1) The weight of a glass bowl and the marbles it contains is 50 ounces. If the number of marbles in the bowl is doubled, the total weight of the bowl and marbles is 92 ounces. What is
the weight of the bowl? (Assume that each of the marbles has the same weight).
2) Square ABCD and rectangle AEFG each have an area of 36 square meters. E is the midpoint of AB. What is the perimeter of rectangle AEFG?
3) A slow clock loses 3 minutes every hour. Suppose the slow clock and a correct clock both show the correct time at 9 A.M. What time will the slow clock show when the correct clock
shows 10 o'clock the evening of the same day?
1) Method 1: Since the weight of one bowl and the marbles it contains is 50 ounces, the weight of two bowls and twice as many marbles is 100 ounces. It is given that the weight of one
bowl and twice as many marbles is 92 ounces. Then the difference of weights between 100 ounces and 92 ounces is the weight of the bowl, or 8 ounces.
Method 2: The increase in weight from 50 ounces to 92 ounces occurs because the weight of the marbles is doubled. That increase therefore is the weight of the original set of
marbles, 42 ounces. If follows that the bowl weighs 8 ounces.
2) The length of a side of the square is 6m. The dimensions of rectangle AEFG are 3m and 12m. The perimeter of the rectangle is 2 x (3m + 12m) which is equivalent to 30 m.
3) There are 13 hours between 9 AM and 10 PM. Thus the slow clock will lose 13 x 3 or 39 minutes and show 9:21 or 21:21 on a 24-hour clock
1) Suppose the time is now 2 o'clock on a twelve-hour which runs continuously. What time will it show 1,000 hours from now?
2) When a natural number is multiplied by itself, the result is a perfect square. For example, 1, 4, and 9 are perfect squares because 1 x 1 = 1, 2 x 2 = 4 and 3 x 3 = 9. How many perfect
squares are less than 10,000?
3) A work team of four people completes half of a job in 15 days. How many days will it take a team of ten people to complete the remaining half of the job? (Assume that each person of
both teams works at the same rate as each of the other people).
1) The time 2 o'clock is repeated every twelve hours. There are 83 twelves in 1,000 plus a remainder of 4. Therefore the clock will show a time of 6 o'clock 1,000 hours from now.
2) Since 10,000 = 100 x 100, each of the perfect squares 1 x 1, 2 x 2, 3 x 3, . . . , 99 x 99 is less than 100 x 100. There are 99 numbers in the above sequence.
3) Four people working 15 days is equivalent to one person working 60 days. To complete the other half of the job, ten people would have to work 6 days which is also equivalent to one
person working 60 days
1) A motorist made a 60-mile trip averaging 20 miles per hour. On the return trip, he averaged 30 miles per hour. What was the motorist's average speed for the entire trip?
2) 100 pounds of chocolate is packaged into boxes each containing 1 1/4 pounds of chocolate. Each box is then sold for $1.75. What is the total selling price for all of the boxes of chocolate?
3) In the multiplication problem below, A and B stand for different digits. Find A and B.
x B A
1 1 4
3 0 4
3 1 5 4
1) The average speed for any trip is the total distance divided by the total time spent in traveling. The total distance was 120 miles and the total time was 5 hours. The average speed equals
(120 miles)/(5 hours) or 24 miles/hour or 24 mph.
2) We need to know the number of boxes that were packaged in order to find the total selling price. The number of boxes is obtained by dividing 100 by 1 1/4: 100 divided by 5/4 = 100 x 4/5 = 80.
3) The first partial product 114 is equal to the product of AB and A. The second partial product 304 is equal to the product of AB and B. Then A must be less than B.
Method 1: Since the product of AB and A is 114, A is a divisor of 114. Therefore A may be 2, 3, or 6. Since AB x A = 114, A cannot be 2 because AB x A would then be less 60. Similarly, A
cannot be 6 since AB x A would then be greater than 360. Therefore A must be 3 and AB must be 114/3 or 38. Thus A = 3 and B = 8.
Method 2: From the first partial product, observe that B x A must end in 4. Since A is less than B, A = 2 and B = 7, or A = 3 and B = 8, or A = 4 and B = 6. But 27 x 2 = 54, 46 x 4 = 184, and
38 x 3 = 114. Only the third equation satisfies the given conditions. So A = 3 and B = 8.
Method 3: From the second partial product 304, we see that B x B ends in 4. Then B = 2 or 8. If B =2, then A must be 1 because A is less than B. But 12 x 21 = 252. If B = 8 and AB x B = 304,
then AB = 304/8 or 38 and 38 x 83 = 3154. Therefore A = 3 and B = 8.