Logical Reasoning & DI (LRDI) Test

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Logical Reasoning & DI (LRDI) Test - 27

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Question 1

3 / -1

Directions For Questions

A student appears for a management exam which is conducted in multiple-choice question type. The exam pattern has five sections in it as English, Quant, Verbal, General Aptitude, and LRDI. The following table shows the points scored by the candidate in each of the sections. Note that some details are absent from the table:

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50
QUANTS	60	58
VERBAL	80	45		22
GK	70				120
LRDI	80		50

It is also known that:

1. For every correct answer, 5 marks are awarded and for every incorrect attempt, 2 marks are deducted. Also, 1 mark is lost for every unattempted problem.

2. Total marks scored in the Verbal section is ½ of the marks scored in the English section

3. The number of incorrect attempts in GK is 1/4 of his incorrect attempts in all the sections and is twice that in English.

4. In all the sections together, the total attempts are 260 and the total marks scored is 660.

Using the above information given answer the following questions:

...view full instructions

Find the total marks scored by the candidate in English section?

72

82

36

232 Solution

EXPLANATION:

According to statement 4, total attempt is 260 and the total marks scored is 660.

⇒ From the table, total no of questions = 50 + 60 + 80 + 70 + 80 = 340

⇒ Total no of unattempted questions= 340 - 260 = 80

Important Points

Let’s assume x number of incorrect attempts

∴ (260 - x) × 5 - 2x - 80 × 1 = 660

⇒ 1300 - 5x - 2x - 80 = 660

⇒ 1220 - 7x = 660

⇒ 7x = 560

⇒ x = 80

∴ Total no of incorrect attempts = 80

⇒ Total no of correct attempts = 260 - 80 = 180

From statement 3, we calculate the total no of incorrect attempts in Section 4 = 1/4 × 80 = 20

⇒ Total no of incorrect attempts in English = 20/2 = 10

Also, total correct answers in Verbal section = (45 - 22) = 23

⇒ Unattempted = 80 - 45 = 35

⇒ Total score scored in Verbal section= 23 × 5 - 22 × 2 - 35 × 1 = 115 - 44 - 35 = 115 - 79 = 36

∴ From statement 2, total marks scored in English section = 36 × 2 = 72.

Assume total no of correct attempts in English section = a

⇒ Total no of attempted questions = a + 10( as 10 is incorrect)

⇒ Total no of unattempted answers = 50 - (a +10) = 40 - a

∴ a × 5 - (10 × 2) - (40 - a) × 1 = 72

⇒ 5a - 20 - 40 + a = 72

⇒ 6a - 60 = 72

⇒ 6a = 132

⇒ a = 22

Therefore, correct attempts in English section = 22

⇒ Unattempted questions = 40 - 22 = 18

and total attempted questions = 22 + 10 = 32

Now, take no of correct questions in GK section as m

⇒ Attempted questions = m + 20

⇒ Unattempted questions = 70 - (m + 20) = 50 - m

⇒ Total marks = m × 5 - 20 × 2 - (50 - m) × 1 = 120

⇒ 5m - 40 - 50 + m = 120

⇒ 6m - 90 = 120

⇒ 6m = 210

⇒ m = 35

∴ Total correct questions in GK section = 35

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80		50
TOTAL	340	260	180	80	660

From the table, we can now fill the remaining spaces

⇒ Correct attempts in Quants = 180 - (22 + 23 + 35 + 50) = 50

∴ Incorrect attempts in Quants = 58 - 50 = 8

⇒ Attempted questions in LRDI = 260 - (32 + 58 + 45 + 55) = 70

∴ Incorrect attempts in LRDI = 70 - 50 = 20

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20
TOTAL	340	260	180	80	660

Total marks in Quants = 50 × 5 - 8 × 2 - (60 - 58) × 1

⇒ 250 - 16 -2

⇒ 250 - 18

⇒ 232

Total marks in LRDI = 50 × 5 - 20 × 2 - (80 - 70) × 1

⇒ 250 - 40 -10

⇒ 250 - 50

⇒ 200

Total marks in LRDI = 660 - (72 + 232 + 36 + 120)

⇒ 660 - 460

⇒ 200

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8	232
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20	200
TOTAL	340	260	180	80	660

∴ The total marks scored by the candidate in the English section is 72.

