## Logical Reasoning – Set – Each visitor to an amusement park ## DILR – Logical Reasoning – Set

The following set contains four questions related to Logical Reasoning. Choose the best answer to each question.

Each visitor to an amusement park needs to buy a ticket. Tickets can be Platinum, Gold, or Economy. Visitors are classified as Old, Middle-aged, or Young. The following facts are known about visitors and ticket sales on a particular day:

1. 140 tickets were sold.
2. The number of Middle-aged visitors was twice the number of Old visitors, while the number of Young visitors was twice the number of Middle-aged visitors.
3. Young visitors bought 38 of the 55 Economy tickets that were sold, and they bought half the total number of Platinum tickets that were sold.
4. The number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.

Q 1. If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Platinum tickets, then which among the following could be the total number of Platinum tickets sold?

a) 38

b) 32

c) 36

d) 34

Q 2. If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Economy tickets, then the number of Old visitors buying Gold tickets was

Q 3. If the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, then the number of Middle-aged visitors buying Gold tickets was

Q 4. Which of the following statements MUST be FALSE?

a) The numbers of Gold and Platinum tickets bought by Young visitors were equal

b) The numbers of Middle-aged and Young visitors buying Gold tickets were equal

c) The numbers of Old and Middle-aged visitors buying Economy tickets were equal

d) The numbers of Old and Middle-aged visitors buying Platinum tickets were equal

Solution: Let the number of old visitors = x so number of middle aged and young visitors will be 2x and 4x respectively. So x+2x+4x=140 or x=20

Let total number of platinum ticket sold = p (even)

Now on the basis of information given following table can be made:

 Old Middle-aged young Total Platinum p/2 p Gold a 43 – p/2 – a 42 – p/2 85 – p Economy a 17 – a 38 55 Total 20 40 80 140

1. Given , the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Platinum tickets, So

 Old Middle-aged young Total Platinum p/4 p/4 p/2 p Gold a 43 – p/2 – a 42 – p/2 85 – p Economy a 17 – a 38 55 Total 20 40 80 140

Now old = 40 so 2a + p/4 = 20

Or a + p/8 = 10

As p/8 must be an integer so p must be a multiple of 8. Thus option a) 38, c) 36 and d)34 get eliminated.

Only possible option is b) 32

2. Given, the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Economy tickets So

 Old Middle-aged young Total Platinum x p/2 p Gold a 42 – p/2 Economy a x 38 55 Total 20 40 80 140

So

2a + x = 20—–1) and

a+x + 38 =55——-2)

By solving above equations a = 3

So the number of Old visitors buying Gold tickets = 3

3. The number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, so 42 – p/2 < a

Or a + p/2 > 42

So minimum value of a + p/2 = 43

Number of Middle-aged visitors buying Gold tickets = 43 – (p/2 +a) = 0

4. Since Old – Economy + Middle age – economy = 17 which is odd so these two can never be equal. Hence, the statement that “The numbers of Old and Middle-aged visitors buying Economy tickets were equal” is false.

Answer: c) The numbers of Old and Middle-aged visitors buying Economy tickets were equal

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## Logical Reasoning – Set – Seven candidates, Akil, Balaram ## DILR – Logical Reasoning – Set

The following set contains four questions related to Data Interpretation. Choose the best answer to each question.

Seven candidates, Akil, Balaram, Chitra, Divya, Erina, Fatima, and Ganeshan, were invited to interview for a position. Candidates were required to reach the venue before 8 am. Immediately upon arrival, they were sent to one of three interview rooms: 101, 102, and 103. The following venue log shows the arrival times for these candidates. Some of the names have not been recorded in the log and have been marked as ‘?’.

Additionally here are some statements from the candidates: Balaram: I was the third person to enter Room 101.

Chitra: I was the last person to enter the room I was allotted to.

Erina: I was the only person in the room I was allotted to.

Fatima: Three people including Akil were already in the room that I was allotted to when I entered it.

