Mathematics Grade 10 Paper 1 North West June 2018 Memo

Downloaded from hlayiso.com GRADE 10 MATHEMATICS PAPER 1 MID YEAR EXAMINATION MEMORANDUM 2018 MARKS: 75 This memorandum consists of 6 pages Demo NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics Memo/P1 2 NW/June 2018 Grade 10 QUESTION 1[10] 1.1   1  0,4 3 0,4 3 and 3  27 1  3  27 (2) 1.2 9,87 answer (1) 1.3  0,4 3  0,433333333...  let x  0,4 3  equation 1 10x  4, 3        (1) equation 2  100x  43, 3        (2) equation (2) - (1)  90x  39 90x  39 answer 13 (4) x  30 1.4 7 2  98  7 2  98 81  98  100  81  98  100 9  98  10 answer  it lies between integers 9 and 10 (3) QUESTION 2[16] 2.1 (2a  3b)(3a 2  2ab  3b 2 )  2a(3a 2  2ab  3b 2 )  3b(3a 2  2ab  3b 2 )  6a 3  4a 2b  6ab2  6a  4a b  6ab  9a b  6ab  9b 3 2 2 2 2 3   9a 2b  6ab2  9b 3  6a 3  5a 2 b  12ab 2  9b 3 answer (3) 2.2.1 3x  6 y  ax  2ay  3( x  2 y)  a( x  2 y) common factor  ( x  2 y)(3  a) answer (2) 2.2.2 3x 2  17 x  6  (3x  1)( x  6)  (3 x  1)  ( x  6) (2) Demo NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics Memo/P1 3 NW/June 2018 Grade 10 2.3.1 2 3n  2.8 n 3 2 3n 2 2.2 3( n 3)  4 3n  2 2 2 (3n  2) simplify numerator 2 3n 2 2.2 3n 2 9  2 6n 2 4 simplify  2 3n 3n 6 n.2 29 4 denominator  2 0.2 3 simplification  1 1  2 3 8 1 answer  (5) 8 2.3.2 p q 3 3 3 p  3q ( p  q)( p 2  pq  q 2 ) 3( p  q) factor sum of two    cubes 6p 2 p q 2 2 6p 2 ( p  q)( p  q) common factor p 2  pq  q 2 factor diff. of two  2 p2 squares answer (4) QUESTION 3[22] 3.1.1 3x  5  2(3x  1) 3x  5  6 x  2 removing brackets 3x  6 x  2  5 simplify  3x  3  x  1 answer (3) 3.1.2 x( x  2)  8 x2  2 x  8 removing brackets standard form x2  2 x  8  0 factors ( x  4)( x  2)  0  x  4 or x  2 both answers (4) 3.1.3 1 32 x  2  27 1 32 x  2  3 3 2x2  3 3 3  3 3 equating exponents  2 x  2  3 2 x  5 simplification 5 answer x   2 (4) Demo NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics Memo/P1 4 NW/June 2018 Grade 10 3.2 1 x 1 1    3 6 18 1  2  x 1  simplification 3 1  2 1  x  1 3 4 critical values 1  x  notation 3 graphical representation -1 1.33 (4) 3.3.1 let price of fruit yoghurt be x and of plain yoghurt be y x  y4 each equation 5 x  3 y  84 (2) 3.3.2 let fruit yoghurt be x and plain yoghurt be y x  y4 5 x  3 y  84 substitute substitution 5( y  4)  3 y  84 removing brackets 5 y  20  3 y  84 simplification 8 y  64 y 8 y 8 x  84  12  x  12 (5)  each fruit yoghurt cost R12 and each plain yoghurt cost R8 QUESTION 4[9] 4.1.1 4 7 10 pattern 3 3 next term is 13 answer  the fourth diagram will have 13 blocks Answer only: full marks (2) 4.1.2 Pattern is linear, Tn  dn  c To solve for c , substitute n  1 and T1  4 substitution 4  3(1)  c  value of c c  1 answer  Tn  3n  1 (3) Demo NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics Memo/P1 5 NW/June 2018 Grade 10 4.1.3 Tn  3n  1 70  3n  1 substitution 3n  69  n  23 answer rd The 23 diagram will have 70 blocks (2) 4.2 8888888 987654 9  2 987654 +2 (2) QUESTION 5[12] 5.1.1 C is y - intercept of f , x  0  subst. x  0  y  0  9 2 9  C( 0;9)  C( 0;9) E is y - intercept of g  y  0  3  3  E(0;3)  E(0;3) (3) 5.1.2 CE = CO + OE method = 9 + 3 = 12 units answer (2) 5.1.3 A is point of intersection  f ( x)  g ( x) equating  x2  9  x  3 x 2  x  12  0 standard form ( x  4)( x  3)  0 x  4 or x  3  x  4 y  4  3  7 each coordinate  A(-4;-7) (4) Demo NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics Memo/P1 6 NW/June 2018 Grade 10 5.2 shape increasing function asymptote below x – axis (3) QUESTION 6[6] 6.1 2 h( x )  1 x x - intercept, y  0 subst. y  0 2 0  1 x 2 1 x x2 x  2 (2) 6.2 y  1 answer (1) 6.3 x – intercept asymptotes shape (3) Demo NW/JUNE/MATH/ EMIS/6*******