If y is the number on the number line between 10 and 40 that is three × as far from 10 as from 40, then what is the value of y?
y is equal to 32 1/3.
To find the value of y, we need to establish the distances from y to 10 and from y to 40. Setting up the equation based on the problem statement, we find that y must be positioned 10/3 units away from 40, leading directly to the solution of 32 1/3.
This value does not satisfy the condition that y is three times as far from 10 as it is from 40. If we check the distances, the ratio would not hold, which means this choice is incorrect.
While 32 is a number within the specified range, it does not maintain the necessary ratio of distances between y, 10, and 40. The distances calculated from 32 do not yield the required relationship that y is three times further from 10 than from 40.
This value correctly represents y as it maintains the required distance ratio. The distance from 10 is 22 1/3, while the distance from 40 is 7 1/3. The ratio of these distances is 22 1/3 to 7 1/3, which simplifies to 3:1, confirming y’s correct placement.
This choice also fails to meet the distance condition specified in the problem. The calculated distances from both 10 and 40 do not reflect the necessary 3:1 ratio, making this option incorrect.
While 33 is within the range and appears reasonable, it does not satisfy the ratio condition either. The distances from 10 and 40 for this choice do not yield a 3:1 relationship, which is essential for the solution.
The value of y found to be 32 1/3 adheres to the requirement of being three times as far from 10 as it is from 40. By confirming the distance ratios with this value, we establish that it uniquely satisfies the conditions of the problem, while all other options fail to do so.
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