The average of an evenly spaced set = the average of the endpoints. The
endpoints of the set are 50 and 60. Their average is 50 – 60 2 = 55. You know
from earlier that the number of items in the set is 11. Plug these values into
the formula:
average × number of items = sum
↓ ↓ ↓
55 × 11 = 605
Subtract the sum from the number of items: 605 – 11 = 594.- B Let x be the first integer in the set and x + 1 be the next integer. You know:
x + (x + 1) = 37
2 x + 1 = 37
2 x = 36
x = 18
To find the sum of the last three integers, determine the value of the last
integer and work from there to get the value of the previous two integers.
If the first integer is 18, then the 20th term in the set will be 18 + 19 = 37.
The values of the 2nd-to-last and 3rd-to-last terms are 36 and 35, respectively.
35 + 36 + 37 = 108.
Quantitative Comparison Questions
- A If a series of consecutive integers has more positive than negative numbers,
the positive numbers must all offset the negative numbers. Thus the sum of the
series will be positive. You can also see this by testing numbers. Try –2, –1, 0,
1, 2, 3. The sum is greater than zero. Try –4, –3, –2, –1, 0, 1, 2, 3, 4, 5. The sum
is still greater than zero. - B Don’t calculate! Determine how the two columns differ.
Column A can be rewritten as (6 × 7 × 8... 60) × 61.
Column B can be rewritten as: 2 × 3 × 4 × 5 × (6 × 7 × 8... 60).
Divide both columns by (6 × 7 × 8... 60). The new comparison is 61 versus
(2 × 3 × 4 × 5). 2 × 3 × 4 × 5 = 120. 120 > 61. Thus the value in Column B
is greater. - B If given the endpoints of an evenly spaced set, the fastest way to determine
the average of the set is to take the average of the endpoints. The average of the
integers from 2–20, inclusive is (20 + 2)/2 = 11. Quantity B is greater. - B Don’t calculate! Determine how the two quantities differ.
Quantity A can be rewritten as: (8 + 9 + 10... 24) + 25.
Quantity B can be rewritten as: 1 + 2 + 3 + 4 + 5 + 6 + 7 + ( 8 + 9 +
10... 24).
206 PART 4 ■ MATH REVIEW03-GRE-Test-2018_173-312.indd 206 12/05/17 11:51 am