PE 1 - Mathematical Reasoning | LR - Mathematical Reasoning

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.

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).

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.

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.

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.

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.

Answer: 22653

Explanation :

Consider the solution to the first question. 

The total of the chits picked by Cabra is 22653.

Hence, 22653.

Answer: 9687

Explanation :

Consider the solution to the first question. 

The highest chit picked in round 3 is 9687.

Hence, 9687.

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.