Logical Reasoning & DI (LRDI) Test

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Logical Reasoning & DI (LRDI) Test - 14

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Question 1
3 / -1

Directions For Questions

King Porus was defending his last fort against the army of Alexander. He had only 40 soldiers left in his army. He arranged his army in all directions in a square frame(as given below), in 8 flanks, with 11 soldiers always defending each side of the fort. The number of soldiers in any flank is a whole number, and the set containing the number of soldiers in each flank can have at most 3 distinct numbers.

Porus lost his 4 soldiers during each of the first, second, third, and fourth attacks and 2 soldiers during the fifth attack. The rest of the soldiers were taken captive after the sixth attack.

Alexander lauded Porus' war acumen after the last attack, as he noticed that the number of soldiers defending each side of the square remained the same in every attack.

...view full instructions

In how many ways can Porus deploy his army for the first three attacks?

A
35

B
10

C
105

D
70

Solution

Total number of soldiers = 40

The possible ways to arrange the 40 soldiers are:

After the first attack, the total number of soldiers left = $$40-4\ =\ 36$$

The possible deployments at any side of the square will be 3-7-1, 2-7-2, and 4-7-0.

The possible arrangements of 7-3-1 will be :

Thus, the total number of arrangements will be 2.

Similarly, for 4-7-0, the total arrangements will be 2.

For 2-7-2, the total number of arrangements will be 1.

After the second attack, the total number of soldiers left = $$40-\left(2\times\ 4\right)\ =\ 32$$

The possible deployments at any side of the square will be 5-5-1, 4-5-2, 3-5-3, and 5-6-0.

For 5-5-1, 4-5-2, and 5-6-0, the number of arrangements will be 2.

For 3-5-3, the number of arrangements will be 1.

Thus, the total number of arrangements = $$3\times\ \left(2\times\ 2+1\right)\times\ \left(2\times\ 3+1\right)\ =\ 105$$

Thus, the correct answer is C.
Question 2
3 / -1

Directions For Questions

King Porus was defending his last fort against the army of Alexander. He had only 40 soldiers left in his army. He arranged his army in all directions in a square frame(as given below), in 8 flanks, with 11 soldiers always defending each side of the fort. The number of soldiers in any flank is a whole number, and the set containing the number of soldiers in each flank can have at most 3 distinct numbers.

Porus lost his 4 soldiers during each of the first, second, third, and fourth attacks and 2 soldiers during the fifth attack. The rest of the soldiers were taken captive after the sixth attack.

Alexander lauded Porus' war acumen after the last attack, as he noticed that the number of soldiers defending each side of the square remained the same in every attack.

...view full instructions

In how many ways could Porus have arranged his soldiers for the fifth attack?

A
1

B
11

C
13

D
7

Solution

Number of soldiers left after the fourth attack = $$40-\left(4\times\ 4\right)\ =\ 24$$

The possible arrangement for the soldiers per side will be 5-1-5, 4-1-6, 3-1-7, 2-1-8, 1-1-9, and 0-1-10.

For 5-1-5,

Thus, the number of arrangements for 5-1-5 will be 1.

For 6-1-4,

Thus, the number of arrangements for 6-1-4 will be 2.

Similarly, for 3-1-7, 2-1-8, 1-1-9, and 0-1-10, the number of arrangements will be 2.

Thus, the total number of arrangements will be 11.

Thus, the correct option will be B.
Question 3
3 / -1

Directions For Questions

King Porus was defending his last fort against the army of Alexander. He had only 40 soldiers left in his army. He arranged his army in all directions in a square frame(as given below), in 8 flanks, with 11 soldiers always defending each side of the fort. The number of soldiers in any flank is a whole number, and the set containing the number of soldiers in each flank can have at most 3 distinct numbers.

Porus lost his 4 soldiers during each of the first, second, third, and fourth attacks and 2 soldiers during the fifth attack. The rest of the soldiers were taken captive after the sixth attack.

Alexander lauded Porus' war acumen after the last attack, as he noticed that the number of soldiers defending each side of the square remained the same in every attack.

