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Sample Test: LR & DI- 1 - CAT MCQ


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20 Questions MCQ Test - Sample Test: LR & DI- 1

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*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 1

Directions: Read the given information carefully and answer the question that follows.

Amelia bought six different types (Type-1 to Type-6) of cookie packets from a bakery for her birthday party, each at a different price. The quantity of no two types of cookies bought by her is the same. Further, the quantity of each cookie packet that she bought is a natural number. She bought 5 packets of Type-1 and 3 packets of Type-2 cookies. The price of Type-3 cookies is more than that of Type-4 cookies, but less than that of Type-5 cookies. She bought a total of 21 packets of cookies, of which 6 packets were of the costliest type. Type-1 cookies are cheaper than Type-3 cookies, while Type-6 cookies are cheaper than Type-4 cookies, of which she bought 2 packets. Type-1 cookies are costlier than at least two types of cookies. Type-6 cookies are not the cheapest type of cookies that she bought.

Q. How many packets of the cheapest type of cookies did Amelia buy?


Detailed Solution for Sample Test: LR & DI- 1 - Question 1

Given:
Amelia bought 6 types of cookies.
She bought a total of 21 packets and with no two types having the same number of packets, the number of packets of each type must be among 1, 2, 3, 4, 5 and 6.
The quantity of the costliest type is 6 packets.
As she bought 5 packets of Type-1, 3 packets of Type-2 and 2 packets of Type-4 cookies, these are not the costliest.
Also, Type-6 is cheaper than Type-4 cookies and Type-3 are cheaper than Type-5.
Therefore, Type-5 is the costliest.
Given that Type-1 is costlier than at least 2 types of cookies, but cheaper than Type-3, Type-6 is cheaper than Type-4, but not the cheapest cookies.
So, Type-1, Type-3, Type-4 and Type-6 are not the cheapest cookies.
So, Type-2 is the cheapest type of cookies.
Also, Type-1, Type-4 and Type-6 are cheaper than Type-3 cookies.
Type-3 is the second costliest type of cookies.
As Type-1 is costlier than at least two other types of cookies, it must either be third or fourth costliest type of cookies.
As Type-4 is costlier than Type-6 cookies, it must also be either the third or fourth costliest type of cookies.
So, Type-6 is the second cheapest type of cookies.

The order according to price is as follows.
(i) Type-5 > Type-3 > Type-1 > Type-4 > Type-6 > Type-2 (or)
(ii) Type-5 > Type-3 > Type-4 > Type-1 > Type-6 > Type-2

The number of packets purchased by Amelia of Type-5 is 6, Type-1 is 5, Type-2 is 3 and Type-4 is 2 packets.
Thus, she bought 4 packets and 1 packet of Type-3 and Type-6, in any order.

Thus, the cookies can be arranged as follows.

She purchased 3 packets of the cheapest cookies.

Sample Test: LR & DI- 1 - Question 2

Directions: Read the given information carefully and answer the question that follows.

Amelia bought six different types (Type-1 to Type-6) of cookie packets from a bakery for her birthday party, each at a different price. The quantity of no two types of cookies bought by her is the same. Further, the quantity of each cookie packet that she bought is a natural number. She bought 5 packets of Type-1 and 3 packets of Type-2 cookies. The price of Type-3 cookies is more than that of Type-4 cookies, but less than that of Type-5 cookies. She bought a total of 21 packets of cookies, of which 6 packets were of the costliest type. Type-1 cookies are cheaper than Type-3 cookies, while Type-6 cookies are cheaper than Type-4 cookies, of which she bought 2 packets. Type-1 cookies are costlier than at least two types of cookies. Type-6 cookies are not the cheapest type of cookies that she bought.

Q. If Amelia bought 5 packets of the third cheapest type of cookies, then what is the third costliest type of cookies?

Detailed Solution for Sample Test: LR & DI- 1 - Question 2

Given:
Amelia bought 6 types of cookies.
She bought a total of 21 packets and with no two types having the same number of packets, the number of packets of each type must be among 1, 2, 3, 4, 5 and 6.
The quantity of the costliest type is 6 packets.
As she bought 5 packets of Type-1, 3 packets of Type-2 and 2 packets of Type-4 cookies, these are not the costliest.
Also, Type-6 is cheaper than Type-4 cookies and Type-3 are cheaper than Type-5.
Therefore, Type-5 is the costliest.
Given that Type-1 is costlier than at least 2 types of cookies, but cheaper than Type-3, Type-6 is cheaper than Type-4, but not the cheapest cookies.
So, Type-1, Type-3, Type-4 and Type-6 are not the cheapest cookies.
So, Type-2 is the cheapest type of cookies.
Also, Type-1, Type-4 and Type-6 are cheaper than Type-3 cookies.
Type-3 is the second costliest type of cookies.
As Type-1 is costlier than at least two other types of cookies, it must either be third or fourth costliest type of cookies.
As Type-4 is costlier than Type-6 cookies, it must also be either the third or fourth costliest type of cookies.
So, Type-6 is the second cheapest type of cookies.

The order according to price is as follows.
(i) Type-5 > Type-3 > Type-1 > Type-4 > Type-6 > Type-2 (or)
(ii) Type-5 > Type-3 > Type-4 > Type-1 > Type-6 > Type-2

The number of packets purchased by Amelia of Type-5 is 6, Type-1 is 5, Type-2 is 3 and Type-4 is 2 packets.
Thus, she bought 4 packets and 1 packet of Type-3 and Type-6, in any order.

Thus, the cookies can be arranged as follows.

She bought 5 packets of Type-1.
Thus, if she bought 5 packets of the third cheapest cookies, then the third costliest type of cookies would be Type-4.

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Sample Test: LR & DI- 1 - Question 3

Directions: Read the given information carefully and answer the question that follows.

Amelia bought six different types (Type-1 to Type-6) of cookie packets from a bakery for her birthday party, each at a different price. The quantity of no two types of cookies bought by her is the same. Further, the quantity of each cookie packet that she bought is a natural number. She bought 5 packets of Type-1 and 3 packets of Type-2 cookies. The price of Type-3 cookies is more than that of Type-4 cookies, but less than that of Type-5 cookies. She bought a total of 21 packets of cookies, of which 6 packets were of the costliest type. Type-1 cookies are cheaper than Type-3 cookies, while Type-6 cookies are cheaper than Type-4 cookies, of which she bought 2 packets. Type-1 cookies are costlier than at least two types of cookies. Type-6 cookies are not the cheapest type of cookies that she bought.

