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  • GETSenPhaseMaths display tagshide tags

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Breuke - Deling met breuke

Module by: Siyavula Uploaders. E-mail the author

WISKUNDE

Gewone Breuke

OPVOEDERS AFDELING

Memorandum

  • b)
Table 1
10 10
1 10 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {} 1 10 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {}

Answers is the same

(i) = 3434 size 12{ { { size 8{3} } over { size 8{4} } } } {} x 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {}

x = 3838 size 12{ { { size 8{3} } over { size 8{8} } } } {}

y = 18 2323 size 12{ { { size 8{2} } over { size 8{3} } } } {}

(ii) = 7 x 8383 size 12{ { { size 8{8} } over { size 8{3} } } } {}

= 563563 size 12{ { { size 8{"56"} } over { size 8{3} } } } {}

  1. (i) = 6161 size 12{ { { size 8{6} } over { size 8{1} } } } {} x 5454 size 12{ { { size 8{5} } over { size 8{4} } } } {}

= 304304 size 12{ { { size 8{"30"} } over { size 8{4} } } } {}

m = 7 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {}

(iv) = 2727 size 12{ { { size 8{2} } over { size 8{7} } } } {} x 1919 size 12{ { { size 8{1} } over { size 8{9} } } } {}

n = 263263 size 12{ { { size 8{2} } over { size 8{"63"} } } } {}

  • b)

(i) x = 3838 size 12{ { { size 8{3} } over { size 8{8} } } } {}924924 size 12{ { { size 8{9} } over { size 8{"24"} } } } {}

= 3838 size 12{ { { size 8{3} } over { size 8{8} } } } {} x 249249 size 12{ { { size 8{"24"} } over { size 8{9} } } } {}

x = 1

(ii) k = 15181518 size 12{ { { size 8{"15"} } over { size 8{"18"} } } } {}456456 size 12{ { { size 8{"45"} } over { size 8{6} } } } {}

= 15181518 size 12{ { { size 8{"15"} } over { size 8{"18"} } } } {} x 645645 size 12{ { { size 8{6} } over { size 8{"45"} } } } {}

k = 1919 size 12{ { { size 8{1} } over { size 8{9} } } } {}

(iii) c = 7979 size 12{ { { size 8{7} } over { size 8{9} } } } {}5656 size 12{ { { size 8{5} } over { size 8{6} } } } {}

= 7979 size 12{ { { size 8{7} } over { size 8{9} } } } {} x 6565 size 12{ { { size 8{6} } over { size 8{5} } } } {}

c = 14151415 size 12{ { { size 8{"14"} } over { size 8{"15"} } } } {}

(iv) f = 11121112 size 12{ { { size 8{"11"} } over { size 8{"12"} } } } {}6565 size 12{ { { size 8{6} } over { size 8{5} } } } {}

= 11121112 size 12{ { { size 8{"11"} } over { size 8{"12"} } } } {} x 5656 size 12{ { { size 8{5} } over { size 8{6} } } } {}

= 55725572 size 12{ { { size 8{"55"} } over { size 8{"72"} } } } {}

23.3 c)

(i) b = 214214 size 12{2 { { size 8{1} } over { size 8{4} } } } {}3232 size 12{ { { size 8{3} } over { size 8{2} } } } {}

= 9494 size 12{ { { size 8{9} } over { size 8{4} } } } {} x 2323 size 12{ { { size 8{2} } over { size 8{3} } } } {}

b = 1 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {}

(ii) e = 3 4545 size 12{ { { size 8{4} } over { size 8{5} } } } {}  2 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {}

= 195195 size 12{ { { size 8{"19"} } over { size 8{5} } } } {} x 2525 size 12{ { { size 8{2} } over { size 8{5} } } } {}

e = 38253825 size 12{ { { size 8{"38"} } over { size 8{"25"} } } } {}

e = 1 13251325 size 12{ { { size 8{"13"} } over { size 8{"25"} } } } {}

  1. (i) g = 3 4747 size 12{ { { size 8{4} } over { size 8{7} } } } {}  1 2727 size 12{ { { size 8{2} } over { size 8{7} } } } {}

= 257257 size 12{ { { size 8{"25"} } over { size 8{7} } } } {} x 7979 size 12{ { { size 8{7} } over { size 8{9} } } } {}

= 259259 size 12{ { { size 8{"25"} } over { size 8{9} } } } {}

g = 2 7979 size 12{ { { size 8{7} } over { size 8{9} } } } {}

(iv) r = 15 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {}  5 1414 size 12{ { { size 8{1} } over { size 8{4} } } } {}

= 312312 size 12{ { { size 8{"31"} } over { size 8{2} } } } {} x 421421 size 12{ { { size 8{4} } over { size 8{"21"} } } } {}

= 62216221 size 12{ { { size 8{"62"} } over { size 8{"21"} } } } {}

r = 2 20212021 size 12{ { { size 8{"20"} } over { size 8{"21"} } } } {}

Leerders Afdeling

Inhoud

AKTIWITEIT: Deling met breuke [LU 1.7.3, LU 2.1.5]

23. Kom ons kyk nou na DELING MET BREUKE!

23.1 Deling van heelgetalle deur breuke en andersom:

a) Werk saam met ’n maat en kyk goed na die volgende probleme.