Question 2

3 / -1

Directions For Questions

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50
QUANTS	60	58
VERBAL	80	45		22
GK	70				120
LRDI	80		50

It is also known that:

1. For every correct answer, 5 marks are awarded and for every incorrect attempt, 2 marks are deducted. Also, 1 mark is lost for every unattempted problem.

2. Total marks scored in the Verbal section is ½ of the marks scored in the English section

3. The number of incorrect attempts in GK is 1/4 of his incorrect attempts in all the sections and is twice that in English.

4. In all the sections together, the total attempts are 260 and the total marks scored is 660.

Using the above information given answer the following questions:

...view full instructions

Find the total marks deducted in the Quants section because of incorrect questions?

14

13

12

16 Solution

EXPLANATION:

According to statement 4, total attempt is 260 and the total marks scored is 660.

⇒ From the table, total no of questions = 50 + 60 + 80 + 70 + 80 = 340

⇒ Total no of unattempted questions= 340 - 260 = 80

Important Points

Let’s assume x number of incorrect attempts

∴ (260 - x) × 5 - 2x - 80 × 1 = 660

⇒ 1300 - 5x - 2x - 80 = 660

⇒ 1220 - 7x = 660

⇒ 7x = 560

⇒ x = 80

∴ Total no of incorrect attempts = 80

⇒ Total no of correct attempts = 260 - 80 = 180

From statement 3, we calculate the total no of incorrect attempts in Section 4 = 1/4 × 80 = 20

⇒ Total no of incorrect attempts in English = 20/2 = 10

Also, total correct answers in Verbal section = (45 - 22) = 23

⇒ Unattempted = 80 - 45 = 35

⇒ Total score scored in Verbal section= 23 × 5 - 22 × 2 - 35 × 1 = 115 - 44 - 35 = 115 - 79 = 36

∴ From statement 2, total marks scored in English section = 36 × 2 = 72.

Assume total no of correct attempts in English section = a

⇒ Total no of attempted questions = a + 10( as 10 is incorrect)

⇒ Total no of unattempted answers = 50 - (a +10) = 40 - a

∴ a × 5 - (10 × 2) - (40 - a) × 1 = 72

⇒ 5a - 20 - 40 + a = 72

⇒ 6a - 60 = 72

⇒ 6a = 132

⇒ a = 22

Therefore, correct attempts in English section = 22

⇒ Unattempted questions = 40 - 22 = 18

and total attempted questions = 22 + 10 = 32

Now, take no of correct questions in GK section as m

⇒ Attempted questions = m + 20

⇒ Unattempted questions = 70 - (m + 20) = 50 - m

⇒ Total marks = m × 5 - 20 × 2 - (50 - m) × 1 = 120

⇒ 5m - 40 - 50 + m = 120

⇒ 6m - 90 = 120

⇒ 6m = 210

⇒ m = 35

∴ Total correct questions in GK section = 35

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80		50
TOTAL	340	260	180	80	660

From the table, we can now fill the remaining spaces

⇒ Correct attempts in Quants = 180 - (22 + 23 + 35 + 50) = 50

∴ Incorrect attempts in Quants = 58 - 50 = 8

⇒ Attempted questions in LRDI = 260 - (32 + 58 + 45 + 55) = 70

∴ Incorrect attempts in LRDI = 70 - 50 = 20

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20
TOTAL	340	260	180	80	660

Total marks in Quants = 50 × 5 - 8 × 2 - (60 - 58) × 1

⇒ 250 - 16 -2

⇒ 250 - 18

⇒ 232

Total marks in LRDI = 50 × 5 - 20 × 2 - (80 - 70) × 1

⇒ 250 - 40 -10

⇒ 250 - 50

⇒ 200

Total marks in LRDI = 660 - (72 + 232 + 36 + 120)

⇒ 660 - 460

⇒ 200

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8	232
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20	200
TOTAL	340	260	180	80	660

∴ The total marks deducted due to incorrect questions attempted in the Quants section = 8 × 2 = 16.

Question 3

3 / -1

Directions For Questions

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50
QUANTS	60	58
VERBAL	80	45		22
GK	70				120
LRDI	80		50

It is also known that:

1. For every correct answer, 5 marks are awarded and for every incorrect attempt, 2 marks are deducted. Also, 1 mark is lost for every unattempted problem.

2. Total marks scored in the Verbal section is ½ of the marks scored in the English section

3. The number of incorrect attempts in GK is 1/4 of his incorrect attempts in all the sections and is twice that in English.

