Problem Solving

If Jeff's current calorie intake is x, we want to find how long it will take for him to reach 0.5x.

Statement 1
We know that Jeff started at 2500 calories. But we have no idea how quickly he's going to cut down on his calories. Maybe he'll eat 1% fewer calories every week. Maybe he'll eat 30% fewer calories every week. Without this rate, we can't solve. Insufficient.

Statement 2
We now know our rate - 10% each week. It's tempting to think that we need to combine this info with Statement 1 to determine how long it takes to reach 1225 calories (half of 2500). However, we actually don't need it.

If Jeff's intake drops by 10% each week, his intake will be 0.9 * his intake the week before. So we multiply by 0.9 for each week, or $$0.9^nx$$ where n is the number of weeks and x is his current calorie intake. We want to find how long it take for him to reach 0.5x. So we can set our expressions equal to each other $$0.9^nx=0.5x$$ $$0.9^n=0.5$$
And solve for n . We don't actually need to solve for n (and we shouldn't since it'll take a long time) - we just need to know that we can. So Statement 2 is sufficient to solve for the number of weeks it will take for Jeff to reduce his calorie intake by half.

swerve wrote:Jeff plans to lose weight by reducing his daily calorie intake each week. How many weeks does it take for his calorie intake to fall below half of its original level?

A. Jeff currently consumes 2,500 calories per day.
B. Jeff plans to reduce his daily calorie intake by 10 percent each week.

Statement One Only:

Jeff currently consumes 2,500 calories per day.

Knowing his calories intake per day does not allow us to determine the time it takes for his calorie intake to fall below half of its original level. Statement one alone is not sufficient.

Statement Two Only:

Jeff plans to reduce his daily calorie intake by 10 percent each week.

Knowing his calorie intake is reduced by 10 percent each week does allow us to determine the number of weeks it takes for his calorie intake to fall below half of its original level. For example, we can let his original intake be 2000, n be the number of weeks it takes for his calorie intake to fall below half of its original level and create the equation:

2000(0.9)^n = 1000

We don't need to actually solve the equation, but we can see that it's solvable. Statement two alone is sufficient.

Answer: B