...view full instructions

If the sum of the soldiers occupying the N and N-W flank after the second attack is the maximum possible, what is the maximum sum of any 5 flanks?

A
25

B
26

C
27

D
29

Solution

The number of soldiers left after the second attack = $$40-\left(2\times\ 4\right)\ =\ 32$$

The possible deployments at any side of the square will be 5-5-1, 4-5-2, 3-5-3, and 5-6-0.

For the maximum sum case, the deployment of the soldiers will be :

Thus, the maximum possible sum of any 5 flanks = $$5\times\ 3+6\times\ 2\ =\ 27$$

Thus, the correct option is C.
Question 4
3 / -1

Directions For Questions

King Porus was defending his last fort against the army of Alexander. He had only 40 soldiers left in his army. He arranged his army in all directions in a square frame(as given below), in 8 flanks, with 11 soldiers always defending each side of the fort. The number of soldiers in any flank is a whole number, and the set containing the number of soldiers in each flank can have at most 3 distinct numbers.

Porus lost his 4 soldiers during each of the first, second, third, and fourth attacks and 2 soldiers during the fifth attack. The rest of the soldiers were taken captive after the sixth attack.

Alexander lauded Porus' war acumen after the last attack, as he noticed that the number of soldiers defending each side of the square remained the same in every attack.

...view full instructions

In how many ways can Porus arrange his army for the next attack after the third attack?

A
9

B
7

C
13

D
4

Solution

The number of soldiers left after the second attack = $$40-\left(3\times\ 4\right)\ =\ 28$$

The possible deployments at any side of the square will be 4-3-4, 5-3-3, 6-3-2, 7-3-1, 8-3-0.

Case 4-3-4:

The number of possible arrangements will be 1.

Case 5-3-3:

The number of possible arrangements will be 2.

Similarly, for 6-3-2, 7-3-1, and 8-3-0, the number of cases will be 2.

Thus, the total number of cases = $$1+2\times\ 4=9$$

Thus, the correct option will be A.
Question 5
3 / -1

Directions For Questions

King Porus was defending his last fort against the army of Alexander. He had only 40 soldiers left in his army. He arranged his army in all directions in a square frame(as given below), in 8 flanks, with 11 soldiers always defending each side of the fort. The number of soldiers in any flank is a whole number, and the set containing the number of soldiers in each flank can have at most 3 distinct numbers.

Porus lost his 4 soldiers during each of the first, second, third, and fourth attacks and 2 soldiers during the fifth attack. The rest of the soldiers were taken captive after the sixth attack.

Alexander lauded Porus' war acumen after the last attack, as he noticed that the number of soldiers defending each side of the square remained the same in every attack.

...view full instructions

Porus received a tip about the fourth attack that Alexander's army would try to infiltrate from the North-West side.

In how many ways can he arrange his army without breaking the formation, such that he should deploy maximum possible soldiers on the nearest three flanks(of the attack)?

A
1

B
2

C
3

D
Data Inconsistent

Solution

After the third attack, the total number of soldiers left = $$40-\left(4\times\ 3\right)\ =\ 28$$

For the maximum number of soilders on the nearest 3 flanks, the pattern of soldiers deployed on any flank is 0-8-3.

The nearest 3 flanks of the attacks are N-W, N, and W.

Total soldiers in these three flanks are $$\left(8+3+3\right)=14$$

Since only one case is possible for the aforementioned condition,

The correct option is A.
Question 6
3 / -1

Directions For Questions

Ashok is the cricket coach of V.S.R public school. For an upcoming mini-cricket tournament, he has to select a team of six consisting of 2 opening batsmen, 2 Regular bowlers, an all-rounder and a wicket keeper. The following are the players in contention for the selection:
Opening Batsman - Right Handed - A, B; Left Handed - C, D
Regular Bowler - E - right- spinner, F - right - fast, G - left - spinner, H - left-fast
All-rounder - P right-handed batsman and left handed fast, Q - left-handed batsman and left - spinner
Wicketkeeper - R - right-handed batsman, S-left handed batsman
The selection is also subject to the following conditions:
(i) The team cannot consist of more than three right or left handed regular batsman
(ii) The team must consist of at least two left-handed bowler
(iii) The team must consist of exactly one spinner
(iv) C cannot play in a team with a left-handed spinner and D cannot play in a team with a right-handed spinner

Note: Opening batsmen are considered regular batsmen. The bowlers are not considered as regular batsman and the all-rounders are not considered as regular bowlers but are considered regular batsmen. Wicket keepers are considered as regular batsmen.

