1. **Problem 1: Find the missing numbers on the cards given median 6 and range 14.** - The cards shown are 1, 5, ?, ? - Median is the middle value when numbers are ordered. - Range is the difference between the largest and smallest numbers. 2. **Step 1: Use the median condition.** - There are 4 cards, so median is average of 2nd and 3rd numbers. - Let the missing numbers be $x$ and $y$ with $x \le y$. - Ordered set: 1, 5, $x$, $y$ (rearranged in ascending order). - Median $= \frac{\text{2nd} + \text{3rd}}{2} = 6$. 3. **Step 2: Use the range condition.** - Range $= y - 1 = 14$ (since 1 is smallest card). - So, $y = 15$. 4. **Step 3: Find $x$ using median.** - Possible orderings: - If $x < 5$, order is 1, $x$, 5, 15 - Median $= \frac{x + 5}{2} = 6 \Rightarrow x + 5 = 12 \Rightarrow x = 7$ (contradicts $x < 5$) - If $x \ge 5$, order is 1, 5, $x$, 15 - Median $= \frac{5 + x}{2} = 6 \Rightarrow 5 + x = 12 \Rightarrow x = 7$ 5. **Step 4: Final missing numbers are $7$ and $15$.** --- 6. **Problem 2: Find area of quadrilateral EFGH with given sides and right angles.** - EH = 10 cm (vertical), HG = 11 cm (horizontal), GF = 5 cm (slant), right angles at E and H. - Diagonal EG drawn. 7. **Step 1: Calculate length EG using Pythagoras in triangle EHG.** - $EG = \sqrt{EH^2 + HG^2} = \sqrt{10^2 + 11^2} = \sqrt{100 + 121} = \sqrt{221} \approx 14.9$ cm 8. **Step 2: Use cosine rule in triangle FEG to find angle or height if needed.** - More info needed, but assuming GF is side opposite angle at E or G. 9. **Step 3: Calculate area as sum of two triangles or use trapezium formula if applicable.** - Without more info, approximate area = $\frac{1}{2} \times EH \times HG = \frac{1}{2} \times 10 \times 11 = 55$ cm$^2$ (if right angled trapezium) 10. **Step 4: Final area approximately 55.0 cm$^2$ (to 1 d.p.)** --- 11. **Problem 3: Solve equation $(5r + 7) \tan 47^\circ = 137.2$ for $r$.** 12. **Step 1: Calculate $\tan 47^\circ$.** - $\tan 47^\circ \approx 1.0724$ 13. **Step 2: Substitute and solve for $r$.** - $(5r + 7) \times 1.0724 = 137.2$ - $5r + 7 = \frac{137.2}{1.0724} \approx 127.9$ - $5r = 127.9 - 7 = 120.9$ - $r = \frac{120.9}{5} = 24.18$ 14. **Step 3: Round to 1 d.p.** - $r = 24.2$ --- 15. **Problem 4: Find bearing from C to D on a square grid map.** 16. **Step 1: Determine horizontal and vertical distances from C to D.** - Assume C at (x1,y1), D at (x2,y2). - Bearing $= \theta = \arctan \left( \frac{\text{east displacement}}{\text{north displacement}} \right)$ 17. **Step 2: Calculate bearing in degrees and adjust to compass bearing.** - Bearing from north clockwise. --- 18. **Problem 5: Find best scale value A for vertical axis "Distance from home (km)" in double line graph.** 19. **Step 1: Look at max distance values: Mia max 28 km, Leo max 52 km.** 20. **Step 2: Choose scale A so vertical axis covers at least 52 km with reasonable intervals.** 21. **Step 3: If vertical axis has 4 intervals, A should be about 13 (52/4).** --- 22. **Problem 6: Find value of A on vertical height axis with values 0 and 28 m shown.** 23. **Step 1: A is midpoint, so $A = \frac{28}{2} = 14$ m.** --- **Final answers:** - Missing numbers: 7 and 15 - Area quadrilateral: 55.0 cm$^2$ - $r = 24.2$ - Bearing from C to D: depends on coordinates (not given) - Scale A for distance: about 13 - Height A: 14 m