Q. Which of the following statements is definitely false?

Detailed Solution for Sample Test: LR & DI- 1 - Question 3

Given:
Amelia bought 6 types of cookies.
She bought a total of 21 packets and with no two types having the same number of packets, the number of packets of each type must be among 1, 2, 3, 4, 5 and 6.
The quantity of the costliest type is 6 packets.
As she bought 5 packets of Type-1, 3 packets of Type-2 and 2 packets of Type-4 cookies, these are not the costliest.
Also, Type-6 is cheaper than Type-4 cookies and Type-3 are cheaper than Type-5.
Therefore, Type-5 is the costliest.
Given that Type-1 is costlier than at least 2 types of cookies, but cheaper than Type-3, Type-6 is cheaper than Type-4, but not the cheapest cookies.
So, Type-1, Type-3, Type-4 and Type-6 are not the cheapest cookies.
So, Type-2 is the cheapest type of cookies.
Also, Type-1, Type-4 and Type-6 are cheaper than Type-3 cookies.
Type-3 is the second costliest type of cookies.
As Type-1 is costlier than at least two other types of cookies, it must either be third or fourth costliest type of cookies.
As Type-4 is costlier than Type-6 cookies, it must also be either the third or fourth costliest type of cookies.
So, Type-6 is the second cheapest type of cookies.

The order according to price is as follows.
(i) Type-5 > Type-3 > Type-1 > Type-4 > Type-6 > Type-2 (or)
(ii) Type-5 > Type-3 > Type-4 > Type-1 > Type-6 > Type-2

The number of packets purchased by Amelia of Type-5 is 6, Type-1 is 5, Type-2 is 3 and Type-4 is 2 packets.
Thus, she bought 4 packets and 1 packet of Type-3 and Type-6, in any order.

Thus, the cookies can be arranged as follows.

Amelia purchased 2 packets of Type-4 cookies and 3 packets of Type-2 cookies, but the price of Type-2 is less than that of Type-4 cookies. Hence, we cannot tell the total amount spent on which of the cookies is the highest. Hence, option (1) may be correct or incorrect.

The price of Type-3 cookies is more than the price of Type-6 cookies, the quantities of each of these cannot be determined. Hence, no conclusion can be drawn on total money spent.
Hence, option (2) may be correct or incorrect.

The maximum quantity of Type-6 cookies that Amelia could have purchased is 4 packets. She bought 5 packets of Type-1 cookies and the price of Type-1 cookies is more than that of Type-6 cookies. More money was spent on Type-1 cookies than on Type-6 cookies.
Hence, option (3) is definitely false.
Thus, this is the correct option.

As the quantity of Type-3 cookies cannot be determined, so no conclusion can be drawn about the total money spent.
Hence, option (4) may be correct or incorrect.

Sample Test: LR & DI- 1 - Question 4

Directions: Read the given information carefully and answer the question that follows.

Amelia bought six different types (Type-1 to Type-6) of cookie packets from a bakery for her birthday party, each at a different price. The quantity of no two types of cookies bought by her is the same. Further, the quantity of each cookie packet that she bought is a natural number. She bought 5 packets of Type-1 and 3 packets of Type-2 cookies. The price of Type-3 cookies is more than that of Type-4 cookies, but less than that of Type-5 cookies. She bought a total of 21 packets of cookies, of which 6 packets were of the costliest type. Type-1 cookies are cheaper than Type-3 cookies, while Type-6 cookies are cheaper than Type-4 cookies, of which she bought 2 packets. Type-1 cookies are costlier than at least two types of cookies. Type-6 cookies are not the cheapest type of cookies that she bought.

Q. The maximum amount of money that Amelia spent was on

Detailed Solution for Sample Test: LR & DI- 1 - Question 4

Given:
Amelia bought 6 types of cookies.
She bought a total of 21 packets and with no two types having the same number of packets, the number of packets of each type must be among 1, 2, 3, 4, 5 and 6.
The quantity of the costliest type is 6 packets.
As she bought 5 packets of Type-1, 3 packets of Type-2 and 2 packets of Type-4 cookies, these are not the costliest.
Also, Type-6 is cheaper than Type-4 cookies and Type-3 are cheaper than Type-5.
Therefore, Type-5 is the costliest.
Given that Type-1 is costlier than at least 2 types of cookies, but cheaper than Type-3, Type-6 is cheaper than Type-4, but not the cheapest cookies.
So, Type-1, Type-3, Type-4 and Type-6 are not the cheapest cookies.
So, Type-2 is the cheapest type of cookies.
Also, Type-1, Type-4 and Type-6 are cheaper than Type-3 cookies.
Type-3 is the second costliest type of cookies.
As Type-1 is costlier than at least two other types of cookies, it must either be third or fourth costliest type of cookies.
As Type-4 is costlier than Type-6 cookies, it must also be either the third or fourth costliest type of cookies.
So, Type-6 is the second cheapest type of cookies.

The order according to price is as follows.
(i) Type-5 > Type-3 > Type-1 > Type-4 > Type-6 > Type-2 (or)
(ii) Type-5 > Type-3 > Type-4 > Type-1 > Type-6 > Type-2

The number of packets purchased by Amelia of Type-5 is 6, Type-1 is 5, Type-2 is 3 and Type-4 is 2 packets.
Thus, she bought 4 packets and 1 packet of Type-3 and Type-6, in any order.

Thus, the cookies can be arranged as follows.

Amelia purchased the maximum quantity of the costliest cookies i.e. Type-5 cookies. Hence, she would have spent the maximum amount of money on Type-5 cookies.

Sample Test: LR & DI- 1 - Question 5

Directions: Read the given information carefully and answer the question that follows.