Ganeshan: I was one among the two candidates allotted to Room 102

Q 1. What best can be said about the room to which Divya was allotted?

a) Definitely Room 102

b) Definitely Room 101

c) Either Room 101 or Room 102

d) Definitely Room 103

Q 2. Who else was in Room 102 when Ganeshan entered?

a) No one

b) Chitra

c) Akil

d) Divya

Q 3. When did Erina reach the venue?

a) 7:10 am

b) 7:15 am

c) 7:45 am

d) 7:25 am

Q 4. If Ganeshan entered the venue before Divya, when did Balaram enter the venue?

a) 7:15 am

b) 7:25 am

c) 7:45 am

d) 7:10 am

Solution:

From Chitra’s statement it is clear that Chitra and Fatima can not be in the same room. And also there will be 4 people in Fatima’s room and at least two people in Chitra’s room.

As there are only 7 people and 3 rooms . So the only possible case is that there must be exactly 1, 2 and 4 people in each room in any order.

Now from Balram’s statement it is clear that there were more than 2 people in room number 101. So number of people in room 101 = 4

As Ganeshan was in room number 102 so he/she must be there with Chitra as chitra was last one to enter his room. And E would be the last one to come.

These information can be tabulated as bellow

 Time (am) 7:10 7:15 7:25 7:30 7:40 7:45 Candidate Akil, G/D G/D/B B/G Chitra Fatima Erina

And

 Room no. 101 102 103 Candidate Akil, , Divya (D) Balram (B), Fatima Ganeshan , Chitra Erina

Now all questions can be answered

1. b) Definitely Room 101

2. a) no one

3. c) 7 : 45 am

4. b) 7 :25 am

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## Logical Reasoning – Set – The base exchange rate of a currency ## DILR – Logical Reasoning – Set

The following set contains four questions related to Data Interpretation. Choose the best answer to each question.

The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.

A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly with another. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates.

The following facts are known about the outlet on a particular day:

1. The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
2. The amounts of L used by the outlet to buy A and B are in the ratio 5:3.
3. The amounts of L the outlet received from the sales of A and B are in the ratio 5:9.
4. The outlet received 88000 units of L by selling A during the day.
5. The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
6. The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.

Q 1. How many units of currency A did the outlet buy on that day?

Q 2. How many units of currency C did the outlet sell on that day?

a) 6000

b) 19000

c) 22000

d) 3000

Q 3. What was the base exchange rate of currency B with respect to currency L on that day?

Q 4. What was the buying exchange rate of currency C with respect to currency L on that day?

a) 90

b) 10

c) 20

d) 95

Solution:

Given,

The base exchange rates (BER) of currencies A, B and C with respect to L is in the ratio 100 : 120 : 1. Also The buying exchange rates (BR) of each of A, B, and C with respect to L will be in the ratio  = 95 : 114: 0.95 and corresponding selling rates (SR) will be 110, 132 and 1.1 respectively as it is 10% more than base exchange rates.

The given information can be tabulated as follows:

 A B C BER 100x 120x x BR 95x 114x 0.95x SR 110x 132x 1.1x Net Addition 800 0 3000

From point 4) The outlet received 88,000 units of L by selling A.

From point 2) & 3)  the ratio of amounts of L used to by A and B are in the ratio 5 : 3 and from the sales of A and B are in the ratio 5 : 9.

Let the base exchange rates for A, B and C 100x,120x and x .

Units sold of A = 88,000/110x = 800/x.

As the net addition is 800, the units of A bought is 800 + 800/x

Amount of L used in buying 800 + 800/x units is (800 +800/x)×0.95 × 100x = 76000(x+1) .

