The mathematics club in a school has 53 members 5/7 of the girl members and 3/5 of the boys members are from lower secondary. The total numbers of lower secondary members is 35. How many members of the club are boys?
b+g = 53
5/7 g + 3/5 b = 35
5/7 (53-b) + 3/5 b = 35
b = 25
check:
g = 53-25 = 28
5/7 * 28 + 3/5 * 25 = 20 + 15 = 35
right on.
To find out how many members of the club are boys, we need to determine the number of boys in the club first. Let's break down the information given:
Total number of members = 53
Let's assume the number of girls in the club is g, and the number of boys is b.
Given:
5/7 of the girl members are from lower secondary: (5/7)g
3/5 of the boy members are from lower secondary: (3/5)b
The total number of lower secondary members is 35:
(5/7)g + (3/5)b = 35
Now, let's solve this equation to find the values of g and b:
To simplify the equation, we can simplify the fractions by finding a common denominator. The least common multiple of 7 and 5 is 35. So we can multiply both sides of the equation by 35:
35 * [(5/7)g + (3/5)b] = 35 * 35
Simplifying further, we get:
(5/7) * 35g + (3/5) * 35b = 35 * 35
Now, let's simplify both sides of the equation:
25g + 21b = 35 * 35
Since we know that g + b = 53 (total number of members), we can solve these two equations simultaneously to find the value of b (number of boys).
We can use substitution method or elimination method to solve this system of equations. Let's use elimination.
Multiply the first equation by 21, and the second equation by 35 to eliminate g:
(25 * 21)g + (21 * 21)b = 21 * (35 * 35)
(25g + 21b) * 21 = 35 * (35 * 35)
Simplifying further:
525g + 441b = 735 * 35
Now we have two equations:
25g + 21b = 35 * 35
525g + 441b = 735 * 35
Let's subtract the first equation from the second equation:
(525g + 441b) - (25g + 21b) = (735 * 35) - (35 * 35)
Simplifying further:
525g + 441b - 25g - 21b = 25725 - 1225
Combining like terms:
500g + 420b = 24500
Now, we have the following equation:
500g + 420b = 24500
25g + 21b = 35 * 35
To solve this system of equations, we can use substitution or elimination method again. Let's use elimination.
Multiply the second equation by 20 to eliminate g:
(25g + 21b) * 20 = (35 * 35) * 20
Simplifying further:
500g + 420b = 49000
Now we have two equations:
500g + 420b = 24500
500g + 420b = 49000
Since both equations are the same, there are infinite solutions. This means that the number of boys (b) can be any value.
Therefore, the number of members of the club who are boys cannot be determined with the given information.