Amelia bought six different types (Type-1 to Type-6) of cookie packets from a bakery for her birthday party, each at a different price. The quantity of no two types of cookies bought by her is the same. Further, the quantity of each cookie packet that she bought is a natural number. She bought 5 packets of Type-1 and 3 packets of Type-2 cookies. The price of Type-3 cookies is more than that of Type-4 cookies, but less than that of Type-5 cookies. She bought a total of 21 packets of cookies, of which 6 packets were of the costliest type. Type-1 cookies are cheaper than Type-3 cookies, while Type-6 cookies are cheaper than Type-4 cookies, of which she bought 2 packets. Type-1 cookies are costlier than at least two types of cookies. Type-6 cookies are not the cheapest type of cookies that she bought.

Q. If Amelia spent the same amount of money on purchasing Type-2 and Type-6 cookies, how many packets of the second costliest cookies did she buy?

Detailed Solution for Sample Test: LR & DI- 1 - Question 5

Given:
Amelia bought 6 types of cookies.
She bought a total of 21 packets and with no two types having the same number of packets, the number of packets of each type must be among 1, 2, 3, 4, 5 and 6.
The quantity of the costliest type is 6 packets.
As she bought 5 packets of Type-1, 3 packets of Type-2 and 2 packets of Type-4 cookies, these are not the costliest.
Also, Type-6 is cheaper than Type-4 cookies and Type-3 are cheaper than Type-5.
Therefore, Type-5 is the costliest.
Given that Type-1 is costlier than at least 2 types of cookies, but cheaper than Type-3, Type-6 is cheaper than Type-4, but not the cheapest cookies.
So, Type-1, Type-3, Type-4 and Type-6 are not the cheapest cookies.
So, Type-2 is the cheapest type of cookies.
Also, Type-1, Type-4 and Type-6 are cheaper than Type-3 cookies.
Type-3 is the second costliest type of cookies.
As Type-1 is costlier than at least two other types of cookies, it must either be third or fourth costliest type of cookies.
As Type-4 is costlier than Type-6 cookies, it must also be either the third or fourth costliest type of cookies.
So, Type-6 is the second cheapest type of cookies.

The order according to price is as follows.
(i) Type-5 > Type-3 > Type-1 > Type-4 > Type-6 > Type-2 (or)
(ii) Type-5 > Type-3 > Type-4 > Type-1 > Type-6 > Type-2

The number of packets purchased by Amelia of Type-5 is 6, Type-1 is 5, Type-2 is 3 and Type-4 is 2 packets.
Thus, she bought 4 packets and 1 packet of Type-3 and Type-6, in any order.

Thus, the cookies can be arranged as follows.

Type-2 is the cheapest and she purchased 3 packets of Type-2. Amelia could have purchased 1 packet or 4 packets of Type-6. If she purchased 4 packets of Type-6 at a price costlier than that of Type-2, she must have spent more amount on buying Type-6. Hence, Amelia must have purchased 1 packet of Type-6. Therefore, Amelia would have purchased 4 packets of Type-3, the second costliest cookies.

Sample Test: LR & DI- 1 - Question 6

Directions: Study the following information and answer the question that follows.

Four friends P, Q, R and S were playing a game in which they had to randomly pick a ball out of five balls of five different colours from a bag. They played the game for four rounds. The balls were of Red, Blue, Yellow, Green and Purple colours. Whenever they picked a ball, they got points 1, 3, 5, 7 and 11 for picking Purple, Yellow, Blue, Red and Green balls, respectively.

Further, the following information is known:
i. Each one of them picked different balls in each round except the last.
ii. Purple ball was picked more times than Green ball but less times than Blue ball.
iii. Each ball was picked at least once.
iv. R picked the ball which was picked only in round 2.
v. Red ball was picked in all the rounds and Q picked it in the second round.
vi. P, R and S picked different balls in all the four rounds.
vii. Total points earned because of Yellow ball were 18.
viii. P picked a Yellow ball in Round 2.

Q. Who picked a Purple ball in round 3?

Detailed Solution for Sample Test: LR & DI- 1 - Question 6

The total number of times balls were picked is 4*4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

Therefore, it cannot be determined who picked a Purple ball in round 3.
Hence, option 4 is the correct answer.

Sample Test: LR & DI- 1 - Question 7

Directions: Study the following information and answer the question that follows.

Four friends P, Q, R and S were playing a game in which they had to randomly pick a ball out of five balls of five different colours from a bag. They played the game for four rounds. The balls were of Red, Blue, Yellow, Green and Purple colours. Whenever they picked a ball, they got points 1, 3, 5, 7 and 11 for picking Purple, Yellow, Blue, Red and Green balls, respectively.

Further, the following information is known:
i. Each one of them picked different balls in each round except the last.
ii. Purple ball was picked more times than Green ball but less times than Blue ball.
iii. Each ball was picked at least once.
iv. R picked the ball which was picked only in round 2.
v. Red ball was picked in all the rounds and Q picked it in the second round.
vi. P, R and S picked different balls in all the four rounds.
vii. Total points earned because of Yellow ball were 18.
viii. P picked a Yellow ball in Round 2.

Q. Winner is the person scoring the highest points at the end of all the rounds. Who is the winner?

Detailed Solution for Sample Test: LR & DI- 1 - Question 7

The total number of times balls were picked is 4*4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

Points scored by P = 1 + 5 + 3 + 7 = 16
Points scored by Q = 3 × 3 + 7 = 16
Points scored by R = 5 + 9 + 7 + 3 = 24
Points scored by S = 7 + 5 + 1 + 3 = 16
Therefore, R is the winner.
Hence, option 3 is the correct answer.

*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 8

Directions: Study the following information and answer the question that follows.

Four friends P, Q, R and S were playing a game in which they had to randomly pick a ball out of five balls of five different colours from a bag. They played the game for four rounds. The balls were of Red, Blue, Yellow, Green and Purple colours. Whenever they picked a ball, they got points 1, 3, 5, 7 and 11 for picking Purple, Yellow, Blue, Red and Green balls, respectively.

Further, the following information is known:
i. Each one of them picked different balls in each round except the last.
ii. Purple ball was picked more times than Green ball but less times than Blue ball.
iii. Each ball was picked at least once.
iv. R picked the ball which was picked only in round 2.
v. Red ball was picked in all the rounds and Q picked it in the second round.
vi. P, R and S picked different balls in all the four rounds.
vii. Total points earned because of Yellow ball were 18.
viii. P picked a Yellow ball in Round 2.