Now As the amount used to buy A and B are in the ratio 5 : 3,

The amount used to buy B is 76000(x+1) /5 X 3 = 45,600(x+1)

Number of units of B bought = 45,600(x+1)/114x = 400(x+1)/x

As the net addition of B is zero, number of units of B sold = 400(x+1)/x

The amount received form selling A = 88,000

As given 88,000 : 52800(x+1) = 5 : 9

880/528(x+1) = 5/9

(x+1) = 880*9/558*5 = 3

x = 2

1. The units of A bought is 800 + 800/x = 800+ 800/2 = 1200

2. As the net addition in the number of units of C is 3,000 and the buying and selling rates are in the ratio 0.95 and 1.1, assuming x units are sold 0.95 (x + 3000) = 1.1 (x) 0.15x = 2850 X= 19000

3. The base exchange rate of currency B with respect to L is 120x = 120*2 = 240

4. The buying exchange rate of currency C with respect to L on that day was 0.95x = 1.9

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## Logical Reasoning – Set – According to a coding scheme the sentence ## DILR – Logical Reasoning – Set

The following set contains four questions related to Data Interpretation. Choose the best answer to each question.

According to a coding scheme the sentence

“Peacock is designated as the national bird of India” is coded as

5688999 35 1135556678 56 458 13666689 1334 79 13366

This coding scheme has the following rules:

1. The scheme is case-insensitive (does not distinguish between upper case and lower case letters).
2. Each letter has a unique code which is a single digit from among 1,2,3, …, 9.
3. The digit 9 codes two letters, and every other digit codes three letters.
4. The code for a word is constructed by arranging the digits corresponding to its letters in a non-decreasing sequence.

Q 1. What best can be concluded about the code for the letter L?

a) 6

b) 8

c) 1 or 8

d) 1

Q 2. What best can be concluded about the code for the letter B?

a) 1 or 3 or 4

b) 3 or 4

c) 3

d) 1

Q 3. For how many digits can the complete list of letters associated with that digit be identified?

a) 1

b) 3

c) 2

d) 0

Q 4. Which set of letters CANNOT be coded with the same digit?

a) S,U,V

b) X,Y,Z

c) I,B,M

d) S,E,Z

Solution:

Given ‘peacock is designated as the national bird of India’ is coded as ‘ 5688999 35 1135556678 56 458 13666689 1334 79 13366’

So from first and 2nd last word of sentence that is from the coding of peacock and of  we can say  9 is the code for o and c .  It means F is coded as 7 from the word of.

Now from 2nd word (is ) and the 3rd last word (bird) we can say the code for ‘I’ is 3. So S is coded as 5 from the word is and A is coded as 6 from the word ‘as’.

Now N is coded as 6 from the word national. Thus the only alphabet left in the word India ‘D ‘ is coded as 1. T is coded as 8 from the word ‘the’ and ‘National’ similarly E is coded as 5 from the word designated. Thus H is coded as 4 from the word ‘the’. Also P and K are coded as 8 from the word ‘peacock’.

B and R are coded as 3 and 4 many order from the word ‘bird’.

G is coded as 7. L is coded as 1 from the word ‘National’. P and K are coded as 8 from the word ‘peacock’.

So following table can be made for the letters and their code digit

 Code Letter 1 D,L 2 3 I 4 H 5 S,E 6 A,N 7 F,G 8 T,P,K 9 O,C

Now all the questions can be answered.

1. d) 1

2. b) 3 or 4

3. c) 2 , for 8 and 9

4. a) S,U,V , as S and E are already coded with 5 and one digit can not be coded for more than 3 letters.

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## Logical Reasoning – Set – Fun Sports (FS) provides training in ## DILR – Logical Reasoning – Set

The following set contains four questions related to Data Interpretation. Choose the best answer to each question.

Fun Sports (FS) provides training in three sports – Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 39 students each of whom is enrolled in at least one of the three sports. The following details are known:

1. The number of students enrolled only in L is double the number of students enrolled in all the three sports.
2. There are a total of 17 students enrolled in G.
3. The number of students enrolled only in G is one less than the number of students enrolled only in L.
4. The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
5. The maximum student enrollment is in L.
6. Ten students enrolled in G are also enrolled in at least one more sport.

Q 1. What is the minimum number of students enrolled in both G and L but not in K?

Q 2. If the numbers of students enrolled in K and L are in the ratio 19:22, then what is the number of students enrolled in L?

a) 19

b) 22

c) 17

d) 18

Q 3. Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and K?