...view full instructions

If the team consists of A, R and P, then how many different combinations of teams are possible?

A
1

B
2

C
3

D
4

Solution

Since, A, R and P are right-handed batsman, the 4th regular batsman should be either C or D. Also, from (iii) exactly one among E, G or Q must be selected. Since an all-rounder is selected already, the spinner must be either G or E. Thus, the second bowler should be from among - F/H.
The different team combinations possible are given
Question 7
3 / -1

Directions For Questions

Ashok is the cricket coach of V.S.R public school. For an upcoming mini-cricket tournament, he has to select a team of six consisting of 2 opening batsmen, 2 Regular bowlers, an all-rounder and a wicket keeper. The following are the players in contention for the selection:
Opening Batsman - Right Handed - A, B; Left Handed - C, D
Regular Bowler - E - right- spinner, F - right - fast, G - left - spinner, H - left-fast
All-rounder - P right-handed batsman and left handed fast, Q - left-handed batsman and left - spinner
Wicketkeeper - R - right-handed batsman, S-left handed batsman
The selection is also subject to the following conditions:
(i) The team cannot consist of more than three right or left handed regular batsman
(ii) The team must consist of at least two left-handed bowler
(iii) The team must consist of exactly one spinner
(iv) C cannot play in a team with a left-handed spinner and D cannot play in a team with a right-handed spinner

Note: Opening batsmen are considered regular batsmen. The bowlers are not considered as regular batsman and the all-rounders are not considered as regular bowlers but are considered regular batsmen. Wicket keepers are considered as regular batsmen.

...view full instructions

If the team consists of A, R and P, then who must be present in the team if C is also selected?

A
F

B
H

C
G

D
none of these

Solution

Since, A, R, and P are right-handed batsman, the 4th regular batsman should be either C or D. Also, from (iii) exactly one among E, G or Q must be selected. Since an all-rounder is selected already, the spinner must be either G or E. Thus, the second bowler should be from among - F/H.
The different team combinations possible are given

From the table we can see that, H must also be present in the team is C is selected.
Question 8
3 / -1

Directions For Questions

Ashok is the cricket coach of V.S.R public school. For an upcoming mini-cricket tournament, he has to select a team of six consisting of 2 opening batsmen, 2 Regular bowlers, an all-rounder and a wicket keeper. The following are the players in contention for the selection:
Opening Batsman - Right Handed - A, B; Left Handed - C, D
Regular Bowler - E - right- spinner, F - right - fast, G - left - spinner, H - left-fast
All-rounder - P right-handed batsman and left handed fast, Q - left-handed batsman and left - spinner
Wicketkeeper - R - right-handed batsman, S-left handed batsman
The selection is also subject to the following conditions:
(i) The team cannot consist of more than three right or left handed regular batsman
(ii) The team must consist of at least two left-handed bowler
(iii) The team must consist of exactly one spinner
(iv) C cannot play in a team with a left-handed spinner and D cannot play in a team with a right-handed spinner

Note: Opening batsmen are considered regular batsmen. The bowlers are not considered as regular batsman and the all-rounders are not considered as regular bowlers but are considered regular batsmen. Wicket keepers are considered as regular batsmen.

...view full instructions

If H, D, and Q are selected, then who among the following must be selected in the team?