Q. How many more minimum conditions are required to find who picked what in all the rounds?


Detailed Solution for Sample Test: LR & DI- 1 - Question 8

The total number of times balls were picked is 4*4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

It is clear that even if we know which coloured ball P, R or S picked in round 1 or 3, we'll know who picked what in each round.
Therefore, 1 more condition is required.

Sample Test: LR & DI- 1 - Question 9

Directions: Study the following information and answer the question that follows.

Four friends P, Q, R and S were playing a game in which they had to randomly pick a ball out of five balls of five different colours from a bag. They played the game for four rounds. The balls were of Red, Blue, Yellow, Green and Purple colours. Whenever they picked a ball, they got points 1, 3, 5, 7 and 11 for picking Purple, Yellow, Blue, Red and Green balls, respectively.

Further, the following information is known:
i. Each one of them picked different balls in each round except the last.
ii. Purple ball was picked more times than Green ball but less times than Blue ball.
iii. Each ball was picked at least once.
iv. R picked the ball which was picked only in round 2.
v. Red ball was picked in all the rounds and Q picked it in the second round.
vi. P, R and S picked different balls in all the four rounds.
vii. Total points earned because of Yellow ball were 18.
viii. P picked a Yellow ball in Round 2.

Q. If points scored by R in round 3 were 5, what was the colour of the ball picked by S in round 1?

Detailed Solution for Sample Test: LR & DI- 1 - Question 9

The total number of times balls were picked is 4*4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

If R scored 5 points in round 3, then he must have picked Blue in round 3.
Therefore, S must have picked a Purple coloured ball in round 1.
Hence, option 1 is the correct answer.

*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 10

Directions: Read the following passage and answer the question.

Mr. Joshua, owner of a group of industries, has five factories - Factory-1 to Factory-5. The distance that he travels in any week depends on the factory that he goes to during the week. In the weeks that he goes to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, he travels a distance longer than when he does not go to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, in any week in which case he travels exactly 30 km in that week. However, during the week, if he goes to Factory-1, he travels an additional 10 km; if he goes to Factory-4, he travels an additional 20 km; if he goes to Factory-5, he travels an additional 30 km; if he goes to Factory-3, he travels an additional 40 km; if he goes to Factory-2, he travels an additional 50 km. In any week, if he goes to more than one of the five types of factories mentioned above, he travels the additional distances corresponding to all the types of factories that he goes to in that week. He does not travel more after he goes to his factories in a particular week.
The table below provides the distance that Mr. Joshua travelled for nine weeks, from Week 1 through Week 9. Further, it is known that he did not travel to the same type of factory in any two consecutive weeks during the nine weeks.

Q. During the nine weeks, for how many pairs of consecutive weeks did Mr. Joshua go to Factory-2 in the first week and Factory-5 in the next week?


Detailed Solution for Sample Test: LR & DI- 1 - Question 10

It is given that Mr. Joshua travels 30 km irrespective of his factory visits.
In week 1, he must have travelled an additional 30 km. This can happen only if he went to Factory-5 OR to Factory-4 and Factory-1.
In Week 2, he could have travelled to Factory-3 and Factory-4 OR to Factory-2 and Factory-1 OR to Factory-5, Factory-4 and Factory-1.
Similarly, we can find the possibilities for each week. This is presented in the table below:

In Week 9, if he went to Factory-2 and Factory-1, he cannot have any of the three options in Week 8. In Week 9, if he went to Factory-5, Factory-4 and Factory-1, he cannot have any of the three options in Week 8. Hence, he must have gone to Factory-3 and Factory-4 in Week 9. Since he had travelled to Factory-3 and Factory-4 in Week 9, he must have travelled to Factory-2 and Factory-5 in Week 8. Since he travelled to Factory-2 and Factory-5 in Week 8, he must have travelled to Factory-3 and Factory-1 in Week 7. Since he travelled to Factory-3 and Factory-1 in Week 7, he must have travelled to Factory-2 and Factory-4 in Week 6. Since he travelled to Factory-2 and Factory-4 in Week 6, he must have travelled to Factory-3 and Factory-1 in Week 5. Since he travelled to Factory-3 and Factory-1 in Week 5, he must have travelled to Factory-2 and Factory-4 in Week 4. Since he travelled to Factory-2 and Factory-4 in Week 4, he must have travelled to Factory-3 and Factory-5 in Week 3. Since he travelled to Factory-3 and Factory-5 in Week 3, he must have travelled to Factory-2 and Factory-1 in Week 2. Since he travelled to Factory-2 and Factory-1 in Week 2, he must have travelled to Factory-5 in Week 1.

The following table provides the factories that he travelled to in each week:

Only one instance satisfies the given condition - Week 2 and Week 3.

Sample Test: LR & DI- 1 - Question 11

Directions: Read the following passage and answer the question.

Mr. Joshua, owner of a group of industries, has five factories - Factory-1 to Factory-5. The distance that he travels in any week depends on the factory that he goes to during the week. In the weeks that he goes to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, he travels a distance longer than when he does not go to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, in any week in which case he travels exactly 30 km in that week. However, during the week, if he goes to Factory-1, he travels an additional 10 km; if he goes to Factory-4, he travels an additional 20 km; if he goes to Factory-5, he travels an additional 30 km; if he goes to Factory-3, he travels an additional 40 km; if he goes to Factory-2, he travels an additional 50 km. In any week, if he goes to more than one of the five types of factories mentioned above, he travels the additional distances corresponding to all the types of factories that he goes to in that week. He does not travel more after he goes to his factories in a particular week.
The table below provides the distance that Mr. Joshua travelled for nine weeks, from Week 1 through Week 9. Further, it is known that he did not travel to the same type of factory in any two consecutive weeks during the nine weeks.