Q 4. Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and L?

a) 7

b) 5

c) 8

d) 6

Solution:

Let the number of students enrolled in all the three sports = g and

The number of students enrolled only in K = a

So from point 1) The number of students enrolled only in L = 2g

And from point 3) The number of students enrolled only in  G = 2g – 1

So on the  basis of information given, following Venn Diagram can be made, From point 6) , b+c+g=10——1)

From point 2) (2g-1)+b+c+g=17——2)

From eq 1) & eq 2)

(2g-1)=7 or g=4

As total number of students = 39

So (2g-1)+b+c+g +(2g+a-g+a )=39
17+(2g+a-g+a )=39
17+4+2a=39
a = 9

So final Venn diagram will look like, As number of students in L is maximum , so 8+6-b>9+b or b<2.5

So b can be either 0, 1 or 2 .

1. Minimum number of students enrolled in both G and L but not K = 6 –b

So it will be minimum ,when b is maximum which is 2. So

Minimum number of students enrolled in both G and L but not K = 6 -2 = 4

2. As per above question maximum and minimum possible number of student in L are 23 and 21 respectively . So required ratio will be 19 : 22 if there is 22 students in L.

3. From g = 4, one person moves to (6-b), one person to b and two persons to (a-g) . after the withdrawal 4. d) 6

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## Logical Reasoning – Set – Fuel contamination levels at each of 20 ## DILR – Logical Reasoning – Set

The following set contains four questions related to Logical Reasoning. Choose the best answer to each question.

Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low.

1. Contamination levels at three pumps among P1 – P5 were recorded as high.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

1) Which of the following MUST be true?

a) The contamination level at P20 was recorded as medium.

b) The contamination level at P10 was recorded as high.

c) The contamination level at P12 was recorded as high.

d) The contamination level at P13 was recorded as low.

2) What best can be said about the number of pumps at which the contamination levels were recorded as medium?

a) At least 8

b) More than 4

c) Exactly 8

d) At most 9

3) If the contamination level at P11 was recorded as low, then which of the following MUST be true?

a) The contamination level at P14 was recorded as medium.

b) The contamination level at P18 was recorded as low.

c) The contamination level at P15 was recorded as medium.

d) The contamination level at P12 was recorded as high.

4) If contamination level at P15 was recorded as medium, then which of the following MUST be FALSE?

a) Contamination level at P14 was recorded to be higher than that at P15.

b) Contamination levels at P13 and P17 were recorded as the same.

c) Contamination levels at P11 and P16 were recorded as the same.

d) Contamination levels at P10 and P14 were recorded as the same.

Solution:

From point 2) we can say that P6 was the only pump among P1 – P10 where the contamination level was recorded as low and in other pipe it can be either medium or high .

From point 3) only P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded

And From point 1) Contamination levels at three pumps among P1 – P5 were recorded as high it means P1 , P3 and P5 have high Contamination level and P2 and P4 will have medium Contamination level. Thus we get two possible cases for P1 to P10 .

Case a )

 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 H M H M H L H H M H

Case b)

 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 H M H M H L M M H M

From point 5)

H = 2L

Thus H + L = 3L must be multiple of 3.

Case 1) From point 4)  High contamination levels were not recorded at any of the pumps P16 – P20.

So maximum possible number of pipe with High contamination levels = 8 (P1,P3,P5,P7,P8 ,P10 , P12 and P14)

In this case number of pipes with Low contamination levels = 8/2 = 4

So medium (M) = 8

The possible arrangement will be

 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 P16 P17 P18 P19 P20 H M H M H L H H M H M H M H M/L L/M M/L L/M M/L L/M

Or

 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 P16 P17 P18 P19 P20 H M H M H L H H M H L M H M H M L M L M

Case 2) if Low contamination levels pipes were minimum , which can be 3,then number of pipes with high contamination level = 6

Number of pipes with medium contamination level = 11 which is not possible .

So only case 1) is possible

Now all the questions can be answered.

