Question

There are two positive integers x and y. When x is divided by 237, the remainder is 192. When y is divided by 117, the quotient is same but the remainder is 108. The remainder when the sum of x and y is divided by 118 is

Answer 1

Answer:x=237q+192----------eq-1

Y=117q+108----------- eq 2

Add eq 1&2

We get X+Y=354q+300

Divide X+Y/118

THEN WE GET REMAIMDER =64

(354/118=3, 300=236+64

236/118= 2 SO REMAINDER=64

Step-by-step explanation:

Answer 2

Step by step explanation:

Dividend = Divisor × Quotient(q) + Remainder

So from the question we can write :

x = 237 × Q + 192 .. (1)

y = 117 × Q + 108 ..(2)

After adding equation (1) and (2) :

$x + y = 354 \times q + 300$

$\frac{x + y}{118 } = \frac{354 \times q}{118} + \frac{300}{118}$

Now we can write 300 ÷ 118 as

$300 = 118 \times 3 + 64$

$\frac{x + y}{118} = 3 \times q + \frac{236 + 64}{118}$

$\frac{x + y}{118} = (3 \times q + 3) + 64$

So quotient is 3q + 3 and,

remainder is 64

I hope you get your answer,

Thank you