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

Solution:
[tex]34+x=4(10+x) \Rightarrow 34+x=40+4x \Rightarrow 3x=-6 \Rightarrow x=-2[/tex]
The negative answer means that this was the case [tex]2[/tex] years ago: the father was [tex]32[/tex] and the son [tex]8[/tex]. From now on the ratio of their ages only decreases, so it will never happen again.

Solution:
Work backwards. If [tex]y[/tex] horses remain after a sale from [tex]x[/tex] horses, then [tex]y=\frac{x}{2}-\frac{1}{2}[/tex], i.e. [tex]x=2y+1[/tex]:
[tex]5\to 2\times 5+1=11\to 2\times 11+1=23\to 2\times 23+1=47[/tex]
Check: [tex]47-24=23[/tex], [tex]23-12=11[/tex], [tex]11-6=5[/tex].

Solution:
After the first day [tex]\frac{4}{9}[/tex] of the book remains. On the second day he read [tex]\frac{3}{8}\times \frac{4}{9}=\frac{1}{6}[/tex] of the book. The last part is:
[tex]\dfrac{4}{9}-\dfrac{1}{6}=\dfrac{8-3}{18}=\dfrac{5}{18}[/tex]
[tex]\dfrac{5}{18}x=40 \Rightarrow x=144[/tex]

Solution:
Let me be [tex]h[/tex] and my wife [tex]w[/tex] years old. I was [tex]w[/tex] years old [tex]h-w[/tex] years ago; then she was [tex]w-(h-w)=2w-h[/tex]:
[tex]h=2(2w-h) \Rightarrow 3h=4w[/tex]
[tex]h+w=63 \Rightarrow \dfrac{4w}{3}+w=63 \Rightarrow w=27, h=36[/tex]
Check: [tex]9[/tex] years ago I was [tex]27[/tex] and she was [tex]18[/tex], and [tex]36=2\times 18[/tex].

Solution:
[tex]\overline{ab}+63=\overline{ba} \Rightarrow 10a+b+63=10b+a \Rightarrow 9(b-a)=63 \Rightarrow b-a=7[/tex]
Together with [tex]a+b=11[/tex] we get [tex]a=2[/tex], [tex]b=9[/tex]. The number is [tex]29[/tex] (check: [tex]29+63=92[/tex]).

Solution:
Let there be [tex]x[/tex] coins of [tex]1[/tex] and [tex]51-x[/tex] coins of [tex]0.1[/tex]:
[tex]x+0.1(51-x)=33 \Rightarrow 0.9x=27.9 \Rightarrow x=31[/tex]
Check: [tex]31+0.1\times 20=31+2=33[/tex]

Solution:
Let the thicker candle be [tex]L[/tex] long; the thinner one is [tex]L-4[/tex]. In [tex]2[/tex] hours the thicker one burns [tex]\frac{2}{5}[/tex] of its length and the thinner one [tex]\frac{2}{3}[/tex], so the remaining parts are [tex]\frac{3}{5}L[/tex] and [tex]\frac{1}{3}(L-4)[/tex]:
[tex]\dfrac{3}{5}L=2\times \dfrac{L-4}{3} \Rightarrow 9L=10L-40 \Rightarrow L=40[/tex]
The candles are [tex]40[/tex] and [tex]36[/tex] long.

Solution:
Pages [tex]1[/tex] to [tex]9[/tex] use [tex]9[/tex] digits, and pages [tex]10[/tex] to [tex]99[/tex] use [tex]90\times 2=180[/tex] digits, together [tex]189[/tex]. The remaining digits belong to three-digit page numbers:
[tex](2322-189)\div 3=2133\div 3=711[/tex]
[tex]99+711=810[/tex]

Solution:
If there are [tex]n[/tex] students, each gives [tex]n-1[/tex] photos:
[tex]n(n-1)=992=32\times 31 \Rightarrow n=32[/tex]

Solution:
The number of matches is [tex]\frac{4\times 3}{2}=6[/tex]. A match with a winner gives [tex]3[/tex] points in total, and a draw gives [tex]2[/tex]. If [tex]x[/tex] matches ended in a draw:
[tex]3(6-x)+2x=16 \Rightarrow 18-x=16 \Rightarrow x=2[/tex]

Solution:
Let there be [tex]m[/tex] men, [tex]w[/tex] women and [tex]c[/tex] children:
[tex]m+w+c=13, 2m+\dfrac{w}{2}+\dfrac{c}{4}=13[/tex]
Multiply the second equation by [tex]4[/tex] and subtract the first:
[tex]8m+2w+c=52 \Rightarrow 7m+w=39 \Rightarrow w=39-7m[/tex]
From [tex]w\ge 0[/tex] we get [tex]m\le 5[/tex], and from [tex]m+w=39-6m\le 13[/tex] we get [tex]m\ge 5[/tex]. So [tex]m=5[/tex], [tex]w=4[/tex], [tex]c=4[/tex].
Check: [tex]5\times 2+4\times 0.5+4\times 0.25=10+2+1=13[/tex]

Solution:
Let the numbers be [tex]n, n+1, \dots, n+8[/tex]. Their sum is [tex]9n+36[/tex], so the erased number is [tex]9n+36-166=9n-130[/tex]. It is one of the nine numbers:
[tex]n\le 9n-130\le n+8 \Rightarrow 130\le 8n\le 138 \Rightarrow n=17[/tex]
[tex]9\times 17-130=23[/tex]

Solution:
Let the average age of the [tex]10[/tex] players be [tex]a[/tex] (so their total age is [tex]10a[/tex]) and let the captain be [tex]c[/tex] years old:
[tex]\dfrac{10a+c}{11}=a+1 \Rightarrow 10a+c=11a+11 \Rightarrow c=a+11[/tex]
The team average is [tex]a+1[/tex], so the captain is older than it by [tex](a+11)-(a+1)=10[/tex] years.

Solution:
In [tex]43+48+44[/tex] every student with two top grades is counted twice and every student with three top grades three times. Subtracting the pairs removes the students with three top grades completely, so we add them back once:
[tex]43+48+44-14-17-15+6=135-46+6=95[/tex]

Solution:
By the clock, from [tex]9{:}25[/tex] p.m. to [tex]6{:}31[/tex] a.m. there are [tex]9[/tex] h [tex]6[/tex] min, i.e. [tex]546[/tex] min. In every real hour ([tex]60[/tex] min) the clock moves [tex]63[/tex] min, so the real time that passes is:
[tex]546\times \dfrac{60}{63}=520[/tex]
[tex]520[/tex] min [tex]=8[/tex] h [tex]40[/tex] min, so the alarm rings at [tex]6{:}05[/tex] a.m. real time. Ivo can lie in for [tex]26[/tex] minutes.