LR - Mathematical Reasoning - Previous Year CAT/MBA Questions
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Answer the next 5 questions based on the information given below:
Adhara, Bithi, Chhaya, Dhanavi, Esther, and Fathima are the interviewers in a process that awards funding for new initiatives. Every interviewer individually interviews each of the candidates individually and awards a token only if she recommends funding. A token has a face value of 2, 3, 5, 7, 11, or 13. Each interviewer awards tokens of a single face value only.
Once all six interviews are over for a candidate, the candidate receives a funding that is Rs.1000 times the product of the face values of all the tokens. For example, if a candidate has tokens with face values 2, 5, and 7, then they get a funding of Rs.1000 × (2 × 5 × 7) = Rs.70,000.
Pragnyaa, Qahira, Rasheeda, Smera, and Tantra were five candidates who received funding. The funds they received, in descending order, were Rs.390,000, Rs.210,000, Rs.165,000, Rs.77,000, and Rs.66,000.
The following additional facts are known:
1. Fathima awarded tokens to everyone except Qahira, while Adhara awarded tokens to no one except Pragnyaa.
2. Rashida received the highest number of tokens that anyone received, but she did not receive one from Esther.
3. Bithi awarded a token to Smera but not to Qahira, while Dhanavi awarded a token to Qahira but not to Smera.
From (1): Fathima awarded tokens to everyone except Qahira,
∴ Fatima must have awarded token number 3 to everyone and Qahira received tokens whose product is 77.
From (1): Adhara awarded tokens to no one except Pragnyaa
∴ Adhara must have awared token 13 and Pragnyaa received tokens whose product is 390.
From (2): Rashida received the highest number of tokens that anyone received, but she did not receive one from Esther.
∴ Rashida must have received 4 tokes whose product can only be 210. Also, since Esther did not give her the token, hence Esther must have distributed token number 11.
From (3): Dhanavi awarded a token to Qahira but not to Smera.
∴ Dhanavi must have awared token number 7.
From (3): Bithi awarded a token to Smera but not to Qahira
∴ Bithi/Chhaya could have awared token 2/5 in any order.
If Bithi gives token number 2 to S/T, then Chhaya will given token number 5 to T/S and vice-a-versa.
Answer the next 4 questions based on the information given
Each of the bottles mentioned in this question contains 50 ml of liquid. The liquid in any bottle can be 100% pure content (P) or can have certain amount of impurity (I). Visually it is not possible to distinguish between P and I. There is a testing device which detects impurity, as long as the percentage of impurity in the content tested is 10% or more.
For example, suppose bottle 1 contains only P, and bottle 2 contains 80% P and 20% I. If content from bottle 1 is tested, it will be found out that it contains only P. If content of bottle 2 is tested, the test will reveal that it contains some amount of I. If 10 ml of content from bottle 1 is mixed with 20 ml content from bottle 2, the test will show that the mixture has impurity, and hence we can conclude that at least one of the two bottles has I. However, if 10 ml of content from bottle 1 is mixed with 5 ml of content from bottle 2. the test will not detect any impurity in the resultant mixture.
A.
5 ml of content from bottle A is mixed with 5 ml of content from bottle B. The resultant mixture, when tested, detects the presence of I. If it is known that bottle A contains only P, what BEST can be concluded about the volume of I in bottle B?
B.
10 ml or more
C.
Less than 1 ml
D.
10 ml
E.
1 ml
Answer: Option
A
Explanation :
Since impurity is detected in the final mixture, hence the mixture has at least 10% impurities.
∴ % impurity in the mixture = ≥ 10
⇒ 5P ≥ 100
⇒ P ≥ 20%
∴ Volume of impurities in bottle B is at least 20% of 50 ml i.e., ≥ 10 ml.
There are four bottles. Each bottle is known to contain only P or only I. They will be considered to be “collectively ready for despatch” if all of them contain only P. In minimum how many tests, is it possible to ascertain whether these four bottles are “collectively ready for despatch”?
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Answer:1
Explanation :
The bottles contain only P or only I.
Let us mix equal quantities from each of these bottles and check for impurities.
Case 1: All bottles contain only P
Concentration of I in final mixture = = 0%
∴ Impurities will not be detected.
Case 2: 3 bottles contain P and 1 contains I
Concentration of I in final mixture = = 25%
∴ Impurities will be detected.
Case 3: 2 bottles contain P and 2 contain I
Concentration of I in final mixture = = 50%
∴ Impurities will be detected.
Case 4: 1 bottle contains P and 3 contain I
Concentration of I in final mixture = = 75%
∴ Impurities will be detected.
Case 5: All bottles contain only I
Concentration of I in final mixture = = 100%
∴ Impurities will be detected.
There is only one case where impurities are not detected and that is when all the bottles have only P.
⇒ If we mix equal quantities of all 4 bottles and test for impurities, if impurity is not detected then we can safely accept all 4 bottles.
There are four bottles. It is known that three of these bottles contain only P, while the remaining one contains 80% P and 20% I. What is the minimum number of tests required to definitely identify the bottle containing some amount of I?
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Answer:2
Explanation :
Let the bottles be A, B, C and D.
We first mix equal quantities of A and B and test the mixture.
Case 1: If any one of A or B contains 20% impurities, test will detect the impurity.
Then we check A for impurity. If impurity is detected then A contains I else B contains I.
∴ 2 tests are required to detect I.
Case 2: If none of A or B contains impurities, test will not detect the impurity.
This means one of C or D contains I.
Then we check C for impurity. If impurity is detected then C contains I else D contains I.
∴ 2 tests are required to detect I.
