Higher June 2018 Paper 2 Q27
27 Prove algebraically that \(\; 2.7\dot{5} \;\) converts to the fraction \(\dfrac{124}{45}\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Working with 2.75….. | ||
| \(10x = 27.5\)… or \(100x = 275.5\)… | M1 | oe multiplication by a power of 10 eg \(1000x = 2755.5\)… any letter |
| \(10x - x = 27.5\)… \(-\) 2.75… or \(9x = 24.8\) with \(10x = 27.5\)… seen or \(100x - 10x = 275.5\)… \(-\) 27.5… or \(90x = 248\) with \(100x = 275.5\)… and \(10x = 27.5\)… seen or \(100x - x = 275.5\)… \(-\) 2.75… or \(99x = 272.8\) with \(100x = 275.5\)… seen | M1dep | oe subtraction to eliminate recurring digits eg \(1000x - 10x = 2755.5\)… \(-\) 27.5… or \(990x = 2728\) with \(1000x = 2755.5\)… and \(10x = 27.5\)… seen numbers must all be correct |
| \(x = 2.75\)… stated and M2 scored and \(9x = 24.8\) and \(x = \dfrac{24.8}{9} = \dfrac{124}{45}\) or \(x = 2.75\)… stated and M2 scored and \(90x = 248\) and \(x = \dfrac{248}{90} = \dfrac{124}{45}\) or \(x = 2.75\)… stated and M2 scored and \(99x = 272.8\) and \(x = \dfrac{272.8}{99} = \dfrac{124}{45}\) | A1 | oe eg \(x = 2.75\)… stated and M2 scored and \(990x = 2728\) and \(x = \dfrac{2728}{990} = \dfrac{124}{45}\) |
| Alternative method 2 Working with 0.75….. | ||
| \(10x = 7.5\)… or \(100x = 75.5\)… | M1 | oe multiplication by a power of 10 eg \(1000x = 755.5\)… any letter |
| \(10x - x = 7.5\)… \(-\) 0.75… or \(9x = 6.8\) with \(10x = 7.5\)… seen or \(100x - 10x = 75.5\)… \(-\) 7.5… or \(90x = 68\) with \(100x = 75.5\)… and \(10x = 7.5\)… seen or \(100x - x = 75.5\)… \(-\) 0.75… or \(99x = 74.8\) with \(100x = 75.5\)… seen | M1dep | oe subtraction to eliminate recurring digits eg \(1000x - 10x = 755.5\)… \(-\) 7.5… or \(990x = 748\) with \(1000x = 755.5\)… and \(10x = 7.5\)… seen numbers must all be correct |
| \(x = 0.75\)… stated and M2 scored and \(9x = 6.8\) and \(x = \dfrac{6.8}{9}\) and \(2\dfrac{6.8}{9} = \dfrac{124}{45}\) or \(x = 0.75\)… stated and M2 scored and \(90x = 68\) and \(x = \dfrac{68}{90}\) and \(2\dfrac{68}{90} = \dfrac{124}{45}\) or \(x = 0.75\)… stated and M2 scored and \(99x = 74.8\) and \(x = \dfrac{74.8}{99}\) and \(2\dfrac{74.8}{99} = \dfrac{124}{45}\) | A1 | oe eg \(x = 0.75\)… stated and M2 scored and \(990x = 748\) and \(x = \dfrac{748}{990}\) and \(2\dfrac{748}{990} = \dfrac{124}{45}\) |
| Alternative method 3 Working with 0.05….. | ||
| \(10x = 0.5\)… or \(100x = 5.5\)… | M1 | oe multiplication by a power of 10 eg \(1000x = 55.55\)… any letter |
| \(10x - x = 0.5\)… \(-\) 0.05… or \(9x = 0.5\) with \(10x = 0.5\)… seen or \(100x - 10x = 5.5\)… \(-\) 0.5… or \(90x = 5\) with \(100x = 5.5\)… and \(10x = 0.5\)… seen or \(100x - x = 5.5\)… \(-\) 0.05… or \(99x = 5.5\) with \(100x = 5.5\)… seen | M1dep | oe subtraction to eliminate recurring digits eg \(1000x - 10x = 55.5\)… \(-\) 0.5… or \(990x = 55\) with \(1000x = 55.5\)… and \(10x = 0.5\)… seen numbers must all be correct |
| \(x = 0.05\)… stated and M2 scored and \(9x = 0.5\) and \(x = \dfrac{0.5}{9}\) and \(2.7 + \dfrac{0.5}{9} = \dfrac{124}{45}\) or \(x = 0.05\)… stated and M2 scored and \(90x = 5\) and \(x = \dfrac{5}{90}\) and \(2.7 + \dfrac{5}{90} = \dfrac{124}{45}\) or \(x = 0.05\)… stated and M2 scored and \(99x = 5.5\) and \(x = \dfrac{5.5}{99}\) and \(2.7 + \dfrac{5.5}{99} = \dfrac{124}{45}\) | A1 | oe eg \(x = 0.05\)… stated and M2 scored and \(990x = 55\) and \(x = \dfrac{55}{990}\) and \(2.7 + \dfrac{55}{990} = \dfrac{124}{45}\) |
Additional guidance
| \(124 \div 45 = 2.75\)… | M0M0A0 |
| Alt 1 M1dep oe subtraction to eliminate recurring decimals includes \(100x - 10x = 248\) with \(100x = 275.5\)… and \(10x = 27.5\)… seen or \(90x = 275.5\)… \(-\) 27.5… with \(100x = 275.5\)… and \(10x = 27.5\)… seen (apply same principle in Alts 2 and 3) | |
| Alt 2 equivalents for final part of A1 eg For \(2\dfrac{68}{90} = \dfrac{124}{45}\) allow \(2 + \dfrac{68}{90} = \dfrac{124}{45}\) | |
| Alt 3 equivalents for final part of A1 eg For \(2.7 + \dfrac{5}{90} = \dfrac{124}{45}\) allow \(2 + \dfrac{7}{10} + \dfrac{5}{90} = \dfrac{124}{45}\) |