Gr 3 Mathematics (IsiNdebele) 2023 Possible Answers

UKUHLOLWA KWESIFUNDA SINYIKHABA 2023 IBANGA 3 INDLELA YOKUMAKA IIMBALO: ISINDEBELE Amakhasi asi 6

INDLELA YOKUMAKHA Ibanga 3 IIMBALO 1 C 1 imaki 2 B 1 imaki 3 D 1 imaki 4 A 1 imaki 5 B 1 imaki 6 C 1 imaki 7 C 1 imaki 8 D 1 imaki 9 B 1 imaki 10 D 1 imaki 11 7:40 ekuseni nanyana imizuzu ematjhumi amabili  1 imaki Ngaphambi kwe iri lobunane. 12 Amakota amabili  1 imaki 13 Imizuzu ema 30  1 imaki 14 iPentagoni  1 imaki 15 Amakhilomitha  1 imaki 16 A 500  Imaki linye B 1 500  ependulweni (2 amamaki) 2

INDLELA YOKUMAKHA Ibanga 3 IIMBALO 17 1 imaki  Yokulandelanisa ekungikho 18 706; 705; 701; 659; 607  1 imaki lokulandelanisa (1 imaki) 19 A B – ikhiyubhi  1 imaki lependulo B C – ikhoni  (2 amamaki) 20 Amakota amathathu azeleko  1 imaki 21 A C3 nanyana 3C  4 amamaki B D4 nanyana 4D  Imaki linye lependulo ekungiyo C C2 nanyana 2C  (4 amamaki) D iphiramidi  22 1 bala uhlanganise zoke iinkwere = 16  1 imaki linye lependulo nanyana ekungiyo 2 ireyi x ikholomu = 16  (2 amamaki) nanyana 3 4 x 4 = 16 (enyenenye indlela ingasetjenziswa) 3

INDLELA YOKUMAKHA Ibanga 3 IIMBALO 23 6 x 4 = 24  1 imaki lependulo OR ekungiyo 12 + 12 = 24 (2 amamaki) OR 24 amaqanda  (enyenenye indlela ingasetjenziswa) 24 360 ÷ 12 = 30 amagremu  1 imaki lependulo 2 x 30 amagremu = 60 amagremu  ekungiyo (2 amamaki) (enyenenye indlele ingasetjenziswa) 25 uSara mude ngama cm ali 13 kuno Jay  1 imaki lependulo 151 cm – 138 cm = 13 cm  ekungyo (2 amamaki) 26 12 ÷ 4 = 3  1 imaki lependulo nanyana ekungiyo 12 ÷ 3 = 4 (1 imaki) Enyenenye indlela ivumelekile 27 5 x 6 = 30  1 imaki lendlela Amastikha ama 30 ekungiyo 1 imaki Enyenenye indlela ingasetjenziswa  lependulo (2 amamaki) 4

INDLELA YOKUMAKHA Ibanga 3 IIMBALO 28 36 ÷ 3 = 12  1 imaki lependulo 1 imaki lomeqo ekungiwo (2 amamaki)  0 3 6 9 12 15 18 21 24 27 30 33 36 39 Yamukela enyenenye indlela 29 8 x 4 = 32  1 imaki lependulo 1 imaki lomeqo Usebenzise amalamune ama 32 ukwenza inamaneyidi  (2 amamaki) 30 15 + 6 = 21  1 imaki lendlela ekungiyo ne 1 Amahabhula ama 21 sele awoke  imaki lependulo (2 amamaki) 31 242 + 353 1 imaki lendlela yokwenza 1 imaki = (200 + 40 + 2) + (300 + 50 + 3)  lependulo (200 + 300) (40 + 50) (2 + 3) (2 amamaki) 500 + 90 + 5 595  5

INDLELA YOKUMAKHA Ibanga 3 IIMBALO 32 598 – 75 =  1 imaki lendlela yokwenza 1 imaki = (500 and 90 and 8) – (70 and 5) lependulo (500 – 0) (90 – 70) (8 – 5) (2 amamaki) = 500 – 20 – 3 = 523  Yamukela enyenenye indlela yokwenza 33 A iincwadi zeenlwana  1 imaki lependulo B 23  (3 amamaki) C abafundi aba 2  INANI: 50 amamaki 6