# PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning

**Answer the next 3 questions based on the information given below:**

In a class of 8 students students were ranked according to their marks in 'Quant' section as given in the table.

Now, each student recieved zero or more than zero marks in the LRDI section.

Following observations were made based :

- Total marks scored in LRDI section by all students did not exceed 60.
- The rank of any student after adding marks for LRDI section was not the same as that before.
- The marks of no two students were same.

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

If exactly 2 students scored non-zero marks in LRDI section, what is the minimum total marks scored by all of them in LRDI section?

Answer: 47

**Explanation** :

Arranging marks in descending order

Difference in marks between consecutively ranked students:

E – C = 5

H – E = 10

B – H = 9

F – B = 5

D – F = 15

G – D = 6

A – G = 10

Since rank of all students changed after adding marks of LRDI section, hence C who had least marks in Quant section will not have least overall marks.

⇒ C will definitely get non-zero marks in LRDI section.

If exactly 2 students scored non-zero in LRDI one of the two will be C.

Difference in marks of D and F is highest. Hence, if C get 30 marks in LRDI he will move above F in ranking, while ranks of F, B, H and E will slip by 1 position.

Now we are left with A, G and D.

If D gets 17 marks in LRDI, he will move above A in raking while ranks of A and G will slip by 1 position.

Total marks after above changes.

Now, ranks of none of the students is same as that earlier while all of them have different total marks.

∴ Least total marks scored in LRDI section by all students = 30 + 17 = 47.

Hence, 47.

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

What can be the least possible total marks scored by all of them together in LRDI section.

Answer: 38

**Explanation** :

Arranging marks in descending order

Difference in marks between consecutively ranked students:

E – C = 5

H – E = 10

B – H = 9

F – B = 5

D – F = 15

G – D = 6

A – G = 10

Since rank of all students changed after adding marks of LRDI section, hence C who had least marks in Quant section will not have least overall marks.

⇒ C will definitely get non-zero marks in LRDI section.

Since we have to minimize the total marks in LRDI section, we will ignore the highest difference in the marks.

For example: We have to increase C’s marks. We can either increase it by 6 marks to take C above E or increase it by 16 marks to take C above H. In the second case increase is very high, hence we will consider only 1st case.

∴ C gets 6 marks in LRDI section.

If H gets 15 marks in LRDI section he will move above F.

While, if D gets 17 marks in LRDI section he will move above A.

Total marks after above changes.

∴ Minimum total marks obtained in LRDI section = 6 + 15 + 17 = 38.

Hence, 38.

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

If exactly 5 students scored zero marks in LRDI section, what is the maximum possible total marks scored by any student in LRDI section?

Answer: 53

**Explanation** :

Arranging marks in descending order

Difference in marks between consecutively ranked students:

E – C = 5

H – E = 10

B – H = 9

F – B = 5

D – F = 15

G – D = 6

A – G = 10

Since rank of all students changed after adding marks of LRDI section, hence C who had least marks in Quant section will not have least overall marks.

⇒ C will definitely get non-zero marks in LRDI section.

If exactly 5 students scored zero in LRDI, it means only 3 students scored non-zero marks in LRDI, one of whom is C.

Maximum total score of all students in LRDI can be 60.

If C gets enough marks in LRDI such that he gets the first rank, ranks of all other students will slip by one position. But for this C needs to score at least 61 marks which is not possible.

Suppose C score just enough to change his own rank i.e., C gets 6 marks in LRDI, while one more student gets 1 mark in LRDI, then the third student can get (60 – 6 – 1 =) 53 marks in LRDI section.

This third student can be H. If H gets 53 marks in LRDI, he will get first rank and ranks of A, G, D, F and B will slip by one position.

∴ Maximum marks scored by a student can be 53.

Hence, 53.

Workspace:

**Answer the next 4 questions based on the information given below.**

Company XYZ has two types of products – A & B. Every year, the company sells all the products that it manufactures in that particular year. The production of ‘A’ commenced in 2016. Exactly 10% of the quantity of ‘A’ sold in a year is discarded exactly two years later; another 10% (of the total) is discarded in the next year and so on, so that the entire quantity of A manufactured in a year is completely discarded exactly at the end of eleven years. It is also known that the company manufactured at least 100 units of ‘A’ each year and did not manufacture the same quantity in any two years.

The production of ‘B’ started in 2018. 20% of the units of ‘B’ that are sold in a year are discarded exactly two years later; another 20% (of the total) is discarded in the next year and so on, so that the entire quantity of B manufactured in a year is completely discarded exactly at the end of six years. The bar graph below gives the number of units of these products that were in use by the citizens in the given years.

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many units of A were manufactured in 2016?

- (a)
100

- (b)
150

- (c)
None of these

- (d)
Cannot be determined

Answer: Option C

**Explanation** :

**Product B**:

**2018**: 80 units of B were produced in 2018.

**2019**: Since 125 units were in use in 2019, hence (125 – 80 =) 45 units of B were produced in 2019.

Also, 20% of 80 (units produced in 2018) will be discarded at the end of 2019. Hence, number of units left for next year is 125 – 16 = 109

**2020**: There were 184 units in use during 2020. Hence, number of units produced in 2020 = 184 – 109 = 75.

Also, 10% of 125 (80 + 25 units produced in 2018 & 2019) will be discarded at the end of 2020. Hence, number of units left for next year is 184 – 25 = 159.

