Logical Reasoning & DI (LRDI) Test

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Logical Reasoning & DI (LRDI) Test - 32

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Question 1
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

The ratio distribution of total marks of five students (Parul, Qasim, Rahul, Sonia, and Tarun) in English for the last three class tests is as follows:

Parul : Qasim = 4:5, Qasim : Rahul = 5:3, Rahul : Sonia = 15:22 and Sonia : Tarun = 11:9

It is also known that:

I. The total marks scored by the five students in Test 1 were 20 less than the total marks of Test 2, which in turn is 40 less than total marks of Test 3

II. The marks of all the students in each of the three tests were integers. The maximum marks for each of the tests were 50

III. None of the students scored less than 20 marks in any of the tests

IV. The marks scored by both Rahul and Tarun in three tests were in AP with a common difference of 5. Both of them scored the least marks in Test 1 and the highest marks in Test 3

V. The marks scored by Parul and Sonia are the same in Test 1, which is 20% more than the marks scored by Rahul in Test 2

VI. The ratio of Sonia’s marks in Test 2 to that in Test 3 is 2:3

VII. The ratio of Qasim’s marks in Test 2 to that in Test 3 is 8:7.

...view full instructions

Who scored the maximum marks in Test 3?

A
Parul

B
Qasim

C
Sonia

D
Tarun

Solution

The combined ratio of total marks of five students in English for the last three tests can be calculated as follows,

Sonia : Tarun = 11 : 9 = 22 : 18 Rahul : Sonia = 15 : 22

Qasim : Rahul = 5 : 3 = 25 : 15

Parul : Qasim = 4 : 5 = 20 : 25

So, Parul : Qasim : Rahul : Sonia : Tarun = 20 : 25 : 15 : 22 : 18

Let total marks of five students in English for last three tests = 100x Parul = 22x,

Qasim = 25x, Rahul = 15x, Sonia = 22x and Tarun = 18x

Let the total marks scored by the five students in Test 3 = y

So, total marks scored by the five students in Test 2 = y – 40 and in Test 1 = y – 60 So, 3y – 100 = 100x

Given, the maximum marks for each of the test is 50 and no one scored less than 20 marks in any of the tests

So, 300 ≤ 100x ≤ 750 Considering the marks of all the students in each of the three tests as integers,

Thus, total marks of five students for last three tests = 500

Now, let the score of Rahul in Test 1 = a, Test 2 = a + 5 and Test 3 = a + 10

Solving, a + a + 5 + a + 10 = 75 => a = 20

Similarly, let the score of Tarun in Test 1 = b, Test 2 = b + 5 and Test 3 = b + 10

Solving, b + b + 5 + b + 10 = 90 => b = 25

The marks scored by Parul and Sonia each in Test 1 = 1.2 × 25 = 30

Let Sonia’s marks in Test 2 = 2c and Test 3 = 3c

Solving, 30 + 2c + 3c = 110 => c = 16

Also, Qasim’s marks in Test 1 = 140 – (30 + 20 + 30 + 25) = 35

Let Qasim’s marks in Test 2 = 8d and Test 3 = 7d

Solving, 35 + 8d + 7d = 125 => d = 6

Similarly, Parul’s marks in Test 2 = 160 – (48 + 25 + 32 + 30) = 25

And,

Parul’s marks in Test 3 = 200 – (42 + 30 + 48 + 35) = 45

The rest of the information can be gathered as follows

50% weightage scores from tests 1, 2 and 3

Parul = 10% of 30 + 15% of 25 + 25% of 45 = 18

Qasim = 10% of 35 + 15% of 48 + 25% of 42 = 21.2

Sonia = 10% of 30 + 15% of 32 + 25% of 48 = 19.8

Tarun = 10% of 25 + 15% of 30 + 25% of 35 = 15.75

Sonia scored the maximum marks in Test 3 = 48
Question 2
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

The ratio distribution of total marks of five students (Parul, Qasim, Rahul, Sonia, and Tarun) in English for the last three class tests is as follows:

Parul : Qasim = 4:5, Qasim : Rahul = 5:3, Rahul : Sonia = 15:22 and Sonia : Tarun = 11:9

It is also known that:

I. The total marks scored by the five students in Test 1 were 20 less than the total marks of Test 2, which in turn is 40 less than total marks of Test 3

II. The marks of all the students in each of the three tests were integers. The maximum marks for each of the tests were 50

III. None of the students scored less than 20 marks in any of the tests

IV. The marks scored by both Rahul and Tarun in three tests were in AP with a common difference of 5. Both of them scored the least marks in Test 1 and the highest marks in Test 3

V. The marks scored by Parul and Sonia are the same in Test 1, which is 20% more than the marks scored by Rahul in Test 2

VI. The ratio of Sonia’s marks in Test 2 to that in Test 3 is 2:3

VII. The ratio of Qasim’s marks in Test 2 to that in Test 3 is 8:7.

...view full instructions

Who scored the least marks in Test 2?

