1001 Algebra Problems.PDF

Set 3 (Page 5)

33. a.Express both fractions using the least com-
    mon denominator, then add:^59 – ^14 =
    ^59  44 – ^14 ^99 = ^2306 –  396 = ^203 – 6 ^9 = ^1316 


34. b.First, rewrite all fractions using the least
    common denominator, which is 30. Then, add:

 12  5 + ^15 +  61 +  13  0 =  125 ^22 +  51 ^66 + ^16  55 +  130 ^33

=  340 +  360 +  350 +  390 

=

= ^2340 

= ^45 

35. d.The square is divided into 8 congruent
    parts, 3 of which are shaded. Thus,^38 of the
    figure is shaded.


36. c.First, rewrite both fractions using the least
    common denominator, which is 60. Then,
    subtract:

^1270 – ^56 = ^1270 ^33 – ^56 ^1100 = ^56  01 –  6500 = ^516 – 0 ^50 =  610 

37. c.Rewrite this as a multiplication problem,
    cancel factors that are common to the numer-
    ator and denominator, and then multiply:

^158  290 = ^158 ^290 =  = 8

38. c.A reasonable strategy is to begin with one of
    the fractions, say ^58 , and compare it to the next
    one in the list. Discard whichever is smaller
    and compare the remaining one with the next
    in the list. Repeat this until you reach the end
    of the list. Doing so results in the following
    three comparisons:

Comparison 1:^58 ^23 Cross multiplying yields

the false statement 15 > 16.

This implies that ^23 is larger.

Comparison 2:^23   181 Cross multiplying yields

the false statement 22 > 24. This implies that

 181 is larger.

Comparison 3: 181  140 Cross multiplying

yields the true statement 80 > 44. This implies

that  181 is larger.

Thus, we conclude that  181 is the largest of the

choices.

39. a.The fact that ^14 < ^58 < ^23 is evident from the
    following two comparisons:

Comparison 1:^14 ^58 Cross multiplying yields

the true statement 8 < 20, so the original

inequality is true.

Comparison 2:^58 ^23 Cross multiplying yields

the true statement 15 < 16, so the original

inequality is true.

40. a.Since^35 (360) = ^35 (5 72) = 216, we conclude
    that Irma has read 216 pages.


41. a.Cancel factors that are common to the
    numerator and denominator, then multiply:

^58 ^47 =  4

5  2  =  154 

42. d.The reciprocal of^2412 is ^4221 , which is equiva-
    lent to 2.


43. b.Two real numbers are in a ratio of 4:5 if
    the second number is ^54 times the value of the
    first number. Observe that ^14 is ^54 times the
    value of^15 .


44. c.The remaining 28 (of 42) envelopes need to
    be addressed. Thus, the fraction of envelopes
    that needs to be addressed is ^2482 = =  32 
    1414
     32 .

^4

7

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4.5^

9



^9 ^2

5

4 + 6 + 5 + 9

30

ANSWERS & EXPLANATIONS–