Solutions

1. 7

   To minimize the number of hiragana that Velma will have to learn on Saturday, consider the extreme case in which she learns as many hiragana as possible on the other days. She learns 4 on Friday, leaving 71 − 4 = 67 for the other six days of the week. If Velma learns the maximum of 12 hiragana on the other five days (besides Saturday), then she will have 67 − 5(12) = 7 left for Saturday.

2. 25 square inches

   First, figure the area of each pizza: the small is 100 square inches, and the large is 225 square inches. If the two friends pool their money, they can buy three large pizzas, which have a total area of 675 square inches. If they buy individually, though, then each friend will have to buy one large pizza and one small pizza, so they will only have a total of 2(100 + 225) = 650 square inches of pizza.

3. (D)

   This problem is a grouping problem. You have some integer number of 5’s and some integer number of 7’s. Which of the answer choices cannot be the sum? One efficient way to eliminate choices is first to cross off any multiples of 7 and/or 5; this eliminates choice (E). Now, any other possible sums must have at least one 5 and one 7 in them. So you can subtract off 5’s one at a time until you reach a multiple of 7. (It is easier to subtract 5’s than 7’s, because our number system is base-10.) Choice (A): 31 − 5 = 26; 26 − 5 = 21, a multiple of 7; this eliminates (A). (In other words, 31 = 3 × 7 + 2 × 5.) Choice (B): 29 − 5 = 24; 24 − 5 = 19; 19 − 5 = 14, a multiple of 7; this eliminates (B). Choice (C): 26 − 5 = 21, a multiple of 7; this eliminates (C). So the answer must be choice (D), 23. You can check by successively subtracting 5 and looking for multiples of 7: 23 − 5 = 18, not a multiple of 7; 18 − 5 = 13, also not a multiple of 7; 13 − 5 = 8, not a multiple of 7; and no smaller result will be a multiple of 7 either.

4. $0.17

   The first step is to compute the value of a complete “Collector’s Coin Set”: $1.00 + $0.50 + $0.25 + $0.10 + $0.05 + $0.01 = $1.91. Now, you need to divide $1.91 into $25. A natural first move is to multiply by 10: for $19.10, Colin can buy 10 complete sets. Now add $1.91 successively. Colin can buy 11 sets for $21.01, 12 sets for $22.92, and 13 sets for $24.83. There is $0.17 left over.

5. 1,550

   You can use the formula Total = a + b − e + f. Because all of the tickets will either be premium seating or allow backstage access, f will equal 0. Therefore:

   Total = 1,200 + 500 − 150 = 1,550

6. (B)

   In a week consisting of 5 workdays and 2 weekend days, Susan can write:

   5 × 10 + 2 × 20 = 90 pages

   Therefore, in 10 consecutive full weeks (i.e., 70 consecutive days), she can write 900 pages of her novel, leaving another 50 pages to be written. The least number of days it would take Susan to write 50 pages is 3: 2 weekend days and 1 weekday. Thus, it is possible for Susan to finish her novel in 73 days. (This assumes that Susan chooses her start day appropriately, so as to take advantage of as many weekends as possible.) Therefore, Quantity B is greater.

7. (B)

   Jared can achieve any amount from 1 cent to 19 cents: 1 to 4 cents using the pennies, 5 cents with the nickel, 6 to 9 cents using the nickel along with the pennies, 10 cents using the dime, 11 to 14 cents using the dime along with the pennies, 15 cents using the dime and the nickel, and 16 to 19 cents using the dime and nickel along with the pennies. Notice that 19 cents requires every coin Jared possesses, meaning that 19 is the largest possible value. That makes 19 possible values. Therefore, Quantity B is greater.

8. (C)

   Minimizing the length of the longest piece is equivalent to maximizing the lengths of the remaining pieces, as long as they are shorter than the longest piece. Suppose that the longest piece were 14 inches long (a choice motivated by wanting to be less than the 15 in Quantity B). That would leave 40 − 14 = 26 inches to be accounted for by the other two pieces.

   Because each piece must be a different number of inches long, those pieces cannot each be 13 inches long. This, in turn, implies that one of the two remaining pieces would have to be more than 13 inches long—but then, that piece would be 14 inches long, again violating the constraint that each piece be of a different length. Thus, the longest piece must be at least 15 inches long, and the shorter pieces could then be 12 and 13 inches long, for a total of 40 inches. Thus, the two quantities are equal.

9. (A)

   Once again using the formula Total = a + b − e + f:

   

   Therefore, there will be 25 more customers that purchased both zucchini and cauliflower than those who purchased neither. Thus, Quantity A is greater.