4. In all the sections together, the total attempts are 260 and the total marks scored is 660.

Using the above information given answer the following questions:

...view full instructions

Find the total marks scored by the candidate in section LRDI?

120

72

200 none of these

Solution

EXPLANATION:

According to statement 4, total attempt is 260 and the total marks scored is 660.

⇒ From the table, total no of questions = 50 + 60 + 80 + 70 + 80 = 340

⇒ Total no of unattempted questions= 340 - 260 = 80

Important Points

Let’s assume x number of incorrect attempts

∴ (260 - x) × 5 - 2x - 80 × 1 = 660

⇒ 1300 - 5x - 2x - 80 = 660

⇒ 1220 - 7x = 660

⇒ 7x = 560

⇒ x = 80

∴ Total no of incorrect attempts = 80

⇒ Total no of correct attempts = 260 - 80 = 180

From statement 3, we calculate the total no of incorrect attempts in Section 4 = 1/4 × 80 = 20

⇒ Total no of incorrect attempts in English = 20/2 = 10

Also, total correct answers in Verbal section = (45 - 22) = 23

⇒ Unattempted = 80 - 45 = 35

⇒ Total score scored in Verbal section= 23 × 5 - 22 × 2 - 35 × 1 = 115 - 44 - 35 = 115 - 79 = 36

∴ From statement 2, total marks scored in English section = 36 × 2 = 72.

Assume total no of correct attempts in English section = a

⇒ Total no of attempted questions = a + 10( as 10 is incorrect)

⇒ Total no of unattempted answers = 50 - (a +10) = 40 - a

∴ a × 5 - (10 × 2) - (40 - a) × 1 = 72

⇒ 5a - 20 - 40 + a = 72

⇒ 6a - 60 = 72

⇒ 6a = 132

⇒ a = 22

Therefore, correct attempts in English section = 22

⇒ Unattempted questions = 40 - 22 = 18

and total attempted questions = 22 + 10 = 32

Now, take no of correct questions in GK section as m

⇒ Attempted questions = m + 20

⇒ Unattempted questions = 70 - (m + 20) = 50 - m

⇒ Total marks = m × 5 - 20 × 2 - (50 - m) × 1 = 120

⇒ 5m - 40 - 50 + m = 120

⇒ 6m - 90 = 120

⇒ 6m = 210

⇒ m = 35

∴ Total correct questions in GK section = 35

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80		50
TOTAL	340	260	180	80	660

From the table, we can now fill the remaining spaces

⇒ Correct attempts in Quants = 180 - (22 + 23 + 35 + 50) = 50

∴ Incorrect attempts in Quants = 58 - 50 = 8

⇒ Attempted questions in LRDI = 260 - (32 + 58 + 45 + 55) = 70

∴ Incorrect attempts in LRDI = 70 - 50 = 20

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20
TOTAL	340	260	180	80	660

Total marks in Quants = 50 × 5 - 8 × 2 - (60 - 58) × 1

⇒ 250 - 16 -2

⇒ 250 - 18

⇒ 232

Total marks in LRDI = 50 × 5 - 20 × 2 - (80 - 70) × 1

⇒ 250 - 40 -10

⇒ 250 - 50

⇒ 200

Total marks in LRDI = 660 - (72 + 232 + 36 + 120)

⇒ 660 - 460

⇒ 200

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8	232
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20	200
TOTAL	340	260	180	80	660

∴ Total marks scored by the candidate in section LRDI = 200

Question 4

3 / -1

Directions For Questions

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50
QUANTS	60	58
VERBAL	80	45		22
GK	70				120
LRDI	80		50

It is also known that:

1. For every correct answer, 5 marks are awarded and for every incorrect attempt, 2 marks are deducted. Also, 1 mark is lost for every unattempted problem.

2. Total marks scored in the Verbal section is ½ of the marks scored in the English section

3. The number of incorrect attempts in GK is 1/4 of his incorrect attempts in all the sections and is twice that in English.

4. In all the sections together, the total attempts are 260 and the total marks scored is 660.

Using the above information given answer the following questions:

...view full instructions

Suppose the passing marks in all of the sections of the test is 130, then calculate in how many sections was the candidate able to pass?

3

4

2

1 Solution

EXPLANATION:

According to statement 4, total attempt is 260 and the total marks scored is 660.

