# PE 4 - Puzzles | LR - Puzzles

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

The following table gives the number of students across six different classes of BVN Higher Secondary School in the years 2020 & 2021.

It was known that:

- Students who fail in a class in a year will study in the same class next year.
- New students are admitted only in class 1.
- No student leaves the school before passing out class 6.

**PE 4 - Puzzles | LR - Puzzles**

What is the minimum possible number of students who could have joined the school in 2020?

Answer: 48

**Explanation** :

To minimise the number of students who could have joined the school in 2021, we need to minimise the number of students that exit the school in 2021.

∴ We need to maximise the number of students failing in class 6.

If all 33 students in class 6 fail in 2020, these 33 students will remain in class 6 in 2021.

Hence, out of 48 students enrolled in class 6 in 2021, 33 of them were those who failed previous year and hence remained in class 6.

It means (48 - 33 =) 15 students of class 6 in 2021 were those who passed in class 5 in 2020.

Similarly, we can draw the following chart.

Hence, minimum number of students who could have joined the school in 2021 is 48.

Hence, 48.

Workspace:

**PE 4 - Puzzles | LR - Puzzles**

What is the maximum possible number of students who could have joined the school in 2020?

Answer: 77

**Explanation** :

To maximise the number of students who could have joined the school in 2021, we need to maximise the number of students that exit the school in 2021.

∴ We need to minimise the number of students failing in class 6.

There are 48 students in class 6 in 2021. Out of these 48, maximum 44 could have passed in class 5 and moved to class 6. Hence, minimum 4 students need to fail in class 6.

Now, since all 44 students passed from class 5 to class 6, no student failed in class 5 in 2020.

Now in 2021 all 49 students in class 5 could have come from passing class 4.

Hence, out of 59 students in class 4 in 2020, at most 49 passed to next class and hence at least 10 failed and stayed in class 4.

Similarly, we can draw the following chart.

Hence, maximum number of students who could have joined the school in 2021 is 77.

Hence, 77.

Workspace:

**PE 4 - Puzzles | LR - Puzzles**

In 2020, which of the following was not a possible pass percentage of Class 3?

- A.
27.54%

- B.
33.33%

- C.
69.57%

- D.
None of these

Answer: Option D

**Explanation** :

Consider the solution to first 2 questions of this set.

Minimum students that could have passed class 3 in 2020 = 19

∴ Minimum pass percentage = 19/69 × 100% = 27.54%

Maximum students that could have passed class 3 in 2020 = 48

∴ Maximum pass percentage = 48/69 × 100% = 69.57%

∴ Pass percentage for class 3 can be between 27.54% and 69.57%.

Hence, option (d).

Workspace:

**PE 4 - Puzzles | LR - Puzzles**

In 2020, if the pass percentage in Class 2 was 50%, then what was the pass percentage in Class 5?

- A.
78.39%

- B.
85%

- C.
86.36%

- D.
44.44%

Answer: Option C

**Explanation** :

Since 50% students passed in class 2, hence out of 64 students in class 3 next year 37 came from class 2.

It means (64 - 37 =) 27 students must have failed and remaind in class 3.

Since 27 students failed in class 3, hence (69 - 27 =) 42 students passed from class 3 to class 4.

Out of 58 students in class 4 in 2021, 42 came from class 3, hence (58 - 42 =) 16 failed in class 4 in 2020.

Since 16 students failed in class 4 in 2020, hence (59 - 16 =) 43 students passed from class 4 to class 5.

Out of 49 students in class 5 in 2021, 43 came from class 4, hence (49 - 43 =) 6 failed in class 5 in 2020.

Since 6 students failed in class 5 in 2020, hence (44 - 6 =) 38 students passed from class 4 to class 5.

Similarly, we can draw the following chart.

∴ pass % for class 5 = 38/44 × 100% = 86.36%.

Hence, option (c)..

Workspace:

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

Five bakeries in Bhopal made 100 cakes each during the month of October. Each cake was decorated using at least one of the following add-ons: Fruits, Chocolate, Fondant, Sprinkles and Frosting.

Each cake was supposed to be designed by certain guidelines as follows:

1. If Chocolate was used, Sprinkles was not used.

2. If Fondant was used, Chocolate was not used.

3. If a Sprinkles was used, Fondant was also used.

4. If Fruits were used, Chocolate was also used.

5. If Frosting was used, Fruits were also used.

Each bakery, except 1, violated exactly one of the above-mentioned guidelines. Also, each bakery violated a different guideline. The table below gives the details of the number of cakes and add-ons used by each of the 5 bakeries.

**PE 4 - Puzzles | LR - Puzzles**

Which bakery did not violate any guidelines?

- A.
A

- B.
C

- C.
D

- D.
E

Answer: Option C

**Explanation** :

Let us look at the guidelines.

Guideline 1 says: If Chocolate was used, Sprinkles was not used. It means Chocolate and Sprinkles cannot be used together.

G1 ⇒ Chocolate × Sprinkles

G2 ⇒ Chocolate × Fondant

G3 ⇒ Sprinkles → Fondant

G4 ⇒ Fruits → Chocolate

G5 ⇒ Frosting → Fruits → Chocolate

Possible combination of add-ons are:

1. Sprinkles + Fondant

2. Fondant

3. Frosting + Fruits + Chocolate

4. Fruits + Chocolate

5. Chocolate

⇒ Each cake will have one of Fondant or Chocolate.

∴ Number of cakes with Fondant + Number of cakes with Chocolate = Total number of cakes = 100.

We see that only Bakery D satisfies the above equation, hence Bakery D must have followed all the rules.

**Bakery A**: Cakes with Fondant + Cakes with Chocolate > 100. It means there must be some cakes which have both Fondant and Chocolate and hence Bakery A violated guideline number 2.

100 = Cakes with Chocolate + Cakes with Fondant – Cakes with both Fondant and Chocolate

⇒ 100 = 55 + 55 - Cakes with both Fondant and Chocolate

⇒ Cakes with both Fondant and Chocolate = 110 – 100 = 10

Here, we will have a 6^{th} type of cake which contains both fondant and chocolates.

**Bakery B: **

There are 60 cakes with chocolate, hence there must be 40 cake with fondant on it. But that is not the case.

To make 100 cakes Bakery B did not use Fondant whenever it used Sprinkles, hence it violated guideline number 3.

**Bakery C:**

Bakery C did not use Chocolates whenever it used Fruits, hence it violated guideline number 4.

**Bakery D:**

Bakery D has not violated any guideline.

**Bakery E:**

Bakery E did not use Fruits whenever it used Frosting, hence it violated guideline number 5.

Hence, option (c).

Workspace:

**PE 4 - Puzzles | LR - Puzzles**

Which guideline did C break?

- A.
5

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Consider the solution to the first question of this set.

Bakery C violated 4^{th} guideline.

Hence, option (d).

Workspace:

**PE 4 - Puzzles | LR - Puzzles**

How many cakes had only Chocolate on them from bakery C?

Answer: 32

**Explanation** :

Consider the solution to the first question of this set.

32 cakes had only chocolate on them from bakery C.

Hence, 32.

Workspace:

**PE 4 - Puzzles | LR - Puzzles**

How many cakes had at least 3 add-ons from bakery E?

Answer: 0

**Explanation** :

Consider the solution to the first question of this set.

No cake had 2 add-ons on them from bakery E.

Hence, 0.

Workspace:

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