A
Parul

B
Rahul

C
Tarun

D
Both Parul and Rahul

Solution

The combined ratio of total marks of five students in English for the last three tests can be calculated as follows,

Sonia : Tarun = 11 : 9 = 22 : 18 Rahul : Sonia = 15 : 22

Qasim : Rahul = 5 : 3 = 25 : 15

Parul : Qasim = 4 : 5 = 20 : 25

So, Parul : Qasim : Rahul : Sonia : Tarun = 20 : 25 : 15 : 22 : 18

Let total marks of five students in English for last three tests = 100x Parul = 22x,

Qasim = 25x, Rahul = 15x, Sonia = 22x and Tarun = 18x

Let the total marks scored by the five students in Test 3 = y

So, total marks scored by the five students in Test 2 = y – 40 and in Test 1 = y – 60 So, 3y – 100 = 100x

Given, the maximum marks for each of the test is 50 and no one scored less than 20 marks in any of the tests

So, 300 ≤ 100x ≤ 750 Considering the marks of all the students in each of the three tests as integers,

Thus, total marks of five students for last three tests = 500

Now, let the score of Rahul in Test 1 = a, Test 2 = a + 5 and Test 3 = a + 10

Solving, a + a + 5 + a + 10 = 75 => a = 20

Similarly, let the score of Tarun in Test 1 = b, Test 2 = b + 5 and Test 3 = b + 10

Solving, b + b + 5 + b + 10 = 90 => b = 25

The marks scored by Parul and Sonia each in Test 1 = 1.2 × 25 = 30

Let Sonia’s marks in Test 2 = 2c and Test 3 = 3c

Solving, 30 + 2c + 3c = 110 => c = 16

Also, Qasim’s marks in Test 1 = 140 – (30 + 20 + 30 + 25) = 35

Let Qasim’s marks in Test 2 = 8d and Test 3 = 7d

Solving, 35 + 8d + 7d = 125 => d = 6

Similarly, Parul’s marks in Test 2 = 160 – (48 + 25 + 32 + 30) = 25

And,

Parul’s marks in Test 3 = 200 – (42 + 30 + 48 + 35) = 45

The rest of the information can be gathered as follows

50% weightage scores from tests 1, 2 and 3

Parul = 10% of 30 + 15% of 25 + 25% of 45 = 18

Qasim = 10% of 35 + 15% of 48 + 25% of 42 = 21.2

Sonia = 10% of 30 + 15% of 32 + 25% of 48 = 19.8

Tarun = 10% of 25 + 15% of 30 + 25% of 35 = 15.75

Sonia scored the maximum marks in Test 3 = 48
Question 3
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

The ratio distribution of total marks of five students (Parul, Qasim, Rahul, Sonia, and Tarun) in English for the last three class tests is as follows:

Parul : Qasim = 4:5, Qasim : Rahul = 5:3, Rahul : Sonia = 15:22 and Sonia : Tarun = 11:9

It is also known that:

I. The total marks scored by the five students in Test 1 were 20 less than the total marks of Test 2, which in turn is 40 less than total marks of Test 3

II. The marks of all the students in each of the three tests were integers. The maximum marks for each of the tests were 50

III. None of the students scored less than 20 marks in any of the tests

IV. The marks scored by both Rahul and Tarun in three tests were in AP with a common difference of 5. Both of them scored the least marks in Test 1 and the highest marks in Test 3

V. The marks scored by Parul and Sonia are the same in Test 1, which is 20% more than the marks scored by Rahul in Test 2

VI. The ratio of Sonia’s marks in Test 2 to that in Test 3 is 2:3

VII. The ratio of Qasim’s marks in Test 2 to that in Test 3 is 8:7.

...view full instructions

How many students scored at least 64% marks in at least 2 tests?

