The mathematics club in a school has 53 members. 5/7 of the girls members and 3/5 of the boys members are from lower secondary. The total number of lower secondary members is 35. How many members of the club are boys?
b+g=53
3/5 b + 5/7 g = 35
or, knowing that
g is a multiple of 7
b is a multiple of 5
how many such multiple pairs add to 53?
To find the number of boys in the mathematics club, we need to determine the number of boys in the lower secondary section. Let's break down the given information:
Let's assume that there are x girls and y boys in the club.
According to the problem, 5/7 of the girls members are from lower secondary. So, the number of lower secondary girls can be calculated as (5/7) * x.
Similarly, 3/5 of the boys members are from lower secondary. So, the number of lower secondary boys can be calculated as (3/5) * y.
The problem states that the total number of lower secondary members is 35. So, the equation becomes:
(5/7) * x + (3/5) * y = 35 (Equation 1)
We also know that the total number of members in the club is 53. Therefore:
x + y = 53 (Equation 2)
Now, we have two equations with two unknowns (x and y). We can solve these equations simultaneously.
First, let's multiply Equation 1 by 7 and Equation 2 by 5 to eliminate the fractions:
5 * (5/7) * x + 7 * (3/5) * y = 5 * 35
35x/7 + 21y/5 = 175
7 * x + 5 * y = 7 * 53
7x + 5y = 371
Now, we can solve the two resulting equations:
35x/7 + 21y/5 = 175 (Equation 3)
7x + 5y = 371 (Equation 4)
Rearranging Equation 4, we get:
7x = 371 - 5y
x = (371 - 5y)/7
Substituting this value of x into Equation 3, we have:
35(371 - 5y)/7 + 21y/5 = 175
Now, we can solve for y:
5(371 - 5y) + 49y/5 = 175
1855 - 25y + 49y/5 = 175
5(1855 - 25y) + 49y = 175 * 5
9275 - 125y + 49y = 875
-76y = 875 - 9275
-76y = -8400
y = -8400 / -76
y = 110
Therefore, the number of boys in the mathematics club is 110.