Answer the following question based on the information given below.
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:
The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2 : 1 : 1.
The number of satellites serving all three of B, C, and S is 100.
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.
The number of satellites serving O is the same as the number of satellites serving both Cand S but not B.
1.
CAT 2018 LRDI Slot 1
| LR - Venn Diagram
What best can be said about the number of satellites serving C?
Let number of satellites exclusively serving B be ‘10x’ and number of satellites serving O be ‘z’.
∴ number of satellites exclusively serving C = number of satellites exclusively serving S = 3a
As the number of satellites serving C = number of satellites serving S, the number of satellites serving C and B but not S = number of satellites exclusively serving S and B but not C = y (assume).
∴ 10x + 2y + 100 + 6x + 2z
= 1600 ⇒ 8x + y + z
= 750 … (i)
From (1), 10x + 2y + 100
= 6x + 2y + 2z + 200
∴ z = 2x – 50 … (ii)
Thus, the minimum value that ‘x’ can takes is 25.
From (i) and (ii),
10x + y = 800 ⇒ y = 800 – 10x
Thus, the maximum value that ‘x’ can takes is 80.
The number of satellites serving
C = 3x + y + z + 100 = 3x + (800 – 10x) + (2x – 50) + 100 = 850 – 5x
As value of ‘x’ is between 25 and 80, value of (850 – 5x) must be between 450 and 725.
Answer the following question based on the information given below.
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:
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.
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.
60% of the politicians are in the administration committee, and 20% are in the teachingcommittee.
5.
CAT 2018 LRDI Slot 1
| LR - Selection & Distribution
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 administration 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 teaching committee
D.
In the administration committee the number of bureaucrats is equal to the number of educationalists
As each committee has at least one bureaucrat, one educationalist and one politician, the members of each type must be in the range [3, 18].
From (1), ratio of the number of bureaucrats in the research, teaching and administration committees = 3 : 3 : 4
Therefore, the number of bureaucrats = 10
From (2), If there are x educationists in the research committee, there must be (3x) educationists in all.
From (3), ratio of the number of politicians in the research, teaching and administration committees = 1 : 1 : 3
Therefore, there can be 5 or 10 politicians. If there are 10 politicians, number of educationists = 4, which is not a multiple of 3.
Therefore there are 5 politicians and 9 educationists.
Thus, we have
[1]: The administration committee has 11 or 12 members. Research committee has 7 members which is definitely less than the members of administration committee. Therefore, this statement is true.
[2]: The teaching committee has 1 or 2 educationists and 1 politician. Therefore, this statement can be true.
[3]: The teaching committee has 5 or 6 members. The research committee has 7 members which is definitely more than members of the teaching committee. Therefore, this statement is definitely false.
[4]: The administration committee has 4 bureaucrats and either 4 or 5 educationists. Therefore, this statement can be true.
Answer the following question based on the information given below.
A company administers a written test comprising of three sections of 20 marks each – Data Interpretation (DI), Written English (WE) and General Awareness (GA), for recruitment. A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections. Candidates who score less than 70% marks in two or more sections are disqualified. From among the rest, the four with the highest composite scores are recruited. If four or less candidates qualify, all who qualify are recruited.
Ten candidates appeared for the written test. Their marks in the test are given in the table below. Some marks in the table are missing, but the following facts are known:
No two candidates had the same composite score.
Ajay was the unique highest scorer in WE.
Among the four recruited, Geeta had the lowest composite score.
Indu was recruited.
Danish, Harini, and Indu had scored the same marks the in GA.
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
Qualified candidate score 14 or more marks in two or more sections.
Composite scores of few students:
Chetna: 2(19) + 4 + 12 = 54
Ester: 2(12) + 18 + 16 = 58
Falak: 2(15) + 7 + 10 = 47
From (6), Jatin scored 20 marks in DI. Therefore, his Composite score is 2(20) + 16 + 14 = 70
Therefore, Indu’s Composite score = 60
If Indu scored 20 marks in DI, her score in GA = 60 – 2(20) – 8 = 12.
In this case, Indu scored less than 70% in WE and GA. But from (4) Indu was recruited. Therefore, this case is not valid. This means she must have scored 100% marks in GA i.e., 20 marks in GA.
∴ indu's marks (out of 20) in DI =
= 16
From (5), Danish and Harini also scored 20 marks in GA.
From (2), Ajay must have scored more than 18 marks. Assuming his score in WE as 19, composite score would be 2(8) + 19 + 16 = 51. This cannot be true as Danish’s composite score was 51. Thus, Ajay’s score in WE = 20 and his composite score 2(8) + 20 + 16 = 52.
Bala, Chetna and Falak scored less than 70% marks in at least two sections and hence were disqualified.
From (4), Geeta had the lowest score. Maximum composite score of Harini = 2(5) + 20 + 20 = 50
Ajay(52), Danish(51), Ester(58), Indu(60) and Jatin(70) are definitely qualified.
