Answer: The answer is 
Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.
Let, AC = 3x and CM = 4x.
In the right-angled triangle ACM, we have

Now,

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.
So,

Comparing equations (A) and (B), we have

Thus,

First you have to distribute. -3y +12=18. -3y=6. y= -2.
If the sample is random, it is not valid.
4:2 can represented as 4 / 2. Or also 2.
30 / 2 = 15