# PE 2 - Puzzles | LR - Puzzles

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

In Tokyo Olympics 2020 athletes from 4 different countries – USA, China, Japan and India won certain number of medals.

The graph below shows the number of athletes from each country.

Each athlete won either 1, 2, 3, 4, or 5 medals. The table below shows the distribution of athletes and number of medals.

No 2 athletes from a country won same number of medals.

**PE 2 - Puzzles | LR - Puzzles**

If athletes from India win the maximum possible number of medals, what is the total number of medals won by all athletes from India?

Answer: 7

**Explanation** :

12 athletes won at least 1 medal and there are exactly 12 athletes from the given countries hence, every athlete won at least 1 medal.

There are 5 athletes winning at least 2 medals.

There are 11 athletes winning at most 4 medals hence, there will be only 1 athlete winning 5 medals.

There are 7 athletes winning at least 3 medals, hence there will be 6 athletes winning either 3 or 4 medals.

We know there are 5 athletes from USA, and they all won different number of medals.

Also, there is only 1 athlete who wins 5 medals. Hence, the only athlete to win 5 medals is from USA.

Hence, we get,

Since India wins maximum possible number of medals hence, 2 athletes from India will win 3 and 4 medals i.e., total 7 medals.

Hence, 7.

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**PE 2 - Puzzles | LR - Puzzles**

If it is known that China wins the maximum possible medals and maximum possible number of athletes got 4 medals, then how many athletes won 3 medals?

Answer: 2

**Explanation** :

Consider the solution to the first question of this set.

If China wins the maximum possible medals and maximum possible number of athletes got 4 medals, then we get the following table.

Since there are 6 athletes winning either 3 or 4 medals hence, at max 4 athletes will win 4 medals each and 2 athletes will win 3 medals each.

Hence, 2.

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**PE 2 - Puzzles | LR - Puzzles**

In the previous question if minimum possible athletes won 2 medals, then how medals did athletes from Japan win?

- (a)
6

- (b)
5

- (c)
4

- (d)
Cannot be determined

Answer: Option B

**Explanation** :

Consider the solution to previous question.

Since minimum number of athletes win 2 medals, hence only 2 athletes will win 2 medals each and 3 athletes will win 1 medal each.

∴ Total number of medals won by athletes from Japan = 1 + 4 = 5.

Hence, option (b).

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**PE 2 - Puzzles | LR - Puzzles**

What is the total number of medals that athletes from these 4 countries win?

- (a)
36

- (b)
35

- (c)
34

- (d)
Cannot be determined

Answer: Option D

**Explanation** :

Consider the solution to first question of this set.

Since we can only draw the following table from the information given, we cannot determine the total number of medals won by athletes from these 4 countries.

Hence, option (d).

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

Harappa was a tribe in the northwestern regions of South Asia. It heavily depended on trade of gold, silver, and bronze coins. Any member of the tribe could exchange coins amongst gold or silver or bronze for any of the other 2 types of coins. The coins are traded as a whole and can’t be cut for the purpose of exchange.

One gold coin can be exchanged for 2 silver and 2 bronze coins. 3 silver coins can be exchanged for 1 gold and 2 bronze coins.

**PE 2 - Puzzles | LR - Puzzles**

One of the tribesman had certain number of coins with him. He exchanged one coin and ended up with same number of coins of each type. What is the minimum number of coins with him initially.

Answer: 3

**Explanation** :

Let the value of each gold, silver and bronze coin be ‘g’, ‘s’ and ‘b’ respectively.

Given,

g = 2s + 2b …(1) &

3s = g + 2b …(2)

Solving (1) and (2) we get

g = 10b and s = 4b.

The tribesman exchanged 1 coin. Since coins have to be exchange as a whole without cutting them, he could’ve exchanged either 1 gold or 1 silver coin and ends up with same number of coins.

**Case 1**: He could’ve exchanged 1 silver coin with 4 bronze coins, hence at the end he has at least 4 bronze coins.

Hence, he has at least 4 coins of each type at the end.

∴ Initially he had at least 4 gold, and 5 silver coins i.e., a total of 9 coins.

**Case 2 (a)**: He could’ve exchanged 1 gold coin for 10 bronze coins, hence at the end he has at least 10 bronze coins.

Hence, he has at least 10 coins of each type at the end.

∴ Initially he had at least 11 gold, and 10 silver coins i.e., a total of 21 coins.

**Case 2 (b)**: He could’ve exchanged 1 gold coin for 2 silver and 2 bronze coins, hence at the end he has at least 2 silver and 2 bronze coins.

Hence, he has at least 2 coins of each type at the end.

∴ Initially he had at least 3 gold coins.

Hence, 3.

Workspace:

**PE 2 - Puzzles | LR - Puzzles**

One of the tribesmen had 3 gold, 5 silver and 8 bronze coins. He made some exchanges and ended up with least number of total coins. How many coins does he have now?

Answer: 7

**Explanation** :

Consider the solution to first question of this set.

