# LR - Mathematical Reasoning - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic LR - Mathematical Reasoning. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

**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 = $\frac{5\times 0\%+5\times P\%}{5+5}$ ≥ 10

⇒ 5P ≥ 100

⇒ P ≥ 20%

∴ Volume of impurities in bottle B is at least 20% of 50 ml i.e., ≥ 10 ml.

Hence, option (a).

Workspace:

**CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

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”?

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 = $\frac{0+0+0+0}{4}$ = 0%

∴ Impurities will not be detected.

**Case 2**: 3 bottles contain P and 1 contains I

Concentration of I in final mixture = $\frac{0+0+0+100}{4}$ = 25%

∴ Impurities will be detected.

**Case 3**: 2 bottles contain P and 2 contain I

Concentration of I in final mixture = $\frac{0+0+100+100}{4}$ = 50%

∴ Impurities will be detected.

**Case 4**: 1 bottle contains P and 3 contain I

Concentration of I in final mixture = $\frac{0+100+100+100}{4}$ = 75%

∴ Impurities will be detected.

**Case 5**: All bottles contain only I

Concentration of I in final mixture = $\frac{100+100+100+100}{4}$ = 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.

Only one test is required is such a situation.

Hence, 1.

Workspace:

**CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

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?

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.

In both cases 2 tests are required to detect I.

Hence, 2.

Workspace:

**CAT 2021 LRDI Slot 3 | LR - Mathematical Reasoning**

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 = $\frac{0+15+15+15}{4}$ = 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 = $\frac{0+0+15+15}{4}$ = 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.

Hence, option (d).

Workspace:

**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.

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

What is the minimum possible number of different types of prizes?

Answer: 2

**Explanation** :

There is exactly one prize of type a.

There can be 99 items of type b. Thus, there can be only two types of items.

Answer: 2.

Workspace:

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

What is the maximum possible number of different types of prizes?

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.

Answer: 6.

Workspace:

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

Which of the following is not possible?

- A.
There are exactly 30 items of type b.

- B.
There are exactly 75 items of type e.

- C.
There are exactly 60 items of type d.

- D.
There are exactly 45 items of type c.

Answer: Option D

**Explanation** :

There is exactly one prize of type a.

[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.

Hence, option 4.

Workspace:

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

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

So number of remaining items = 25

Now, 25 = (1 + 2 + 4 + 17) or (1 + 3 + 6 + 16) ,… etc

So there can be 5 types of different items.

Hence, option 1.

Workspace:

**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.

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

Which digit does the letter A represent?

Answer: 1

**Explanation** :

Units place of (F + F) is F; F = 0 or 5 Units place of (H + H) or (H + H + 1) is also F. This is possible only when H is 5 and there is no carry over.

So, F = 0 and H = 5

∴ (1 + B + A) = (10A + A) ⇒ A = 1 and B = 9

As units place of (G + K) = 1, actual value of G + K = 11

So, 1 + F + A = 1 + 1 + 0 = 2 = C

Now, 1 + J = G and G + K = 11

Also, G, J and K ∈ {3, 4, 6, 7, 8}

Therefore, (G, J, K) = (8, 3, 7) or (7, 4, 6) or (4, 7, 3), accordingly D and E can be represented by (4,6) or (3, 8) or (6, 8) in some order.

The letter A represents digit 1.

Answer: 1.

Workspace:

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

Which digit does the letter B represent?

Answer: 9

**Explanation** :

Consider the solutoin to first question of this set.

B represents the digit 9.

Answer: 9.

Workspace:

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?

Answer: 7

**Explanation** :

Consider the solutoin to first question of this set.

The digit 7 cannot be represented by the letter D.

Answer: 7.

Workspace:

**CAT 2019 LRDI Slot 1 | LR - Mathematical Reasoning**

Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?

Answer: 6

**Explanation** :

Consider the solutoin to first question of this set.

The digit 6 cannot be represented by the letter G.

Answer: 6.

Workspace:

**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.

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

Which of the following statements MUST be true?

- 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 = $\frac{60-8-20}{2}$

$=\frac{32}{2}$ = 16

From (5), Danish and Harini also scored 20 marks in GA.

Therefore, his Composite score = 2(8) + 15 + 20 = 51

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.

Jatin’s composite score = 70 and Danish’s composite score = 51

Thus, statement 1 is true.

Indu’s score in DI = 16 and Chetna’s score in DI = 19

Thus, statement 2 is true.

Jatin’s score in GA = 14 and Indu’s score in GA = 20

Thus, statement 3 is false.

Hence, option 1.

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

Which of the following statements MUST be FALSE?

- A.
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.

Hence, statement 2 must be false.

Hence, option 2.

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

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: 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.