A
2

B
3

C
4

D
None

Solution

The combined ratio of total marks of five students in English for the last three tests can be calculated as follows,

Sonia : Tarun = 11 : 9 = 22 : 18 Rahul : Sonia = 15 : 22

Qasim : Rahul = 5 : 3 = 25 : 15

Parul : Qasim = 4 : 5 = 20 : 25

So, Parul : Qasim : Rahul : Sonia : Tarun = 20 : 25 : 15 : 22 : 18

Let total marks of five students in English for last three tests = 100x Parul = 22x,

Qasim = 25x, Rahul = 15x, Sonia = 22x and Tarun = 18x

Let the total marks scored by the five students in Test 3 = y

So, total marks scored by the five students in Test 2 = y – 40 and in Test 1 = y – 60 So, 3y – 100 = 100x

Given, the maximum marks for each of the test is 50 and no one scored less than 20 marks in any of the tests

So, 300 ≤ 100x ≤ 750 Considering the marks of all the students in each of the three tests as integers,

Thus, total marks of five students for last three tests = 500

Now, let the score of Rahul in Test 1 = a, Test 2 = a + 5 and Test 3 = a + 10

Solving, a + a + 5 + a + 10 = 75 => a = 20

Similarly, let the score of Tarun in Test 1 = b, Test 2 = b + 5 and Test 3 = b + 10

Solving, b + b + 5 + b + 10 = 90 => b = 25

The marks scored by Parul and Sonia each in Test 1 = 1.2 × 25 = 30

Let Sonia’s marks in Test 2 = 2c and Test 3 = 3c

Solving, 30 + 2c + 3c = 110 => c = 16

Also, Qasim’s marks in Test 1 = 140 – (30 + 20 + 30 + 25) = 35

Let Qasim’s marks in Test 2 = 8d and Test 3 = 7d

Solving, 35 + 8d + 7d = 125 => d = 6

Similarly, Parul’s marks in Test 2 = 160 – (48 + 25 + 32 + 30) = 25 And,

Parul’s marks in Test 3 = 200 – (42 + 30 + 48 + 35) = 45 The rest of the information can be gathered as follows

50% weightage scores from tests 1, 2 and 3

Parul = 10% of 30 + 15% of 25 + 25% of 45 = 18

Qasim = 10% of 35 + 15% of 48 + 25% of 42 = 21.2

Sonia = 10% of 30 + 15% of 32 + 25% of 48 = 19.8

Tarun = 10% of 25 + 15% of 30 + 25% of 35 = 15.75

Sonia scored the maximum marks in Test 3 = 48
Question 4
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

The ratio distribution of total marks of five students (Parul, Qasim, Rahul, Sonia, and Tarun) in English for the last three class tests is as follows:

Parul : Qasim = 4:5, Qasim : Rahul = 5:3, Rahul : Sonia = 15:22 and Sonia : Tarun = 11:9

It is also known that:

I. The total marks scored by the five students in Test 1 were 20 less than the total marks of Test 2, which in turn is 40 less than total marks of Test 3

II. The marks of all the students in each of the three tests were integers. The maximum marks for each of the tests were 50

III. None of the students scored less than 20 marks in any of the tests

IV. The marks scored by both Rahul and Tarun in three tests were in AP with a common difference of 5. Both of them scored the least marks in Test 1 and the highest marks in Test 3

V. The marks scored by Parul and Sonia are the same in Test 1, which is 20% more than the marks scored by Rahul in Test 2

VI. The ratio of Sonia’s marks in Test 2 to that in Test 3 is 2:3

VII. The ratio of Qasim’s marks in Test 2 to that in Test 3 is 8:7.

...view full instructions

In the final evaluation, an annual exam was conducted for 50% marks. The remaining 50% weightage was given to the scores from tests 1, 2, and 3 in the following manner: 10% from Test 1 + 15% from Test 2 and 25% from Test 3. Which student can score the minimum marks in the final exam and still be the top scorer in the final evaluation?

