tawdry tendon
There are 25 students in Class 1A, with more boys than girls. It is known that the number of boys wearing glasses is 3 times the number of girls without glasses; and that the number of girls wearing glasses is 2 times the number of boys without glasses. Find the number of boys in Class 1A.
show step please.
manic robinBOT
quartz crystal
okay first have you written out all of these givens as equations
@tawdry tendon
(ping me if you answer)
tawdry tendon
yeah but it doesnt work
by just putting numbers into it, i can get the answer is 15
but that doesnt suffice as a step
so i need help with that
toxic barn
yeah but it doesnt work
how does it not work
quartz crystal
okay ill give it a shot:
we have b as number of boys, g as number of girls
b + g = 25
b > g
-> b > 12, g < 13
(let x be # of girls with glasses and y be # of boys with glasses)
y = 3(g - x)
x = 2(b - y)
okay so y = 3g - 3x, x = 2b - 2y
so x = 2b - 2(3g - 3x) = 2b - 6g + 6x
so -5x = 2b - 6g
but b = 25 - g
so -5x = 2(25 - g) - 6g = 50 - 8g
-> -x = 10 - 8g/5
-> x = 8g/5 - 10
is that all right up to this point? let me check
seems right
okay so clearly x is an integer, and its at least 0
so g ≥ 50/8 = 6.25
and g ≤ 12
g has to be (5n)/8 for some n since x is a positive integer
and g has to be (8m)/8 for some m since g is a positive integer-> thus g is 40k/8 for some k -> g is divisible by 5
since 6.25 ≤ g ≤ 12 and g divisible by 5, we know that g = 10 and b = 15. furthermore, x = 6 and y = 12. so there are 15 boys and 10 girls, and 6 of the girls have glasses and 12 of the boys have glasses. i'll let you clean this up @tawdry tendon
(there is almost certainly a more elegant way, please look for it. but use some of the ideas i used in my proof: mostly, divisibility and narrowing down what range a number can be in. no "easy"/"normal" methods will work because we have a system of equations with 3 equations and 4 variables)
tawdry tendon
ok