Q. In which of the following weeks did Mr. Joshua go to Factory-1?

Detailed Solution for Sample Test: LR & DI- 1 - Question 11

It is given that Mr. Joshua travels 30 km irrespective of his factory visits.
In week 1, he must have travelled an additional 30 km. This can happen only if he went to Factory-5 OR to Factory-4 and Factory-1.
In Week 2, he could have travelled to Factory-3 and Factory-4 OR to Factory-2 and Factory-1 OR to Factory-5, Factory-4 and Factory-1.
Similarly, we can find the possibilities for each week. This is presented in the table below:

In Week 9, if he went to Factory-2 and Factory-1, he cannot have any of the three options in Week 8. In Week 9, if he went to Factory-5, Factory-4 and Factory-1, he cannot have any of the three options in Week 8. Hence, he must have gone to Factory-3 and Factory-4 in Week 9. Since he had travelled to Factory-3 and Factory-4 in Week 9, he must have travelled to Factory-2 and Factory-5 in Week 8. Since he travelled to Factory-2 and Factory-5 in Week 8, he must have travelled to Factory-3 and Factory-1 in Week 7. Since he travelled to Factory-3 and Factory-1 in Week 7, he must have travelled to Factory-2 and Factory-4 in Week 6. Since he travelled to Factory-2 and Factory-4 in Week 6, he must have travelled to Factory-3 and Factory-1 in Week 5. Since he travelled to Factory-3 and Factory-1 in Week 5, he must have travelled to Factory-2 and Factory-4 in Week 4. Since he travelled to Factory-2 and Factory-4 in Week 4, he must have travelled to Factory-3 and Factory-5 in Week 3. Since he travelled to Factory-3 and Factory-5 in Week 3, he must have travelled to Factory-2 and Factory-1 in Week 2. Since he travelled to Factory-2 and Factory-1 in Week 2, he must have travelled to Factory-5 in Week 1.

The following table provides the factories that he travelled to in each week:

Mr. Joshua went to Factory-1 in week 2.

Sample Test: LR & DI- 1 - Question 12

Directions: Read the following passage and answer the question.

Mr. Joshua, owner of a group of industries, has five factories - Factory-1 to Factory-5. The distance that he travels in any week depends on the factory that he goes to during the week. In the weeks that he goes to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, he travels a distance longer than when he does not go to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, in any week in which case he travels exactly 30 km in that week. However, during the week, if he goes to Factory-1, he travels an additional 10 km; if he goes to Factory-4, he travels an additional 20 km; if he goes to Factory-5, he travels an additional 30 km; if he goes to Factory-3, he travels an additional 40 km; if he goes to Factory-2, he travels an additional 50 km. In any week, if he goes to more than one of the five types of factories mentioned above, he travels the additional distances corresponding to all the types of factories that he goes to in that week. He does not travel more after he goes to his factories in a particular week.
The table below provides the distance that Mr. Joshua travelled for nine weeks, from Week 1 through Week 9. Further, it is known that he did not travel to the same type of factory in any two consecutive weeks during the nine weeks.

Q. Which of the following factories did Mr. Joshua go to in the maximum number of weeks during the given period?

Detailed Solution for Sample Test: LR & DI- 1 - Question 12

It is given that Mr. Joshua travels 30 km irrespective of his factory visits.
In week 1, he must have travelled an additional 30 km. This can happen only if he went to Factory-5 OR to Factory-4 and Factory-1.
In Week 2, he could have travelled to Factory-3 and Factory-4 OR to Factory-2 and Factory-1 OR to Factory-5, Factory-4 and Factory-1.
Similarly, we can find the possibilities for each week. This is presented in the table below:

In Week 9, if he went to Factory-2 and Factory-1, he cannot have any of the three options in Week 8. In Week 9, if he went to Factory-5, Factory-4 and Factory-1, he cannot have any of the three options in Week 8. Hence, he must have gone to Factory-3 and Factory-4 in Week 9. Since he had travelled to Factory-3 and Factory-4 in Week 9, he must have travelled to Factory-2 and Factory-5 in Week 8. Since he travelled to Factory-2 and Factory-5 in Week 8, he must have travelled to Factory-3 and Factory-1 in Week 7. Since he travelled to Factory-3 and Factory-1 in Week 7, he must have travelled to Factory-2 and Factory-4 in Week 6. Since he travelled to Factory-2 and Factory-4 in Week 6, he must have travelled to Factory-3 and Factory-1 in Week 5. Since he travelled to Factory-3 and Factory-1 in Week 5, he must have travelled to Factory-2 and Factory-4 in Week 4. Since he travelled to Factory-2 and Factory-4 in Week 4, he must have travelled to Factory-3 and Factory-5 in Week 3. Since he travelled to Factory-3 and Factory-5 in Week 3, he must have travelled to Factory-2 and Factory-1 in Week 2. Since he travelled to Factory-2 and Factory-1 in Week 2, he must have travelled to Factory-5 in Week 1.

The following table provides the factories that he travelled to in each week:

Mr. Joshua went to Factory-2 on four weeks. Mr. Joshua did not go to any of the other factories given in the options in more than 3 weeks.

*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 13

Directions: Read the following passage and answer the question.

Mr. Joshua, owner of a group of industries, has five factories - Factory-1 to Factory-5. The distance that he travels in any week depends on the factory that he goes to during the week. In the weeks that he goes to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, he travels a distance longer than when he does not go to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, in any week in which case he travels exactly 30 km in that week. However, during the week, if he goes to Factory-1, he travels an additional 10 km; if he goes to Factory-4, he travels an additional 20 km; if he goes to Factory-5, he travels an additional 30 km; if he goes to Factory-3, he travels an additional 40 km; if he goes to Factory-2, he travels an additional 50 km. In any week, if he goes to more than one of the five types of factories mentioned above, he travels the additional distances corresponding to all the types of factories that he goes to in that week. He does not travel more after he goes to his factories in a particular week.
The table below provides the distance that Mr. Joshua travelled for nine weeks, from Week 1 through Week 9. Further, it is known that he did not travel to the same type of factory in any two consecutive weeks during the nine weeks.