1) b) The contamination level at P10 was recorded as high

2) c) Exactly 8

3) a) The contamination level at P14 was recorded as medium

4) c) Contamination levels at P11 and P16 were recorded as the same.

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## Logical Reasoning – Set – Twenty four people are part of three ## DILR – Logical Reasoning – Set

The following set contains four questions related to Logical Reasoning. Choose the best answer to each question.

Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:

1. The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2. The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3. 60% of the politicians are in the administration committee, and 20% are in the teaching committee.

1) Based on the given information, which of the following statements MUST be FALSE?

a) The size of the research committee is less than the size of the teaching committee

b) In the teaching committee the number of educationalists is equal to the number of politicians

c) The size of the research committee is less than the size of the administration committee

d) In the administration committee the number of bureaucrats is equal to the number of educationalists

2) What is the number of bureaucrats in the administration committee?

3) What is the number of educationalists in the research committee?

4) Which of the following CANNOT be determined uniquely based on the given information?

a) The size of the teaching committee

b) The size of the research committee

c) The total number of educationalists in the three committees

d) The total number of bureaucrats in the three committees

Solution:

Let the number of bureaucrats in the research committee = x and total number of politicians = y

From point 1) Then number of bureaucrats in the teaching and administration committees will be x and 4x/3 respectively.

From point 3) number of the politicians in the administration committee teaching committee will be 0.6y and 0.2y respectively.

Now let Number of educationalist in research , teaching and administration department are a, b and z   such that a > b and a = (b+z)/2 thus z > a > b and a+b+z = 3a

From the given information following table can be made :

 Research Teaching Administration Total Bureaucrats x x 4x/3 10x/3 Politicians 0.2y 0.2y 0.6y y Educationalist a > b b z 3a

Now as all the values in the table must be integers so minimum value of x = 3 and y = 5 also

10x/3 + y + 3a = 24

So the only possible value for a, x and y are 3, 3 and 5 . Thus the table with required number of people in each committees will be

 Research Teaching Administration Total Bureaucrats 3 3 4 10 Politicians 1 1 3 5 Educationalist 3 2 4 9 Total 7 6 11

Or

 Research Teaching Administration Total Bureaucrats 3 3 4 10 Politicians 1 1 3 5 Educationalist 3 1 5 9 Total 7 5 12 24

Now all the question can be answered :

1) a) The size of the research committee is less than the size of the teaching committee

2) 4

3) 3

4) a) The size of the teaching committee

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## Logical Reasoning – Set – Adriana, Bandita, Chitra, and Daisy ## DILR – Logical Reasoning – Set

The following set contains four questions related to Logical Reasoning. Choose the best answer to each question.

Adriana, Bandita, Chitra, and Daisy are four female students, and Amit, Barun, Chetan, and Deb are four male students. Each of them studies in one of three institutes – X, Y, and Z. Each student majors in one subject among Marketing, Operations, and Finance, and minors in a different one among these three subjects. The following facts are known about the eight students:

1. Three students are from X, three are from Y, and the remaining two students, both female, are from Z.
2. Both the male students from Y minor in Finance, while the female student from Y majors in Operations.
3. Only one male student majors in Operations, while three female students minor in Marketing.
4. One female and two male students major in Finance.
5. Adriana and Deb are from the same institute. Daisy and Amit are from the same institute.
6. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
7. Daisy minors in Operations.

1) Who are the students from the institute Z?

a) Bandita and Chitra

d) Chitra and Daisy

2) Which subject does Deb minor in?

a) Cannot be determined uniquely from the given information

b) Marketing

c) Finance

d) Operations

3) Which subject does Amit major in?

a) Finance

b) Operations

c) Cannot be determined uniquely from the given information

d) Marketing

4) If Chitra majors in Finance, which subject does Bandita major in?

a) Finance

b) Operations

c) Cannot be determined uniquely from the given information

d) Marketing

Solution:

Let Daisy and Amit are from some institute A . Adriana and Ded be from the some institute B. So Bandita and Chitra must be from same institute (let’s say C) but as only two females are from Z so Bandita and Chitra must be from Z. Now the given information can be tabulated as bellow –

 Name Gender Institute Major Minor Adriana Female B Bandita Female Z Chitra Female Z Daisy Female A O Amit Male A Barun Male Y O Chetan Male X F Deb Male B

From point 3) three female students minor in Marketing and Daisy minors in operations (O) so other three must have minored in Marketing (M)

From point 2) Female student from Y majors in operations so Daisy cannot be from Y so Daisy is from X so is Amit. So Adriana and Deb are form Y. Also Barun and Deb both will miner in Finance (F) and Adrina will major in Operation (O).

