Skip to content

Mathematics Grade 10 March Test 2020 Paper & Memo

Question paper and memoMathematicsGrade 10202013 pages

Grade 10 Mathematics question paper and memo from the March Test 2020, 13 pages. Read online or get the free PDF download.

Free to read online, no sign-up needed

Loading document…

Loading document…

Document textSearch extracted text and jump to a page.
Downloaded from hlayiso.com testpapers.co.za
Downloaded from hlayiso.com Mathematics 2 Common Test March 2020 NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. - . This question paper consists of 6 questions. Answer ALL the questions. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. Answers only will NOT necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. If necessary, round off answers correct to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. Write neatly and legibly. Copyright reserved Please Turn Over
Downloaded from hlayiso.com Mathematics 3 Common Test March 2020 NSC QUESTION 1 1.1 Factorise the following expressions fully: LLL xy? +43x7y (1) 1.1.2 x°-7x-18 (2) 11.30 x y-164+4y—4x? (3) 1.2 Simplify the following expressions fully: 12.1 (2x-1)(x? -3x+1) (3) 12.2 ee (4) (x+2)tx(x+2) 2x44 Deen ee 12.30 ————_ 4 gn gre! 9-3" ¢ ) {17] QUESTION 2 21 Solve for x: 2.1.1 x(2x-5)=0 (2) 2.12 3x’-2x-8=0 GB) 213. 5221-0 (2) 2.14 x=ytxy (3) 8x° -1 2.1.5 =1 4 2x-1 ® 2.2 The following inequality is given: -11<-2x+1<-9; wherex ER. 2.2.1 Solve for x. (3) 2.2.2 Hence, and without the use of a calculator, show that x = 29 would satisfy the above inequality. (2) [19] Copyright reserved Please Turn Over
Downloaded from hlayiso.com Mathematics 4 Common Test March 2020 NSC QUESTION 3 31 Solve for x and y simultaneously: 2x-y=3 [ 6) 3x+2y =8 3.2 Giventhat M=2°? and M’ =16, determine the value of b. QB) [8] QUESTION 4 Various options are provided as possible answers to the following questions. Write down the question number (4.1 — 4.5) and choose the answer by writing the letter (A—D) next to the question number (4.1 — 4.5) in your answer book, for example: 4.6) D 4.1 Which description below does NOT guarantee that a quadrilateral is a square? A. Quadrilateral is both a rectangle and a rhombus B. Quadrilateral is a parallelogram with perpendicular diagonals Cc Quadrilateral has all sides equal and all angles equal D. Quadrilateral has all right angles and has all sides equal qd) 4.2. Which of the following statements is true? A. All quadrilaterals are rectangles B. All quadrilaterals are squares Cc. All rectangles are quadrilaterals Dz. All quadrilaterals are parallelograms (1) 43 In the diagram below rectangle KLMN has KM = 6x + 16 and LN = 49. Find the value of x. K L N M A. x=5,5 B. x= 33 Cc, ee D x =6,5 Q) Copyright reserved Please Turn Over
Downloaded from hlayiso.com Mathematics 5 Common Test March 2020 NSC 44 A quadrilateral with only one pair of opposite sides parallel is called a: A. Trapezium B. Square e Kite D. Rhombus (1) 4.5 Inquadrilateral ABCD, ACD =2x+4 and ACB=5x-11. For what value of x is ABCD a rhombus? " GVaAW>S> He HOM OH ll ADO P (1) [5] Give reasons for your statements in the answers to QUESTIONS 5 and 6. Copyright reserved Please Turn Over
Downloaded from hlayiso.com Mathematics 6 Common Test March 2020 NSC QUESTION 5 5.1 In the diagram below, straight lines DEF and TMB are parallel to each other. It is also given that EM = EB and BEF = 64°, D 64° Calculate the size of MEB. (4) 5.2 In the diagram below, PQRS is a parallelogram. PT = 4x -2, TR=x + 28, ST=4y-7 and TQ=y+ 14. Ss Determine, with reasons, the values of x and y. (4) Copyright reserved Please Turn Over
Downloaded from hlayiso.com Mathematics i Common Test March 2020 NSC 33 In the diagram below, ABCD and BECD are parallelograms with common base DC. BC LBD and DAB=40° . A B E 40° D C Determine the size of BEC . (4) [12] Copyright reserved Please Turn Over