⇒ From the table, total no of questions = 50 + 60 + 80 + 70 + 80 = 340

⇒ Total no of unattempted questions= 340 - 260 = 80

Important Points

Let’s assume x number of incorrect attempts

∴ (260 - x) × 5 - 2x - 80 × 1 = 660

⇒ 1300 - 5x - 2x - 80 = 660

⇒ 1220 - 7x = 660

⇒ 7x = 560

⇒ x = 80

∴ Total no of incorrect attempts = 80

⇒ Total no of correct attempts = 260 - 80 = 180

From statement 3, we calculate the total no of incorrect attempts in Section 4 = 1/4 × 80 = 20

⇒ Total no of incorrect attempts in English = 20/2 = 10

Also, total correct answers in Verbal section = (45 - 22) = 23

⇒ Unattempted = 80 - 45 = 35

⇒ Total score scored in Verbal section= 23 × 5 - 22 × 2 - 35 × 1 = 115 - 44 - 35 = 115 - 79 = 36

∴ From statement 2, total marks scored in English section = 36 × 2 = 72.

Assume total no of correct attempts in English section = a

⇒ Total no of attempted questions = a + 10( as 10 is incorrect)

⇒ Total no of unattempted answers = 50 - (a +10) = 40 - a

∴ a × 5 - (10 × 2) - (40 - a) × 1 = 72

⇒ 5a - 20 - 40 + a = 72

⇒ 6a - 60 = 72

⇒ 6a = 132

⇒ a = 22

Therefore, correct attempts in English section = 22

⇒ Unattempted questions = 40 - 22 = 18

and total attempted questions = 22 + 10 = 32

Now, take no of correct questions in GK section as m

⇒ Attempted questions = m + 20

⇒ Unattempted questions = 70 - (m + 20) = 50 - m

⇒ Total marks = m × 5 - 20 × 2 - (50 - m) × 1 = 120

⇒ 5m - 40 - 50 + m = 120

⇒ 6m - 90 = 120

⇒ 6m = 210

⇒ m = 35

∴ Total correct questions in GK section = 35

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80		50
TOTAL	340	260	180	80	660

From the table, we can now fill the remaining spaces

⇒ Correct attempts in Quants = 180 - (22 + 23 + 35 + 50) = 50

∴ Incorrect attempts in Quants = 58 - 50 = 8

⇒ Attempted questions in LRDI = 260 - (32 + 58 + 45 + 55) = 70

∴ Incorrect attempts in LRDI = 70 - 50 = 20

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20
TOTAL	340	260	180	80	660

Total marks in Quants = 50 × 5 - 8 × 2 - (60 - 58) × 1

⇒ 250 - 16 -2

⇒ 250 - 18

⇒ 232

Total marks in LRDI = 50 × 5 - 20 × 2 - (80 - 70) × 1

⇒ 250 - 40 -10

⇒ 250 - 50

⇒ 200

Total marks in LRDI = 660 - (72 + 232 + 36 + 120)

⇒ 660 - 460

⇒ 200

SECTIONS	TOTAL NO OF QUESTIONS	ATTEMPTED QUESTIONS	CORRECT ATTEMPTS	INCORRECT ATTEMPTS	TOTAL
ENGLISH	50	32	22	10	72
QUANTS	60	58	50	8	232
VERBAL	80	45	23	22	36
GK	70	55	35	20	120
LRDI	80	70	50	20	200
TOTAL	340	260	180	80	660

From the table, it is clear that the candidate passed only in the Quants and LRDI section only.

∴ The candidate has passed only in 2 sections.

Question 5

3 / -1

Directions For Questions

A birthday party is going on at three different places and Crispy Burger was required to supply the burgers to these parties. The total number of burgers it had to deliver was 800, 70% of which were to be delivered to Third birthday party and the rest equally divided between First birthday party and Second birthday party. Burgers could be of King Size (K) or Large (L) variety and come in either with Cheese (C) or without Cheese (WC) versions.

Therefore, there are four types of burgers: K-C, K-WC, L-C and L-WC. Some other information about proportions of K and C burgers ordered is given below:

	King Size (K)	Cheese (C)
First birthday party	0.6
Second birthday party	0.55	0.3
Third birthday party		0.65
Total	0.375	0.52

...view full instructions

How many King size burgers are to be delivered to the third birthday party?

398

162

96

364 Solution

Explanation:

As per the question, there are 3 birthday parties and the Crispy Burger has been given the orders of sending the burgers to all these 3 parties.

These burgers are of two types and are either with cheese(C) or without cheese(WC); this means that there are total four combinations of the burgers, i.e.