A
A

B
R

C
S

D
F

Solution

Since, Q is a spinner, from (iii), E and G cannot be selected. Thus, both F and H must be selected as regular bowlers. Thus, D is the right choice.
Question 9
3 / -1

Directions For Questions

Ashok is the cricket coach of V.S.R public school. For an upcoming mini-cricket tournament, he has to select a team of six consisting of 2 opening batsmen, 2 Regular bowlers, an all-rounder and a wicket keeper. The following are the players in contention for the selection:
Opening Batsman - Right Handed - A, B; Left Handed - C, D
Regular Bowler - E - right- spinner, F - right - fast, G - left - spinner, H - left-fast
All-rounder - P right-handed batsman and left handed fast, Q - left-handed batsman and left - spinner
Wicketkeeper - R - right-handed batsman, S-left handed batsman
The selection is also subject to the following conditions:
(i) The team cannot consist of more than three right or left handed regular batsman
(ii) The team must consist of at least two left-handed bowler
(iii) The team must consist of exactly one spinner
(iv) C cannot play in a team with a left-handed spinner and D cannot play in a team with a right-handed spinner

Note: Opening batsmen are considered regular batsmen. The bowlers are not considered as regular batsman and the all-rounders are not considered as regular bowlers but are considered regular batsmen. Wicket keepers are considered as regular batsmen.

...view full instructions

If C is selected , then how many people are definitely not selected

A
4

B
2

C
3

D
5

Solution

Since C is selected we cannot select any left handed spinner i.e. Q & G . The team must contain at least 1 spinner and there is only one right handed spinner - E . Since E is selected we cannot select D as well. The team must also have at least 2 left handed bowlers , Since we have already selected E , The other two bowlers must be left handed . Hence , we cannot select F as well.
Finally - We cannot select D,F,G and Q
Question 10
3 / -1

Directions For Questions

Ashok is the cricket coach of V.S.R public school. For an upcoming mini-cricket tournament, he has to select a team of six consisting of 2 opening batsmen, 2 Regular bowlers, an all-rounder and a wicket keeper. The following are the players in contention for the selection:
Opening Batsman - Right Handed - A, B; Left Handed - C, D
Regular Bowler - E - right- spinner, F - right - fast, G - left - spinner, H - left-fast
All-rounder - P right-handed batsman and left handed fast, Q - left-handed batsman and left - spinner
Wicketkeeper - R - right-handed batsman, S-left handed batsman
The selection is also subject to the following conditions:
(i) The team cannot consist of more than three right or left handed regular batsman
(ii) The team must consist of at least two left-handed bowler
(iii) The team must consist of exactly one spinner
(iv) C cannot play in a team with a left-handed spinner and D cannot play in a team with a right-handed spinner

Note: Opening batsmen are considered regular batsmen. The bowlers are not considered as regular batsman and the all-rounders are not considered as regular bowlers but are considered regular batsmen. Wicket keepers are considered as regular batsmen.

...view full instructions

What is the probability of selecting a right handed spinner if A, B and P are already selected?

A
$$\frac{1}{2}$$

B
$$\frac{1}{3}$$

C
$$\frac{2}{5}$$

D
$$\frac{3}{5}$$

Solution

It is given,

A, B and P are already selected. This implies three right hand batsmen are already selected and we cannot further select a right hand batsmen. Therefore, S will be selected as wicket-keeper.

It is given, team should consist of exactly one spinner. As P is selected as all-rounder, Q cannot be selected. Either E or G should be selected as spinner.

Case 1: E is selected as spinner.

It is given, team consists of at least two left hand bowler.

E is right spinner and P is left fast bowler.

This implies the other player should be a left fast bowler.

From the remaining players, only H is a left fast bowler.

Team - A, B, P, S, E, H

Case 2: G is selected as spinner

It is given, team consists of at least two left hand bowler.

G is left hand spinner and P is left fast bowler.

This implies the other player can be either right fast bowler or left fast bowler.

From the remaining players, F is right fast bowler and H is a left fast bowler.