Now the table will look like ,

 Name Gender Institute Major Minor Adriana Female Y O M Bandita Female Z F/O M Chitra Female Z F/O M Daisy Female X F/M O Amit Male X F O/M Barun Male Y O F Chetan Male X F O/M Deb Male Y M F

Now all the questions can be answered.

1. a) Bandita and Chitra

2. c) Finance

3. a) Finance

4. b) operation

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## Logical Reasoning – Set – 1600 satellites were sent up by a ## DILR – Logical Reasoning – Set

The following set contains four questions related to Logical Reasoning. Choose the best answer to each question.

1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O.

The following facts are known about the satellites:

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

1) What best can be said about the number of satellites serving C?

a) Must be between 450 and 725

b) Must be at least 100

c) Cannot be more than 800

d) Must be between 400 and 800

2) What is the minimum possible number of satellites serving B exclusively?

a) 500

b) 100

c) 200

d) 250

3) If at least 100 of the 1600 satellites were serving O, what can be said about the number of satellites serving S?

a) No conclusion is possible based on the given information

b) Exactly 475

c) At most 475

d) At least 475

4) If the number of satellites serving at least two among B, C, and S is 1200, which of the following MUST be FALSE?

a) All 1600 satellites serve B or C or S

b) The number of satellites serving B is more than 1000

c) The number of satellites serving C cannot be uniquely determined

d) The number of satellites serving B exclusively is exactly 250

Solution:

It is given that the satellites serving either B, C or S do not serve O.

From point (1), let the number of satellites serving B, C and S be 2K, K, K respectively.

Let the number of satellites exclusively serving B be x.

From point (3), the number of satellites exclusively serving C and exclusively serving S will each be 0.3x

From point (4), the number of satellites serving O is same as the number of satellites serving only C and S. Let that number be y.

Since the number of satellites serving C is same as the number of satellites serving S,

we can say that (number of satellites serving only B and C) + 0.3x + 100 + y = (number of satellites serving only B and S) + 0.3x + 100 + y

Let the number of satellites serving only B and C = the number of satellites serving only B and S = Z

Therefore, the venn diagram will be as follows: Given that there are a total of 1600 satellites So

x + z + 0.3x + z + 100 + y + 0.3x + y = 1600

1.6x + 2y + 2z = 1500—————(1)

Also K = 0.3x + z + y +100

Satellites serving B = 2K = x + 2z + 100

Or 2(0.3x + z + y + 100)= x + 2z + 100

0.4x = 2y + 100

x = 5y + 250 ——————-(2)

Substituting (2) in (1), we will get

1.6 (5y + 250)+ 2y + 2z = 1500

10y + 2z = 1100

z= 550 – 5y ———— (3)

1) The number of satellites serving C = z + 0.3x + 100 + y = (550 – 5y) + 0.3(5y + 250) + 100 + y = 725 – 2.5y This number will be maximum when y is minimum. Minimum value of y is 0. So the maximum number of satellites serving C will be 725.

From equation 3), z = 550 – 5y, Since the number of satellites cannot be negative

z ≥ 0

or 550 – 5y ≥ 0

y ≤ 110.

Maximum value of y is 110.

When y = 110, the number of satellites serving C will be 725 – 2.5 × 110 = 450.

This will be the minimum number of satellites serving C. The number of

satellites serving C must be between 450 and 725.

Answer: a) Must be between 450 and 725

2) From equation 2) , the number of satellites serving B exclusively is x = 5y + 250

This is minimum when y is minimum. Minimum value of y = 0. The minimum number of satellites serving B exclusively = 5 × 0 + 250 = 250.