1) King size- with cheese- K-C

2) King size- without cheese- K-WC

3) Large- with cheese- L-C

4) Large- without cheese- L-WC

Important Point:

Also, it is told that the total number of burgers being ordered by the three parties is 800 and that 70% of these are to be delivered to the third birthday party and the remaining ones have to be equally distributed between the first and the second birthday party.

This means that the number of burgers with the respective parties is:

⇒ Third birthday party- 70/100 × 800 = 560.

Remaining burgers to be distributed equally, i.e.

⇒ 800 - 560 = 240.

So, 240 burgers to be divided equally between the first and the second birthday party. This means the number of burgers with the parties is:

⇒ First birthday party - 120.

⇒ Second birthday party- 120.

Key Points

From the given table, we know that the total proportion of King Size burgers is 0.375.

Therefore, the total number of King Size (K) burgers is 0.375 × 800 = 300.

Thus, the total number of Large (L) burgers is 800 - 300 = 500.

Key Points

It is also told that the proportion of King Size burgers received by the first birthday party is 0.6, i.e. 0.6 × 120 = 72.

And, by the second birthday party is 0.55, i.e. 0.55 × 120 = 66.

Thus, the number of King Size burgers ordered by the third birthday party is:

⇒ 300 - (72 + 66) = 162.

Now, the total number of Large Size burgers ordered by the parties is:

⇒ First birthday party = 120 - 72 = 48

⇒ Second birthday party = 120 - 66 = 54

⇒ Third birthday party = 560 - 162 = 398.

Let us put all this information in a table.

	Total number of burgers	King Size - C	King Size-WC	King Size (K)-Total	Large-C	Large- WC	Large (L)-Total
First birthday party	120			72			48
Second birthday party	120			66			54
Third birthday party	560			162			398
Total	800			300			500

Hence, from the table, we can see that the total number of King size burgers ordered by the third birthday party is 162.

Question 6

3 / -1

Directions For Questions

Therefore, there are four types of burgers: K-C, K-WC, L-C and L-WC. Some other information about proportions of K and C burgers ordered is given below:

	King Size (K)	Cheese (C)
First birthday party	0.6
Second birthday party	0.55	0.3
Third birthday party		0.65
Total	0.375	0.52

...view full instructions

How many Large size burgers were ordered by the First birthday party?

36

66

48

72 Solution

Explanation:

As per the question, there are 3 birthday parties and the Crispy Burger has been given the orders of sending the burgers to all these 3 parties.

These burgers are of two types and are either with cheese(C) or without cheese(WC); this means that there are a total of four combinations of the burgers, i.e.

1) King size- with cheese- K-C

2) King size- without cheese- K-WC

3) Large- with cheese- L-C

4) Large- without cheese- L-WC

Important Point:

Also, it is told that the total number of burgers being ordered by the three parties is 800 and that 70% of these are to be delivered to the third birthday party and the remaining ones have to be equally distributed between the first and the second birthday party.

This means that the number of burgers with the respective parties is:

⇒ Third birthday party- 70/100 × 800 = 560.

Remaining burgers to be distributed equally,i.e.

⇒ 800 - 560 = 240.

So, 240 burgers to be divided equally between the first and the second birthday party. This means the number of burgers with the parties is:

⇒ First birthday party - 120.

⇒ Second birthday party- 120.

Key Points

From the given table, we know that the total proportion of King Size burgers is 0.375.

Therefore, the total number of King Size (K) burgers is 0.375 × 800 = 300.

Thus, the total number of Large (L) burgers is 800 - 300 = 500.

Key Points

It is also told that the proportion of King Size burgers received by the first birthday party is 0.6, i.e. 0.6 × 120 = 72.

And, by the second birthday party is 0.55, i.e. 0.55 × 120 = 66.

Thus, the number of King Size burgers ordered by the third birthday party is:

⇒ 300 - (72 + 66) = 162.

Now, the total number of Large Size burgers ordered by the parties is:

⇒ First birthday party = 120 - 72 = 48

⇒ Second birthday party = 120 - 66 = 54

⇒ Third birthday party = 560 - 162 = 398.

Let us put all this information on a table.

	Total number of burgers	King Size - C	King Size-WC	King Size (K)-Total	Large-C	Large- WC	Large (L)-Total
First birthday party	120			72			48
Second birthday party	120			66			54
Third birthday party	560			162			398
Total	800			300			500

Hence, from the table, we can see that the total number of Large size burgers ordered by the first birthday party is 48.