For d = 13, the composite score = 46

Thus, Bala could have scored 13 marks in DI.

Therefore, the required answer is 13.

Answer: 13

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

If all the candidates scored different marks in WE then what is the maximum marks that Harini could have scored in WE?

Answer: 14

**Explanation** :

Marks scored in WE are 20, 19, 18, 16, 15, 9, 8, 7, 4 and ‘a’.

As all scored different marks, ‘a’ can take one of the following values: 17, 14, 13, 12, 11, 10, 6, 5, 3, 2, 1

For a = 17, Harini’s composite score i.e., 47 would be same as Falak’s. Hence, a ≠ 17

For a = 14, Harini’s composite score i.e., 44 which is different from others.

Therefore, the required answer is 14.

Answer: 14

Workspace:

**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.

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

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

Answer: 7

**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)

(iii) 500 × 9 = 4500, 200 × 1 = 200 and 100 × 3 = 300

(Notes of Rs. 500 denomination = 9 and notes of Rs. 200 and Rs. 100 denominations = 1 + 3 = 4, 9 > 4)

(iv) 500 × 9 = 4500, 100 × 5 = 500

(Notes of Rs. 500 denomination = 9 and notes of Rs. 200 and Rs. 100 denominations = 0 + 5 = 5, 9 > 5) If the ATM machine dispenses 8 notes of Rs. 500;

(v) 500 × 8 = 4000 and 200 × 5 = 1000

(Notes of Rs. 500 denomination = 8 and notes of Rs. 200 and Rs. 100 denominations = 5 + 0 = 5, 8 > 5)

(vi) 500 × 8 = 4000, 200 × 4 = 800 and 100 × 2 = 200

(Notes of Rs. 500 denomination = 8 and notes of Rs. 200 and Rs. 100 denominations = 4 + 2 = 6, 8 > 6)

(vii) 500 × 8 = 4000, 200 × 3 = 600 and 100 × 4 = 400

(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.

Therefore, the required answer is 7.

Answer: 7

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

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?

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)

(iii) 500 × 9 = 4500, 200 × 1 = 200 and 100 × 3 = 300

(Notes of Rs. 500 denomination = 9 and notes of Rs. 200 and Rs. 100 denominations = 1 + 3 = 4, 9 > 4)

(iv) 500 × 9 = 4500, 100 × 5 = 500

(Notes of Rs. 500 denomination = 9 and notes of Rs. 200 and Rs. 100 denominations = 0 + 5 = 5, 9 > 5) If the ATM machine dispenses 8 notes of Rs. 500;

(v) 500 × 8 = 4000 and 200 × 5 = 1000

(Notes of Rs. 500 denomination = 8 and notes of Rs. 200 and Rs. 100 denominations = 5 + 0 = 5, 8 > 5)

(vi) 500 × 8 = 4000, 200 × 4 = 800 and 100 × 2 = 200

(Notes of Rs. 500 denomination = 8 and notes of Rs. 200 and Rs. 100 denominations = 4 + 2 = 6, 8 > 6)

(vii) 500 × 8 = 4000, 200 × 3 = 600 and 100 × 4 = 400

(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.

Therefore, the required answer is 6.

Answer: 6

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

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.

Hence, option 2.

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Mathematical Reasoning**

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

Hence, option 3.

Workspace:

**Answer the following question based on the information given below.**

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 withanother. 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:

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

**CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

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

Answer: 1200

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

Answer: 1200

Workspace:

**CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

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

- A.
3000

- B.
6000

- C.
19000

- D.
22000

Answer: Option C

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

The outlet sold 19000 units of currency C on that day.

Hence, option 3.

Workspace:

**CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

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

Answer: 240

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

The base exchange rate of B w.r.t. L on that day was 240

Therefore the required answer is 240.

Answer: 240

Workspace:

**CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

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

- A.
0.95

- B.
1.10

- C.
1.90

- D.
2.20

Answer: Option C

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

The buying exchange rate of C w.r.t. L on that day was 1.90

Hence, option 3.

Workspace:

**Answer the following question based on the information given below.**

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 particular day:

- 140 tickets were sold.
- 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.
- 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.
- The number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.

**CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

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.
32

- B.
34

- C.
38

- D.
36

Answer: Option A

**Explanation** :

From points 1 and 2, the number of tickets bought by Young people = 80, the number of tickets bought by the middle aged people = 40 and the number of tickets bought by the old people = 20.

Using points 3 & 4, we get the following

Now all the questions can be answered.

We have, 20 - 2y = x + 2y - 20. Therefore, x + 4y = 40.

Therefore, x = 40 - 4y = 4(10 - y)

or 2x = 8(10 - y). Therefore, x is a multiple of 8. Out of the given options, 32 is a possible answer.

Hence, option 1.

Workspace:

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