, and 45 marks respectively.

∴ Bhama’s minimum marks such that she gets calls from all the colleges = 45 + 45 + 46 + 45 = 181

Hence, option (b).

Note: This is already greater than the highest aggregate cut-off of all colleges (which is 180 for college 5). So, she will get calls from all 6 colleges.

Answer: Option C

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Explanation :

The aggregate cut-off for each college is given in the common data. In order for Charlie to get minimum marks in one of the sections, he should have got maximum marks (i.e. 50) in the other three sections.

For example, the aggregate cut-off in college 1 is 176. Since, we want minimum marks in a section he should have gotten an aggregate of exactly 176. To minimise one of the sections, assume that he got 50 marks in the 3 sections whose cut-off is given in the common data. Then, Charlie will get a call from college 1 if he gets at least 176 – (50 × 3) = 26 marks in section D, provided that the cut-off for this section is also 26.

Now, there is at least one unknown sectional cut-off for each of the colleges, so we can use the same logic as used above for each of the remaining colleges.

For college 2, the minimum marks that Charlie needs to get a call = 175 – 150 = 25

For college 3, the minimum marks that Charlie needs to get a call = 171 – 150 = 21

For college 4, the minimum marks that Charlie needs to get a call = 178 – 150 = 28

For college 5, the minimum marks that Charlie needs to get a call = 180 – 150 = 30

For college 6, the minimum marks that Charlie needs to get a call = 176 – 150 = 26

The question states that Charlie only gets a call from 2 of the colleges. So, Charlie got 25 marks.

Hence, option (c).

Answer: Option C

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Explanation :

The change is growth rate of European subscribers in '07 over '06 = $\frac{500-380}{380}=\frac{6}{19}$

The change is growth rate of European subscribers in '05 over '04 = $\frac{270-180}{180}=\frac{1}{2}$

∴ The percentage change in growth rate of 2007 (over ‘06) relative to 2005 (over ‘04) is,

$\left[\frac{\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{6}{19}}\right)}{\left({\displaystyle \frac{1}{2}}\right)}\right]\times 100=\left[\frac{\left({\displaystyle \frac{7}{38}}\right)}{\left({\displaystyle \frac{1}{2}}\right)}\right]\times 100=\frac{7}{19}$ × 100 = 36.84%

Hence, option (c)

Answer: Option A

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Explanation :

The estimated subscription revenue in Europe in 2008 = 605 million USD

The %growth rate in the revenue from 2006 to 2007 = $\frac{500-380}{380}\times 100=\frac{6}{19}$ × 100 

∴ The compound subscription revenue in Europe in 2008 = 500 × $\left(1+\frac{6}{19}\right)$

= 500 × $\left(\frac{25}{19}\right)=\frac{12500}{19}$ ≈ 658

∴ The difference between the estimated and the computed values = 658 – 605 = 53

The closest value among the given options is 50.

Hence, option (a).

Answer: Option A

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Explanation :

Let the number of men subscribers in 2003 be m, and the number of women subscribers in 2003 be w.

It is given that the women subscribers increase at the rate of 10% per year and the men increase at the rate of 5% per year.

∴ Number of men subscribers in 2007 = m(1 + 0.05)⁷ ≈ m[1 + ⁷C₁ (0.05) + ⁷C₂ (0.05)² + ⁷C₃ (0.05)³]

≈ m(1 + 0.35 + 0.0525 + 0.0043)

≈ 1.4m

Number of women subscribers in 2007 = w(1 + 0.1)⁷ ≈ w [1 + ⁷C₁(0.1) + ⁷C₂ (0.1)² + ⁷C₃ (0.1)³]

≈ w$\left[1+7(0.1)+\frac{7\times 6}{2}{(0.1)}^2+\frac{7\times 6\times 5}{2\times 3}{(0.1)}^3\right]$

≈ w(1 + 0.7 + 0.21 + 0.035)

≈ 1.95w

Now, it is given that in 2003, men are 60% of the total European subscribers. So, women are 40% of the subscribers. Let the total European subscribers be P. Then,

m = 0.6P and

w = 0.4P

∴ The total number of subscribers in 2003 = m + w = 0.6P + 0.4P = P

And the total number of subscribers in 2007 = 1.4m + 1.95w

= 1.4(0.6P) + 1.95(0.4P)

= 0.84P + 0.78P = 1.62P

∴ Percentage growth of subscribers between '03 and '07 = $\frac{(1.62)P-P}{P}$ × 100 = 62%

Hence, option (a).