Question 7

3 / -1

Directions For Questions

Therefore, there are four types of burgers: K-C, K-WC, L-C and L-WC. Some other information about proportions of K and C burgers ordered is given below:

	King Size (K)	Cheese (C)
First birthday party	0.6
Second birthday party	0.55	0.3
Third birthday party		0.65
Total	0.375	0.52

...view full instructions

What is the total number of Large size burgers ordered by all the three parties?

800

500

300

120 Solution

Explanation:

As per the question, there are 3 birthday parties and the Crispy Burger has been given the orders of sending the burgers to all these 3 parties.

These burgers are of two types and are either with cheese(C) or without cheese(WC); this means that there are total of four combinations of the burgers, i.e.

1) King size- with cheese- K-C

2) King size- without cheese- K-WC

3) Large- with cheese- L-C

4) Large- without cheese- L-WC

Important Point:

Also, it is told that the total number of burgers being ordered by the three parties is 800 and that 70% of these are to be delivered to the third birthday party and the remaining ones have to be equally distributed between the first and the second birthday party.

This means that the number of burgers with the respective parties is:

⇒ Third birthday party- 70/100 × 800 = 560.

Remaining burgers to be distributed equally,i.e.

⇒ 800 - 560 = 240.

So, 240 burgers to be divided equally between the first and the second birthday party. This means the number of burgers with the parties is:

⇒ First birthday party - 120.

⇒ Second birthday party- 120.

Key Points

From the given table, we know that the total proportion of King Size burgers is 0.375.

Therefore, the total number of King Size (K) burgers is 0.375 × 800 = 300.

Thus, the total number of Large (L) burgers is 800 - 300 = 500.

Key Points

It is also told that the proportion of King Size burgers received by the first birthday party is 0.6, i.e. 0.6 × 120 = 72.

And, by the second birthday party is 0.55, i.e. 0.55 × 120 = 66.

Thus, the number of King Size burgers ordered by the third birthday party is:

⇒ 300 - (72 + 66) = 162.

Now, the total number of Large Size burgers ordered by the parties is:

⇒ First birthday party = 120 - 72 = 48

⇒ Second birthday party = 120 - 66 = 54

⇒ Third birthday party = 560 - 162 = 398.

Let us put all this information on a table.

	Total number of burgers	King Size - C	King Size-WC	King Size (K)-Total	Large-C	Large- WC	Large (L)-Total
First birthday party	120			72			48
Second birthday party	120			66			54
Third birthday party	560			162			398
Total	800			300			500

Hence, from the table, we can see that the total number of Large Size burgers ordered by all the three birthday parties is 500.

Question 8

3 / -1

Directions For Questions

Therefore, there are four types of burgers: K-C, K-WC, L-C and L-WC. Some other information about proportions of K and C burgers ordered is given below:

	King Size (K)	Cheese (C)
First birthday party	0.6
Second birthday party	0.55	0.3
Third birthday party		0.65
Total	0.375	0.52

...view full instructions

For the second birthday party, if Large Size Burgers with cheese variety are 50% of the Large Size burgers were from Without Cheese variety, what was the difference between the numbers of K-WC and L-WC burgers to be delivered to the second birthday party?

12

20

48

36 Solution

Explanation:

As per the question, there are 3 birthday parties and the Crispy Burger has been given the orders of sending the burgers to all these 3 parties.

These burgers are of two types and are either with cheese(C) or without cheese(WC); this means that there are total of four combinations of the burgers, i.e.

1) King size- with cheese- K-C

2) King size- without cheese- K-WC

3) Large- with cheese- L-C

4) Large- without cheese- L-WC

Important Point:

Also, it is told that the total number of burgers being ordered by the three parties is 800 and that 70% of these are to be delivered to the third birthday party and the remaining ones have to be equally distributed between the first and the second birthday party.

This means that the number of burgers with the respective parties is:

⇒ Third birthday party- 70/100 × 800 = 560.

Remaining burgers to be distributed equally,i.e.

⇒ 800 - 560 = 240.

So, 240 burgers to be divided equally between the first and the second birthday party. This means the number of burgers with the parties is:

⇒ First birthday party - 120.

⇒ Second birthday party- 120.

Key Points

From the given table, we know that the total proportion of King Size burgers is 0.375.

Therefore, the total number of King Size (K) burgers is 0.375 × 800 = 300.

Thus, the total number of Large (L) burgers is 800 - 300 = 500.

