Foundation June 2017 Paper 1 Q13
13
\[2 + 0 + 1 + 7 = 10\]Make the following calculations correct.
Use only the symbols \(\quad +,\ \ -,\ \ \times,\ \ \div\ \) and \(\ (\ )\) [3 marks]
\[\begin{array}{ccccccc} 2 & \qquad & 0 \qquad & 1 \qquad & 7 \qquad & = & -4 \\[4pt] 2 & \qquad & 0 \qquad & 1 \qquad & 7 \qquad & = & 0 \\[4pt] 2 & \qquad & 0 \qquad & 1 \qquad & 7 \qquad & = & 2^4 \end{array}\]| Answer | Mark | Comments |
|---|---|---|
| \(2 + 0 + 1 - 7 = -4\) or \(2 - 0 + 1 - 7 = -4\) | B1 | |
| \(2 \times 0 \times 1 \times 7 = 0\) or \(2 \times 0 \div 1 \times 7 = 0\) or \(2 \times 0 \times 1 \div 7 = 0\) or \(2 \times 0 \div 1 \div 7 = 0\) or \(2 \times 0 \times (1 + 7) = 0\) or \(2 \times 0 \div (1 + 7) = 0\) | B1 | Allow any brackets in pairs for first four Allow – instead of + for last two |
| \((2 + 0) \times (1 + 7) = 2^4\) or \((2 - 0) \times (1 + 7) = 2^4\) or \(2 \times (0 + 1 + 7)\) | B1 |
Additional guidance
| In all cases, allow extra pairs of brackets which do not alter the result of the calculation eg in 3rd calculation \(\quad ((2 + 0) \times (1 + 7)) = 2^4\) | B1 |
| Brackets can be used in the place of a multiplication sign eg in 2nd calculation \(\quad 2 \times 0(1 + 7) = 0\) | B1 |
| Each gap must have a bracket or an operator in | |
| Allow additional + or - signs in any gap, if correct eg in 1st calculation \(\quad 2 + 0 + 1 + -7 = -4\) | B1 |