GRE-Math-Questions-to-Know-by-Test-Day-2nd-Edition_pag165

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25. (B) Eliminate (D) because only numbers are involved. The sum of the whole numbers 
from 1 to n is +n n( 1)
2
. Thus, Quantity A is 50 51
2
1,275.
⋅
= Quantity B is greater. 
Tip: Another way to work this problem is to list the sum of the numbers from 1 to 50 in 
two ways, one in increasing order and one in decreasing order, and then add the pairs of 
numbers as shown here.
 
1 2 3 4 5 46 47 48 49 50
50 49 48 47 46 5 4 3 2 1
51 51 51 51 51 51 51 51 51 51
+ + + + + + + + + +
+ + + + + + + + + +
+ + + + + + + + + +
 
 There are 50 pairs, each with a sum of 51. The double sum is 50(51) 2,550= , so the 
single sum is 2,550 2 1,275÷ = .
26. (C) Because = ⋅ +3 4 0 3, the least positive number that will leave a remainder of 3 when 
divided by 4 is 3 (Quantity A). Because = ⋅ +3 5 0 3, the least positive number that will leave 
a remainder of 3 when divided by 5 is 3 (Quantity B). The two quantities are equal. Select (C).
27. (B) Let x = the number of $5 bills and y = the number of $10 bills. You have two 
equations: (1) + =x y 8, = −y x8 ; and (2) + =x y$5 $10 $50. Substitute (1) = −y x8 into 
(2) and solve for x (omitting units for convenience).
 + =x y5 10 50 
 + − =x x5 10(8 ) 50 
 + − =x x5 80 10 50 
 = x30 5 
=x 6 (Quantity B); = − = − =y x8 8 6 2 (Quantity A)
 Quantity B is greater.
28. (D) Substitute values for n. If n = 1, Quantity A is + = + =n 2 1 2 3, and Quantity 
B is + = + =n
n
2 1 2(1) 1
1
3. In this case, the two quantities are equal. If n = 2, Quantity A 
is + = + =n 2 2 2 4, and Quantity B is + = + =n
n
2 1 2(2) 1
2
2.5. In this case, Quantity A is 
greater. Because you found different results (one where the quantities were equal and one 
where Quantity A was greater), the relationship cannot be determined from the information 
given. Select (D).
29. (C) Eliminate (D) because only numbers are involved. In this question, use the division 
algorithm that states if an integer n is divided by a nonzero integer d, there exist unique 
integers q and r such that n = qd + r, where 0 ≤ r