Q. In how many of the nine weeks did Mr. Joshua go to Factory-3 but not to Factory-5?


Detailed Solution for Sample Test: LR & DI- 1 - Question 13

It is given that Mr. Joshua travels 30 km irrespective of his factory visits.
In week 1, he must have travelled an additional 30 km. This can happen only if he went to Factory-5 OR to Factory-4 and Factory-1.
In Week 2, he could have travelled to Factory-3 and Factory-4 OR to Factory-2 and Factory-1 OR to Factory-5, Factory-4 and Factory-1.
Similarly, we can find the possibilities for each week. This is presented in the table below:

In Week 9, if he went to Factory-2 and Factory-1, he cannot have any of the three options in Week 8. In Week 9, if he went to Factory-5, Factory-4 and Factory-1, he cannot have any of the three options in Week 8. Hence, he must have gone to Factory-3 and Factory-4 in Week 9. Since he had travelled to Factory-3 and Factory-4 in Week 9, he must have travelled to Factory-2 and Factory-5 in Week 8. Since he travelled to Factory-2 and Factory-5 in Week 8, he must have travelled to Factory-3 and Factory-1 in Week 7. Since he travelled to Factory-3 and Factory-1 in Week 7, he must have travelled to Factory-2 and Factory-4 in Week 6. Since he travelled to Factory-2 and Factory-4 in Week 6, he must have travelled to Factory-3 and Factory-1 in Week 5. Since he travelled to Factory-3 and Factory-1 in Week 5, he must have travelled to Factory-2 and Factory-4 in Week 4. Since he travelled to Factory-2 and Factory-4 in Week 4, he must have travelled to Factory-3 and Factory-5 in Week 3. Since he travelled to Factory-3 and Factory-5 in Week 3, he must have travelled to Factory-2 and Factory-1 in Week 2. Since he travelled to Factory-2 and Factory-1 in Week 2, he must have travelled to Factory-5 in Week 1.

The following table provides the factories that he travelled to in each week:

Mr. Joshua went to Factory-3 but not to Factory-5 in three weeks (Week 5, Week 7 and Week 9).

Sample Test: LR & DI- 1 - Question 14

Directions: Read the following passage and answer the question.

Mr. Joshua, owner of a group of industries, has five factories - Factory-1 to Factory-5. The distance that he travels in any week depends on the factory that he goes to during the week. In the weeks that he goes to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, he travels a distance longer than when he does not go to Factory-1 or Factory-2 or Factory-3 or Factory-4 or Factory-5, in any week in which case he travels exactly 30 km in that week. However, during the week, if he goes to Factory-1, he travels an additional 10 km; if he goes to Factory-4, he travels an additional 20 km; if he goes to Factory-5, he travels an additional 30 km; if he goes to Factory-3, he travels an additional 40 km; if he goes to Factory-2, he travels an additional 50 km. In any week, if he goes to more than one of the five types of factories mentioned above, he travels the additional distances corresponding to all the types of factories that he goes to in that week. He does not travel more after he goes to his factories in a particular week.
The table below provides the distance that Mr. Joshua travelled for nine weeks, from Week 1 through Week 9. Further, it is known that he did not travel to the same type of factory in any two consecutive weeks during the nine weeks.

Q. Which factories were visited by Mr. Joshua in the week immediately preceded by the one which witnessed his visit to the same factories as the one that immediately succeeded that week?

Detailed Solution for Sample Test: LR & DI- 1 - Question 14

It is given that Mr. Joshua travels 30 km irrespective of his factory visits.
In week 1, he must have travelled an additional 30 km. This can happen only if he went to Factory-5 OR to Factory-4 and Factory-1.
In Week 2, he could have travelled to Factory-3 and Factory-4 OR to Factory-2 and Factory-1 OR to Factory-5, Factory-4 and Factory-1.
Similarly, we can find the possibilities for each week. This is presented in the table below:

In Week 9, if he went to Factory-2 and Factory-1, he cannot have any of the three options in Week 8. In Week 9, if he went to Factory-5, Factory-4 and Factory-1, he cannot have any of the three options in Week 8. Hence, he must have gone to Factory-3 and Factory-4 in Week 9. Since he had travelled to Factory-3 and Factory-4 in Week 9, he must have travelled to Factory-2 and Factory-5 in Week 8. Since he travelled to Factory-2 and Factory-5 in Week 8, he must have travelled to Factory-3 and Factory-1 in Week 7. Since he travelled to Factory-3 and Factory-1 in Week 7, he must have travelled to Factory-2 and Factory-4 in Week 6. Since he travelled to Factory-2 and Factory-4 in Week 6, he must have travelled to Factory-3 and Factory-1 in Week 5. Since he travelled to Factory-3 and Factory-1 in Week 5, he must have travelled to Factory-2 and Factory-4 in Week 4. Since he travelled to Factory-2 and Factory-4 in Week 4, he must have travelled to Factory-3 and Factory-5 in Week 3. Since he travelled to Factory-3 and Factory-5 in Week 3, he must have travelled to Factory-2 and Factory-1 in Week 2. Since he travelled to Factory-2 and Factory-1 in Week 2, he must have travelled to Factory-5 in Week 1.

The following table provides the factories that he travelled to in each week:

Mr. Joshua visited F-3 and F-1 in Week 5 (immediately before Week 6) and the same factories in Week 7 (immediately after Week 6), so he visited F-2 and F-4 in Week 6.

*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 15

Directions: Answer the given question based on the following data:

In a group of 200 people, number of people having at least primary education : number of people having at least middle school education: number of people having at least high school education are in the ratio 7 : 3 : 1. Out of these, 90 play football and 60 play hockey. Also, 5 in Category III (people having high school education) and one-fourth each in Categories I and II (people having primary school education only and people having middle school education but not high school education, respectively) do not play any game. In each of the above categories, the number of people who play only hockey equals the number of people who play only football. Two persons each in Categories I and II and one person in Category III play both the games. Two persons who play both the games are uneducated (Category IV). Five persons in Category III play only hockey.
Assume middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under one of the four categories described above.

Q. How many people have middle school education?


Detailed Solution for Sample Test: LR & DI- 1 - Question 15

Let 'x' people have high school education.
∴ 3x have middle school education and 7x have primary school education.
Also, as all middle school educational people have primary school education and all high school educated people have middle school education, number of people in Category I = 4x, Category II = 2x, Category III = x.