A
Parul

B
Qasim

C
Sonia

D
Tarun

Solution

The combined ratio of total marks of five students in English for the last three tests can be calculated as follows,

Sonia : Tarun = 11 : 9 = 22 : 18 Rahul : Sonia = 15 : 22

Qasim : Rahul = 5 : 3 = 25 : 15

Parul : Qasim = 4 : 5 = 20 : 25

So, Parul : Qasim : Rahul : Sonia : Tarun = 20 : 25 : 15 : 22 : 18

Let total marks of five students in English for last three tests = 100x Parul = 22x,

Qasim = 25x, Rahul = 15x, Sonia = 22x and Tarun = 18x

Let the total marks scored by the five students in Test 3 = y

So, total marks scored by the five students in Test 2 = y – 40 and in Test 1 = y – 60 So, 3y – 100 = 100x

Given, the maximum marks for each of the test is 50 and no one scored less than 20 marks in any of the tests

So, 300 ≤ 100x ≤ 750 Considering the marks of all the students in each of the three tests as integers,

Thus, total marks of five students for last three tests = 500

Now, let the score of Rahul in Test 1 = a, Test 2 = a + 5 and Test 3 = a + 10

Solving, a + a + 5 + a + 10 = 75 => a = 20

Similarly, let the score of Tarun in Test 1 = b, Test 2 = b + 5 and Test 3 = b + 10

Solving, b + b + 5 + b + 10 = 90 => b = 25

The marks scored by Parul and Sonia each in Test 1 = 1.2 × 25 = 30

Let Sonia’s marks in Test 2 = 2c and Test 3 = 3c

Solving, 30 + 2c + 3c = 110 => c = 16

Also, Qasim’s marks in Test 1 = 140 – (30 + 20 + 30 + 25) = 35 Let Qasim’s marks in Test 2 = 8d and Test 3 = 7d

Solving, 35 + 8d + 7d = 125 => d = 6

Similarly, Parul’s marks in Test 2 = 160 – (48 + 25 + 32 + 30) = 25 And,

Parul’s marks in Test 3 = 200 – (42 + 30 + 48 + 35) = 45 The rest of the information can be gathered as follows

50% weightage scores from tests 1, 2 and 3

Parul = 10% of 30 + 15% of 25 + 25% of 45 = 18

Qasim = 10% of 35 + 15% of 48 + 25% of 42 = 21.2

Sonia = 10% of 30 + 15% of 32 + 25% of 48 = 19.8

Tarun = 10% of 25 + 15% of 30 + 25% of 35 = 15.75

Sonia scored the maximum marks in Test 3 = 48
Question 5
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

Ten friends A, B, C, D, E, F, G, H, I and J meet a trade analyst seeking advice on the share market. The trade analyst advises to buy any two stocks among five listed companies P, Q, R, S and T. The price of one share of P, Q, R, S and T are Re.1, Rs. 10, Rs. 20, Rs. 100 and Rs. 200 in the same order.

Each friend selects two stocks out of five advised and buys one share of each. No two friends buy the shares of the same pair of companies.

I. Total amount invested by A, B and C is Rs. 161

II. The total amount invested by F, H and J is Rs. 540

III. G buys 1 share of P

IV. E buys 1 share of R

V. I buys 1 share of P.

...view full instructions

Who among the following definitely has one share of P?

A
A

B
E

C
C

D
B

Solution

Since all ten friends bought 1 share of 2 stocks, with all pair being different.

Thus the possible pairs would be

P + Q = Rs 11, P + R = Rs 21, P + S = Rs 101, P + T = Rs 201,

Q + R = Rs 30, Q + S = Rs 110, Q + T = Rs 210,

R + S = Rs 120, R + T = Rs 220 and S + T = Rs 300

Total amount invested by A + B + C = Rs 161

Possible combinations, A + B + C = (11 + 30 + 120) or (21 + 30 + 110) in any order

Total amount invested by F + H + J = Rs 540

Possible combinations, F + H + J = (30 + 210 + 300) or (110 + 210 + 220) in any order

Since, the pair having price = Rs 30 is there in the possible combinations of A + B + C, thus F + H + J = (110 + 210 + 220) is the only possible combination left in any order.

Also, Rs 110 being used among F + H + J, thus the only possible combination left for A + B + C = (11 + 30 + 120) in any order

The remaining combinations of possible pairs left for D, E, G and I are (in any order) P + R = Rs 21, P + S = Rs 101, P + T = Rs 201 and S + T = Rs 300

Given, E buys 1 share of R, so only possible pair for E is P + R = Rs 21 Also given, G and I buys 1 share of P, so they bought P + S = Rs 101 and P + T = Rs 201 in any order.

Thus D buys the pair of S + T = Rs 300

The rest of the information can be gathered as follows-
Question 6
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

Ten friends A, B, C, D, E, F, G, H, I and J meet a trade analyst seeking advice on the share market. The trade analyst advises to buy any two stocks among five listed companies P, Q, R, S and T. The price of one share of P, Q, R, S and T are Re.1, Rs. 10, Rs. 20, Rs. 100 and Rs. 200 in the same order.

