The average of three distinct positive integers x, y, and z is 64. If a new number d is add, will the percent increase exceed 13%?
1) d is twice the value of one of the original numbers
2) d is half the value of the largest of the original numbers
B
I actually got the answer right, but I am not sure if my math is correct or if I just got lucky.
Here is how I solved it:
1. (x+y+z)/3 = 64 -> x+y+z = 192
2. 13% of 192 is approximately 23. So, in order for d to be at least a 13% increase, d > 23
3. The most extreme number combinations are 189+1+2, in which case (for statement 2) d=94.5, and 63+64+65, in which case d=32.5. Since both are greater than 23, the percent increase is greater than 13%.
Does this make sense or am I missing something?
Thank you
1) d is twice the value of one of the original numbers
2) d is half the value of the largest of the original numbers
B
I actually got the answer right, but I am not sure if my math is correct or if I just got lucky.
Here is how I solved it:
1. (x+y+z)/3 = 64 -> x+y+z = 192
2. 13% of 192 is approximately 23. So, in order for d to be at least a 13% increase, d > 23
3. The most extreme number combinations are 189+1+2, in which case (for statement 2) d=94.5, and 63+64+65, in which case d=32.5. Since both are greater than 23, the percent increase is greater than 13%.
Does this make sense or am I missing something?
Thank you
