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Session 9: Solutions
 
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Solutions for Session 9 Homework

See solutions for Problems: H1 | H2 | H3 | H4


Problem H1

a. 

The product must be positive but smaller than either c or d; therefore, c • d = b.

b. 

The quotient must be larger than 1, since d is larger than c; therefore, d c = e.

c. 

The result must be negative, since c is smaller than d; therefore, c - d = a.

d. 

The result must be larger than d, since both c and d are positive; therefore, there are two possible answers: c + d = 1 or c + d = e.

<< back to Problem H1


 

Problem H2

a. 

The product is smaller, because b is less than 1 and will therefore produce a smaller result: a • b < a.

b. 

There is not enough information, because c may be less than 1 or greater than 1: b • c ? b.

c. 

There is not enough information, because the product of a and c may be less than 1 or greater than 1: a • b • c ? b.

d. 

The quotient is larger, because b is less than 1 (see the area model for division in Problem A3 for more details): ab > a.

e. 

There is not enough information, because c may be less than 1 (yielding a larger quotient) or greater than 1 (yielding a smaller quotient): ac ? a.

f. 

As in question (e), there is not enough information, because c may be less than 1 (yielding a larger quotient) or greater than 1 (yielding a smaller quotient): bc ? b.

g. 

The answer to bb will always be 1, and we are told that b is less than 1. Therefore, bb > b.

h. 

As in question (a), the product of b2 will be smaller than b, since b is less than 1: b2 < b.

<< back to Problem H2


 

Problem H3

This is not the same, because the additional 50% off is coming from the sale price rather than the original price. The sale price is only 0.8 times as large as the original, so the additional discount is really 40% (0.8 • 50%) of the original price.

For example, if an item cost $100 originally, it would be $80 with the sale price. With the additional sale, 50% is taken from the $80 sale price, not $100, and the final price is $40.

Put another way, the resulting price is 0.5 • 0.8 = 0.40 of the original, which is a 60% discount, not a 70% discount.

<< back to Problem H3


 

Problem H4

Since we can choose any two one-digit numbers as our starting pair, there are 100 possible two-digit combinations, so the total number of digits in the bracelets will be exactly 100. There are six total bracelets. Here they are:

112358314594370774156178538190998752796516730336954932572910 [60 digits]

134718976392 [12 digits]

22460662808864044820 [20 digits]

2684 [4 digits]

550 [3 digits]

0 [1 digit: 0, 0, 0, . . .]

There are a lot of patterns in the 60-digit bracelet. See what you can find!

<< back to Problem H4


 

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