Question

The sum of four consecutive even numbers is 107 more than the sum of three consecutive odd numbers. If the sum of the smallest odd number and the smallest even number is 55, then what is the smallest even number?

• A. 36
• B. 40
• C. 32
• D. 38

Hint:

When dealing with consecutive numbers, express them algebraically and use the relationships given in the problem to form equations. Solve these equations systematically to find the unknowns.

Answer 1

The Correct Option is D

Let the smallest odd number be \( x \). Then, the three consecutive odd numbers will be \( x, x+2, x+4 \). The sum of the three consecutive odd numbers is:

x + (x + 2) + (x + 4) = 3x + 6

Let the smallest even number be \( y \). Then, the four consecutive even numbers will be \( y, y+2, y+4, y+6 \). The sum of the four consecutive even numbers is:

y + (y + 2) + (y + 4) + (y + 6) = 4y + 12

We are told that the sum of the four consecutive even numbers is 107 more than the sum of the three consecutive odd numbers. Therefore, we can write the equation:

4y + 12 = (3x + 6) + 107

Simplifying:

4y + 12 = 3x + 113
4y = 3x + 101  {(Equation 1)}

We are also told that the sum of the smallest odd number and the smallest even number is 55. Therefore:

x + y = 55  {(Equation 2)}

Now, solve these two equations simultaneously. From Equation 2:

y = 55 - x

Substitute this into Equation 1:

4(55 - x) = 3x + 101
220 - 4x = 3x + 101
220 - 101 = 3x + 4x
119 = 7x
x = 119 / 7 = 17

Substitute \( x = 17 \) into Equation 2:

17 + y = 55
y = 55 - 17 = 38

Thus, the smallest even number is \( \boxed{38} \).