A round archery target of diameter 1 m is marked with four scoring regions from the centre outwards as red, blue, yellow and white. The radius of the red band is 0.20 m. The width of all the remaining bands is equal. If archers throw arrows towards the target, what is the probability that the arrows fall in the red region of the archery target ?

Okay, this looks like one of those questions where there's a simple way hiding behind the details. The question asks for the probability of hitting the red region. For a target like this, probability is simply the ratio of the area you are aiming for to the total area. So, Probability = (Area of Red Region) / (Total Area of Target). As we know from basic math, the area of a circle is "r "r "². Because the constant 'pi' is in both the numerator and denominator, it cancels out. This means we don't need to do any complex calculations with 3.14. This is a common shortcut in aptitude tests. The problem just becomes a ratio of the squares of the radii: Probability = (radius of red region) "² / (radius of total target) "². 1. **Get the radii:** - Radius of the red region is given: 0.20 m. - Total diameter of the target is 1 m, so its radius is half of that: 0.50 m. 2. **Form the ratio of the radii:** - Ratio = (0.20) / (0.50). This is the same as 20/50, which simplifies to 2/5. Being good with simplifying ratios is a key skill tested in papers from previous years. [15] 3. **Square the ratio:** - The probability is the square of this ratio: (2/5) "² = 4/25. 4. **Convert to decimal:** - To turn 4/25 into a decimal, we can multiply the top and bottom by 4 to make the denominator 100. So, (4 * 4) / (25 * 4) = 16/100, which is 0.16. The information about the blue, yellow, and white bands is extra information, a common pattern in CSAT questions where unnecessary details are added to test if you can focus on what's important, as seen in a 2015 question about average speed where the distance was irrelevant. [32] Checking the options, 0.16 matches option (C).