Answer: Option D

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Explanation :

The gap between subscription revenues in US and Europe in 2003 = 420 – 105 = 315

The gap between subscription revenues in the US and Europe in 2004 = 525 – 180 = 345

∴ The percentage change in the gap between subscription revenues in the US and Europe in the period 2003-04 is

$\frac{345-315}{315}=\frac{30}{315}=\frac{2}{21}$

Similarly, the percentage change in the gap between subscription revenues in the US and Europe in the period 2005-06 is

$\frac{270-320}{320}=\frac{-50}{320}$

Hence, the absolute change is $\frac{5}{32}$

The percentage change in the gap between subscription revenues in the US and Europe in the period 2006-07 is

$\frac{220-270}{270}=\frac{-50}{270}$

Hence, the absolute change is $\frac{5}{27}$

The percentage change in the gap between subscription revenues in the US and Europe in the period 2008-09 is

$\frac{110-185}{185}=\frac{-75}{185}$

Hence, the absolute change is $\frac{15}{37}$

The percentage change in the gap between subscription revenues in the US and Europe in the period 2009-10 is

$\frac{100-110}{110}=\frac{-10}{110}$

Hence, the absolute change is $\frac{1}{11}$

The highest value among $\frac{2}{21},\frac{6}{33},\frac{5}{27},\frac{15}{37}$ and $\frac{1}{11}$ is clearly $\frac{15}{37}$

∴ The absolute value of change was the highest in the period 2008-09.

 Hence, option (d).

Answer: Option C

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Explanation :

Let the age of the employee being transferred from the

1. Marketing department to the Finance department be x.

2. Finance department to the Marketing department be y.

3. Marketing department to the HR department be z.

Now,

The sum of the ages of all employees in Finance originally was 30 × 20 = 600. Later, an employee of x years of age joined the department and one of y years of age left it.

So, the new average age for the Finance department = $\frac{600+x-y}{20}$ = 31 (given)

∴ 600 + x – y = 620

∴ x – y = 20                                            … (i)

The sum of the ages of all employees in Marketing originally was 35 × 30 = 1050.

Later, two employees of x years and z years of age left the department and one of y years of age joined it. Since 2 employees left and 1 joined, hence the number of employees currently in this department is 29

So, the new average age for the Marketing department = .$\frac{1050-x+y-z}{29}$ = 35 (given)

∴ 1050 – x + y – z = 1015

∴ x – y + z = 35                                       … (ii)

From equations (i) and (ii), we get,

20 + z = 35

∴ z = 15

The sum of the ages of all employees in HR originally was 45 × 5 = 225.

Later, one employee of z years of age joined the department. Also, the number of employees increases by one to 6.

So, the new average age for the HR department = $\frac{225+z}{6}=\frac{225+15}{6}=\frac{240}{6}=40$

Hence, option (c).

Answer: Option C

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Explanation :

The average age of the Marketing department is 35 years and that of the HR department is 45 years. So, the employee is being transferred from a department with a lower average age to one with a higher average age, which means that he gets an additional allowance of 10% of basic pay over his current allowance.

His current allowance = 80% of 8000 = 6400

Therefore, his new allowance = 6400 + 10% of 8000 = 6400 + 800 = 7200

After the transfer, his gross pay = 8000 + 7200 = 15200

Initially, the average gross pay of the HR department = 5000 + 70% of 5000 = 8500

The new average gross pay of the HR department (i.e. after the transfer of the 40-yr old)

$=\frac{8500\times 5+15200}{6}=\frac{57700}{6}\approx 9617$

∴ The percentage change in the average gross pay of the HR department

$=\frac{9617-8500}{8500}$ × 100 ≈ 13.13%

Hence, option (c).

Answer: Option B

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Explanation :

Note that in this question, the percentage change in basic pay is asked. According to the common data, only the allowances (and hence the gross pay) is affected when a person is transferred. The basic pay of a person remains unaltered.

∴ The average basic pay after the transfers have taken place

$$\begin{gathered} =\frac{(5000\times 5)+(6000\times 2)+\left(8000\right)}{8} \\ =\frac{45000}{8}=5625 \end{gathered}$$

∴ The percentage change in the average basic pay of the HR department

$=\frac{5625-5000}{5000}$ × 100 = 12.5%

Hence, option (b).