Each friend selects two stocks out of five advised and buys one share of each. No two friends buy the shares of the same pair of companies.

I. Total amount invested by A, B and C is Rs. 161

II. The total amount invested by F, H and J is Rs. 540

III. G buys 1 share of P

IV. E buys 1 share of R

V. I buys 1 share of P.

...view full instructions

The pair of shares bought by G are

A
P and Q

B
P and S

C
P and T

D
Cannot be determined

Solution

Since all ten friends bought 1 share of 2 stocks, with all pair being different.

Thus the possible pairs would be

P + Q = Rs 11, P + R = Rs 21, P + S = Rs 101, P + T = Rs 201,

Q + R = Rs 30, Q + S = Rs 110, Q + T = Rs 210

R + S = Rs 120, R + T = Rs 220 and S + T = Rs 300

Total amount invested by A + B + C = Rs 161

Possible combinations, A + B + C = (11 + 30 + 120) or (21 + 30 + 110) in any order Total amount invested by F + H + J = Rs 540

Possible combinations, F + H + J = (30 + 210 + 300) or (110 + 210 + 220) in any order

Since, the pair having price = Rs 30 is there in the possible combinations of A + B + C, thus F + H + J = (110 + 210 + 220) is the only possible combination left in any order.

Also, Rs 110 being used among F + H + J, thus the only possible combination left for A + B + C = (11 + 30 + 120) in any order

The remaining combinations of possible pairs left for D, E, G and I are (in any order) P + R = Rs 21, P + S = Rs 101, P + T = Rs 201 and S + T = Rs 300

Given, E buys 1 share of R, so only possible pair for E is P + R = Rs 21 Also given, G and I buys 1 share of P, so they bought P + S = Rs 101 and P + T = Rs 201 in any order. Thus D buys the pair of S + T = Rs 300

The rest of the information can be gathered as follows-

The pair of shares bought by G are either P and S or P and T. Hence, cannot be determined
Question 7
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

Ten friends A, B, C, D, E, F, G, H, I and J meet a trade analyst seeking advice on the share market. The trade analyst advises to buy any two stocks among five listed companies P, Q, R, S and T. The price of one share of P, Q, R, S and T are Re.1, Rs. 10, Rs. 20, Rs. 100 and Rs. 200 in the same order.

Each friend selects two stocks out of five advised and buys one share of each. No two friends buy the shares of the same pair of companies.

I. Total amount invested by A, B and C is Rs. 161

II. The total amount invested by F, H and J is Rs. 540

III. G buys 1 share of P

IV. E buys 1 share of R

V. I buys 1 share of P.

...view full instructions

If A does not buy a share of R, then A buys shares of

A
Q and S

B
P and S

C
P and Q

D
Cannot be determined

Solution

Since all ten friends bought 1 share of 2 stocks, with all pair being different.

Thus the possible pairs would be

P + Q = Rs 11, P + R = Rs 21, P + S = Rs 101, P + T = Rs 201,

Q + R = Rs 30, Q + S = Rs 110, Q + T = Rs 210,

R + S = Rs 120, R + T = Rs 220 and S + T = Rs 300

Total amount invested by A + B + C = Rs 161

Possible combinations, A + B + C = (11 + 30 + 120) or (21 + 30 + 110) in any order Total amount invested by F + H + J = Rs 540

Possible combinations, F + H + J = (30 + 210 + 300) or (110 + 210 + 220) in any order

Since, the pair having price = Rs 30 is there in the possible combinations of A + B + C, thus F + H + J = (110 + 210 + 220) is the only possible combination left in any order.

Also, Rs 110 being used among F + H + J, thus the only possible combination left for A + B + C = (11 + 30 + 120) in any order

The remaining combinations of possible pairs left for D, E, G and I are (in any order) P + R = Rs 21, P + S = Rs 101, P + T = Rs 201 and S + T = Rs 300

Given, E buys 1 share of R, so only possible pair for E is P + R = Rs 21 Also given, G and I buys 1 share of P, so they bought P + S = Rs 101 and P + T = Rs 201 in any order. Thus D buys the pair of S + T = Rs 300

The rest of the information can be gathered as follows-

If A does not buy a share of R, then A buys shares of P and Q
Question 8
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

Ten friends A, B, C, D, E, F, G, H, I and J meet a trade analyst seeking advice on the share market. The trade analyst advises to buy any two stocks among five listed companies P, Q, R, S and T. The price of one share of P, Q, R, S and T are Re.1, Rs. 10, Rs. 20, Rs. 100 and Rs. 200 in the same order.

