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Dan pays $2.00 for the first fifth mile and $7.60 for the rest of the ride. Each fifth of a mile costs 40¢, and 760 ÷ 40 = 19. The $7.60 pays for a ride of 19 fifths of a mile. Adding 1/5 mile for the first $2.00,

The cost increases by $.40 for each fifth of a mile. When the desired amount is reached, the taxi will have traveled a length of 20/5 or 4 miles.

Let N = the number of fifths of a mile after the first fifth mile that Dan can ride.

2.00 + .40N = 9.60

.40N = 7.60 (Subtract 2.00 from each side)

N = 19 (Divide each side by .40)

Add the first fifth of a mile, bringing the length of the ride to 20/5 or 4 miles.

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In 45 consecutive days there are 6 weeks and 3 days. Each of the 6 weeks contains one Monday. In order to have the greatest number of Mondays, one of the 3 days left must also be a Monday.

Suppose day 1 is a Monday. Mondays will occur on days 1, 8, 15, 22, 29, 36, 43. The next Monday would be after day 45. The greatest number of Mondays in 45 consecutive days is 7.

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N is composite, so it is the product of two or more (not necessarily different) primes. None of the prime factors can be 2, 3, or 5, for then at least one of the fractions given would not be in lowest terms. The possible prime factors of N are 7, 11, 13, 17, etc. The composites using these factors are 7 × 7, 7 × 11, 7 × 13, 11 × 11, etc. Because only the first two composites are between 20 and 80,

Start with the numbers 21, 22, 23, . . ., 79. 2/N , 3/N , and 5/N are in lowest terms, so eliminate multiples of 2, 3, and 5. This leaves 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, and 79. N is composite, so eliminate primes, leaving only 49 and 77 The possible values of N are 49 and 77.

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Suppose 2/N , 3/N, and 5/N are three fractions in lowest terms. What are all the possible composite whole number values for N between 20 and 80?

]]>If the average of the five numbers is 8, then the sum of those five numbers is 5 × 8, or 40. Subtract the sum of the two known numbers, 7, from 40. The sum of the other three numbers is 33.

The simplest set of 5 numbers whose average is 8 is 8, 8, 8, 8, and 8. Replacing one of the 8's with 2 drops the total by 6, and replacing another with 5 drops the total still further by 3. The total of the five numbers must be increased by 9 (since 6 + 3 = 9), so each of the three remaining 8's must be increased by 3 (three 3's = 9). That means the numbers are 2, 5, and three 11's. The value of any one of the three equal numbers is 11.

(2) What is its smallest possible value? [10]

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The average of five numbers is 8. Two of the numbers are 2 and 5. The other three numbers are equal. What is the value of one of the three equal numbers?

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