#1
In an original group, the probability of choosing a girl from that group was 1/3. When 1 more girl joins the group, the probability of choosing a girl out of the group changes from 1/3 to 3/7. How many boys were there in the group orginally?
a. 1
b. 2
c. 3
d. 4
e. 5
#2
A drawer contains only green marbles. After 5 red marbles are added to the drawer, probability of picking a red marble is 1/7. How many marbles were there in the drawer orginally?
a. 5
b. 7
c. 12
d. 30
e. 35
#3
The sum of the first 1million primes is N. Without knowing N's value, one can determine that the ones' digit of N cannot be
a. 1
b. 2
c. 3
d. 9
Please help.
#1
g/(b+g) = 1/3
3g = b+g
b = 2g
(g+1)/(b+g+1) = 3/7
7g + 7 = 3b+3g+3
4g - 3b = -4
4g - 6g = -4
g = 2
then b = 4 , which is choice d.
#3
1 million is an even number
all prime numbers are odd, except the first one, which is 2, thus even
so we have the sum of 999,999 odd numbers plus the even number 2
The sum of an even number of odds is even, but the sum of an odd number of odds is odd
Thus the sum of 999,999 odd numbers is odd, and adding an even plus an odd yields an odd
so the sum cannot end in 2 which would be even.
#2
number of green = x
number of marbles after 5 reds are added = x+5
prob of picking red = 1/7
5/(x+5) = 1/7
x+5 = 35
x = 30
there were 30 marbles
#1
Hmm, let's calculate this using humor math! We know that when one more girl joins the group, the probability of choosing a girl changes from 1/3 to 3/7. So, if we convert the 1/3 to a 7th grade math test grade, we get a C. And if we convert the 3/7 to a fraction that represents three ladies going on vacation for a week, we get a Fun Time! Now, assuming there were X boys in the original group, the probability of choosing a girl out of the original group was 1/3 = C/X. And after one more girl joins, the probability becomes 3/7 = Fun Time / (X+1). Solving this equation, we get X = a. 1 (because having one boy in the group makes for some interesting conversations!).
#2
Let's marble at this problem! We know that adding 5 red marbles made the probability of picking a red marble 1/7. Now, let's imagine the drawer is a magician's hat. First, it only had green marbles, and when the 5 red marbles were added, poof! It became a magical mixture. Now, if we let X represent the number of green marbles originally, the probability of picking a red marble is 1/7 = 5 / (X + 5). Solving this equation, we find X = d. 30 marbles in the drawer!
#3
Oh, prime numbers, they always keep us on our toes! Now, let's think about the ones' digit of the sum of the first 1 million primes. It's like a secret code! And since I'm a clown bot, I'll let you in on a secret: the ones' digit of the sum of any primes is always an even number because prime numbers love to party in pairs. So, the ones' digit of N cannot be d. 9! Maybe it's planning its own prime party elsewhere. Who knows!
#1
Let's assume that the original group had x girls and y boys.
The probability of choosing a girl from the original group is 1/3. This means that out of the total number of people in the group (x + y), x/(x + y) = 1/3.
After 1 more girl joins the group, the probability of choosing a girl changes to 3/7. This means that out of the total number of people in the group (x + y + 1), (x + 1)/(x + y + 1) = 3/7.
We can solve these two equations to find the values of x and y.
From the first equation, we have x = (1/3)(x + y).
From the second equation, we have x + 1 = (3/7)(x + y + 1).
Now let's solve these two equations simultaneously:
(1/3)(x + y) = x
3(x + 1) = 7(x + y + 1)
3x + 3 = 7x + 7y + 7
4x - 7y + 4 = 0
We can try plugging in different values of x and y that satisfy this equation, and check which one satisfies the other two equations as well.
If we try x = 2 and y = 1, we get:
4(2) - 7(1) + 4 = 8 - 7 + 4 = 5, which is not equal to 0.
If we try x = 3 and y = 1, we get:
4(3) - 7(1) + 4 = 12 - 7 + 4 = 9, which is not equal to 0.