Answer: Option A

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Explanation :

Firstly, let us try to understand the way the investments of the three traders behave.

Abdul buys shares at 10 am everyday and sells them at a particular price at 3 pm. So his return is determined by the difference in the share price at these two times.

Bikram and Chetan buy shares at equal intervals. But since Chetan buys them in equal amount he would end up buying more when the price is less and less when the price is more.

Whether the prices are continuously rising or continuously falling down or in a fluctuating market, Chetan always has a higher proportion of lower priced shares as compared to Bikram. This increases his profit in a rising market and reduces his loss in a falling market. Therefore Chetan never has return lower than that of Bikram.

We have explained this concept by taking examples. For more depth we have also provided the theoretical explanation. The theoretical explanation is only for better understanding and may not be suitable in a test environment.

Consider the scenario when the share price keeps rising throughout the day.

Let the share price at 10 am be Rs. 100, 11 am be Rs. 110, 12 noon be Rs. 140, 1 pm be Rs. 150, 2 pm be Rs. 180, and finally at 3 pm be Rs. 200.

​​​​​​​

Abdul buys shares at Rs. 100 at 10 am and sells them at Rs. 200 at 3 pm.

∴ Abdul’s return is 100%.

Let Bikram buy one share at each interval. So, at 10 am, he buys a share for Rs. 100; at 11 am, he buys a share for Rs. 110; at 12 noon, he buys a share for Rs. 140; at 1 pm, he buys a share for Rs. 150; and at 2 pm, he buys a share for 180 × 1 = Rs. 180.

Thus, he buys a total of 5 shares for 100 + 110 + 140 + 150 + 180 = Rs. 680

At 3 pm, he sells all 5 shares for 200 × 5 = Rs. 1,000. Thus, his profit will be 1,000 − 680 = Rs. 320

Hence, Bikram's return is $\frac{320}{800}$ × 100 ≈ 47%

Let Chetan invest Rs. 415,800 at each interval. So, at 10 am, he buys 415800/100 = 4158 shares; at 11 am, he buys 415800/110 = 3780 shares; at 12 noon, he buys 415800/140 = 2970 shares; at 1 pm, he buys 415800/150 = 2772 shares; at 2 pm, he buys 415800/180 = 2310 shares.

Thus, he buys 4158 + 3780 + 2970 + 2772 + 2310 = 15990 shares for 415800 × 5 = Rs. 20,79,000. He sells these shares for 200 × 15990 = Rs. 31,98,000. His profit will be 3198000 − 2079000 = Rs. 11,19,000.

Hence Chetan's returns = $\frac{1119000}{2079000}\times 100=\frac{373}{693}$ × 100 ≈ 53%

From the above example, we see that in case of continuously rising share prices,

Abdul’s return > Chetan’s return > Bikram’s return

Thus Bikram gets the minimum return on a “boom” day.

Hence, option (a).

Note: Theoretical Explanation:

Let x1, x2, … , x6 be the share prices at 10 am, 11 am, 12 noon, 1 pm, 2 pm and 3 pm respectively.

For Abdul:

Abdul buys shares at Rs. x1 and sells them at Rs. x6.

∴ Abdul's returns = $\frac{x_6-x_1}{x_1}$

For Bikram:

Let Bikram have bought n shares at each hourly interval.

His investment amount = nx₁ + nx₂ + nx₃ + nx₄ + nx₅

= n(x₁ + x₂ + x₃ + x₄ + x₅)

$=n\times \underset{i=1}{\overset{5}{\sum x_t}}$

At 3 pm, he sells his shares for (5n × x₆)

Hence, his profit/loss = (n × 5x₆) - n × $\underset{i=1}{\overset{5}{\sum x_t}}$

= n × $\left(5x_6-\underset{i=1}{\overset{5}{\sum x_t}}\right)$

∴ Bikram's returns = $\frac{n\times (5x_6-{\displaystyle \underset{i=1}{\overset{5}{\sum x_t}}}}{n\times \underset{i=1}{\overset{5}{\sum x_t}}}=\frac{5x_6}{\underset{i=1}{\overset{5}{\sum x_t}}}-1=\frac{x_6}{{\displaystyle \frac{\underset{i=1}{\overset{5}{\sum x_t}}}{5}}}-1$

Hence, Bikram's returns = $\frac{x_6}{(\mathrm{Arithmetic} \mathrm{mean} \mathrm{of} x_1,x_2,....,x_5)}-1$

For Chetan:

Let Chetan invest Rs. P at each hourly interval.