So, Ester(58), Indu(60) and Jatin(70) were top three scorers. Geeta must have scored more than 52 marks. Therefore marks scored by her in WE ≥ 53 – 2(14) − 6 i.e., marks scored by her in WE ≥ 19
If she had scored 20 marks in WE, her composite marks would be 2(14) + 20 + 6 = 54
But Chetna’s composite score = 54. So, Geeta’s marks in WE = 19 and her composite score = 53
Thus we have
‘a’ denotes marks of Harini in WE while ‘d’ denote marks of Bala in DI.
Harini’s composite score = (30 + a) and Falak’s composite score = 47
Thus, statement 1 is may not be true for values of ‘a’ more than or equal to 17.
In DI, if Bala’s score is 19 or 20, his composite score would be 58 (or 60). But, no two candidates had the same composite scores. Therefore, Bala’s score must be less than or equal to 18 and hence his composite score must be less than or equal to 56.
Therefore, Chetna scored more than Bala in DI and Bala’s composite score was less than that of Ester. Thus, statements 2 and 3 are true.
If Bala scores same as Jatin in DI, Bala’s composite score would be 60. This is not valid.
If all the candidates except Ajay and Danish had different marks in DI, and Bala's composite score was less than Chetna's composite score, then what is the maximum marks that Bala could have scored in DI?
Answer the following question based on the information given below.
The multi-layered pie-chart below shows the sales of LED television sets for a big retail electronics outlet during 2016 and 2017. The outer layer shows the monthly sales during this period, with each label showing the month followed by sales figure of that month. For some months, the sales figures are not given in the chart. The middle-layer shows quarter wise aggregate sales figures (in some cases, aggregate quarter-wise sales numbers are not given next to the quarter). The innermost layer shows annual sales. It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression, as do the three monthly sales figures in the fourth quarter (October, November, December) of that year.
13.
CAT 2018 LRDI Slot 1
| DI - Tables & Graphs
What is the percentage increase in sales in December 2017 as compared to the sales in December 2016?
The sales figures during April, May and June of 2016 form an arithmetic progression. Assuming ‘d’ as common difference, 40 + (40 + d) + (40 + 2d) = 150 ⇒ d = 10
Therefore, sales in May was 50 and that in June was 60.
The sales figures during October, November and December of 2016 form an arithmetic progression. Assuming ‘f’ as common difference, 100 + (100 + f) + (100 + 2f) = 360 ⇒ f = 20
Therefore, sales in November was 120 and that in December was 140.
Let sales in August 2017 and December 2017 be ‘a’ and ‘b’.
∴ 60 + a + 70 = 220 and 150 + 170 + b = 500
∴ a = 90 and b = 180
The values for quarter-wise aggregate sales can be found by simply adding sales values of the corresponding months and for values of for annual sales we need to add either values for all quarter-wise sales of that year or values for all the months of that year.
Thus, we have
Sales in December 2017 = 180 and Sales in December 2016 = 140
Answer the following question based on the information given below.
You are given an n × n square matrix to be filled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the first or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.
17.
CAT 2018 LRDI Slot 1
| LR - Board Games
What is the minimum number of different numerals needed to fill a 3×3 square matrix?
A number say ‘x’ can be filled in four corner cells. Another number say ‘y’ can be used in the central cell. Now the remaining cells are middle cells of top row, bottom row, left column and right column. Two more numerals can be used to fill these cells. Assume that ‘z’ and ‘w’ are those numerals.
A number say ‘x’ can be filled in four corner cells. Another number say ‘y’ can be used in the central cell. Now the remaining cells are middle cells of top row, bottom row, left column and right column. Two more numerals can be used to fill these cells. Assume that ‘z’ and ‘w’ are those numerals.
Thus,
So, at least 4 different numerals are required for a 3 × 3 matrix.
Top row i.e., the first row of the 5 × 5 matrix can be (x, z, x, z, x). Similarly, other cells from the second and the third row can be filled.
For the first column i.e., the leftmost column, the fourth cell from top can be filled with the number from the second cell from top i.e. w and the bottom cell can be filled with the number from the top most cell of the column i.e., x. Similarly, remaining cells of the remaining columns can be filled.
Thus, four different numerals are sufficient to fill a 5 × 5 matrix.
Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to fill a 5×5 matrix?
A number say ‘x’ can be filled in four corner cells. Another number say ‘y’ can be used in the central cell. Now the remaining cells are middle cells of top row, bottom row, left column and right column. Two more numerals can be used to fill these cells. Assume that ‘z’ and ‘w’ are those numerals.
Thus,
4 different numerals are required to fill a 5 × 5 matrix. As one is allowed to make only one mistake, it is possible to change exactly one entry such that all other conditions are not violated.
Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a 5×5 square matrix?
Assuming that the particular cell is not a cell at one of the corner, it has 8 adjacent cells.
Hence, 8 + 1 = 9 different numerals are required to fill the particular cell and the adjacent cells.
Earlier we have seen that 4 different numerals are required to fill a 5 × 5 matrix. Hence, 9 different numerals will be definitely sufficient to fill the 5 × 5 matrix.
Answer the following question based on the information given below.
