In a hall, there are 151 boys and girls altogether. '1/3 of the boys and '1/5 of the girls wear glasses. The total number of boys and girls who wear glasses is 45. How many boys are there in the Hall?
More data needed.
1/3b + 1/5g = 45
1/3b + 1/5g = 45
b+g = 151
or,
5b+3g = 675
3b+3g = 453
--------------------
2b = 222
b = 111
Let's solve this step-by-step.
First, let's assign variables to the unknowns:
Let B = number of boys in the hall
Given:
Total number of boys and girls = 151
We can write this as the equation: B + G = 151, where G represents the number of girls.
1/3 of the boys wear glasses:
The number of boys wearing glasses = 1/3 * B = B/3
1/5 of the girls wear glasses:
The number of girls wearing glasses = 1/5 * G = G/5
The total number of boys and girls who wear glasses is 45:
B/3 + G/5 = 45
We can simplify the equation by multiplying through by the least common multiple (LCM) of 3 and 5, which is 15:
15(B/3) + 15(G/5) = 15(45)
5B + 3G = 675
Now we can solve the system of equations:
B + G = 151
5B + 3G = 675
Using the method of substitution, we can solve for G in terms of B by rearranging the first equation:
G = 151 - B
Substituting this expression for G in the second equation:
5B + 3(151 - B) = 675
5B + 453 - 3B = 675
2B + 453 = 675
2B = 675 - 453
2B = 222
B = 222/2
B = 111
Therefore, there are 111 boys in the hall.
To solve this problem, you can set up equations based on the given information.
Let's assume that the number of boys in the hall is "b" and the number of girls is "g." We are trying to find the value of "b."
According to the given information, 1/3 of the boys wear glasses and 1/5 of the girls wear glasses. This means:
Number of boys wearing glasses = (1/3) * b
Number of girls wearing glasses = (1/5) * g
The total number of boys and girls wearing glasses is 45, so we can write the equation:
(1/3) * b + (1/5) * g = 45
We also know that the total number of boys and girls in the hall is 151, so we can write another equation:
b + g = 151
Now we have a system of two equations with two variables. We can solve this system to find the value of "b."
To eliminate the fractions, you can multiply the first equation by 15 to get rid of the denominators:
15 * [(1/3) * b] + 15 * [(1/5) * g] = 15 * 45
5b + 3g = 675
Now we have the following system of equations:
5b + 3g = 675
b + g = 151
You can solve this system using any method you prefer, such as substitution or elimination. I'll solve it for you using the elimination method.
Multiply the second equation by 3:
3 * (b + g) = 3 * 151
3b + 3g = 453
Now subtract the second equation from the first equation to eliminate "g":
(5b + 3g) - (3b + 3g) = 675 - 453
2b = 222
Divide both sides by 2 to solve for "b":
b = 222 / 2
b = 111
Therefore, there are 111 boys in the hall.