Question

In the hall, how many chairs are available?
(a) 407
(b) 143
(c) 539
(d) 209
If one chair is removed, which arrangements are possible now?
(a) 2
(b) 3
(c) 4
(d) 5
If one chair is added to the total number of chairs, how many chairs will be left when arranged in 11's.
(a) 1
(b) 2
(c) 3
How many chairs will be left in original arrangement if same number of chairs will be arranged in 7's?
(b) 1
(c) 2
(d) 3​

Answer 1

Given:

Refer to the attached image in the question.

To find :

The required answers.

Solution:

1) In the hall, how many chairs are available?

(a) 407

(b) 143

(c) 539

(d) 209

• Divide all the given options by 2, 3, 4, 5, 6, and 11.
• When we divide the number 539 by 2, 3, 4, 5, 6, and 11, it will leave the remainder 1, 2, 3, 4, 5, 0 respectively.

So, Option (C) => 539 is the required answer.

_________________________________________

2) If one chair is removed, which arrangements are possible now?

(a) 2

(b) 3

(c) 4

(d) 5

• There are a total of 539 chairs. if 1 removed then there will be 538 chairs
• On arranging chairs in 3's, 4's and 5's then 1, 2, 3 chairs are left respectively.
• On arranging chairs in 2's then no chair will be left.

So, option (A) => 2 is the required answer.

__________________________________________

3) If one chair is added to the total number of chairs, how many chairs will be left when arranged in 11's.

(a) 1

(b) 2

(c) 3

(d) 4

• The total number of chairs are 539
• On arranging in 11's then no chair will be left.
• so, if we add 1 extra chair then that added 1 chair will be left.

So, option (A) => 1 is the required answer.

__________________________________________

4) How many chairs will be left in original arrangement if same number of chairs will be arranged in 7's?

(a) 0

(b) 1

(c) 2

(d) 3

• On dividing 539 by 7, we will get a remainder 0.
• On arranging 539 chairs in 7's, no chair will be left.

So, option (A) => 0 is the required answer.

Answer 2

Solution :-

given that,

→ when arrange in 2 = 1 chair left

→ when arrange in 3 = 2 chair left

→ when arrange in 4 = 3 chair left

→ when arrange in 5 = 4 chair left

→ when arrange in 6 = 5 chair left

as we can see that,

• 2 - 1 = 3 - 2 = 4 - 3 = 5 - 4 = 6 - 5 = 1 => Difference is same .

so,

→ Required number = LCM (2,3,4,5,6) - 1

prime factors of 2,3,4,5,6 are :-

→ 2 = 1 * 2

→ 3 = 1 * 3

→ 4 = 2 * 2

→ 5 = 1 * 5

→ 6 = 2 * 3

so,

→ LCM = 2 * 2 * 3 * 5 = 60

now, given that, when arrange in 11's no chair is left .

then,

→ Required number will be in the form of = (60*k - 1)/11 = Remainder 0 .

putting values of k now, we get :-

• k = 1 => (60*1 - 1)/11 = 59/11 ≠ Remainder 0 .
• k = 2 => (60*2 - 1)/11 = 119/11 ≠ Remainder 0 .
• k = 3 => (60*3 - 1)/11 = 179/11 ≠ Remainder 0 .
• k = 4 => (60*4 - 1)/11 = 239/11 ≠ Remainder 0 .
• k = 5 => (60*5 - 1)/11 = 299/11 ≠ Remainder 0 .
• k = 6 => (60*6 - 1)/11 = 359/11 ≠ Remainder 0 .
• k = 7 => (60*7 - 1)/11 = 419/11 ≠ Remainder 0 .
• k = 8 => (60*8 - 1)/11 = 479/11 ≠ Remainder 0 .
• k = 9 => (60*9 - 1)/11 = 539/11 = Remainder 0 .
• Here, we can use divisibility rule of 11 = Difference between sum of alternate numbers is 0,11,22,___

therefore,

→ Required smallest number = 539

hence,

→ Total available chairs = 539 (c) (Ans.)

now,

→ when one chair is removed = 539 - 1 = 538

checking 538 is divisible by which number 2,3,4,5,6 :-

• 538 => 8 in last => Divisible by 2 .
• 538 => sum of digits is 16 => 16/3 ≠ remainder 0 => Not divisible by 3 .
• 538 => Last two digits 38 is not divisible by 4 .
• 538 => Last digits are not 0 or 5 . so not divisible by 5 .
• 538 => Not divisible by 3 . so not divisible by 6 .

Therefore , when one chair is removed only arrangement of 2 (a) is possible .

now,

→ when 1 chair is added , total chair = 539 + 1 = 540

then,

→ when arranged in 11's = 540/11 = 1 remainder = 1 chair left (a)

now,

→ when arranged in 7's = 539/7 = 0 remainder = 0 chair left .

[Note :- In competitive exams we can solve through options also but we must know basic method to solve the problem . ]

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