Higher November 2020 Paper 3 Q26
26 Prove algebraically that \(\quad 3.4\dot{7} = \dfrac{313}{90}\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Working with \(3.47\ldots\) | ||
| \(10x = 34.7\ldots\) or \(100x = 347.7\ldots\) | M1 | oe multiplication by a power of 10 eg \(1000x = 3477.7\ldots\) any letter |
| \(10x - x = 34.7\ldots - 3.47\ldots\) or \(9x = 31.3\) with \(10x = 34.7\ldots\) seen or \(100x - 10x = 347.7\ldots - 34.7\ldots\) or \(90x = 313\) with \(100x = 347.7\ldots\) and \(10x = 34.7\ldots\) seen or \(100x - x = 347.7\ldots - 3.47\ldots\) or \(99x = 344.3\) with \(100x = 347.7\ldots\) seen | M1dep | oe subtraction to eliminate recurring digits eg \(1000x - 10x = 3477.7\ldots - 34.7\ldots\) or \(990x = 3443\) with \(1000x = 3477.7\ldots\) and \(10x = 34.7\ldots\) seen numbers must all be correct |
| \(x = 3.47\ldots\) stated and M2 scored and \(9x = 31.3\) and \((x =)\ \dfrac{31.3}{9}\) and \(\dfrac{313}{90}\) or \(x = 3.47\ldots\) stated and M2 scored and \(90x = 313\) and \((x =)\ \dfrac{313}{90}\) or \(x = 3.47\ldots\) stated and M2 scored and \(99x = 344.3\) and \((x =)\ \dfrac{344.3}{99}\) and \(\dfrac{313}{90}\) | A1 | oe eg \(x = 3.47\ldots\) stated and M2 scored and \(990x = 3443\) and \((x =)\ \dfrac{3443}{990}\) and \(\dfrac{313}{90}\) |
| Alternative method 2 Working with \(0.47\ldots\) | ||
| \(10x = 4.7\ldots\) or \(100x = 47.7\ldots\) | M1 | oe multiplication by a power of 10 eg \(1000x = 477.7\ldots\) any letter |
| \(10x - x = 4.7\ldots - 0.47\ldots\) or \(9x = 4.3\) with \(10x = 4.7\ldots\) seen or \(100x - 10x = 47.7\ldots - 4.7\ldots\) or \(90x = 43\) with \(100x = 47.7\ldots\) and \(10x = 4.7\ldots\) seen or \(100x - x = 47.7\ldots - 0.47\ldots\) or \(99x = 47.3\) with \(100x = 47.7\ldots\) seen | M1dep | oe subtraction to eliminate recurring digits eg \(1000x - 10x = 477.7\ldots - 4.7\ldots\) or \(990x = 473\) with \(1000x = 477.7\ldots\) and \(10x = 4.7\ldots\) seen numbers must all be correct |
| \(x = 0.47\ldots\) stated and M2 scored and \(9x = 4.3\) and \((x =)\ \dfrac{4.3}{9}\) and \(3\dfrac{4.3}{9}\) and \(\dfrac{313}{90}\) or \(x = 0.47\ldots\) stated and M2 scored and \(90x = 43\) and \((x =)\ \dfrac{43}{90}\) and \(3\dfrac{43}{90}\) and \(\dfrac{313}{90}\) or \(x = 0.47\ldots\) stated and M2 scored and \(99x = 47.3\) and \((x =)\ \dfrac{47.3}{99}\) and \(3\dfrac{47.3}{99}\) and \(\dfrac{313}{90}\) | A1 | oe eg \(x = 0.47\ldots\) stated and M2 scored and \(990x = 473\) and \((x =)\ \dfrac{473}{990}\) and \(3\dfrac{473}{990}\) and \(\dfrac{313}{90}\) |
| Alternative method 3 Working with \(0.07\ldots\) | ||
| \(10x = 0.7\ldots\) or \(100x = 7.7\ldots\) | M1 | oe multiplication by a power of 10 eg \(1000x = 77.7\ldots\) any letter |
| \(10x - x = 0.7\ldots - 0.07\ldots\) or \(9x = 0.7\) with \(10x = 0.7\ldots\) seen or \(100x - 10x = 7.7\ldots - 0.7\ldots\) or \(90x = 7\) with \(100x = 7.7\ldots\) and \(10x = 0.7\ldots\) seen or \(100x - x = 7.7\ldots - 0.07\ldots\) or \(99x = 7.7\) with \(100x = 7.7\ldots\) seen | M1dep | oe subtraction to eliminate recurring digits eg \(1000x - 10x = 77.7\ldots - 0.7\ldots\) or \(990x = 77\) with \(1000x = 77.7\ldots\) and \(10x = 0.7\ldots\) seen numbers must all be correct |
| \(x = 0.07\ldots\) stated and M2 scored and \(9x = 0.7\) and \((x =)\ \dfrac{0.7}{9}\) and \(3.4 + \dfrac{0.7}{9}\) and \(\dfrac{313}{90}\) or \(x = 0.07\ldots\) stated and M2 scored and \(90x = 7\) and \((x =)\ \dfrac{7}{90}\) and \(3.4 + \dfrac{7}{90}\) and \(\dfrac{313}{90}\) or \(x = 0.07\ldots\) stated and M2 scored and \(99x = 7.7\) and \((x =)\ \dfrac{7.7}{99}\) and \(3.4 + \dfrac{7.7}{99}\) and \(\dfrac{313}{90}\) | A1 | oe eg \(x = 0.07\ldots\) stated and M2 scored and \(990x = 77\) and \((x =)\ \dfrac{77}{990}\) and \(3.4 + \dfrac{77}{990}\) and \(\dfrac{313}{90}\) |
Additional guidance
| \(313 \div 90 = 3.47\ldots\) | M0M0A0 |
| Alt 1 M1dep oe subtraction to eliminate recurring decimals includes \(100x - 10x = 313\) with \(100x = 347.7\ldots\) and \(10x = 34.7\ldots\) seen or \(90x = 347.7\ldots - 34.7\ldots\) with \(100x = 347.7\ldots\) and \(10x = 34.7\ldots\) seen (apply same principle in Alt 2 and Alt 3) | |
| Alt 2 equivalents for final part of A1 eg For \(3\dfrac{43}{90}\) and \(\dfrac{313}{90}\) allow \(3 + \dfrac{43}{90}\) and \(\dfrac{313}{90}\) | |
| Alt 3 equivalents for final part of A1 eg For \(3.4 + \dfrac{7}{90}\) and \(\dfrac{313}{90}\) allow \(3 + \dfrac{4}{10} + \dfrac{7}{90}\) and \(\dfrac{313}{90}\) |