Ma bak vyf koeke en wil graag vir jou en jou maats elkeen ’n halwe ( 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {}) stuk gee. Hoeveel maats kan van die koek eet?

  • Op ’n getallelyn lyk dit so:
Figure 1
Figure 1 (graphics1.png)

Dus: 5 ÷ 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} = 1010 kinders kan elkeen 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} koek kry.

Ma bak weer, maar hierdie keer net een reghoekige koek. Sy besluit om die helfte daarvan tussen haar vyf kinders te verdeel. Watter breuk kry elkeen?

  • Kom ons maak ’n skets daarvan!

12345

Kan jy sien dat elke kind een tiende ( 110110 size 12{ { { size 8{1} } over { size 8{"10"} } } } {}) van die koek sal kry?Dus: 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} ÷ 5 = 110110 size 12{ { { size 8{1} } over { size 8{"10"} } } } {}

b) Voltooi die tabel:

Table 2
   
5 ÷ 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} = ............ 5 × 2121 size 12{ { { size 8{2} } over { size 8{1} } } } {} = ............
1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} ÷ 5151 size 12{ { { size 8{5} } over { size 8{1} } } } {} = ............ 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} × 1515 size 12{ { { size 8{1} } over { size 8{5} } } } {} = ............

Wat merk jy op? ___________________________________________________

_____________________________________________________________________

c) Het jy geweet?

Enige deelsom met breuke kan in ’n vermenigvuldigingsom verander word! Ons doen dit deur die deler in sy resiprook te verander. Ons “keer dus die deler om”!

Dus:

Figure 2
Figure 2 (graphics2.wmf)
÷ 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} =
Figure 3
Figure 3 (graphics3.wmf)
´ 2121 size 12{ { { size 8{2} } over { size 8{1} } } } {} = 10

d) Verbind kolom A met die korrekte antwoord in kolom B:

Table 3
A   B
÷ deur 5   × met 4343 size 12{ { { size 8{4} } over { size 8{3} } } } {}
     
÷ deur 3434 size 12{ { { size 8{3} } over { size 8{4} } } } {}   × met 3
     
÷ deur 7878 size 12{ { { size 8{7} } over { size 8{8} } } } {}   × met 5
     
÷ deur 1313 size 12{ { { size 8{1} } over { size 8{3} } } } {}   × met 1515 size 12{ { { size 8{1} } over { size 8{5} } } } {}
     
÷ deur 1515 size 12{ { { size 8{1} } over { size 8{5} } } } {}   × met 8787 size 12{ { { size 8{8} } over { size 8{7} } } } {}

e) Bereken die volgende:

i) x=34÷2x=34÷2 size 12{x= { { size 8{3} } over { size 8{4} } } div 2} {}

___________________________________________________

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ii) y=7÷38y=7÷38 size 12{y=7 div { { size 8{3} } over { size 8{8} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

iii) m=6÷45m=6÷45 size 12{m=6 div { { size 8{4} } over { size 8{5} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

iv) n=27÷9n=27÷9 size 12{n= { { size 8{2} } over { size 8{7} } } div 9} {}

___________________________________________________

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23.2 Deling van breuke deur breuke:

a) Werk weer saam met ’n maat en bestudeer die volgende:

x = 6 25 ÷ 3 5 x = 6 25 ÷ 3 5 size 12{x= { { size 8{6} } over { size 8{"25"} } } div { { size 8{3} } over { size 8{5} } } } {}
(1)

Ek weet ek moet die volgende stappe volg:

1. Verander die ÷ in ×

2. Draai die breuk na die ÷ (deler) om – kry dus resiprook

3. Vermenigvuldig soos gewoonlik: teller × teller noemer × noemer teller × teller noemer × noemer size 12{ { { ital "teller" times ital "teller"} over { ital "noemer" times ital "noemer"} } } {}

Dus: 625÷35=625×53625÷35=625×53 size 12{ { { size 8{6} } over { size 8{"25"} } } div { { size 8{3} } over { size 8{5} } } = { { size 8{6} } over { size 8{"25"} } } times { { size 8{5} } over { size 8{3} } } } {}

