Mathematics Grade 10 Mpumalanga March Test 2025 Paper & Memo

Testpapers.co.za ce@ty Starynorephysics.com education MPUMALANGA PROVINCE REPUBLIC OF SOUTH AFRICA . SENIOR CERTIFICATE MARKS: 50 TIME: 1 HOUR This question paper consists of 5 pages including the cover page Please turn over

Testpapers.co.za wanownloadedifrsm Stanmorephysics.com 24 March 2025 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions: L This question paper consists of 3 questi. Answer all questions An approved calcula c grammable and non-graphical) may be used, unless stated otherwise. Number the answers a mbering system used in this question paper It is in your own interest to write legibly and to present the work neatly. If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise Diagrams are not drawn according to scale. Please turn over

Testpapers.co.za matownoadediinsm Stanmorephysies.com QUESTION 1 1.1 Indicate between which two consecutive integers V47 lie 1.2 Write 0,13 as a fraction, clearly show all your steps. 1.3 1.4 Simplify fully: 14.1 (2x-5)(x? -2x+3) 142 x2-x-2 x+l1 143 3, 7.9% 4a 24 March 2025 (2) (3) (2) (2) (2) (3) (2) (2) (2) [20] Please turn over

Testpapers.co.za matownoadediinsm Stanmorephysies.com QUESTION 2 2.2 2.2 23 2.4 Solve for x in each of the following equation: 211 x(5-x)=0 242 48 x mvs om Solve the folloy 23,1 -15<1-4x<5 andxeR 2.3.2 Represent your answer above on a number line. Solve for x and y simultaneously: x+3y=12 ax+y=9 24 March 2025 (2) (3) (3) (4) (3) (2) (5) [22] Please turn over

Testpapers.co.za matownoadediinsm Stanmorephysics.com 24 March 2025 QUESTION 3 3.1 In APRQ, OR=4cm, PR=5cm andQP =3cm. PQ is perpendicular to QRandP=0. PR = ysics.com 3.1.1 sind (1) 3.1.2 seca GB) 3.13 tan@ () 3.2 Calculate the size of @ (3) [8] TOTAL MARKS: 50 Please turn over

Testpapers.co.za Downloaded from Stanmorephysics.com Education MPUMALANGA PROVINCE REPUBLIC OF SOUTH AFRICA MARKS: 50 TIME: 1 HOUR This marking guidelines consists of 5 pages. Please turn over

Testpapers.co.za Bownloadeatafsom Stanmorephysics.com March 2025 QUESTION 1 Il | 36 < 47 < V49 v 36 < V47 < J49 ” 6< 13 <7 Y answer Between 6 and 7 1.2 ff Letst= 0131313 Y 100x = 13,1313... (3) 100x = 13,1313 Y 100x-x = 13,1313 -0,1313 100x —x = 13,1317 v answer 99x =13 13 99 1.3.1} x?-~3x-10 ¥ ¥ factors (2) (x+2)(x—5) 1.3.2 | 25x" —4y" ¥ (58° -Qy*l @) = (Sx’)? -(2y*)? ¥ factors = (5x? —2y*)(5x? + 2y*) 1.3.3 | 125x3 +27 vv factors (2) = (5x) +3° = (5x + 3)(25x* - 15x +9) OxY —2y +h ¥ 3x(1-2y) Q) t Y \(-2y+l) Y answer Y simplification (2) v answer =2x° -9x? +16x-15 14.2| .?-y-2 _ (x= 2x41) ¥ factors (2) x41 x41 v answer =x-2

Testpapers.co.za Bownloadeatafsom Stanmorephysics.com March 2025 1.4.3 | 36bmpatd 252 350, 750, 950 Y (2.3.7)™ 300 = 300 (4) 42 (3:72) 75003500 7500 Se ae 2502-500 = 3500 7500 5500 sila ¥ answer [20] Y Y answer (2) 2.12 x-2-8 Y standard form * ¥ three factors @) x? -2x=8 Y answer x? -2x-8=0 (x-4)(x+2)=0 x=4or x=-2 2.1.3 | x4 =27x vx =27 3 5. nf (3) x= 27 Vx =3 gaat v answer x=3 2.2 | 4*-3.2"-18 2**-3.2..-18 VOR (4) 2° +3 2* +3 ¥ ¥ factors _ (2° +3)(2"-6). Y answer 2°43 =?" 6: 2.3.1 | -15<1-4x<5 ¥ -16<-4x<4

Testpapers.co.za Bownloaded:firom Stanmorephysics.com March 2025 -16<-4x <4 Y end points (3) 4>x>-1 ¥ notation mae 4 2.3.2 7 a x > vv answer (2) 246 X+3y =12 ¥ equation 3 ax+y=9 ¥ substitution x=12-3y v simplification Substitute (3) into vx-value 2(12-3y)+ y=9 % vy-value GS) 24-6y+y=9 fe ~Sy=-15 y= Subst. y =3 into (3) x=12-3(3) x=3 [22] QUESTION 3

Testpapers.co.za Bownloadeatafsom Stanmorephysics.com March 2025 3.1.1 | siné ¥ answer qd) = oR PR aye 5. seca rae; Jeune B) __! cosa - 2 ¥ substitution = OR Y answer PR eo 4 5 > 4 3.1.3 | tand v answer _QR qd) “OF ot 3 3.2.2 ¥ correct ratio (3) v 7 sin 0 = OR substitution PR ¥ answer sin 0 = ha 4 Gm iar sin ( 3 0 =53,13° [8] TOTAL MARKS: 50