∴ Number of people in Category I and II, who do not play any game = x and x/2 respectively.
∴ Number of people playing only hockey = Number of people playing only football = (3x - 2)/2 and (3x/2 - 2)/2 = (3x - 4)/4 respectively

Also, number of people in Category III playing only hockey = (x - 6)/2 = 5 => x = 16

Now, 3x = 48 people have middle school education.
Number of high school educated people who do not play football = (x - 6)/2 + 5 = 10
Number of people having middle school education but not high school education who play only football = (x - 6)/2 = 11
Number of such people = Number of people having primary school education - Number of people having middle school education
= 7x - 3x = 4x = 64
Number of educated people playing football only = 39
Number of educated people playing hockey only = 39
∴ Number of uneducated people playing football = 90 - (39 + 5) = 46
and number of uneducated people playing hockey = 60 - (39 + 5) = 16
Out of these, 2 play both games.
∴ Number of uneducated people playing at least one game = 46 + 16 - 2 = 60
Number of uneducated people = 200 - 7x = 200 - 112 = 88
∴ 88 - 60 = 28 uneducated people do not play any game.

As per the given information, following four Venn diagrams can be drawn:




People having middle school education = 48

*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 16

Directions: Answer the given question based on the following data:

In a group of 200 people, number of people having at least primary education : number of people having at least middle school education: number of people having at least high school education are in the ratio 7 : 3 : 1. Out of these, 90 play football and 60 play hockey. Also, 5 in Category III (people having high school education) and one-fourth each in Categories I and II (people having primary school education only and people having middle school education but not high school education, respectively) do not play any game. In each of the above categories, the number of people who play only hockey equals the number of people who play only football. Two persons each in Categories I and II and one person in Category III play both the games. Two persons who play both the games are uneducated (Category IV). Five persons in Category III play only hockey.
Assume middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under one of the four categories described above.

Q. How many people have middle school education?


Detailed Solution for Sample Test: LR & DI- 1 - Question 16

Let 'x' people have high school education.
∴ 3x have middle school education and 7x have primary school education.
Also, as all middle school educational people have primary school education and all high school educated people have middle school education, number of people in Category I = 4x, Category II = 2x, Category III = x.

∴ Number of people in Category I and II, who do not play any game = x and x/2 respectively.
∴ Number of people playing only hockey = Number of people playing only football = (3x - 2)/2 and (3x/2 - 2)/2 = (3x - 4)/4 respectively

Also, number of people in Category III playing only hockey = (x - 6)/2 = 5 => x = 16

Now, 3x = 48 people have middle school education.
Number of high school educated people who do not play football = (x - 6)/2 + 5 = 10
Number of people having middle school education but not high school education who play only football = (x - 6)/2 = 11
Number of such people = Number of people having primary school education - Number of people having middle school education
= 7x - 3x = 4x = 64
Number of educated people playing football only = 39
Number of educated people playing hockey only = 39
∴ Number of uneducated people playing football = 90 - (39 + 5) = 46
and number of uneducated people playing hockey = 60 - (39 + 5) = 16
Out of these, 2 play both games.
∴ Number of uneducated people playing at least one game = 46 + 16 - 2 = 60
Number of uneducated people = 200 - 7x = 200 - 112 = 88
∴ 88 - 60 = 28 uneducated people do not play any game.

As per the given information, following four Venn diagrams can be drawn:




Solving the above, high school educated people who do not play football = 10.

*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 17

Directions: Answer the given question based on the following data:

In a group of 200 people, number of people having at least primary education : number of people having at least middle school education: number of people having at least high school education are in the ratio 7 : 3 : 1. Out of these, 90 play football and 60 play hockey. Also, 5 in Category III (people having high school education) and one-fourth each in Categories I and II (people having primary school education only and people having middle school education but not high school education, respectively) do not play any game. In each of the above categories, the number of people who play only hockey equals the number of people who play only football. Two persons each in Categories I and II and one person in Category III play both the games. Two persons who play both the games are uneducated (Category IV). Five persons in Category III play only hockey.
Assume middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under one of the four categories described above.

Q. How many people having middle school education, but not high school education, play only football?


Detailed Solution for Sample Test: LR & DI- 1 - Question 17

Let 'x' people have high school education.
∴ 3x have middle school education and 7x have primary school education.
Also, as all middle school educational people have primary school education and all high school educated people have middle school education, number of people in Category I = 4x, Category II = 2x, Category III = x.

∴ Number of people in Category I and II, who do not play any game = x and x/2 respectively.
∴ Number of people playing only hockey = Number of people playing only football = (3x - 2)/2 and (3x/2 - 2)/2 = (3x - 4)/4 respectively

Also, number of people in Category III playing only hockey = (x - 6)/2 = 5 => x = 16

Now, 3x = 48 people have middle school education.
Number of high school educated people who do not play football = (x - 6)/2 + 5 = 10
Number of people having middle school education but not high school education who play only football = (x - 6)/2 = 11
Number of such people = Number of people having primary school education - Number of people having middle school education
= 7x - 3x = 4x = 64
Number of educated people playing football only = 39
Number of educated people playing hockey only = 39
∴ Number of uneducated people playing football = 90 - (39 + 5) = 46
and number of uneducated people playing hockey = 60 - (39 + 5) = 16
Out of these, 2 play both games.
∴ Number of uneducated people playing at least one game = 46 + 16 - 2 = 60
Number of uneducated people = 200 - 7x = 200 - 112 = 88
∴ 88 - 60 = 28 uneducated people do not play any game.

As per the given information, following four Venn diagrams can be drawn:




People having middle school education, but not high school education, who play only football = 16 - 5 = 11.

*Answer can only contain numeric values
Sample Test: LR & DI- 1 - Question 18

Directions: Answer the given question based on the following data:

In a group of 200 people, number of people having at least primary education : number of people having at least middle school education: number of people having at least high school education are in the ratio 7 : 3 : 1. Out of these, 90 play football and 60 play hockey. Also, 5 in Category III (people having high school education) and one-fourth each in Categories I and II (people having primary school education only and people having middle school education but not high school education, respectively) do not play any game. In each of the above categories, the number of people who play only hockey equals the number of people who play only football. Two persons each in Categories I and II and one person in Category III play both the games. Two persons who play both the games are uneducated (Category IV). Five persons in Category III play only hockey.
Assume middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under one of the four categories described above.

