At first the ratio of the number of boys to the number of girls in a Mathematics Club was I : 2 when 5 more boys joined the club , the ratio became 3 : 4 How many members were there in the club at first?
b = number of boys
g = number of girls
At first the ratio of the number of boys to the number of girls in a Mathematics Club was 1 : 2
b / g = 1 / 2
When 5 more boys joined the club , the ratio became 3 : 4
( b + 5 ) / g = 3 / 4
b / g = 1 / 2 Multiply both sides by 2
2 b / g = 1 Multiply both sides by g
2 b = g Divide both sides by 2
b = g / 2
( b + 5 ) / g = 3 / 4 Multiply both sides by 4
4 ( b + 5 ) / g = 3
( 4 b + 4 * 5 ) / g = 3
( 4 b + 20 ) / g = 3 Multiply both sides by g
4 b + 20 = 3 g
Replace b = g / 2 in equation 4 b + 20 = 3 g
4 b + 20 = 3 g
4 * g / 2 + 20 = 3 g
2 g + 20 = 3 g Subtract 2 g to both sides
2 g + 20 - 2 g = 3 g - 2 g
20 = g
g = 20
b = g / 2 = 20 / 2 = 10
10 boys and 20 girls
To solve this question, let's define some variables:
Let's say the initial number of boys in the Mathematics Club is "B" and the initial number of girls is "G".
According to the given information, the initial ratio of boys to girls is 1:2, which can be represented as:
B/G = 1/2
Also, it is mentioned that when 5 more boys join the club, the ratio becomes 3:4. This can be represented as:
(B + 5)/G = 3/4
Now, we have a system of two equations with two variables:
Equation 1: B/G = 1/2
Equation 2: (B + 5)/G = 3/4
We can solve the system of equations to find the values of B and G, which will give us the initial number of boys and girls.
To eliminate G, we can multiply Equation 1 by 4 and Equation 2 by 2:
4 * (B/G) = 4 * (1/2)
2 * ((B + 5)/G) = 2 * (3/4)
This simplifies the equations to:
2B = G
(B + 5) = 3G/2
Now we have two equations:
2B - G = 0
B + 5 - (3G/2) = 0
We can solve these equations by substitution or elimination to find the values of B and G.
Substituting the value of G from equation 1 into equation 2:
2B - (2B) = -5 + (3G/2)
0 = -5 + (3G/2)
5 = (3G/2)
10 = 3G
G = 10/3
Now, substituting the value of G back into equation 1:
2B = G
2B = 10/3
B = (10/3) / 2
B = 5/3
Since we cannot have a fractional number of members, we need to find a common multiple of 3 that will give us whole numbers for G and B. The least common multiple of 3 and 2 is 6.
Multiplying both G and B by 2:
G = (10/3) * 2 = 20/3
B = (5/3) * 2 = 10/3
So, the number of boys in the club at first was 10, and the number of girls was 20.
To find the total number of members in the club at first, we add the number of boys and girls:
Total members = B + G = 10/3 + 20/3 = 30/3 = 10
Therefore, there were initially 10 members in the Mathematics Club.