Possible teams are A, B, P, S, G, F and A, B, P, S, G, H

Therefore, probability of selecting a right hand spinner = $$\frac{1}{3}$$

The answer is option B.
Question 11
3 / -1

Directions For Questions

James Watson is a well-known strategist and mathematician. He entered in a world puzzle-solving championship. The puzzle given to the participants was as follows:

A number ABCDEFGH is divided by LMN, and the remainder obtained was zero. The quotient obtained after the division was PQ8RS. There is no restriction on the variables to have unique values.

The solution of the division through the long division method looks as follows:

X represents the unknown numbers in the calculation.

...view full instructions

What is the value of the resultant quotient of the puzzle?

A
80809

B
808094

C
808097

D
808099

Solution

Since the division is solved in 3 steps, the values of Q and R will be zero. This can also be understood through the understanding of the long division method. When two digits are brought down instead of one, then this means that there is a zero in the quotient.

Thus, the number will be in the form P080S. Now, when LMN is multiplied by 8, the resultant number is a three-digit number. Thus, the value of S will be 9(since the multiplication resulted in a four-digit number). This will also mean that the value of LMN is smaller than 125.

Also, the value of P will be equal to 8. This is because when a number smaller than 125 is multiplied by 7 and subtracted by a four-digit number, the resultant number will be greater than the divisor. Consider the divisor to be 124(the highest possible number) and the dividend to be 1000.

1000 - 124*7 = 132

Thus, the quotient will be greater than 7.

So, the quotient will be 80809.

Now, 80809*124 = 10020316 (an 8-digit number)

And, 80809*123 = 9939507 (a 7-digit number)

Since the dividend is an 8-digit number, the value of the divisor will be 124.
Question 12
3 / -1

Directions For Questions

James Watson is a well-known strategist and mathematician. He entered in a world puzzle-solving championship. The puzzle given to the participants was as follows:

A number ABCDEFGH is divided by LMN, and the remainder obtained was zero. The quotient obtained after the division was PQ8RS. There is no restriction on the variables to have unique values.

The solution of the division through the long division method looks as follows:

X represents the unknown numbers in the calculation.

...view full instructions

Which of the following digits was repeated more than once in the puzzle(excluding the value of Xs)?

A
6

B
4

C
3

D
2

Solution

Since the division is solved in 3 steps, the values of Q and R will be zero. This can also be understood through the understanding of long division method. When two digits are brought down instead of one, then this means that there is a zero in the quotient.

Thus, the number will be in the form P080S. Now, when LMN is multiplied by 8, the resultant number is a three digit number. Thus, the value of S will be 9(since the multiplication resulted in a four digit number). This will also mean that the value of LMN is smaller than 125.

Also, the value of P will be equal to 8. This is because when a number smaller than 125 is multiplied by 7 and subtracted by a four digit number, the resultant number will be greater than the divisor. Consider the divisor to be 124(highest possible number) and the dividend to be 1000.

1000 - 124*7 = 132

Thus, the quotient will be greater than 7.

So, the quotient will be 80809.

Now, 80809*124 = 10020316 (an 8-digit number)

And, 80809*123 = 9939507 (a 7-digit number)

Since the dividend is an 8-digit number, the value of the divisor will be 124.

Thus, the numbers in puzzles are 80809, 124, and 10020316.

Since 2 is repeated, the correct option is D.
Question 13
3 / -1

Directions For Questions

James Watson is a well-known strategist and mathematician. He entered in a world puzzle-solving championship. The puzzle given to the participants was as follows:

A number ABCDEFGH is divided by LMN, and the remainder obtained was zero. The quotient obtained after the division was PQ8RS. There is no restriction on the variables to have unique values.

The solution of the division through the long division method looks as follows:

X represents the unknown numbers in the calculation.

...view full instructions

What is the difference between the numbers PQRS and LMN?

A
7885

B
78854

C
78857

D
78859

Solution

Since the division is solved in 3 steps, the values of Q and R will be zero. This can also be understood through the understanding of long division method. When two digits are brought down instead of one, then this means that there is a zero in the quotient.

Thus, the number will be in the form P080S. Now, when LMN is multiplied by 8, the resultant number is a three digit number. Thus, the value of S will be 9(since the multiplication resulted in a four digit number). This will also mean that the value of LMN is smaller than 125.

