MHB How many girls were there in the class?

  • Thread starter Thread starter Johnx1
  • Start date Start date
  • Tags Tags
    Class
AI Thread Summary
The class initially had 42 pupils, consisting of boys and girls. After 6 boys left, the ratio became three times as many girls as boys remaining. The calculations confirm that there are 27 girls and 9 boys left after the departure of the boys. The initial count included 15 boys and 27 girls, which adds up to the total of 42 students. Therefore, the correct number of girls in the class is 27.
Johnx1
Messages
48
Reaction score
0
There were 42 pupils in a class at first. After 6 boys left the class, there were 3 times as many girls as boys. How many girls were there in the class?

My answer: 42 = (x+ 6) + 3x
=> x = 9the (x+6) is basically (boys + 6)

So there are 27 girls and 9 boys? Did I do my work correctly?
 
Mathematics news on Phys.org
I believe that's correct. Another equation we could use is 36 = 3(b - 6) + b - 6. Solving,

36 = 3b - 18 + b - 6
36 = 4b - 24
60 = 4b
b = 15

Then b - 6 = 9 and g = 27, so there are 27 girls in the class.
 
Johnx said:
There were 42 pupils in a class at first. After 6 boys left the class, there were 3 times as many girls as boys. How many girls were there in the class?

My answer: 42 = (x+ 6) + 3x
=> x = 9the (x+6) is basically (boys + 6)

So there are 27 girls and 9 boys? Did I do my work correctly?
It's hard to tell because you start with an equation in "x" but have not said what "x" represents! I presume that you intended x to be the number of boys in the class after the first 6 boys left. The specific question was "How many girls were there in the class?" Yes, there were 27 girls. There were 9 boys left after the first 6 boys left so that there were intially 27 girls and 15 boys in the class. You can check that 27+ 15= 42, the number of students initially in the class.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...

Similar threads

Replies
1
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
5
Views
5K
Replies
4
Views
5K
Back
Top