Question

Mahesh works as a manager in a hotel. He has to arrange seats in hall for a function. A hall has a certain number of chairs. Guests want to sit in different groups like in pairs, triplets, quadruplets, fives and sixes etc. When Mahesh arranges chairs in such pattern like in 2’s, 3’s, 4’s 5’s and 6’s then 1, 2, 3, 4 and 5 chairs are left respectively. But when he arranges in 11’s, no chair will be left.



(i) In the hall, how many chairs are available?

(a) 407 (b) 143

(c) 539 (d) 209​

Answer 1

Answer:

539

Step-by-step explanation:

539÷2=269(1 remainder)

539÷3=179(2 remainder)

539÷4=134(3 remainder)

539÷5=107(4 remainder)

539÷6=89(5 remainder)

539÷11=49(0 remainder)

So 539 leaves no chair.

Answer 2

• Total chairs available in the hall are equal to 539 .

Given :- When Mahesh arranges chairs in such pattern like in 2’s, 3’s, 4’s 5’s and 6’s then 1, 2, 3, 4 and 5 chairs are left respectively. But when he arranges in 11’s, no chair will be left.

To Find :- In the hall, how many chairs are available ?

(a) 407 (b) 143 (c) 539 (d) 209

Concept used :- The least number which when divided by x, y and z leaves the remainders a, b and c respectively where,

• (x - a) = (z - b) = (z - c) = K
• Then, Required number = (LCM of x, y and z) - K .

Solution :-

given that,

→ When Mahesh arranges chairs in pairs = 1 chair left .

→ When Mahesh arranges chairs in triplets = 2 chair left .

→ When Mahesh arranges chairs in quadruplets = 3 chair left .

→ When Mahesh arranges chairs in fives = 4 chair left .

→ When Mahesh arranges chairs in sixes = 5 chair left .

As we can see that,

→ (2 - 1) = (3 - 2) = (4 - 3) = (5 - 4) = (6 - 5) = 1

So,

→ K = 1

then,

→ Least number which when divided by 2,3,4,5 and 6 and gives remainder as 1,2,3,4 and 5 = LCM(2,3,4,5,6) - K

Finding prime factors of 2,3,4,5 and 6 we get,

→ 2 = 1 × 2

→ 3 = 1 × 3

→ 4 = 2 × 2

→ 5 = 1 × 5

→ 6 = 2 × 3

So,

→ LCM = 2 × 2 × 3 × 5 = 60

then,

→ Least number = LCM - K = 60 - 1 = 59

But we have given that, when he arranges in 11’s, no chair will be left. So, we can say that, the least number of chairs must be divisible by 11 .

Then,

→ Required number :- (60n - 1) ÷ 11 = Remainder 0 { where n = natural numbers. }

putting values of n we get :-

• n = 1 => (60×1 - 1) ÷ 11 = (60 - 1) ÷ 11 = 59 ÷ 11 ≠ Remainder 0 .
• n = 2 => (60×2 - 1) ÷ 11 = (120 - 1) ÷ 11 = 119 ÷ 11 ≠ Remainder 0 .
• n = 3 => (60×3 - 1) ÷ 11 = (180 - 1) ÷ 11 = 179 ÷ 11 ≠ Remainder 0 .
• n = 4 => (60×4 - 1) ÷ 11 = (240 - 1) ÷ 11 = 239 ÷ 11 ≠ Remainder 0 .
• n = 5 => (60×5 - 1) ÷ 11 = (300 - 1) ÷ 11 = 299 ÷ 11 ≠ Remainder 0 .
• n = 6 => (60×6 - 1) ÷ 11 = (360 - 1) ÷ 11 = 359 ÷ 11 ≠ Remainder 0 .
• n = 7 => (60×7 - 1) ÷ 11 = (420 - 1) ÷ 11 = 419 ÷ 11 ≠ Remainder 0 .
• n = 8 => (60×8 - 1) ÷ 11 = (480 - 1) ÷ 11 = 479 ÷ 11 ≠ Remainder 0 .
• n = 9 => (60×9 - 1) ÷ 11 = (540 - 1) ÷ 11 = 539 ÷ 11 = Remainder 0 .

Therefore, we can conclude that,

→ Least possible required number = 539 .

Hence, In the hall, (c) 539 chairs are available .

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