3) Given, at least 100 satellites serve O.

We can say in this case that y ≥ 100.

So Number of satellites serving S = 0.3x + z +100 + y. = 725 – 2.5y

This is minimum when y is maximum, i.e. 110, (from eq 3)

Minimum number of satellites serving = 725 – 2.5 × 100 = 450.

This is maximum when y is minimum, i.e., 100 in this case.

Maximum number of satellites serving = 725 – 2.5 × 100 = 475

Therefore, the number of satellites serving S is at most 475

4) The number of satellites serving at least two of B, C or S  =  number of satellites serving exactly two of B, C or S + Number of satellites serving all the three

= z + z + y + 100 = 2(550 – 5y) + y + 100 = 1200 – 9y.

Given that this is equal to 1200 so

1200 – 9y = 1200

Or y = 0

If y = 0, x = 5y + 250 = 250 z = 550 – 5y = 550

No. of satellites serving C = k = z + 0.3x + 100 + y = 550 + 0.3 x 250 + 100 + y = 725

No. of satellites serving B = 2k = 2 x 725 = 1450.

From the given options, we can say that the option “the number of satellites serving C cannot be uniquely determined” must be FALSE.

Answer: c) The number of satellites serving C cannot be uniquely determine

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## Logical Reasoning – Set – An ATM dispenses exactly Rs. 5000 ## DILR – Logical Reasoning – Set

The following set contains four questions related to Logical Reasoning. Choose the best answer to each question.

An ATM dispenses exactly Rs. 5000 per withdrawal using 100, 200 and 500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

1) In how many diﬀerent ways can the ATM serve a customer who gives 500 rupee notes as her preference?

Solution: As it is given that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her So the following cases are possible,

 Cases No. of 500 rupee notes No. of 200 rupee notes No. of 100 rupee notes 1 10 0 0 2 9 2 1 3 9 1 3 4 9 0 5 5 8 5 0 6 8 4 2 7 8 3 4

So total number of required ways = 7

2) If the ATM could serve only 10 customers with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, what is the maximum number of customers among these 10 who could have given 500 rupee notes as their preferences?

Solution: As total number of 500 notes = 50

From the solution of the previous question we can see if the customers could have given 500 rupee notes as their preferences minimum number of 500 rupee notes they will need is 8 . So maximum number of who could have given 500 rupee notes as their preferences = [50/8 ] = 6

3) What is the maximum number of customers that the ATM can serve with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, if all the customers are to be served with at most 20 notes per withdrawal?

a) 13

b) 10

c) 16

d) 12

Solution: To maximize the number of customers we need to minimize the number of 500 rupee notes per customers.

Such that x+y+z ≤ 20——1)
and 500x+200y+100z=5000—–2)

Where x = number of 500 rupee notes

y = number of 200 rupee notes

z = number of 100 rupee notes

So we need to minimize x. and from above two equations minimum value of x =  4.
Thus maximum number of customers that the ATM can serve [50/4] = 12

4.) What is the number of 500 rupee notes required to serve 50 customers with 500 rupee notes as their preferences and another 50 customers with 100 rupee notes as their preferences, if the total number of notes to be dispensed is the smallest possible?

a) 800

b) 900

c) 1400

d) 750

Solution:

Case 1) When with 500 rupee notes is their preference , to minimize the number of notes we need to serve  maximum possible 500 notes, minimum number of notes required = [5000/500] = 10

So total number of 500 notes required to to serve 50 customers = 50*10 = 500

Case 2) When with 100 rupee notes is their preference, to minimize the number of notes we need to serve  maximum possible 500 notes. So

100z+500x=5000———1) and

z>x such that (z+x)isminimum .——-2)

From eq 1) & eq 2) only possible solution is x = 8 and z = 10

Thus total number of 500 notes required to to serve 50 customers = 50*8 = 400

So required number of 500 notes = case 1) + case 2) = 500 + 400 = 900