Question

Sum of 3 consecutive numbers is 2262. What is 41% of the highest number ?

Answer 1

let the numbers be x, x+1 , x+2

3x+3 = 2262

x + 1 = 2262/3

x = 754-1

x = 753

x+2 = 755

41% of 755 = 41×755/100

= 30,955/100

= 30.955


Hope it helps u :-)

Answer 2

Define x:

Let the smallest number be x

The other two number is (x + 1) and (x + 2)

Form the equation:

Given that the sum is 2262

⇒ x + ( x + 1) + (x + 2) = 2262

Solve x:

x + ( x + 1) + (x + 2) = 2262

x + x + 1 + x + 2 = 2262

3x + 3 = 2262

3x = 2259

x = 2259 ÷ 3

x = 753

Find the numbers:

Smallest number = x = 753

2nd number = x + 1 = 753 + 1 = 754

3rd number = x + 2 = 753 + 2 = 755

Find 41% of the highest number:

41% x 755 = 0.41 x 755 = 309.55

Answer: 41% of the highest number is 309.55