Download Quiz 2 Solutions

MAT-103 Quiz 2
Name
Multiply the following fractions and present the answer in simplest form.
(1 pt ea)
1)
    
2)
    
3)
    
4)
   
2∗2∗3
=1
2∗2∗3
5)
   
4∗2∗3∗3
=4∗3=12
3∗2
1 3
2 5
2 7
3 5
4 2
7 3
2 6
4 3
8 9
3 2
=
=
=
=
=
1∗3
3
=
2∗5 10
2∗7 14
=
3∗5 15
4∗2
8
=
7∗3 21


Factor the following numbers completely (1 pt each)
6)
234 -> Even number divides by two
= 2*117 -> 117 is divisible by three
= 2*(3*39) -> 39 is divisible by 3
= 2*(3*3*13) -> Completely factored
7)
177 -> Divisible by three b.c. the sum of digits = 9
= 3*59 -> 59 is prime so we're done
Multiply the following fractions by first factoring each number, canceling like
factors where appropriate, and then multiplying what's left. (2 pts ea)
8)
  
22 −50
−75 −11
=

−
2∗11
3∗25


2∗25 −2∗2 −4
=
=
11
3
3
Why not completely factor 75 and 50? Because we notice that 25 is their
greatest common factor. This saves work so that we can cancel the 25s
and the 11s in between the second and third step.
9)
   
−28 48
16 −14
=
2∗14
16


3∗16
=2∗3=6
14
Again, here we make use of the greatest common factors of 28 and 14 as
well as 48 and 16. This allows us to cancel the 14s and 16s in between
steps 2 and 3.
10)
  
72 −96
32 144
=
 
−

72 3∗32 −3
=
32 2∗72
2
By noticing that 72 goes evenly into 144 (its the greatest common factor)
and that 32 goes evenly into 96, the problem is vastly simplified down to
three steps.