1 gold coin = 10 bronze coins and

1 silver coin = 4 bronze coins.

The tribesman initially had 3 gold + 5 silver + 8 bronze coins

Value of these coins in terms of bronze coins = 3 × 10b + 5 × 4b + 8b = 58 bronze coins.

To minimize the total number of coins such that their value (in terms of bronze coins) is 58 he must have maximum possible number of gold coins and then maximum possible number of silver coins and then remaining coins can be bronze coins.

At the end maximum gold coins, he can have is 5 and then maximum silver coins he can have is 2.

∴ The least number of coins he can have = 5 + 2 = 7 coins.

Hence, 7.

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**PE 2 - Puzzles | LR - Puzzles**

One of the tribeswoman initially had 2 gold, 3 silver and 10 bronze coins. She went to the market to buy a dress for herself. She bought one of the 4 dresses that the shopkeeper had. Which dress she could have bought?

- (a)
Dress 1 costing 1 gold, 5 silver and 15 bronze coins

- (b)
Dress 2 costing 2 gold, 2 silver and 15 bronze coins

- (c)
Dress 3 costing 3 gold, 2 silver and 4 bronze coins

- (d)
Dress 4 costing 2 gold, 4 silver and 7 bronze coins

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

1 gold coin = 10 bronze coins and

1 silver coin = 4 bronze coins.

Number of coins with tribeswoman = 2g + 3s + 10b

Total value of coins with tribeswoman in terms of bronze coins = 2 × 10b + 3 × 4b + 10b = 42b

Now, let us calculate the value of each of the options in terms of bronze coins.

Option (a): 1 × 10b + 5 × 4b + 15b = 45b

Option (b): 2 × 10b + 2 × 4b + 15b = 43b

Option (c): 3 × 10b + 2 × 4b + 4b = 42b

Option (d): 2 × 10b + 4 × 4b + 7b = 43b

Hence, she could’ve only bought dress 3.

Hence, option (c).

Workspace:

**PE 2 - Puzzles | LR - Puzzles**

One of the children in the tribe had 3 gold, 4 silver and 14 bronze coins. He exchanged some of his coins such that he ends up with equal number of coins of each type. How many coins did he exchange?

Answer: 10

**Explanation** :

Consider the solution to first question of this set.

1 gold coin = 10 bronze coins and

1 silver coin = 4 bronze coins.

Let the child ended up with ‘c’ number of coins of each type.

∴ The child has = c gold + c silver + c bronze coins.

Total value of these coins in terms of bronze coins = c × 10b + c × 4b + cb = 15c bronze coins

Total value of coins he initially had = 3 gold + 4 silver + 14 bronze coins

Total value of these coins in terms of bronze coins = 3 × 10b + 4 × 4b + 14b = 60 bronze coins

∴ Initial value of coins (in terms of bronze coins) = Final value of coins (in terms of bronze coins)

⇒ 60 = 15c

⇒ c = 4 coins of each type.

Now, the child has 4 gold + 4 silver and 4 bronze coins

From 3 gold + 4 silver + 14 bronze coins, the child now has 4 gold + 4 silver + 4 bronze coins

⇒ The child exchanged 10 bronze coins for 1 gold coin.

Hence, 10.

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

A die, with the numbers 1 to 6 marked on its six faces, was rolled 300 times and the number which turns up was noted down.

The table below gives the number of times that each individual number on the faces of the die turned up, in the first n rolls, where n = 50, 100, …, 300.

It was also observed that in no two consecutive rolls did the same number turn up.

**PE 2 - Puzzles | LR - Puzzles**

If the number 6 turned up in the 150^{th} roll, then which number/s could not have turned up in the 125^{th} rolled?

- (a)
2

- (b)
1

- (c)
4

- (d)
3

Answer: Option B

**Explanation** :

We can make a table for number of times a number turns up for each set of 50 rolls.

We can see that number 1 does not turn up in any of the rolls from 101st till 150th.

Hence, option (b).

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**PE 2 - Puzzles | LR - Puzzles**

Which of the following numbers must have turned up for the maximum number of times in the first 208 rolls?

- (a)
6

- (b)
5

- (c)
4

- (d)
1

Answer: Option A

**Explanation** :

The number 6 has already turned up for 41 times by the end of the first 200 rolls.

In the next 8 rolls a number can turn up maximum of 4 times. Hence, no other number would turn up for a total of 41 times till 208^{th} roll.

Hence, number 6 turns up for maximum number of times till 208 rolls.

Hence, option (a).

Workspace:

**PE 2 - Puzzles | LR - Puzzles**

What is the least number of times the number 3 could have turned up in the first 280 rolls?

Answer: 43

**Explanation** :

To calculate the minimum possible number of times 3 turned up in 280 rolls = number of times 3 turned up in 300 rolls – maximum possible number of times 3 could have turned up in last 20 rolls

= 50 – 7 = 43

Hence, 43.

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