Q. How many people who completed primary school could not finish middle school?


Detailed Solution for Sample Test: LR & DI- 1 - Question 18

Let 'x' people have high school education.
∴ 3x have middle school education and 7x have primary school education.
Also, as all middle school educational people have primary school education and all high school educated people have middle school education, number of people in Category I = 4x, Category II = 2x, Category III = x.

∴ Number of people in Category I and II, who do not play any game = x and x/2 respectively.
∴ Number of people playing only hockey = Number of people playing only football = (3x - 2)/2 and (3x/2 - 2)/2 = (3x - 4)/4 respectively

Also, number of people in Category III playing only hockey = (x - 6)/2 = 5 => x = 16

Now, 3x = 48 people have middle school education.
Number of high school educated people who do not play football = (x - 6)/2 + 5 = 10
Number of people having middle school education but not high school education who play only football = (x - 6)/2 = 11
Number of such people = Number of people having primary school education - Number of people having middle school education
= 7x - 3x = 4x = 64
Number of educated people playing football only = 39
Number of educated people playing hockey only = 39
∴ Number of uneducated people playing football = 90 - (39 + 5) = 46
and number of uneducated people playing hockey = 60 - (39 + 5) = 16
Out of these, 2 play both games.
∴ Number of uneducated people playing at least one game = 46 + 16 - 2 = 60
Number of uneducated people = 200 - 7x = 200 - 112 = 88
∴ 88 - 60 = 28 uneducated people do not play any game.

As per the given information, following four Venn diagrams can be drawn:




Number of people who completed primary school, but could not finish middle school = 112 - 48 = 64.

Sample Test: LR & DI- 1 - Question 19

Directions: Answer the given question based on the following data:

In a group of 200 people, number of people having at least primary education : number of people having at least middle school education: number of people having at least high school education are in the ratio 7 : 3 : 1. Out of these, 90 play football and 60 play hockey. Also, 5 in Category III (people having high school education) and one-fourth each in Categories I and II (people having primary school education only and people having middle school education but not high school education, respectively) do not play any game. In each of the above categories, the number of people who play only hockey equals the number of people who play only football. Two persons each in Categories I and II and one person in Category III play both the games. Two persons who play both the games are uneducated (Category IV). Five persons in Category III play only hockey.
Assume middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under one of the four categories described above.

Q. How many uneducated people play neither hockey nor football?

Detailed Solution for Sample Test: LR & DI- 1 - Question 19

Let 'x' people have high school education.
∴ 3x have middle school education and 7x have primary school education.
Also, as all middle school educational people have primary school education and all high school educated people have middle school education, number of people in Category I = 4x, Category II = 2x, Category III = x.

∴ Number of people in Category I and II, who do not play any game = x and x/2 respectively.
∴ Number of people playing only hockey = Number of people playing only football = (3x - 2)/2 and (3x/2 - 2)/2 = (3x - 4)/4 respectively

Also, number of people in Category III playing only hockey = (x - 6)/2 = 5 => x = 16

Now, 3x = 48 people have middle school education.
Number of high school educated people who do not play football = (x - 6)/2 + 5 = 10
Number of people having middle school education but not high school education who play only football = (x - 6)/2 = 11
Number of such people = Number of people having primary school education - Number of people having middle school education
= 7x - 3x = 4x = 64
Number of educated people playing football only = 39
Number of educated people playing hockey only = 39
∴ Number of uneducated people playing football = 90 - (39 + 5) = 46
and number of uneducated people playing hockey = 60 - (39 + 5) = 16
Out of these, 2 play both games.
∴ Number of uneducated people playing at least one game = 46 + 16 - 2 = 60
Number of uneducated people = 200 - 7x = 200 - 112 = 88
∴ 88 - 60 = 28 uneducated people do not play any game.

As per the given information, following four Venn diagrams can be drawn:




Number of uneducated people who play neither hockey nor football = 48

Sample Test: LR & DI- 1 - Question 20

Directions: Answer the given question based on the following data:

In a group of 200 people, number of people having at least primary education : number of people having at least middle school education: number of people having at least high school education are in the ratio 7 : 3 : 1. Out of these, 90 play football and 60 play hockey. Also, 5 in Category III (people having high school education) and one-fourth each in Categories I and II (people having primary school education only and people having middle school education but not high school education, respectively) do not play any game. In each of the above categories, the number of people who play only hockey equals the number of people who play only football. Two persons each in Categories I and II and one person in Category III play both the games. Two persons who play both the games are uneducated (Category IV). Five persons in Category III play only hockey.
Assume middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under one of the four categories described above.

Q. How many persons are there who play atleast one game?

Detailed Solution for Sample Test: LR & DI- 1 - Question 20

Let 'x' people have high school education.
∴ 3x have middle school education and 7x have primary school education.
Also, as all middle school educational people have primary school education and all high school educated people have middle school education, number of people in Category I = 4x, Category II = 2x, Category III = x.

∴ Number of people in Category I and II, who do not play any game = x and x/2 respectively.
∴ Number of people playing only hockey = Number of people playing only football = (3x - 2)/2 and (3x/2 - 2)/2 = (3x - 4)/4 respectively

Also, number of people in Category III playing only hockey = (x - 6)/2 = 5 => x = 16

Now, 3x = 48 people have middle school education.
Number of high school educated people who do not play football = (x - 6)/2 + 5 = 10
Number of people having middle school education but not high school education who play only football = (x - 6)/2 = 11
Number of such people = Number of people having primary school education - Number of people having middle school education
= 7x - 3x = 4x = 64
Number of educated people playing football only = 39
Number of educated people playing hockey only = 39
∴ Number of uneducated people playing football = 90 - (39 + 5) = 46
and number of uneducated people playing hockey = 60 - (39 + 5) = 16
Out of these, 2 play both games.
∴ Number of uneducated people playing at least one game = 46 + 16 - 2 = 60
Number of uneducated people = 200 - 7x = 200 - 112 = 88
∴ 88 - 60 = 28 uneducated people do not play any game.

As per the given information, following four Venn diagrams can be drawn:




Number of persons who play atleast one game = 200 - Number of persons who do not play any game
Number of persons who play atleast one game = 200 - (29 + 48) = 200 - 77 = 123

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