Key Points

It is also told that the proportion of King Size burgers received by the first birthday party is 0.6, i.e. 0.6 × 120 = 72.

And, by the second birthday party is 0.55, i.e. 0.55 × 120 = 66.

Thus, the number of King Size burgers ordered by the third birthday party is:

⇒ 300 - (72 + 66) = 162.

Now, the total number of Large Size burgers ordered by the parties is:

⇒ First birthday party = 120 - 72 = 48

⇒ Second birthday party = 120 - 66 = 54

⇒ Third birthday party = 560 - 162 = 398.

Let us put all this information on a table.

	Total number of burgers	King Size - C	King Size-WC	King Size (K)-Total	Large-C	Large- WC	Large (L)-Total
First birthday party	120			72			48
Second birthday party	120			66			54
Third birthday party	560			162			398
Total	800			300			500

Now, the number of Cheeseburgers required to be delivered to the second birthday party is 0.3 × 120 = 36.

As per the question, Large Size Burgers with cheese variety are 50% of the Large Size burgers were from Without Cheese variety

Now, the 4 varieties of burgers required to be delivered to second birthday party are:

Let the number of Large Size burgers were from Without Cheese variety = 2x

⇒ Large (C) + Large (WC) = Large (total) = 54 = x + 2x

⇒ Large (WC) - 54 - 18 = 36

Large (C) = 18

⇒ King Size (C) = Total(c)- Large(c). = 36 - 18 =18

⇒ King Size (WC) = King(total) - King(C) = 66 - 18 = 48.

	Total number of burgers	King Size - C	King Size-WC	King Size (K)-Total	Large-C	Large- WC	Large (L)-Total
First birthday party	120			72			48
Second birthday party	120	18	48	66	18	36	54
Third birthday party	560			162			398
Total	800			300			500

Thus, the difference between the K-WC & L-WC is:

⇒ 48 - 36 = 12.

Question 9

3 / -1

Directions For Questions

Read the given information and answer the following questions:

Seven persons P, Q, R, S, T, U and V like different colors namely blue, black, brown, white, green, orange and red but not necessarily in the same order. They hold different positions in the Indian Army as Captain, Major, Colonel, Brigadier, Major General, General and Field Marshal. The positions given are in increasing order of seniority such that Captain is the junior most position and Field Marshal is the senior most position. P is senior to only three people. Person, who likes brown, is senior to Brigadier. R is junior to the one, who likes black and senior to Colonel. The person, who likes black, is not the most senior. Q, who like white is junior to P but not the least junior. U is senior to one, who likes white and junior to one, who likes brown. T is junior to Brigadier. The one, who likes blue is junior to V and senior to Colonel, who likes green. U does not like black. S likes red.

...view full instructions

Who among the following likes Orange color?

R

T

Brigadier

Major

Solution

Persons: P, Q, R, S, T, U and V.

Positions: Field Marshal, General, Major General, Brigadier, Colonel, Major and Captain.

Color: Blue, Black, Brown, White, Green, Orange and Red

1. P is senior to only three people.

2. Q, who like white is junior to P but not the least junior.

3. Colonel likes green.

So, Q is a Major.

POSITIONS	PERSONS	COLOR
Field Marshal
General
Major General
Brigadier	P
Colonel		Green
Major	Q	White
Captain

4. R is junior to the one, who likes black and senior to Colonel.

5. The person, who likes black, is not the most senior.

So, R must be Major General and the person, who likes black must be General.

POSITIONS	PERSONS	COLOR
Field Marshal
General		Black
Major General	R
Brigadier	P
Colonel		Green
Major	Q	White
Captain

6. U is senior to one, who likes white and junior to one, who likes brown.

7. U does not like black. So, U likes Green.

8. Person, who likes brown, is senior to Brigadier.

	CASE 1		CASE 2
POSITIONS	PERSONS	COLOR	PERSONS	COLOR
Field Marshal				Brown
General		Black		Black
Major General	R	Brown	R
Brigadier	P		P
Colonel	U	Green	U	Green
Major	Q	White	Q	White
Captain

9. T is junior to Brigadier.

10. The one, who likes blue is junior to V and senior to Colonel.

Here in case 1 Blue can go to Brigadier only not Field marshal as blue is Junior to V.In case 2 Blue can be assigned to 2 people.

