Need help with several problems

[Dreamka]

1)R is a set of positive integers less then 50 and S is the set of squeres of the integers in R. How many elements does the interesection of R and S contains?
none
2
4
5
7

The correct answer is 4, but i relaly have no idea how to comme up with this answer.

2) A(2,3,4,5)
B ( 4,5,6,7,8)
Two integers will be randomly selected from the sets above. One integer from the set A and one from the set B. What is the probability that the sum of those 2 integers will wqual 9 ??
0.15
0,20
0.25
0.30
0,33

[Otori Takeo]

1)

less-than-50-squared intergers are
1, 4, 9, 16, 25, 36, and 49.
thus = 7

2)

For normal circumstance, 1 from A = 4C1 = 4, the others from B = 5C1 = 5
Thus = 4 * 5 = 20 cases

sum of two will be equal to 9
then

if you choose 2 from set 1, you must choose 7 = 1
3,6 = 1
4,5 = 1
5,4 = 1
Thus = 4 cases

Prob = 4/20 = 1/5 = 0.2

[Dreamka]

Thank you, but to eb honest it is still not clrar for both problems.
In the first problem, the correct answer is 4 , but the think is i dont really understand what they mean by saying "intersection"
About second problem, i would really appriciate if you explain it more detailed.

And here is one more problem that i can not solve

In the increasing sequesnce of 10 consecutive integers the sum of the first 5 integers is 560 what is the sum of the last 5 integers in the sequence???
585
580
575
570
565
( correct answer is 585)

[Otori Takeo]

the intersected members means the members that are in both sets.

For example

A = (1,2,3,4,5,6)
B = (1,5,7,11,12)
the intersected members of these two sets are (1,5)

When we apply this principle to the 1st answer

R is the set that contains (1,2,......,49)
S is the set that contains (1,4,9,16,.......)

thus the intersection or intersected member are only

1 = 1^2
4 = 2^2
9 = 3^2
16 = 4^2
.
.
.
49 = 7^2
thus 7 is the answer.

2nd problem

All probabilities that you can select two numbers one from Set A and the others from Set B.

1st number = 4C1 = 4
2nd number = 5C1 = 5

Thus all the probabilities = 4x5 = 20

However according to the requirement that both the sum of both numbers should be nine

How many cases that the sum of both numbers can be nine?

(2,7)
(3,6)
(4,5)
(5,4)

Hence, there are four cases.

Therefore the prob = 4/20 = 1/5 ANS

[Otori Takeo]

Originally Posted by Dreamka

And here is one more problem that i can not solve

In the increasing sequesnce of 10 consecutive integers the sum of the first 5 integers is 560 what is the sum of the last 5 integers in the sequence???
585
580
575
570
565
( correct answer is 585)

There are many approaches to nail this problem.

Ans = 560 +1 +2 +3 + 4 + 5 = 585

or 560 / 5 = 112 is the 3rd number in the first 5 numbers.

thus you can come up with the later 5 numbers

110, 111, 112, 113, 114 , 115, 116, 117, 118, 119

[larbgai]

1)R is a set of positive integers less then 50 and S is the set of squeres of the integers in R. How many elements does the interesection of R and S contains?
none
2
4
5
7

First of all, this should read (R is THE set of positive integers less than 50).

Anyway, here is how I solved it.

R = 1,2,3,4,5,6,etc,......50
S = 1^2, 2^2, 3^2, 4^2, etc
so... S = 1,4,9,16,25,36,49,64,81,etc.......

Since R only goes up to 50, we want to disregard everything in S that is below 50.

I already listed all the relevant numbers in S, and then some, above. You can count the numbers below 50 (there are 7). Those will overlap with the numbers in R. The answer is 7.


2) A(2,3,4,5)
B ( 4,5,6,7,8)
Two integers will be randomly selected from the sets above. One integer from the set A and one from the set B. What is the probability that the sum of those 2 integers will wqual 9 ??
0.15
0,20
0.25
0.30
0.33

2+7 = 9
3+6 = 9
4 + 5 = 9
5 + 4 = 9

Clearly, for each number in A, there is only one way to get 9. Each of the numbers required to get to 9 (7,6,5,4), resides in set B.

The total possible combinations of numbers = membersA * membersB = 4 * 5 = 20

The total number of ways to add to 9 = 4 (all listed above).

4/20 = 1/5 = .2

Answer is B.


In the increasing sequesnce of 10 consecutive integers the sum of the first 5 integers is 560 what is the sum of the last 5 integers in the sequence???
585
580
575
570
565

Let's call the first integer x. Since we are dealing with consecutive integers, the first five consecutive integers are:

x, x+1, x+2, x+3, x+4, x+5

Add these together, and you get 5x+10

5x + 10 = 560
5x = 550
x = 110

So our starting number is 10. We then want the next 5 numbers in our sequence. The last number in the first 5 was 110+4 = 114. So the next 5 numbers will be 115+116+117+118+119

Or, a shorter method:
x+5+x+6+x+7+x+8+x+9 = 5x + 35 => 110*5 + 35 = 550 + 35 = 585

Answer is A.

[Dreamka]

Thanks, no it is clear.

[AC Milan]

Originally Posted by Dreamka

1)R is a set of positive integers less then 50 and S is the set of squeres of the integers in R. How many elements does the interesection of R and S contains?
none
2
4
5
7

The correct answer is 4, but i relaly have no idea how to comme up with this answer.

2) A(2,3,4,5)
B ( 4,5,6,7,8)
Two integers will be randomly selected from the sets above. One integer from the set A and one from the set B. What is the probability that the sum of those 2 integers will wqual 9 ??
0.15
0,20
0.25
0.30
0,33

The First problem is POSITIVE INTERGERS meaning R: 0, 2 4 6 8, ....
7 is the cut off line since 7^2= 49
but its positive numbers so 0, 2, 4, 6 can only be in the intersection. which is 4 numbers and therefore B is the correct answer

in the second problem, the answers given are all correct

[Dreamka]

AC Milan
Can you explain it more detailed. I dont understand why we dong count 7 and 1? It seem to be positive too, for me .... kind of confused

[abce_tor]

Originally Posted by AC Milan

The First problem is POSITIVE INTERGERS meaning R: 0, 2 4 6 8, ....
7 is the cut off line since 7^2= 49
but its positive numbers so 0, 2, 4, 6 can only be in the intersection. which is 4 numbers and therefore B is the correct answer

in the second problem, the answers given are all correct

It does not say positive even integers...
positive integers means: all integers greater than zero :: means 1, 2, 3, 4...