3 consecutive even numbers, so let the smallest one be x.

So the middle one is x + 2 and biggest one is x + 4.


When rounded off to1 s.f , their sum is 200


x + x + 2 + x + 4 = 200 (1 s.f)

3x + 6 = 200 (1s.f)

So smallest possible sum is 150.

(Since 50 to 149 will be rounded to 100)


When 3x + 6 = 150,

3x = 150 - 6
3x = 144
x = 144 ÷ 3
x = 48

Smallest possible numbers are 48,50,52.

Now for the largest possible.

The largest possible integer that is rounded to 200(1s.f) would be 249.

But, the sum of 3 even numbers will also be an even number. So we cannot consider 249.


And we cannot consider 248 , the next biggest number (even though 248 is an even number) because :


3x + 6 = 248
3x = 248 - 6
3x = 242
x = 242 ÷ 3
x = 80 R 2

The number is 3x + 6 = 3(x + 2).

So it has to be a multiple of 3. But 248 is not a multiple of 3 and is not divisible by 3.

So we look at the next biggest even number 246.

3x + 6 = 24y
3x = 246 - 6
3x = 240
x = 240 ÷ 3
x = 80


Largest possible numbers are 80,82,84.

Alternatively , you can let x be the middle number.

Then the smallest one will be x - 2 and the largest will be x + 2.

Their sum is

x - 2 + x + x + 2 = 3x


Then consider 3x = 150 (smallest possible sum)

x = 50

and 3x = 246 (largest possible sum)

x = 82



You will get the same results as previously.