# PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning

**Answer the next 5 questions based on the information given below.**

There are 100 beggars standing in a queue outside a temple. Every day 100 people come to pray and while leaving distribute money to these beggars. They distribute the money in the following manner.

First person gives Rs. 1 to each of the 100 beggars. Second person gives Rs. 2 to every 2^{nd} beggar starting with 2^{nd} beggar. Third person gives Rs. 3 to every 3^{rd} beggar starting with 3^{rd} beggar. This continues till the hundredth beggar gives Rs. 100 to 100^{th} beggar.

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the total amount received by 2^{nd}, 4^{th} and 9^{th} beggar?

Answer: 23

**Explanation** :

1^{st} beggar received money from – 1^{st} person

2^{nd} beggar received money from – 1^{st} and 2^{nd} persons

3^{rd} beggar received money from – 1^{st} and 3^{rd} persons

4^{th} beggar received money from – 1^{st}, 2^{nd} and 4^{th} persons

… and so on

We can see that the nth beggar will receive amount from those people whose numbers are factors of n.

Amount received by 1^{st} beggar - 1

Amount received by 2^{nd} beggar – 1 + 2 = Rs. 3

Amount received by 3^{rd} beggar – 1 + 3 = Rs. 4

Amount received by 4^{th} beggar – 1 + 2 + 4 = Rs. 7

… and so on

We can see that the nth beggar will receive amount equal to sum of the factors of n.

∴ Amount received by

Beggar 2 ⇒ 1 + 2 = Rs. 3

Beggar 4 ⇒ 1 + 2 + 4 = Rs. 7

Beggar 9 ⇒ 1 + 3 + 9 = Rs. 13

∴ Amount received by beggar 2, 4 and 9 is 3 + 7 + 13 = Rs. 23

Hence, 23.

Workspace:

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

Which person distributed the least amount?

- A.
51

^{st} - B.
53

^{rd} - C.
100

^{th} - D.
None of these

Answer: Option A

**Explanation** :

Consider the solution to the first question of this set.

All people numbered from 1 to 50 will distribute money to beggars from 100 – 2. Amount distributed by each of them will be greater than Rs. 51.

51^{st} person gives Rs. 51 to only 1 beggar i.e., total Rs. 51.

53^{rd} person gives Rs. 53 to only 1 beggar i.e., total Rs. 53.

100^{th} person gives Rs. 100 to only 1 beggar i.e., total Rs. 100.

∴ Least amount distributed will be Rs. 51 by 51^{st} person.

Hence, option (a).

Workspace:

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the maximum amount distributed by any person?

Answer: 100

**Explanation** :

Consider the solution to the first question of this set.

1^{st} person gives Rs. 1 to all 100 beggars i.e., total Rs. 100.

2^{nd} person gives Rs. 2 to 50 beggars i.e., total Rs. 50.

3^{rd} person gives Rs. 3 to 33 beggars i.e., total Rs. 99.

… and so on.

∴ The maximum amount given be any of the persons will be Rs. 100.

Hence, 100.

Workspace:

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many beggars received money from exactly two people?

Answer: 25

**Explanation** :

Consider the solution to the first question of this set.

n^{th} beggar will receive money from as many people as the number of factors of n.

∴ Those beggars will receive money from 2 people whose number has exactly 2 factors.

Prime numbers have exactly 2 factors.

⇒ Beggars with prime numbers will receive money from 2 people.

There are 25 prime numbers from 1 to 100.

Hence, 25.

Workspace:

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many beggars received money from odd number of people?

Answer: 10

**Explanation** :

Consider the solution to the first question of this set.

n^{th} beggar will receive money from as many people as the number of factors of n.

∴ Those beggars will receive money from odd number of people whose number has odd number of factors.

Only perfect squares have odd number of factors.

⇒ Beggars with perfect square numbers will receive money from odd number of people.

There are 10 perfect squares from 1 to 100.

Hence, 10.

Workspace:

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

Abra, Babra, Cabra, Dabra, Gabra and Kabra play a game where they pick up chits containing number from a box. These numbers however have a specialty i.e., all of them are 4-digit numbers containing different permutations of the digits 6, 7, 8 and 9. Numbers once picked up cannot be picked up again. The table above shows details of some of the numbers picked up at the end of 3 rounds of picking. In the first round, all the 6 pick up a chit one-by-one. The same thing happens for the second and third round.

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

What chit is picked up by Dabra in round 1?

Answer: 6978

**Explanation** :

The chit picked by Gabra in round 3 = 25443 – (9867 + 6897) = 8679

The chit picked by Kabra in round 1 = 25371– (9876+ 6798) = 8697

Now, the various equations that can be formed are:

a + b = 26469 – 7986 = 18,483

This is only possible when the two numbers are of the form 9pqr and 8xyz.

Since last digit of r + z is 3, this is possible when one of r or z is 6 and the other is 7.

Carry over of r + z will be 1.

Now, last digit of 1 + q + y = 8, hence last digit of q + y is 7.

This is possible when one of q or y is 8 and the other is 9.

Carry over of 1 + q + y = 1

Now, last digit of 1 + p + x = 4, hence last digit of q + y is 3.

This is possible when one of p or x is 6 and the other is 7.

∴ a + b can be (9786 + 8697) or (9687 + 8796) in any order.

Since, Kabra picked 8697 in 1st round, Abra cannot pick 8697.

Hence a or b can be 9687 or 8796.

**Case 1**: a = 9687

⇒ f = 51093 – (9687 + 9768 + 6987 + 9867 + 8697) = 6087

This is not possible since the digits should be either 6, 7, 8 or 9.

**Case 2:** a = 8796

⇒ f = 51093 – (8796 + 9768 + 6987 + 9867 + 8697) = 6978

∴ a = 8796, b = 9687 and f = 6978

Now, 149103 = 26469 + 25524 + (6987 + e + 7968) + 23643 + 25443 + 25371

⇒ e = 7698

Also, c = 46935 – (7986 + e + 7689 + 6897 + 9876)

∴ c = 6789

Also, d = 25524 – (9768 + c)

∴ d = 8967

This final table is as follows:

Hence, Dabra picked 6978 in round 1.

Hence, 6978.

Workspace:

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the total of the chits picked up by Cabra?

Answer: 22653

**Explanation** :

Consider the solution to the first question.

The total of the chits picked by Cabra is 22653.

Hence, 22653.

Workspace:

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the highest chit picked by anyone in round 3?

Answer: 9687

**Explanation** :

Consider the solution to the first question.

The highest chit picked in round 3 is 9687.

Hence, 9687.

Workspace:

**PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the largest chit that has never been picked?

Answer: 9786

**Explanation** :

Consider the solution to the first question.

Let us check from highest ticket onwards

Highest ticket is 9876 which was picked by Kabra in round 2.

Next highest ticket is 9867, which was picked by Gabra in round 1.

Next highest ticket is 9786, which has not been picked.

Hence, 9786.

Workspace:

## Feedback

**Help us build a Free and Comprehensive CAT/MBA Preparation portal by providing
us your valuable feedback about Apti4All and how it can be improved.**