Each friend selects two stocks out of five advised and buys one share of each. No two friends buy the shares of the same pair of companies.

I. Total amount invested by A, B and C is Rs. 161

II. The total amount invested by F, H and J is Rs. 540

III. G buys 1 share of P

IV. E buys 1 share of R

V. I buys 1 share of P.

...view full instructions

If F and H buy one share of Q each, then J buys shares of ?

A
Q and R

B
R and S

C
S and T

D
R and T

Solution

Since all ten friends bought 1 share of 2 stocks, with all pair being different.

Thus the possible pairs would be

P + Q = Rs 11, P + R = Rs 21, P + S = Rs 101, P + T = Rs 201,

Q + R = Rs 30, Q + S = Rs 110, Q + T = Rs 210,

R + S = Rs 120, R + T = Rs 220 and S + T = Rs 300

Total amount invested by A + B + C = Rs 161

Possible combinations, A + B + C = (11 + 30 + 120) or (21 + 30 + 110) in any order Total amount invested by F + H + J = Rs 540

Possible combinations, F + H + J = (30 + 210 + 300) or (110 + 210 + 220) in any order

Since, the pair having price = Rs 30 is there in the possible combinations of A + B + C, thus F + H + J = (110 + 210 + 220) is the only possible combination left in any order.

Also, Rs 110 being used among F + H + J, thus the only possible combination left for A + B + C = (11 + 30 + 120) in any order

The remaining combinations of possible pairs left for D, E, G and I are (in any order) P + R = Rs 21, P + S = Rs 101, P + T = Rs 201 and S + T = Rs 300

Given, E buys 1 share of R, so only possible pair for E is P + R = Rs 21 Also given, G and I buys 1 share of P, so they bought P + S = Rs 101 and P + T = Rs 201 in any order. Thus D buys the pair of S + T = Rs 300

The rest of the information can be gathered as follows-

If F and H buys on share of Q each, then J buys shares of R and T
Question 9
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later. Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.

The table below provides partial information about the five firms.

...view full instructions

For which firms can the amounts raised by them be concluded with certainty in each year?

A
Only Drjbna

B
Only Czechy

C
Only Czechy and Drjbna

D
Only Bzygoo and Czechy and Drjbna

Solution

Given that each firm raised Rs 1 crore in its first and last year. Also, each annual increase and decrease was either of Rs 1 crore or by Rs 2 crores.

Let us consider for Alfloo,

First year of existence, 2009 = Rs 1 crore = Last year of existence, 2016

Let amount raised in 2010, 2011, 2012, 2013, 2014, 2015 be a, b, c, d, e and f respectively Solving, 1 + a + b + c + d + e + f + 1 = 21

⇒ a + b + c + d + e + f = 19

Now, even if we consider minimum annual increase and decrease annually, The values can be a = 2, b = 3, c = 4 or 5, d = 5 or 4, e = 3 and f = 2

Now for Bzygoo,

First year of existence, 2012 = Rs 1 crore = Last year of existence, 2015
Question 10
3 / -1

Directions For Questions

Read the passage below and answer the questions that follow

Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later. Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.

The table below provides partial information about the five firms.

...view full instructions

What best can be concluded about the total amount of money raised in 2015?

A
It is either Rs. 8 crores or Rs. 9 crores.

B
It is exactly Rs. 8 crores.

C
It is either Rs. 7 crores or Rs. 8 crores or Rs. 9 crores.

D
It is either Rs. 7 crores or Rs. 8 crores.

Solution

Given that each firm raised Rs 1 crore in its first and last year. Also, each annual increase and decrease was either of Rs 1 crore or by Rs 2 crores.

Let us consider for Alfloo,

First year of existence, 2009 = Rs 1 crore = Last year of existence, 2016

Let amount raised in 2010, 2011, 2012, 2013, 2014, 2015 be a, b, c, d, e and f respectively Solving, 1 + a + b + c + d + e + f + 1 = 21

⇒ a + b + c + d + e + f = 19

Now, even if we consider minimum annual increase and decrease annually, The values can be a = 2, b = 3, c = 4 or 5, d = 5 or 4, e = 3 and f = 2

Now for Bzygoo,

First year of existence, 2012 = Rs 1 crore = Last year of existence, 2015