There are 51 students in a certain class. The number of females is 5 less than 3 times the number of males. find the number of females and the number of males in the class.
Find three consecutive integers such that the sum of the smallest and largest integer is equal to twice the middle number
number of males --- x
number of females --- 3x-5
Can you form the equation from here and solve it?
Please do not post a new question as a reply to somebody else's problem.
Tutors might think it has been answered and not look at it.
I will set it up for you...
1st integer ---- x
2nd integer ----x+1
3rd integer ----x + 2
"smallest plus largest equals twice the middle"
x + (x+2) = 2(x+1)
solve....
ahhh, glancing at this equation it is obvious that you will get
0=0
that means that it will be true for all consecutive even integers.
e.g. let's try 4,5,6
is 4+6 = 2(5) ? YES
To find the number of females and males in the class, let's denote the number of males as "M" and the number of females as "F".
According to the given information, the total number of students in the class is 51, so we can write the equation:
M + F = 51 (Equation 1)
We also know that the number of females is 5 less than 3 times the number of males, which can be expressed as:
F = 3M - 5 (Equation 2)
Now we can solve the equations to find the values of M and F.
First, let's substitute equation 2 into equation 1:
M + (3M - 5) = 51
Combining like terms:
4M - 5 = 51
Now let's isolate the M variable by moving -5 to the other side of the equation:
4M = 51 + 5
4M = 56
Dividing both sides by 4:
M = 56 / 4
M = 14
Now that we have the value of M, we can substitute it back into equation 2 to find the value of F:
F = 3M - 5
F = 3(14) - 5
F = 42 - 5
F = 37
Therefore, there are 37 females and 14 males in the class.