Downloaded from hlayiso.com Mathematics 8 Common Test March 2020 NSC QUESTION 6 6.1 In the diagram below, JKLM is a rectangle. MLPR is a rhombus. JMK = RMP ; JMK =55° and MRP =70° R P Using the diagram, and giving reasons, determine: 6.1.1 MPR (2) 6.1.2 KML () 6.13 KEP 2) 6.2 In the diagram below rectangle ABCD is given with AP = DN . B N E P D 6.2.1 Prove that AABP=ADCN . (4) 6.2.2 Prove that AE = DE. (4) [14] GRAND TOTAL: [75] Copyright reserved
Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 10 MATHEMATICS COMMON TEST MARCH 2020 MARKING GUIDELINES MARKS: 75 TIME: 1½ hours This marking guideline consists of 5 pages. Copyright Reserved Please turn over
Downloaded from hlayiso.com Mathematics 2 March 2020 Common Test NSC – Marking Guidelines QUESTION 1 1.1.1 xy 2  3x 2 y  xy ( y  3x)  answer (1) 1.1.2 x  7 x  18 2  (x – 9)  ( x  9)( x  2)  (x + 2) (2) 1.1.3 x y  16  4 y  4 x 2 2  x 2 y  4 y  4 x 2  16  grouping  y ( x 2  4)  4( x 2  4)  ( x 2  4)( y  4)  common bracket  answer (3) 1.2.1 (2 x  1)( x  3x  1) 2  correct expansion  2 x3  6 x 2  2 x  x 2  3x  1  – 7x2  2 x3  7 x 2  5 x  1  + 5x (3) 1.2.2 x2  1 x 1  factorising of  D.O.T.S ( x  2)  x( x  2) 2 x  4  factorising of ( x  1)( x  1) 2( x  2) common bracket   ( x  2)(1  x) x 1  changing  2  answer (4) 1.2.3 2 2 n .3 3n 22 n .4n  1 .12 3n 22 n.33n  prime bases  22 n.22 n2.(22.3) 3n  raising powers 22 n2 n2 n2.33n  26 n.33n  simplification  26 n26 n.33n3n  22 4  answer (4) [17] 2.1.1 x(2 x  5)  0 5  x=0  x  0 or x  5 2  x 2 (2) 2.1.2 3x  2 x  8  0 2 (3x  4)( x  2)  0  correct factors 4 4  x   or x  2  x 3 3  x2 (3) Copyright Reserved Please turn over
Downloaded from hlayiso.com Mathematics 3 March 2020 Common Test NSC – Marking Guidelines 2.1.3 52 x  1  1  0 52 x 1  50  50 2x 1  0 1 x   answer (2) 2 2.1.4 x  y  xy x  xy  y x(1  y )  y  isolate x terms  factorise x y x 1 y  answer (3) 2.1.5 8x  1 3 1 1 Restriction : x  2x 1 2 (2 x  1)(4 x 2  2 x  1)  1 factorising 2x 1 4 x2  2 x  1  1  simplification 4x  2x  0 2 2 x(2 x  1)  0  factors 1  x  0 or x   2  both x values (4) 2.2.1 11  2 x  1  9 12  2 x  10  12  2 x  10 6   x  5  values 5  x  6  inequality “flip” (3) 2.2.2 25  29  36  creating inequality 25  29  36  25  29  36  5  29  6 Therefore 29 satisfies the inequality in 2.2.1 (2) [19] Copyright Reserved Please turn over
Downloaded from hlayiso.com Mathematics 4 March 2020 Common Test NSC – Marking Guidelines QUESTION 3 3.1 2x  y  3 Eq1 3x  2 y  8 Eq 2  setting up Eq1 2 : 4x  2 y  6 equations Eq 2 : 3x  2 y  8  elimination to Eq1  Eq 2 : 7x  14 solve for x or y x 2  x=2 Sub x  2 into Eq1: 2(2)  y  3  substitution  y  1  y 1 y 1 (5) OR 2x  y  3 Eq1 3x  2 y  8 Eq 2  setting up Eq1  Eq3 : y  2x  3 equation Sub Eq3 into Eq 2 : 3x  2(2 x  3)  8  substitution 3x  4 x  6  8 7 x  14  x=2 x 2 Sub x  2 into Eq3 : y  2(2)  3  substitution y 1  y 1 (5) 3.2 M  16 b  16 = 24 M  2 b 4 but M  20,2  (20,2 )b  24  (20,2 )b  24  0, 2b  4  b  20  answer (3) [8] QUESTION 4 4.1 B  answer (1) 4.2 C  answer (1) 4.3 A  answer (1) 4.4 A  answer (1) 4.5 B  answer (1) [5] Copyright Reserved Please turn over
Downloaded from hlayiso.com Mathematics 5 March 2020 Common Test NSC – Marking Guidelines QUESTION 5 5.1 ˆ  64 BEF ( given)  S ˆ  64 TBE (alt s, DF TB)  R ˆ  64  EMB (s opp  sides)  S/R ˆ  52  MEB ( sum s )  S/R (4) 5.2 4 x  2  x  28 (diagonals of parm)  S/R 3x  30 x  10  x = 10 4 y  7  y  14 (diagonals of parm)  S/R 3 y  21 y3  y=3 (4) 5.3 ˆ  40 EBC (corresponding s; AD BC )  S  R ˆ  90 BCE (alternate s; BD EC )  S/R ˆ  50  BEC ( sum s )  S/R (4) [12] QUESTION 6 6.1.1 ˆ  55 MPR (s opp  sides)  S  R (2) 6.1.2 ˆ  35 KML (adj comp s)  S/R (1) 6.1.3 ˆ  90 MLP ( prop of rect )  S/R  S/R ˆ  70 MLP (opp s rhombus) ˆ  160  KLP  S (3) 6.2.1 In ABP and DCN  S/R 1. Bˆ  Cˆ  90 ( prop of rect ) 2. AP  DN ( given)  S/R 3. AB  DC (opp sides of rect )  S/R  ABP and DCN ( RHS )  S/R (4) 6.2.2 ˆ  NDC BAP ˆ (congruent s proved )  S/R ˆ  ADE  DAE ˆ (adj compl s)  S  R  AE  DE ( sides opp  s )  S/R (4) [14] TOTAL MARKS: 75 Copyright Reserved Please turn over

Study this subject

More from Grade 10 Mathematics

View all