Table 4
Ek kanselleer waar ek kan:
2 6
5 25
×
5 1
3 1

Die antwoord is dus 2×15×1=252×15×1=25 size 12{ { { size 8{2 times 1} } over { size 8{5 times 1} } } = { { size 8{2} } over { size 8{5} } } } {}

b) Probeer die volgende op jou eie:

i) x=38÷924x=38÷924 size 12{x= { { size 8{3} } over { size 8{8} } } div { { size 8{9} } over { size 8{"24"} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

ii) k=1518÷456k=1518÷456 size 12{k= { { size 8{"15"} } over { size 8{"18"} } } div { { size 8{"45"} } over { size 8{6} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

iii) c=79÷56c=79÷56 size 12{c= { { size 8{7} } over { size 8{9} } } div { { size 8{5} } over { size 8{6} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

iv) f=1112÷65f=1112÷65 size 12{f= { { size 8{"11"} } over { size 8{"12"} } } div { { size 8{6} } over { size 8{5} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

23.3 Deling met gemengde getalle:

a) Kan jy die volgende probleem vir ’n maat verduidelik?

’n Gesin eet 1 en ’n halwe ( 112112 size 12{1 { { size 8{1} } over { size 8{2} } } } {}) pizza. As elkeen net een kwart ( 1414 size 12{ { { size 8{1} } over { size 8{4} } } } {}) van die pizza eet, uit hoeveel lede bestaan die gesin?

  • Ek moet 1 1212 size 12{ { { size 8{1} } over { size 8{2} } } } {} ÷ 1414 size 12{ { { size 8{1} } over { size 8{4} } } } {} bereken.
  • Dis makliker as ek dit teken:

It’s easier if I draw it:

Figure 4
Figure 4 (graphics4.png)
  • Die antwoord is dus 6.
  • Wiskundig skryf ek dit so:

y = 1 1 2 ÷ 1 4 3 2 ÷ 1 4 3 2 × 4 1 12 2 6 y = 1 1 2 ÷ 1 4 3 2 ÷ 1 4 3 2 × 4 1 12 2 6 alignl { stack { size 12{y=1 { { size 8{1} } over { size 8{2} } } div { { size 8{1} } over { size 8{4} } } } {} # = { { size 8{3} } over { size 8{2} } } div { { size 8{1} } over { size 8{4} } } {} # = { { size 8{3} } over { size 8{2} } } times { { size 8{4} } over { size 8{1} } } {} # = { { size 8{"12"} } over { size 8{2} } } {} # =6 {} } } {}

  • Ek verkies om ’n getallelyn te gebruik:
Figure 5
Figure 5 (graphics5.png)

b) Het jy geweet?

Ons verander gemengde getalle eers in onegte breuke voordat ons die antwoord bereken.

c) Probeer op jou eie:

i) b=214÷32b=214÷32 size 12{b=2 { { size 8{1} } over { size 8{4} } } div { { size 8{3} } over { size 8{2} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

ii) e=345÷212e=345÷212 size 12{e=3 { { size 8{4} } over { size 8{5} } } div 2 { { size 8{1} } over { size 8{2} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

iii) g=347÷127g=347÷127 size 12{g=3 { { size 8{4} } over { size 8{7} } } div 1 { { size 8{2} } over { size 8{7} } } } {}

___________________________________________________

___________________________________________________

___________________________________________________

iv) r=1512÷514r=1512÷514 size 12{r="15" { { size 8{1} } over { size 8{2} } } div 5 { { size 8{1} } over { size 8{4} } } } {}

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Assessering

Leeruitkomste 1:Die leerder is in staat om getalle en die verwantskappe daarvan te herken, te beskryf en voor te stel, en om tydens probleemoplossing bevoeg en met selfvertroue te tel, te skat, te bereken en te kontroleer.

Assesseringstandaard 1.7: Dit is duidelik wanneer die leerder skat en bereken deur geskikte bewerkings vir probleme wat die volgende behels, kies en gebruik:

1.7.3: optelling, aftrekking en vermenigvuldiging van gewone breuke.

Leeruitkomste 2:Die leerder is in staat om patrone en verwantskappe te herken, te beskryf en voor te stel en probleme op te los deur algebraïese taal en vaardighede te gebruik.

Assesseringstandaard 2.1: Dit is duidelik wanneer die leerder numeriese en meetkundige patrone ondersoek en uitbrei op soek na ‘n verwantskap of reëls, insluitend patrone;

2.1.5: voorgestel in tabelle.

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