Also, the value of P will be equal to 8. This is because when a number smaller than 125 is multiplied by 7 and subtracted by a four digit number, the resultant number will be greater than the divisor. Consider the divisor to be 124(highest possible number) and the dividend to be 1000.

1000 - 124*7 = 132

Thus, the quotient will be greater than 7.

So, the quotient will be 80809.

Now, 80809*124 = 10020316 (an 8-digit number)

And, 80809*123 = 9939507 (a 7-digit number)

Since the dividend is an 8-digit number, the value of the divisor will be 124.

The number formed by PQRS is 8009, and the number formed by LMN is 124.

The required difference = 8009 - 124 = 7885
Question 14
3 / -1

Directions For Questions

James Watson is a well-known strategist and mathematician. He entered in a world puzzle-solving championship. The puzzle given to the participants was as follows:

A number ABCDEFGH is divided by LMN, and the remainder obtained was zero. The quotient obtained after the division was PQ8RS. There is no restriction on the variables to have unique values.

The solution of the division through the long division method looks as follows:

X represents the unknown numbers in the calculation.

...view full instructions

Which one of the following is an odd one out?

A
B

B
G

C
E

D
R

Solution

Since the division is solved in 3 steps, the values of Q and R will be zero. This can also be understood through the understanding of long division method. When two digits are brought down instead of one, then this means that there is a zero in the quotient.

Thus, the number will be in the form P080S. Now, when LMN is multiplied by 8, the resultant number is a three digit number. Thus, the value of S will be 9(since the multiplication resulted in a four digit number). This will also mean that the value of LMN is smaller than 125.

Also, the value of P will be equal to 8. This is because when a number smaller than 125 is multiplied by 7 and subtracted by a four digit number, the resultant number will be greater than the divisor. Consider the divisor to be 124(highest possible number) and the dividend to be 1000.

1000 - 124*7 = 132

Thus, the quotient will be greater than 7.

So, the quotient will be 80809.

Now, 80809*124 = 10020316 (an 8-digit number)

And, 80809*123 = 9939507 (a 7-digit number)

Since the dividend is an 8-digit number, the value of the divisor will be 124.

The values of B, E, and R are equal to 0, and the value of G is 1. Thus, G is the odd one out.

The correct option is B.
Question 15
3 / -1

Directions For Questions

James Watson is a well-known strategist and mathematician. He entered in a world puzzle-solving championship. The puzzle given to the participants was as follows:

A number ABCDEFGH is divided by LMN, and the remainder obtained was zero. The quotient obtained after the division was PQ8RS. There is no restriction on the variables to have unique values.

The solution of the division through the long division method looks as follows:

X represents the unknown numbers in the calculation.

...view full instructions

What is the value of (DEFBC % ABLN)?

A
20

B
204

C
207

D
209

Solution

Since the division is solved in 3 steps, the values of Q and R will be zero. This can also be understood through the understanding of long division method. When two digits are brought down instead of one, then this means that there is a zero in the quotient.

Thus, the number will be in the form P080S. Now, when LMN is multiplied by 8, the resultant number is a three digit number. Thus, the value of S will be 9(since the multiplication resulted in a four digit number). This will also mean that the value of LMN is smaller than 125.

Also, the value of P will be equal to 8. This is because when a number smaller than 125 is multiplied by 7 and subtracted by a four digit number, the resultant number will be greater than the divisor. Consider the divisor to be 124(highest possible number) and the dividend to be 1000.

1000 - 124*7 = 132

Thus, the quotient will be greater than 7.

So, the quotient will be 80809.

Now, 80809*124 = 10020316 (an 8-digit number)

And, 80809*123 = 9939507 (a 7-digit number)

Since the dividend is an 8-digit number, the value of the divisor will be 124.

