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?
Pls help me
smallest 2 digit number divisible by 4 is 12
so we want:
12, 16, 20, 24, 28, 32, 36, 40,.... ,96
this is an arithmetic sequence with
a = 12, d=4
what term number is 96 ?
a + (n-1)d = 96
12 + (n-1)(4) = 96
4n - 4 = 84
4n = 88
n = 22
There would be 22 terms divisible by 4
BUT, we could not have formed 20, 40, 60 and 80
so there are 18 such terms
#2
number of ways with no restrictions
= 4! = 24
consider the 2 girls a single entity
number of ways to arrange with the girls side by side = 3! = 6
So the number of ways the two girls are not together = 18
14 i think
Sure, I can help you with these questions. Let's go through each one step by step.
1. To find the number of 2-digit numbers formed by using the digits 1 to 9 without repetition that are divisible by 4, we need to determine the conditions for divisibility by 4. A number is divisible by 4 if the last two digits of the number are divisible by 4.
The possible 2-digit numbers formed by using the digits 1 to 9 without repetition are: 12, 13, 14, 15, 16, 17, 18, 19, 21, 23, 24, 25, 26, 27, 28, 29, 31, 32, 34, 35, 36, 37, 38, 39, 41, 42, 43, 45, 46, 47, 48, 49, 51, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 64, 65, 67, 68, 69, 71, 72, 73, 74, 75, 76, 78, 79, 81, 82, 83, 84, 85, 86, 87, 89, 91, 92, 93, 94, 95, 96, 97, 98.
Out of these numbers, the ones divisible by 4 are: 12, 16, 24, 28, 32, 36, 42, 48, 52, 56, 64, 68, 72, 76, 84, 88, 92, 96.
So there are 18 two-digit numbers that can be formed using the digits 1 to 9 without repetition that are divisible by 4.
2. To find the number of possible arrangements where 2 boys and 2 girls are arranged in a row such that the girls are not next to each other, we need to consider the different possibilities.
Let's denote the boys as B1 and B2, and the girls as G1 and G2.
The possible arrangements are:
- B1 B2 G1 G2
- B1 G1 B2 G2
- B1 G1 G2 B2
- G1 B1 B2 G2
- G1 B1 G2 B2
- G1 G2 B1 B2
So, there are 6 possible arrangements where the girls are not next to each other.
3. To find the percentage of candidates who cleared one part but failed to clear the other in a selection test with parts A and B, we need to use the information given.
Let's assume the total number of candidates is 100.
75% of the candidates cleared part A, which means 75 candidates cleared part A.
60% of the candidates cleared part B, which means 60 candidates cleared part B.
20% of the candidates failed to clear either part, which means 20 candidates failed to clear both parts.
So, the total number of candidates who cleared at least one part is the sum of the candidates who cleared each part:
75 (part A) + 60 (part B) - 20 (both parts) = 115.
To find the percentage, divide the number of candidates who cleared one part but failed to clear the other (115 - 75 - 60 = 145) by the total number of candidates who cleared at least one part (115) and multiply by 100.
So, the percentage of candidates who cleared one part but failed to clear the other is (145/115) * 100 = 126.09%.
I hope this helped! Let me know if you have any more questions.