General Practice Test 1 Nddc Scholarship

To determine the average for the last 10 innings, we must first calculate the total runs scored in the two different stages of the cricketer's career. First, calculate the total runs in the initial 20 innings: \[ \text{Total runs (initial)} = 20 \times 15 = 300 \text{ runs} \] Next, calculate the total runs after the additional 10 innings (making a total of 30 innings) with the new average of 17: \[ \text{Total runs (final)} = 30 \times 17 = 510 \text{ runs} \] The runs scored specifically during the last 10 innings is the difference between the final total and the initial total: \[ \text{Runs in last 10 innings} = 510 - 300 = 210 \text{ runs} \] Finally, to find the average for those 10 innings, divide the runs by the number of innings: \[ \text{Average} = \frac{210}{10} = 21 \text{ runs} \]

Petroleum production in Africa is dominated by a few key nations with vast natural reserves. Nigeria is consistently the largest producer of oil in Africa, with its economy heavily dependent on the export of crude oil from the Niger Delta. Libya follows closely as another major producer, possessing the largest proven oil reserves on the continent. While Algeria is also a significant producer, the pairing in option B correctly identifies two of the most prolific oil-exporting nations.

The terms in this sequence are the squares of integers in descending order. By observing the numbers, we can see the pattern: \[ 12^2 = 144, \; 11^2 = 121, \; 10^2 = 100 \] The missing term is therefore \( 9^2 = 81 \). The remaining terms confirm the pattern: \( 8^2 = 64 \) and \( 7^2 = 49 \).

Let \( B \) be the number of boys and \( G \) the number of girls. For Daniel: \( B - 1 = G \). For Laura: \( B = 2(G - 1) \). Substitute \( B = G + 1 \) into the second equation: \[ G + 1 = 2(G - 1) \Rightarrow G + 1 = 2G - 2 \Rightarrow G = 3 \] Then \( B = G + 1 = 4 \). So there are 4 boys and 3 girls.

The sequence follows the alphabetical order. After N comes O, which connects to the next group starting with P. So the missing letter is O.

A birthday can fall on any of the 7 days of the week, each equally likely. Therefore, the probability that a randomly selected person was born on a Sunday is \( \frac{1}{7} \). The total number of people is irrelevant.

2 typists type 2 pages in 2 minutes, so 1 typist types 1 page in 2 minutes (rate = 0.5 pages/minute). Let the required number of typists be n. Using Work = Rate × Time × Workers: \[ 18 = n \times 0.5 \times 6 \Rightarrow 18 = 3n \Rightarrow n = 6 \]

To find the final ratio, we must first determine the amount of milk and water in each sample relative to a common total volume for each. Since the quantities used from each sample are equal, let us assume the volume of each sample is the Least Common Multiple (LCM) of the sum of the ratios (3, 5, and 8), which is 120 units.
For Sample 1 (2:1): Total parts = 3. Milk = \( \frac{2}{3} \times 120 = 80 \), Water = \( \frac{1}{3} \times 120 = 40 \).
For Sample 2 (3:2): Total parts = 5. Milk = \( \frac{3}{5} \times 120 = 72 \), Water = \( \frac{2}{5} \times 120 = 48 \).
For Sample 3 (5:3): Total parts = 8. Milk = \( \frac{5}{8} \times 120 = 75 \), Water = \( \frac{3}{8} \times 120 = 45 \).
Now, sum the total milk and total water from all three samples:
Total Milk = \[ 80 + 72 + 75 = 227 \]Total Water = \[ 40 + 48 + 45 = 133 \]Thus, the resulting proportion of milk to water in the final mixture is 227:133.

This problem is a classic application of the Pigeonhole Principle. The principle states that if you have more 'pigeons' than 'pigeonholes', at least one hole must contain more than one pigeon. In this context, the 'pigeonholes' are the two available colours: black and brown. If Mr. Brown picks only 1 glove, he has 1 colour. If he picks 2 gloves, they could be one black and one brown (the worst-case scenario where no pair is formed). However, the moment he picks a 3rd glove, it must be either black or brown. Regardless of which it is, it will inevitably match one of the two gloves already picked, thereby guaranteeing a pair of the same colour. Therefore, the minimum number to ensure a pair is 3.

Let us logically deduce the clothing and hair colour of each girl based on the provided clues:
1. Emma wore a blue dress.
2. The girl in green has black hair.
3. Sharon was not wearing green or blue (Emma), and since Joy has mousey hair, Joy must be the one wearing red or green. However, we know the girl in green has black hair, and Joy has mousey hair, so Joy cannot be wearing green. This leaves Joy wearing the red dress.
4. To verify: If Joy is in red, and Emma is in blue, Sharon must be in the only remaining colour that is not green (since the clue says she isn't in green). If there are four girls (Joy, Tina, Sharon, Emma), and the black-haired girl is in green, then Tina must be the one in green (black hair). Sharon would then be the blonde girl. The blonde (Sharon) was not wearing red, which fits. Thus, Joy is the one wearing the red dress.