The number ABLN = 1014, and the number DEFBC = 20300

ABLN*20 = 20280

Thus, DEFBC % ABLN = 20
Question 16
3 / -1

Directions For Questions

The pie chart shows the break up of production cost of ABC Pvt Ltd which manufactures six products - P, Q, R, S, T, U - in terms of degree made in a pie chart. The total production cost is Rs 250 crore.

Each of the six products is produced in two variants - A & B. For each type, the production cost of one unit of type A and one unit of type B is in the ratio 4:5.

Profit percentage = $$\frac{profit}{cost}*100$$

...view full instructions

For how many of the six products is the profit made on B not more than profit made on A?

A
1

B
2

C
3

D
4

Solution

Production cost of one unit of A : Production cost of one unit of B = 4 : 5

Given the product P contains variants A, B in the ratio 3:2

==> Ratio of total production cost of A, B in product P = 4*3 : 5*2 = 12:10

Production cost of product P = (36/360)*250 = 25 crore.

Therefore production cost of variant A = (12/22)*25 = 13.64.

Production cost of variant B = 11.36.

Similarly, the production cost of variants A, B of other products can be calculated.

The following table breaks down the production cost of each variant.

In Q, S, T, U profit made on B is not more than that on A.
Question 17
3 / -1

Directions For Questions

The pie chart shows the break up of production cost of ABC Pvt Ltd which manufactures six products - P, Q, R, S, T, U - in terms of degree made in a pie chart. The total production cost is Rs 250 crore.

Each of the six products is produced in two variants - A & B. For each type, the production cost of one unit of type A and one unit of type B is in the ratio 4:5.

Profit percentage = $$\frac{profit}{cost}*100$$

...view full instructions

For which product is the total profit to the total production cost the lowest?

A
Q

B
R

C
S

D
U

Solution

The following table breaks down the production cost of each variant.The ratio is the lowest for S which is $$\frac{2.5+2.083}{16.67+20.83} = 0.1444$$
Question 18
3 / -1

Directions For Questions

The pie chart shows the break up of production cost of ABC Pvt Ltd which manufactures six products - P, Q, R, S, T, U - in terms of degree made in a pie chart. The total production cost is Rs 250 crore.

Each of the six products is produced in two variants - A & B. For each type, the production cost of one unit of type A and one unit of type B is in the ratio 4:5.

Profit percentage = $$\frac{profit}{cost}*100$$

...view full instructions

Which product is having the highest absolute difference between the cost of production of Type A and Type B?

A
R

B
U

C
P

D
Q

Solution

The following table breaks down the production cost of each variant. For R, the given difference is the highest which is equal to 22.5-15 = 7.5.
Question 19
3 / -1

Directions For Questions

The pie chart shows the break up of production cost of ABC Pvt Ltd which manufactures six products - P, Q, R, S, T, U - in terms of degree made in a pie chart. The total production cost is Rs 250 crore.

Each of the six products is produced in two variants - A & B. For each type, the production cost of one unit of type A and one unit of type B is in the ratio 4:5.

Profit percentage = $$\frac{profit}{cost}*100$$

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For how many products is the overall profit percentage more than 20%?

A
3

B
34

C
37

D
39

Solution

The following table breaks down the production cost of each variant. For P, Q, T the overall profit percentage is more than 20%.
Question 20
3 / -1

Directions For Questions

The pie chart shows the break up of production cost of ABC Pvt Ltd which manufactures six products - P, Q, R, S, T, U - in terms of degree made in a pie chart. The total production cost is Rs 250 crore.

Each of the six products is produced in two variants - A & B. For each type, the production cost of one unit of type A and one unit of type B is in the ratio 4:5.

Profit percentage = $$\frac{profit}{cost}*100$$

...view full instructions

For the products with lowest total cost of production , Find the ratio of total profit made by all the type A vehicles to total profit made by all the type B vehicles.

A
1:1

B
2:3

C
5:7

D
149:117

Solution

The cost of production is lowest for companies T & P

total profit made by type A of these companies = 2.046 + 3.575 = 5.621

total profit made by type b of these companies = 3.408 + 2.14 = 5.548

Hence ratio = 1:1 ( approximately)