There are four bottles. It is known that either one or two of these bottles contain(s) only P, while the remaining ones contain 85% P and 15% I. What is the minimum number of tests required to ascertain the exact number of bottles containing only P?
A.
3
B.
2
C.
4
D.
1
Answer: Option
D
Explanation :
Case 1: Only bottle contains pure P.
Now if we mix equal quantities from each of the 4 bottles, the concentration of I in the mixture = = 11.25%
∴ Testing this mixture will detect the impurities.
Case 2: 2 bottles contains pure P.
Now if we mix equal quantities from each of the 4 bottles, the concentration of I in the mixture = = 7.5%
∴ Testing this mixture will not detect the impurities.
⇒ Mix equal quantities of all 4 bottles, it I is detected then only 1 bottle contains P else 2 bottles contain P.
Answer the following question based on the information given below.
A new game show on TV has 100 boxes numbered 1, 2, . . . , 100 in a row, each containing a mystery prize. The prizes are items of different types, a, b, c, . . . , in decreasing order of value. The most expensive item is of type a, a diamond ring, and there is exactly one of these. You are told that the number of items at least doubles as you move to the next type. For example, there would be at least twice as many items of type b as of type a, at least twice as many items of type c as of type b and so on. There is no particular order in which the prizes are placed in the boxes.
What is the maximum possible number of different types of prizes?
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Answer:6
Explanation :
There is exactly one prize of type a.
As we need to find maximum possible different types of prizes, number of prizes of type b has to be minimum possible and hence must be 2, number of items of type c = 4 …and so on.
1(type a) + 2(type b) + 4(type c) + 8(type d) + 16(type e) + 32(type e) = 63
Suppose there is prize of type f then number of items has to be at least 64. But then there are more than 100 items, which is not true. So there cannot be prize of type f.
[1]. If there are 30 items of type b, then items of type c = 100 – 30 −1 = 69. So this case is possible.
[2]. There are 75 items of type e, then items of type b, c and d = 100 – 75 −1 = 24.
Some of the values of (b, c, d) are (2, 4, 18) or (2, 5, 17) or (3, 6, 15). So this case is possible.
[3]. If there are 60 items of type d, then items of type b and c = 100 – 60 −1 = 39. So this case is possible as we can find many combinations for (b, c).
[4]. If there are 45 items of type c, then items of type a, b and c in all cannot be more than 1 + 22 + 45 = 68. Now items of type d has to be more than 90. But then total number of items exceed 100. So this case is not possible.
You ask for the type of item in box 45. Instead of being given a direct answer, you are told that there are 31 items of the same type as box 45 in boxes 1 to 44 and 43 items of the same type as box 45 in boxes 46 to 100.
What is the maximum possible number of different types of items?
A.
5
B.
3
C.
6
D.
4
Answer: Option
A
Explanation :
Considering the given options, the maximum number of different types can be 6.
Assume that there are 6 items.
Now number of items of same type as the one in box 45 = 1 + 31 + 43 = 75
So number of remaining items = 25
1 + 2 + 4 + 8 + 16 = 31. If there are 5 types of items, the minimum number of items of 5 types = 31.
31 + 75 >100
So, there cannot be 6 types of items.
Now consider that there are 5 types of items.
Now number of items of same type as the one in box 45 = 1 + 31 + 43 = 75
Answer the following question based on the information given below.
The following table represents addition of two six-digit numbers given in the first and the second rows, while the sum is given in the third row. In the representation, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 has been coded with one letter among A, B, C, D, E, F, G, H, J, K, with distinct letters representing distinct digits.
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.
Jatin's composite score was more than that of Danish.
Indu scored less than Chetna in DI.
Jatin scored more than Indu in GA.
A.
Both 1 and 2
B.
Both 2 and 3
C.
Only 1
D.
Only 2
Answer: Option
A
Explanation :
70% of 20 = 14
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 was less than that of Falak
B.
Bala scored same as Jatin in DI
C.
Bala’s composite score was less than that of Ester
D.
Chetna scored more than Bala in DI
Answer: Option
B
Explanation :
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?
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Answer:13
Explanation :
2d + 20 < 54 ⇒ d < 17
d ≠ 14, 15, 16, 12, 8 and 5
d can take value 13 or 11 or 9 or 7 or 6 or 4 or 3 or 2 or 1.
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?
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Answer:6
Explanation :
Preference was given to Rs. 500 notes.
The ATM machine can dispense 10 notes of Rs. 500.
(i) 500 × 10 = Rs. 5,000
If the ATM machine dispenses 9 notes of Rs. 500;
(ii) 500 × 9 = 4500, 200 × 2 = 400 and 100 × 1= 100
(Notes of Rs. 500 denomination = 9 and notes of Rs. 200 and Rs. 100 denominations = 2 + 1 = 3, 9 > 3)
(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?
A.
13
B.
12
C.
16
D.
10
Answer: Option
B
Explanation :
Here we just need to keep in mind that
> 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?
A.
750
B.
800
C.
900
D.
1400
Answer: Option
C
Explanation :
In order to have the total number of notes to be dispensed is the smallest possible, notes of denomination Rs. 500 has to be maximum.
The ATM machine could have dispensed 10 notes of 500 rupee to the customers who gave 500 rupee notes as their preference.
Therefore, the number of 500 rupee notes required to serve these 50 customers = 10 × 50 = 500
The ATM machine could have dispensed 10 notes of 100 rupee and 8 notes of 500 rupee to the customers who gave 100 rupee notes as their preference.
Therefore, the number of 500 rupee notes required to serve these 50 customers = 8 × 50 = 400
Total number of 500 rupee notes required = 500 + 400 = 900
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