Option 2 : 1

NPCIL Stage-I 2020: Full Mock Test

6327

50 Questions
150 Marks
60 Mins

**Given:**

A number is divided by 11 and remainder = 3.

The quotient is divided by 9, the remainder = 5

**Concept Used:**

Dividend = Divisor × Quotient + Remainder

x = q_{1} × a + r_{1}

q_{1} = q_{2} × b + r_{2}

We assume q_{2} = 1

where, x, a and b are natural numbers.

q_{1} and q_{2} are quotients for x and q_{1} respectively.

r_{1 }and r_{2} are remainders obtained from dividing x and q_{1} by a and b respectively.

**Calculations:**

Let the required number be N.

Let the quotients obtained after dividing N and its quotient be q_{1} and q_{2}.

Let divisors and remainders for N and q_{1} be d_{1}, d_{2} and r_{1}, r_{2} respectively.

Assume q_{2} = 1

q_{1} = q_{2} × d_{2} + r_{2}

⇒ q_{1} = 1 × 9 + 5

⇒ q_{1} = 14

N = q_{1 }× d_{1} + r_{1}

⇒ N = 14 × 11 + 3

⇒ N = 157

On dividing N by 3, we get

⇒ 157/3 = (156 + 1)/3

⇒ Remainder = 1

**∴ The remainder when the number 157 is divided by 3 is 1.**