Word Problems - Proportions, Speed & Time: Very Difficult Problems with Solutions

Solution:
Let him eat [tex]n[/tex] candies a day for [tex]d[/tex] days; the box had [tex]nd[/tex] candies:
[tex](n-1)(d+9)=nd \Rightarrow 9n-d=9[/tex]
[tex](n+1)(d-6)=nd \Rightarrow d-6n=6[/tex]
[tex]3n=15 \Rightarrow n=5, d=36[/tex]
Check: [tex]5\times 36=4\times 45=6\times 30=180[/tex]

Solution:
Let one cow eat [tex]1[/tex] unit of grass per day, let the meadow have [tex]G[/tex] units at the start, and let [tex]r[/tex] units grow every day:
[tex]G+24r=70\times 24=1680, G+60r=30\times 60=1800[/tex]
[tex]36r=120 \Rightarrow r=\dfrac{10}{3}, G=1680-80=1600[/tex]
For [tex]96[/tex] days and [tex]x[/tex] cows:
[tex]96x=G+96r=1600+320=1920 \Rightarrow x=20[/tex]

Solution:
Let the group have [tex]n[/tex] children with average age [tex]a[/tex], so their total age is [tex]na[/tex]:
[tex]\dfrac{na-45}{n-5}=a+1 \Rightarrow na-45=na-5a+n-5 \Rightarrow 5a-n=40[/tex]
[tex]\dfrac{na+85}{n+5}=a+1 \Rightarrow na+85=na+5a+n+5 \Rightarrow 5a+n=80[/tex]
[tex]10a=120 \Rightarrow a=12, n=20[/tex]
Check: [tex]\dfrac{240-45}{15}=\dfrac{240+85}{25}=13[/tex]

Solution:
Let the first sell [tex]x[/tex] lb at price [tex]p[/tex] and the second [tex]100-x[/tex] lb at price [tex]q[/tex]. Then [tex]p(100-x)=72[/tex], [tex]qx=32[/tex], and equal amounts give [tex]px=q(100-x)[/tex]:
[tex]\dfrac{72x}{100-x}=\dfrac{32(100-x)}{x} \Rightarrow 9x^2=4(100-x)^2[/tex]
Both sides are squares of positive numbers, so:
[tex]3x=2(100-x) \Rightarrow 5x=200 \Rightarrow x=40[/tex]
The brothers sold [tex]40[/tex] and [tex]60[/tex] lb. Check: [tex]p=1.2[/tex], [tex]q=0.8[/tex] and [tex]40\times 1.2=60\times 0.8=48[/tex].

Solution:
The total weight is [tex]46\times 233=10718[/tex] lb. If [tex]k[/tex] athletes weigh more than [tex]275[/tex] lb and the others at least [tex]180[/tex] lb:
[tex]10718\gt 275k+180(46-k)=8280+95k \Rightarrow 95k\lt 2438 \Rightarrow k\le 25[/tex]
[tex]25[/tex] is possible: for example, [tex]25[/tex] athletes weighing [tex]276[/tex] lb, [tex]20[/tex] weighing [tex]180[/tex] lb and one weighing [tex]218[/tex] lb.

Solution:
Let him have [tex]x[/tex] dollars and [tex]y[/tex] cents when entering ([tex]y\lt 100[/tex]), i.e. [tex]100x+y[/tex] cents. When leaving he had [tex]\frac{y}{2}[/tex] dollars and [tex]x[/tex] cents, i.e. [tex]50y+x[/tex] cents:
[tex]50y+x=\dfrac{100x+y}{2} \Rightarrow 100y+2x=100x+y \Rightarrow 99y=98x[/tex]
Since [tex]98[/tex] and [tex]99[/tex] have no common divisor greater than [tex]1[/tex], [tex]y[/tex] is a multiple of [tex]98[/tex]. From [tex]0\lt y\lt 100[/tex] we get [tex]y=98[/tex] and [tex]x=99[/tex].
Check: [tex]99.98\div 2=49.99[/tex]

Solution:
Let there be [tex]A[/tex] mushrooms of the first kind, [tex]B[/tex] of the second kind and [tex]n[/tex] pickers. Everyone got [tex]\frac{A}{7}+\frac{B}{9}[/tex] mushrooms:
[tex]n\left(\dfrac{A}{7}+\dfrac{B}{9}\right)=A+B[/tex]
If [tex]n\le 7[/tex], the left side is at most [tex]A+\frac{7B}{9}\lt A+B[/tex]. If [tex]n\ge 9[/tex], it is at least [tex]\frac{9A}{7}+B\gt A+B[/tex]. So [tex]n=8[/tex] (for example, [tex]A=7[/tex], [tex]B=9[/tex] and everyone gets [tex]2[/tex] mushrooms).

Solution:
Let [tex]n[/tex] fishermen catch [tex]F[/tex] fish. The other [tex]n-2[/tex] fishermen caught [tex]F-\frac{F}{5}-\frac{F}{7}=\frac{23F}{35}[/tex] fish, each of them more than [tex]\frac{F}{7}[/tex] and less than [tex]\frac{F}{5}[/tex]:
[tex](n-2)\times \dfrac{F}{7}\lt \dfrac{23F}{35}\lt (n-2)\times \dfrac{F}{5}[/tex]
[tex]5(n-2)\lt 23\lt 7(n-2) \Rightarrow \dfrac{23}{7}\lt n-2\lt \dfrac{23}{5} \Rightarrow n-2=4[/tex]
There were [tex]6[/tex] fishermen (for example, [tex]140[/tex] fish: [tex]28[/tex], [tex]20[/tex] and four catches of [tex]23[/tex]).