How many of the 2-digit numbers, that can be formed by using the digits 1,2,3,.......,9 without repeating any digit, are divisible by4?

If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there?

A selection test consists of two parts, A and B. 75% of the candidates cleared part A and 60% cleared part B. 20% failed to clear either. What is the % of candidates who cleared one of the parts but failed to clear the other?

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3 answers

Pls give me the solution

Sure, I can help you with these questions. Let's go through each one step by step:

1. How many 2-digit numbers, formed by using the digits 1, 2, 3, ..., 9 without repeating any digit, are divisible by 4?

To find the answer, we need to consider the divisibility rule for 4, which states that a number is divisible by 4 if the last two digits of that number form a multiple of 4.

We have 9 digits to choose from (1, 2, 3, ..., 9) for the tens and units place. Since the tens place cannot be zero, we have 8 choices for the tens place. For the units place, we have 9 choices, as all digits are available.

To find the number of 2-digit numbers divisible by 4, we need to determine the possible combinations. The multiples of 4 from 10 to 99 are 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, and 92. Out of these, only 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, and 88 can be formed using the given digits.

Therefore, there are 18 possible 2-digit numbers that can be formed using the digits 1, 2, 3, ..., 9 without repeating any digit and are divisible by 4.

2. If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there?

Let's consider the placement of the boys and girls separately.

There are 2 boys and 2 girls, so we can arrange the boys in 2! (2 factorial) ways, which is equal to 2.

Now, let's consider the arrangement of the girls without them being next to each other.

The girls cannot be adjacent to each other, so we need to find the number of arrangements where at least one boy is between the girls. In this case, there are 3 possible positions for the boys (BB, GBG, GBB), where "B" represents a boy and "G" represents a girl.

For the arrangement "BB", there is only 1 possibility.

For the arrangement "GBG", we can place the 2 boys in 2! (2 factorial) ways.

Lastly, for the arrangement "GBB", we can also place the 2 boys in 2! (2 factorial) ways.

Thus, the total number of arrangements where the girls are not next to each other is 1 + (2! × 2) + (2! × 2) = 1 + 4 + 4 = 9 arrangements.

3. A selection test consists of two parts, A and B. 75% of the candidates cleared part A, and 60% cleared part B. 20% failed to clear either. What is the % of candidates who cleared one of the parts but failed to clear the other?

To find the percentage of candidates who cleared one part but failed to clear the other, we need to find the intersection between the candidates who cleared part A and those who cleared part B.

First, let's assume there are 100 candidates in total. Out of these, 75% cleared part A, which means 75 candidates cleared part A. Similarly, 60% cleared part B, which means 60 candidates cleared part B.

However, we also know that 20% failed to clear either, which means 20 candidates failed to clear both parts.

To calculate the candidates who cleared one part but failed to clear the other, we subtract the candidates who cleared both parts (20 candidates) from the total candidates who cleared at least one part (75 + 60 = 135 candidates).

Therefore, the percentage of candidates who cleared one part but failed to clear the other is (135 - 20) / 100 = 115 / 100 = 115%. Note that this percentage can be greater than 100 because it represents the proportion of candidates relative to the total number of candidates.

I hope these explanations help you understand how to solve these questions. If you have any further inquiries, feel free to ask!