MathRevolution wrote:

Which of the following is satisfied with |x-4|+|x-3|<2?

A. 1<x<5 B. 2<x<5 C. 2.5<x<4.5 D. 2.5<x<4 E. 3<x<4

* A solution will be posted in two days.

Remember that |x-a| represents distance of x from 'a'. As mentioned in the question,

|x-4|+|x-3|<2 simply means that you need to find the range of 'x' such that distance of x from 4 + distance of x from 3 < 2.

Refer to the attached image, you realize that the distance 'y' when it is in between 3 and 4 will always be <2. For regions x<3 or x>4, the sum of the distances of 'y' from 3 and 4 will be <2 only when

y+1+y<2 ---> 2y+1<2 ---> y<0.5 or in other words, range of x is ---> 2.5 < x < 4.5

Attachment:

2016-01-27_20-50-35.jpg [ 6.25 KiB | Viewed 4547 times ]
C is thus the correct answer.

Alternate solution: using POE to arrive at the correct answer.Once you know that the sum of the distances of x from 3 and 4 must be < 2,

A. 1<x<5 . Try with x=1.5. Clearly the distance of 1.5 from 3 and 4 will be >2. Eliminate.B. 2<x<5 . Take x=2.1. Again, the distances of 2.1 from 3 and 4 (=0.9 and 1.9 respectively) will sum to > 2. Hence Eliminate.C. 2.5<x<4.5. Works for all possible values. Keep.D. 2.5<x<4. Take x=4.2. The distances of 4.2 from 3 and 4 (=1.2 and 0.2 respectively) will sum to < 2 but this value is NOT in the given range, thus incomplete range. Hence Eliminate.E. 3<x<4. Take x=2.6. It works for the given sums of the distances < 2 but is not in the given range, thus incomplete range. Hence eliminate.C is thus the correct answer.

Hope this helps.