His investment amount = 5P

Since he invests Rs. P at each interval, he buys $\frac{P}{x_1}$ shares at 10 am; $\frac{P}{x_2}$ at 11 am; and so on until 2 pm.

At 3 pm, he sells each share at x6. So, for all his shares, he receives,

Rs. $\left(\frac{P}{x_1}+\frac{P}{x_2}+\frac{P}{x_3}+\frac{P}{x_4}+\frac{P}{x_5}\right)\times x_6$

= Px₆ × $\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)$

Hence, his profit/loss = Px₆ × $\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)$ - 5P

= P × $\left[x_6\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)-5\right]$

∴ Chetan's returns = $\frac{P\left[x_6\left({\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}+{\displaystyle \frac{1}{x_3}}+{\displaystyle \frac{1}{x_4}}+{\displaystyle \frac{1}{x_5}}\right)-5\right]}{5P}$

$$\begin{gathered} =\frac{x_6\left({\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}+{\displaystyle \frac{1}{x_3}}+{\displaystyle \frac{1}{x_4}}+{\displaystyle \frac{1}{x_5}}\right)-5}{5} \\ =\frac{x_6\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)}{5}-1 \\ =\frac{x_6}{\left[{\displaystyle \frac{5}{\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}\right)}}\right]}-1 \end{gathered}$$

∴ Chetan's returns = $\frac{x_6}{(\mathrm{Harmonic} \mathrm{Mean} \mathrm{of} {\mathrm{x}}_1,{\mathrm{x}}_2,....,{\mathrm{x}}_5)}-1$

Now, let’s compare Bikram’s and Chetan’s returns. Since Arithmetic Mean is always greater than or equal to the Harmonic Mean, Chetan’s returns will be greater than or equal to Bikram’s.

Answer: Option E

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Explanation :

Since Chetan’s return is always higher than or equal to that of Bikram, the trader with the maximum return would be either Abdul or Chetan.

If it is a continuously rising market then Abdul would end up having the highest gain as seen in the example above.

But there might be a scenario when the share price of XYZ would go down after 10 AM and rise in the end at 3 PM to a higher value.

In such a case, if Chetan gets the shares at lower prices than what the price was at 10 AM he would end up making more profit and hence higher return.

​​​​​​​

Here, Abdul’s returns remain unaltered as 100%.

Let Chetan always buy shares worth Rs. 100.

So he would end up buying 1 + 10 + 10 + 10 + 10 = 41 shares.

When he sells the same at Rs. 200 he gets Rs. 8,200 for the same.

∴ Chetan’s profit = 8200 − 500 = 7700

∴ Chetan's returns = $\frac{7700}{500}>100\%$

∴ We cannot say for sure who would have higher returns. 

Hence, option (e).

Answer: Option E

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Explanation :

From the explanation seen till now we can rule out options 1, 3 and 4.

Now option 2 is only partially correct. We have seen that Chetan’s return would be higher than or equal to that of Bikram. It would be equal to Bikram’s return in the scenario when the share price remains at a constant value throughout the day.

∴ Option 2 is not always true.

Hence, option (e).

Answer: Option A

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Explanation :

Let x₁, x₂, … , x₆ be the share prices at 10 am, 11 am, 12 noon, 1 pm, 2 pm and 3 pm respectively.

Now, since Abdul lost money in the transaction,

x₁ > x₆

Combining the above, we have,

x₁ > x₆ > x₅

and x₁ > x₃,

Also, let the money Emily invests at 10 am be Rs. P. Then,

Her investment = Rs. P

And the number of shares she buys = P/x₁

So, after selling these shares at 12 noon, she will get Rs. (P/x₁) × x₃

Now, she invests the money at 1 pm, and the number of shares she buys = $\frac{P{x}_3}{x_1{x}_4}$

So, after selling these shares at 3 pm, she gets Rs. $\frac{P{x}_3}{x_1{x}_4}\times x_6$

So, her returns = $\frac{{\displaystyle \frac{P{x}_3{x}_6}{x_1{x}_4}}-P}{P}=\frac{x_3{x}_6}{x_1{x}_4}-1$

Since she made profit, her returns > 0;

i.e. $\frac{x_3{x}_6}{x_1{x}_4}-1>$ or $\frac{x_3{x}_6}{x_1{x}_4}>1$

Now, we know that x₁ > x₆; so $\frac{x_6}{x_1}$ cannot be > 1.

