(a) John asks Thabo how many learners there are in his class. Thabo answers: "If you take two our number, plus half of our number, plus a quarter of our number, plus 1, the answer is 10. How many learners are there in Thabo's class?"

Let's denote the number of learners in Thabo's class as \( x \). According to Thabo's statement, we can set up the following equation based on the information provided: \[ 2 + \frac{1}{2}x + \frac{1}{4}x + 1 = 10 \] Now, let's simplify this equation step by step. 1. Combine the constant terms on the left side: \[ 2 + 1 = 3 \] So the equation becomes: \[ 3 + \frac{1}{2}x + \frac{1}{4}x = 10 \] 2. Subtract 3 from both sides: \[ \frac{1}{2}x + \frac{1}{4}x = 10 - 3 \] \[ \frac{1}{2}x + \frac{1}{4}x = 7 \] 3. To combine the terms involving \( x \), we need a common denominator. The common denominator for \( \frac{1}{2} \) and \( \frac{1}{4} \) is 4. We can rewrite \( \frac{1}{2}x \) as \( \frac{2}{4}x \): \[ \frac{2}{4}x + \frac{1}{4}x = 7 \] \[ \frac{3}{4}x = 7 \] 4. Now, to solve for \( x \), multiply both sides by \( \frac{4}{3} \): \[ x = 7 \cdot \frac{4}{3} \] \[ x = \frac{28}{3} \] 5. Finally, we can convert \( \frac{28}{3} \) to a mixed number: \[ x = 9 \frac{1}{3} \] Since the number of learners must be a whole number, we can conclude that there is an inconsistency in the problem as stated. However, if we round down, we can say there are approximately 9 learners in Thabo's class. Thus, the answer is that Thabo's class has approximately 9 learners.