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.
(Notes of Rs. 500 denomination = 8 and notes of Rs. 200 and Rs. 100 denominations = 3 + 4 = 7, 8 > 7)
If numbers of notes of denomination Rs. 200 are reduced further, number of notes of the customer’s preferred denomination will not exceed the total number of notes of other denominations.
The ATM machine can dispense 7 notes of Rs. 500, for the minimum number of total notes of the remaining two denominations; one can have 7 notes of Rs. 200 and one note of Rs. 100. But even then, number of notes of the customer’s preferred denomination will be less than the total number of notes of other denominations. As number of notes of denominations reduces, number of notes of the denomination Rs. 500 will be less than the total number of notes of other denominations.
Thus, there are 7 different ways in which the ATM can serve a customer who gives 500 rupee notes as her preference.
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?
(Notes of Rs. 500 denomination = 8 and notes of Rs. 200 and Rs. 100 denominations = 3 + 4 = 7, 8 > 7)
If numbers of notes of denomination Rs. 200 are reduced further, number of notes of the customer’s preferred denomination will not exceed the total number of notes of other denominations.
The ATM machine can dispense 7 notes of Rs. 500, for the minimum number of total notes of the remaining two denominations; one can have 7 notes of Rs. 200 and one note of Rs. 100. But even then, number of notes of the customer’s preferred denomination will be less than the total number of notes of other denominations. As number of notes of denominations reduces, number of notes of the denomination Rs. 500 will be less than the total number of notes of other denominations.
if a customer’s preferred denomination is Rs. 500, the minimum number of notes of Rs. 500 the ATM machine dispenses is 8.
Hence, with fifty 500 rupee notes, the ATM machine can serve 6 customers who could have given 500 rupee notes as their preferences.
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?
> the customers are to be served with at most 20 notes per withdrawal.
> The ATM machine has stock of fifty
> 500 rupee notes and a sufficient number of notes of other denominations.
(i) If the ATM machine dispenses one
(ii) 500 rupee note, for the total number of notes less than or equal to 20, notes of 200 rupee denominations has to be maximum.
500 × 1 + 200 × 22 + 100 × 1 = 5000
i.e., the minimum number of notes = 24
(iii) If the ATM machine dispenses two 500 rupee notes, for the total number of notes less than or equal to 20, notes of 200 rupee denominations has to be maximum.
500 × 2 + 200 × 20 = 5000 i.e., the minimum number of notes = 22
(iv) If the ATM machine dispenses three 500 rupee notes, for the total number of notes less than or equal to 20, notes of 200 rupee denominations has to be maximum.
500 × 3 + 200 × 17 + 100 × 1 = 5000
i.e., the minimum number of notes = 21
(v) If the ATM machine dispenses four 500 rupee notes, for the total minimum number of notes, notes of 200 rupee denominations has to be maximum.
500 × 4 + 200 × 15 = 5000
i.e., the minimum number of notes = 19
Therefore, if the ATM dispenses 4 notes of 500 rupees then it can serve a maximum of 12 customers with 50 notes.
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?
Answer the following question based on the information given below.
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:
Three students are from X, three are from Y, and the remaining two students, both female, are from Z.
Both the male students from Y minor in Finance, while the female student from Y majors in Operations.
Only one male student majors in Operations, while three female students minor in Marketing.
One female and two male students major in Finance.
Adriana and Deb are from the same institute. Daisy and Amit are from the same institute.
Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
Daisy minors in Operations.
25.
CAT 2018 LRDI Slot 1
| LR - Selection & Distribution
From (2), Y has two male and one female student. Hence, from (1), X has two male and one female student. From (2), female student from Y majors in Operations. Daisy minors in Operations. So Daisy (and also Amit) is not from Y.
From (6),
X: Amit, Chetan and Daisy
Y: Barun, Deb and Adriana
Z: Bandita and Chitra
From (3) and (7), Daisy minors in Operations and remaining three girls minors in Marketing.
From (4), the one female student majors in Finance is Chitra. Hence, Bandita majors in Operations.
From (5), the number of pumps where high contamination levels were recorded is even.
From (1) and (2), high contamination levels recorded at 3 petrol pumps among P1 - P6 i.e., P1, P3 and P5 recorded high levels of contamination.
Therefore, using (4) it can be concluded that high contamination levels could have been recorded at pumps P7 – P15. From (3), the maximum number of pumps where the high contamination levels were recorded must be 5. So this number could be 1, 3 or 5 among P7 – P15.
So, we have the following cases.
From (3), the number of pumps with the same levels of contamination cannot be more than 11.
For 11 pumps with same levels of contamination, P1 and P20 should have the same level of contamination, which is not the case as P1 recorded high levels of contamination while P20 recorded either low or medium levels of contamination. Thus, cases (i) and (ii) are not valid.
Considering case iii we have the following cases:
Refer case1 given in the last table.
Statements [1] and [3] are false.
Refer case 6 given in the last table.
Statement [4] is false.
It can be seen that the contamination level at P10 was recorded as high in each case.
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