$\therefore \frac{x_3}{x_4}$ has to be > 1; i.e. x₃ > x₄

∴ The share price at 12 noon is greater than that at 1 pm. 

Hence, option (d) is definitely false.

Also, since in the first half, Emily invests at 10 am and sells at 12 noon, and we know that the share price at 10 am was greater than at 12 noon; hence she must have suffered a loss during this transaction. However, she makes a net profit in the end. So, she must have made profit during the second part of the transaction; i.e. the share price at 1 pm must have been less than that at 3 pm.

i.e. x₄ < x₆,

Also, let Dane buy n shares at 10 am, 11 am and 12 noon.

Hence, her investment = n(x₁ + x₂ + x₃)

And she sells these at 1 pm, 2 pm and 3 pm for n(x₄ + x₅ + x₆)

∴ her returns = $\frac{n(x_4+x_5+x_6)-n(x_1+x_2+x_3)}{n(x_1+x_2+x_3)}=\frac{(x_4+x_5+x_6)}{(x_1+x_2+x_3)}-1$

Since she made profit, her returns are greater than 0;

i.e. $\frac{(x_4+x_5+x_6)}{(x_1+x_2+x_3)}-1>$ or $\frac{(x_4+x_5+x_6)}{(x_1+x_2+x_3)}>1$

Hence, (x₄+ x₅+x₆ )> (x₁+ x₂+x₃ )

Since, x₁ > x₆ and x₃ > x₄, hence x₅ > x₂

So far, we have,

x₁ > x₆ > x₅ > x_2, x₄ < x₆ and x₁ > x₃ > x₄

Now from Dane’s investment, we know that,

(x₄+ x₅+ x₆) - (x₁+ x₂+x₃ ) > 0                         … (i)

Keeping in mind the relationships between the share prices, we have

x₆ = x₁ – b

x₄ = x₁ – b – c

x₃ = x₁ – b – c + a

x₅ = x₁ – d, where a, b, c and d are all positive.

Substituting the above in equation (i), we have,

(x₁ – b – c + x₁ – d + x₁ – b) – (x₁ + x₂ + x₁ – b – c + a) > 0

∴ x₁ - x₂ > b + d + a (which is > 0, since all the variables are positive)

i.e. x₁ > x₂

∴ x₂ < x₁ – b – a – d

∴ x₂ is definitely less than x₆ and x₅.

∴ Although we don’t know when the share price is at its lowest, we do know that x₅ > x₂.

∴ x₅, i.e. the share price at 2 pm is not the lowest.

Hence, option (a) is also definitely false.

Thus there are two options which are correct for this question. This is an ambiguity and therefore, we are not indicating any option as correct.

Answer: Option A

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Explanation :

From the solution of the last question, we can see that,

x₁ > x₆ > x₅ > x₂, x₄ < x₆ and x₁ > x₃ > x₄

∴ x₁,  i.e. the share price at 10 is the highest.

Hence, option (a).

Answer: Option B

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Explanation :

We have to arrange six houses on opposite sides of a road.

From condition (vii), we can say that P, U and S lie on one side of the road as follows:

​​​​​​​

From condition (viii) and (ix) we can further complete the arrangement as follows. We have also used the color of the house P from statement (x).

​​​​​​​

The only left house is definitely T. From conditions (v) and we can complete the arrangement as follows.

​​​​​​​

From condition (vi) it can be deduced that U is the shortest house. Also from the last condition it can be deduced that P is the fourth tallest, R is the fifth tallest and S and Q are second and third tallest not in that order.

Filling all this data we can see the arrangement as follows:

​​​​​​​

​​​​​​​The color of the tallest house (T) is Blue.

Hence, option (b).

Answer: Option D

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Explanation :

The house diagonally opposite to the Yellow coloured house is S which has red colour.

Hence, option (d).

Answer: Option E

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Explanation :

The second tallest house can be either S or Q. We cannot determine for sure which of them is the second tallest.

Hence, option (e).