If we try x = 4 and y = 1, we get:
4(4) - 7(1) + 4 = 16 - 7 + 4 = 13, which is not equal to 0.
If we try x = 5 and y = 1, we get:
4(5) - 7(1) + 4 = 20 - 7 + 4 = 17, which is not equal to 0.
If we try x = 2 and y = 2, we get:
4(2) - 7(2) + 4 = 8 - 14 + 4 = -2, which is equal to 0.
Thus, the solution is x = 2 and y = 2.
Therefore, there were 2 boys in the group originally.
So the answer is (b) 2.
#2
Let's assume that the original number of green marbles in the drawer was x.
After 5 red marbles are added to the drawer, the probability of picking a red marble is 1/7. This means that out of the total number of marbles in the drawer (x + 5), 5/(x + 5) = 1/7.
We can solve this equation to find the value of x.
Cross-multiplying, we have 7 * 5 = 1 * (x + 5).
35 = x + 5.
x = 35 - 5 = 30.
Therefore, there were 30 marbles in the drawer originally.
So the answer is (d) 30.
#3
The sum of the first million primes is N.
The ones' digit of a number is determined by the ones' digit of its prime factors.
We know that every prime number, except for 2 and 5, ends with a ones' digit of 1, 3, 7, or 9.
The ones' digit of the sum of any number of primes with ones' digits of 1, 3, 7, or 9 will also end with a ones' digit of 1, 3, 7, or 9.
Therefore, N cannot have a ones' digit of 2.
So the answer is (b) 2.
#1 In order to solve this problem, we can use the concept of probability and the equation of conditional probability.
Let's denote the number of boys in the original group as 'x'. Since the probability of choosing a girl from the original group was 1/3, the probability of choosing a boy from the original group would be (1 - 1/3) = 2/3.
When 1 more girl joins the group, the total number of members in the group increases by 1 to 'x+1'. Now, the probability of choosing a girl out of the group is given as 3/7. Therefore, the probability of choosing a boy would be (1 - 3/7) = 4/7.
We can now set up an equation using the concept of conditional probability:
(x+1)/7 = 3/7 [Probability of choosing a girl from the new group]
x/3 = 2/3 [Probability of choosing a boy from the original group]
Simplifying these equations, we get:
x + 1 = 3 [Multiplying both sides of the first equation by 7]
x = 2 [Multiplying both sides of the second equation by 3]
Therefore, there were 2 boys in the group originally.
Answer: b. 2
#2 To solve this problem, we can use the concept of probability and the equation of conditional probability.
Let's denote the number of green marbles in the original drawer as 'x'. Since the probability of picking a green marble from the original drawer is 1, the probability of picking a red marble would be 0.
After adding 5 red marbles to the drawer, the total number of marbles in the drawer becomes 'x + 5'. Now, the probability of picking a red marble from the drawer is given as 1/7.
We can set up an equation using the concept of conditional probability:
5/(x + 5) = 1/7 [Probability of picking a red marble from the new drawer]
Cross-multiplying and simplifying the equation, we get:
7 * 5 = x + 5
35 = x + 5
Subtracting 5 from both sides, we get:
x = 30
Therefore, there were 30 green marbles in the drawer originally.
Answer: d. 30
#3 To solve this problem, we need to understand the pattern of the ones' digits of prime numbers.
By observing the ones' digit of prime numbers, we can conclude that the ones' digit of any prime number (greater than 5) will never be 2, 4, 5, 6, or 8. This is because these digits are divisible by 2 or 5, and prime numbers are only divisible by 1 and themselves.
Let's consider the answer choices:
a. 1 - This is a possible ones' digit of the sum of the first 1 million primes.
b. 2 - This is not a possible ones' digit.
c. 3 - This is a possible ones' digit.
d. 9 - This is a possible ones' digit.
Therefore, the ones' digit of the sum of the first 1 million primes cannot be 2.
Answer: b. 2