**UCAT** **Question Analysis - Quantitative Reasoning Question 17**

A field bordering a forest is being regularly used by hunters to shoot game. Its characteristics are as follows:

**Q17.1 What is the area of the field?**

a. 30,101m^2

b. 32,919m^2

c. 36,000m^2

d. 39,370m^2

e.43,055m^2

**Q17.2 The field's owner is nearing retirement and wants to share the field between his four sons. He wishes to allocate:**

**• 1/6 of the field to his first son • 2/7 of the field to his second son • 1/5 of the field to his third son • The remainder to his fourth son.**

**What is the area of the portion of the field allocated to the fourth son?**

a. 12,514yd^2

b. 15,895yd^2

c. 17,657yd^2

d. 18,343yd^2

e. 23,486yd^2

**Q17.3 Two hunters have killed 14 animals in total (pheasants & rabbits). We are told that:**

**• Hunter A killed twice as many rabbits as Hunter B • Hunter A killed three times fewer pheasants than Hunter B • Hunter A killed 2 pheasants.**

**How many rabbits did Hunter B kill?**

a.2

b.3

c.4

d.6

e. 8

**017.4 The field's owner wants to use the field to plant potatoes in rows parallel to the AC edge of the field. The first row will be on line AC and the last row will be on line BD. In-between, he will plant one row every 6 inches. Given that there are 36 inches in a yard. how many rows of potatoes will he be able to plant?**

a. 900

b. 901

c. 1440

d. 1441

e.5400

**Q17.5 The field owner runs along the edge of his field every morning in order to exercise. How long does it take him to run along the full perimeter of the field once, assuming that he runs at a speed of 5 km/h?**

a. 5min 35s

b. 8min 34s

c. 8min 56s

d. 9 min 22s

e. 10min 14s

**Q17.6 The field's owner Is trying to keep hunters out of the field by installing an electric fence. However. for legal reasons, he can only install the fence one metre inside the field along the entire perime-ter. What will be the perimeter of the fence (to the nearest metre)?**

a. 705 m

b. 706 m

c. 709 m

d. 713 m

e. 714 m

**Q17.7 Two hunters walk along the edge of the field. towards the forest. Hunter A walks clockwise from point A. He walks at 60 yd/min. Hunter B walks anticlockwise from point B at a non-specified speed. They meet along the edge of the forest after exactly 5 minutes. At what speed was Hunter B walking?**

a.50 yd/min

b.55 yd/min

c.56 yd/min

d.65 yd/min

e.66 yd/min

**Answer and Explanation**

**Q17.1 — a: 30,101 m^2**

The area of the field is calculated as: (240 / 1.0936) x (150 / 1.0936).

**Q17.2 — a: 12,514 yd^2**

(1 —1/6 — 2/7 —1/5) x (240 x 150) = 12,514 yd2

**Q17.3 — a: 2**

If Hunter A killed 2 pheasants then, since this is 3 times fewer than Hunter B, Hunter B has killed 6 pheasants, making a total of 8 pheasants killed.

Since they killed a total of 14 animals, this means they killed 6 rabbits be-tween them.

We are told that Hunter A killed twice as many rabbits as Hunter B, there-fore Hunter A killed 4 rabbits and Hunter B killed 2 rabbits.

**Q17.4 — b: 901 rows**

This is a simple interval problem. The trap to avoid is not to miss the final row. In 1 yard (i.e. 36 inches) we have 6 intervals of 6 inches each. In 150 yards, we therefore have 900 intervals of 6 inches each. The number of rows (i.e. lines delimiting the intervals) is therefore 901.

**Q17.5 — b: 8min 34s**

The full perimeter of the field measures (150 + 240) x 2 = 780 yd. This is equal to 780/1.0936 = 713.24 m or 0.71324 km. At a speed of 5 km/h, he will run the distance in 0.71324 / 5 = 0.14266 hours, i.e. x 60 = 8.559 minutes. This corresponds to 8 minutes and 0.559 x 60 = 34 seconds.

**Q17.6 — a: 705m**

The length of the field is 240 / 1.0936 = 219.459 m. The length of fence is therefore 2 metres less (1 metre on each side), i.e. 217.459 m

The width of the field is 1501 1.0936 = 137.162 m. The length of fence is therefore 2 metres less (1 metre on each side) i.e. 135.162 m. The perime-ter of the fence is therefore 2 x (217.459 + 135.162) = 705.242 m.

**Q17.7 — e: 66 yd/min**

At the time they meet, the combined distance they will have walked is: 240 + 150 + 240 = 630 yards. Hunter A will have walked 5 x 60 = 300 yards. This means that Hunter B will have walked 630 — 300 = 330 yards during those same 5 minutes. His speed is therefore 330 / 5 — 86 yd/min.

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