**2021**: There were 229 units in use during 2021. Hence, number of units produced in 2021 = 229 – 159 = 70.

We get the following table for product B.

**Product A**:

Since 10% of A produced has to be discarded, hence the number of units of A produced any year will be a multiple of 10.

**2016**: Let 10x units were produced. (x ≥ 10)

**2017**: Let 10y units were produced. (y ≥ 10)

**2018**: Let 10z units were produced. (z ≥ 10)

It is given that at least 100 units of A is produced every year and number of units produced every year is different.

∴ x, y and z are all different.

We can make the following table for product ‘A’ for 2016-2018

**From 2017**: Number of units produced 10y = 300 – 10x

⇒ x + y = 30 …(1)

**From 2018**: Number of units produced = 10z = 390 – (300 - x)

⇒ 10z = 90 + x

Here, the least possible value of x is 10, which gives us z = 10. This is not possible since the number of units produced every year has to be different.

Next possible value of x is 20, which gives z = 11 and y = 10. This is a possible solutions for x, y and z.

For highest values of x, y will not be greater than or equal to 10.

∴ x = 20, y = 10 and z = 11.

Hence, we can complete the table for product ‘A’.

∴ Number of units of A produced in 2016 = 200.

Hence, option (c).

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many units of A were discarded in 2020?

Answer: 41

**Explanation** :

Consider the solution for first question of this set.

Number of units of A discarded in 2020 = 53.

Hence, 53.

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many units of A were manufactured by the end of 2021?

- (a)
790

- (b)
810

- (c)
800

- (d)
None of these

Answer: Option C

**Explanation** :

Consider the solution for first question of this set.

Total number of units of A produced till 2021 = 200 + 100 + 110 + 120 + 140 + 130 = 800.

Hence, option (c).

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many units of B were discarded by the end of 2021?

Answer: 25

**Explanation** :

Consider the solution for first question of this set.

Number of units of B discarded in 2021 = 81.

Hence, 53.

Workspace:

**Answer the next 4 questions based on the information given below.**

The government is planning to reintroduce Cheetahs in the country which earlier went extinct. It plans to introduce 160 Cheetahs equally across 8 national parks. The induction is planned to be completed in 4 phases. Table below shows the number of Cheetahs inducted in each of the 8 parks in phase 1 and 2.

Number of Cheetah’s inducted in any phase in any park is between 1 and 9 (including both). It is known that, in a particular phase, no two parks get an equal number of Cheetahs. The number of Cheetahs inducted in park A in phase 4 is more than that in phase 3.

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many Cheetahs were inducted in park F in phase 3?

Answer: 6

**Explanation** :

Since 160 Cheetahs are equally inducted in each of the 8 national parks, hence, 20 Cheetahs are inducted in each national park.

3 Cheetah’s are inducted in park E in phase 3 and 4 together.

∴ This can be done as (in any order) 1 + 2.

4 Cheetah’s are inducted in park C in phase 3 and 4 together.

∴ This can be done as (in any order) 1 + 3 or 2 + 2.

Now since 2 Cheetah are to be inducted in park E in one of the phases, hence 2 + 2 is not possible, since in a phase each park gets a different number of Cheetahs.

∴ This can be done as (in any order) 1 + 3.

6 Cheetah’s are inducted in park H in phase 3 and 4 together.

Since in a phase each park gets a different number of Cheetahs.

∴ This can be done as (in any order) 2 + 4.

8 Cheetah’s are inducted in park G in phase 3 and 4 together.

Since in a phase each park gets a different number of Cheetahs.

∴ This can be done as (in any order) 2 + 4.

10 Cheetah’s are inducted in park A in phase 3 and 4 together.

Since in a phase each park gets a different number of Cheetahs.

∴ This can be done as (in any order) 6 + 4.

Now, Cheetahs inducted in park A in phase 4 is more than that in phase 3, hence we get.

Case 2 is rejected.

12 Cheetah’s are inducted in park D in phase 3 and 4 together.

Since in a phase each park gets a different number of Cheetahs.

∴ This can be done as 7 + 5.

13 Cheetah’s are inducted in park F in phase 3 and 4 together.

Since in a phase each park gets a different number of Cheetahs.

∴ This can be done as 6 + 7.

17 Cheetah’s are inducted in park B in phase 3 and 4 together.

Since in a phase each park gets a different number of Cheetahs.

∴ This can be done as (in any order) 8 + 9.

⇒ Number of Cheetahs inducted in park F in phase 3 = 6

Hence, 6.

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

Total how many Cheetah’s were inducted in phase 4?

- (a)
35

- (b)
36

- (c)
37

- (d)
Cannot be determined

Answer: Option D

**Explanation** :

Consider the solution to the first question of this set.

Total number of Cheetah’s inducted in phase 4 = 36 or 37.

Hence, option (d).

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

Which park got the least number of Cheetah’s in phase 4?

- (a)
A

- (b)
C

- (c)
E

- (d)
F

Answer: Option B

**Explanation** :

Consider the solution to the first question of this set.

In phase 4 park C gets the least number of Cheetah’s i.e. 1.

Hence, option (b).

Workspace:

**PE 4 - Mathematical Reasoning | LR - Mathematical Reasoning**

If all parks together received more Cheetah’s in phase 3 than phase 4, then how many Cheetah’s were received in park B in phase 4?

Answer: 8

**Explanation** :

Consider the solution to the first question of this set.

In phase 4 park B gets 8 Cheetah’s.

Hence, 8.

Workspace:

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