Answer: Option E

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Explanation :

Let the bold letters denote the teams that have lost.

From condition 3 of stage I,

D lost to A.

D won against C.

D won against F.

These can be represented as:

D -- A

D -- C

D -- F

Similarly, condition 4 of stage I can be represented as:

E -- B

E -- C

E -- F

Since D and E have participated in three matches in stage I, they would not be involved in any other match in stage I.

From the above representations it is clear that all other teams except A have lost at least one match.

∴ From condition 1, of stage I, only A has won all the three matches in stage I.

Also, A will participate in 2 more matches as every team participates in 3 matches in stage I.

∴ A will win in 2 of the remaining 3 matches.

Also A is the top team as it wins all matches in stage I.

From condition 6 of stage I,

F did not play against A.

∴ A won against B and C which can be represented as:

B -- A

A -- C

The only 2 teams which have not won even a single match so far is C and F.

From statement 6 of stage I, F loses in the remaining match against B, which can be

represented as:

F -- B

Stage I can be represented as:

D -- A     B -- A

D -- C     A -- C

D -- F     F -- B

E -- B

E -- C

E -- F

From condition 1 of stage II,

A lost both matches in stage II.

Also, since no team plays against the same team more than once in the event, A plays matches against E and F.

A -- E

A -- F

Since one of the two teams at the bottom after stage I won both matches in stage II, F is the team which has won both the matches in stage II.

Also C lost both matches in stage II.

F -- C

B -- C

The last condition states that one more team lost both matches in stage II.

∴ D lost both matches in stage II.

D -- B

D -- E

Stage II can be represented as:

A -- E

A -- F

F -- C

B -- C

D -- B

D -- E

Now, we can calculate the number of times each team has won.

​​​​​​​

It can be observed from the above table that B and E have most wins in the event.

Hence, option (e).

Answer: Option B

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Explanation :

E and F defeated A.

Hence, option (b).

Answer: Option D

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Explanation :

B, E and F are the three teams that won both matches in stage II.

Hence, option (d).

Answer: Option E

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Explanation :

From the table it is clear that the team that won excatly two matches in the event are D and F.

Hence, option (e).

Answer: Option D

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Explanation :

From the given data we get that,

$\frac{\mathrm{Percentage} \mathrm{of} \mathrm{revenue} \mathrm{from} \mathrm{Data} \mathrm{Transfer}}{100}$ × Total revenue = ARDT × Volume

It is given that the total revenue received is the same for the pairs of countries in the choices.

∴ We need to consider the ratio (% of revenue from Data Transfer/ARDT) for the countries in the given choices

The countries for which this ratio is same would lie on a straight line drawn from the origin.

Consider the choices given. From observation one can see that the UK and Spain lie on a straight line drawn from the origin, whereas all other pairs are not on a straight line.

Hence, option (d).

Answer: Option E

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Explanation :

Let the total revenue for India be USD x and for Singapore be USD y.

$\frac{\mathrm{Percentage} \mathrm{of} \mathrm{revenue} \mathrm{from} \mathrm{Data} \mathrm{Transfer}}{100}$ × Total revenue = ARDT × Volume

From the diagram we can deduce the following values for the quantities.

Percentage of revenue from Data Transfer for India = 9

Percentage of revenue from Data Transfer for Singapore = 20.5

ARDT for India = USD 1

ARDT for Singapore = USD 9

It is given that the volume of data transfer in India is the same as that in Singapore

$\therefore \frac{0.09}{1}\times x=\frac{0.205}{9}\times y$

∴ y = 3.95x

∴ y is about 4 times x

Hence, option (e).

Answer: Option C

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Explanation :

By 2010, Percentage of revenue from transfer of data will triple for India and double for Sweden.

∴ Percentage of revenue from data transfer for India and Sweden will be 27% and 36% respectively.

Let the total revenue in Sweden in 2010 be R.

∴ The total revenue in India in 2010 will be 2R.

It is given that the total volume of data transfer is the same for both the countries.

∴ We can write the following solution for Sweden,

0.36 × R = 6 × Volume                                    ...(i)

Similarly, for India,

0.27 × 2R = ARDT × Volume                          ...(ii)

Dividing (i) by (ii) we get,

ARDT (for India in 2010) = USD 9

ARDT for India now is USD 1.

∴ The increase of ARDT in India is 800%.

Hence, option (c).