Question 24 of 40intermediate🔧 ApplyNumerical4 marks

Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Correct Answer

The required pairs are (6, 8), (8, 10), and (10, 12).

Exercise: EXERCISE 5.1 | Q: 24 | (Page Page 8)
For More Understanding

Explanation

This question is similar to Example 8 in the textbook context. The student needs to set up inequalities for consecutive even positive integers. Let x be the smaller even integer, so the next consecutive even integer is x + 2. The conditions are: (1) both integers larger than 5, meaning x > 5, and (2) their sum less than 23, meaning x + (x + 2) < 23. Solving these inequalities gives 5 < x < 10.5. Since x must be an even positive integer, x can be 6, 8, or 10, giving the three pairs.

Solution Steps

  1. Step 1: Let x be the smaller of the two consecutive even positive integers. Then the other integer is x + 2.

  2. Step 2: Since both integers are larger than 5, we have x > 5 ... (1)

  3. Step 3: Since their sum is less than 23, we have x + (x + 2) < 23

  4. Step 4: Solving: 2x + 2 < 23, which gives 2x < 21, or x < 10.5 ... (2)

  5. Step 5: From (1) and (2), we get 5 < x < 10.5

  6. Step 6: Since x is an even positive integer, x can take the values 6, 8, and 10.

  7. Step 7: The required pairs are (6, 8), (8, 10), and (10, 12).

PreviousNext