	CASE 1		CASE 2		CASE 2A
POSITIONS	PERSONS	COLOR	PERSONS	COLOR	PERSONS	COLOR
Field Marshal				Brown		Brown
General		Black		Black		Black
Major General	R	Brown	R	Blue	R
Brigadier	P	Blue	P		P	Blue
Colonel	U	Green	U	Green	U	Green
Major	Q	White	Q	White	Q	White
Captain	T		T		T

11. S likes red. But in case 2 and case 2A there is no space left for us to assign S and Red as a pair, hence both these cases are eliminated.

	CASE 1
POSITIONS	PERSONS	COLOR
Field Marshal	S	Red
General		Black
Major General	R	Brown
Brigadier	P	Blue
Colonel	U	Green
Major	Q	White
Captain	T

12. The remaining color is orange. So, T likes orange.

13. The remaining person is V. So, V likes Black.

The final arrangement is,

POSITIONS	PERSONS	COLOR
Field Marshal	S	Red
General	V	Black
Major General	R	Brown
Brigadier	P	Blue
Colonel	U	Green
Major	Q	White
Captain	T	Orange

Hence, T likes orange color.

Question 10

3 / -1

Directions For Questions

Read the given information and answer the following questions:

...view full instructions

Who is the immediate senior of Brigadier?

S

V

Q

R

Solution

Persons: P, Q, R, S, T, U and V.

Positions: Field Marshal, General, Major General, Brigadier, Colonel, Major and Captain.

Color: Blue, Black, Brown, White, Green, Orange and Red

1. P is senior to only three people.

2. Q, who like white is junior to P but not the least junior.

3. Colonel likes green.

So, Q is a Major.

POSITIONS	PERSONS	COLOR
Field Marshal
General
Major General
Brigadier	P
Colonel		Green
Major	Q	White
Captain

4. R is junior to the one, who likes black and senior to Colonel.

5. The person, who likes black, is not the most senior.

So, R must be Major General and the person, who likes black must be General.

POSITIONS	PERSONS	COLOR
Field Marshal
General		Black
Major General	R
Brigadier	P
Colonel		Green
Major	Q	White
Captain

6. U is senior to one, who likes white and junior to one, who likes brown.

7. U does not like black. So, U likes Green.

8. Person, who likes brown, is senior to Brigadier.

	CASE 1		CASE 2
POSITIONS	PERSONS	COLOR	PERSONS	COLOR
Field Marshal				Brown
General		Black		Black
Major General	R	Brown	R
Brigadier	P		P
Colonel	U	Green	U	Green
Major	Q	White	Q	White
Captain

9. T is junior to Brigadier.

10. The one, who likes blue is junior to V and senior to Colonel.

Here in case 1 Blue can go to Brigadier only not Field marshal as blue is Junior to V.In case 2 Blue can be assigned to 2 people.

	CASE 1		CASE 2		CASE 2A
POSITIONS	PERSONS	COLOR	PERSONS	COLOR	PERSONS	COLOR
Field Marshal				Brown		Brown
General		Black		Black		Black
Major General	R	Brown	R	Blue	R
Brigadier	P	Blue	P		P	Blue
Colonel	U	Green	U	Green	U	Green
Major	Q	White	Q	White	Q	White
Captain	T		T		T

11. S likes red. But in case 2 and case 2A there is no space left for us to assign S and Red as a pair, hence both these cases are eliminated.

	CASE 1
POSITIONS	PERSONS	COLOR
Field Marshal	S	Red
General		Black
Major General	R	Brown
Brigadier	P	Blue
Colonel	U	Green
Major	Q	White
Captain	T

12. The remaining color is orange. So, T likes orange.

13. The remaining person is V. So, V likes Black.

The final arrangement is,

POSITIONS	PERSONS	COLOR
Field Marshal	S	Red
General	V	Black
Major General	R	Brown
Brigadier	P	Blue
Colonel	U	Green
Major	Q	White
Captain	T	Orange

Hence, R is immediate senior to Brigadier.

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Logical Reasoning & DI (LRDI) Test - 27

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Logical Reasoning & DI (LRDI) Test - 27

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SHARING IS CARING

Question 1

72

82

36

232

Solution

Question 2

14

13

12

16

Solution

Question 3

120

72

200

none of these

Solution

Question 4

3

4

2

1

Solution

Question 5

398

162

96

364

Solution

Question 6

36

66

48

72

Solution

Question 7

800

500

300

120

Solution

Question 8

12

20

48

36

Solution

Question 9

R

T

Brigadier

Major

Solution

Question 10

S

V

Q

R

Solution

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