heureka

(a)

Two candles of equal length are lit at the same time.

One candle takes 6 hours to burn out, and the other takes 9 hours to burn out.

After how much time will the slower burning candle be exactly twice as long as the faster burning one?

\(\begin{array}{|rcll|} \hline \text{Rest fast candle}&=& 1Candle - \dfrac{1candle}{6h}\times t \\ \text{Rest slow candle}&=& 1Candle - \dfrac{1candle}{9h}\times t \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2\times \text{Rest fast candle } &=& \text{Rest slow candle} \\\\ 2\times \left( 1Candle - \dfrac{1candle}{6h}\times t\right) &=& 1Candle - \dfrac{1candle}{9h}\times t \\\\ 2 - \dfrac{2}{6}\times t &=& 1 - \dfrac{1}{9}\times t \\\\ 2 - \dfrac{1}{3}\times t &=& 1 - \dfrac{1}{9}\times t \\\\ \dfrac{1}{3}\times t- \dfrac{1}{9}\times t &=& 1 \\\\ \dfrac{(9-3)}{3*9}\times t &=& 1 \\\\ \dfrac{6}{3*9}\times t &=& 1 \\\\ \dfrac{2}{9}\times t &=& 1 \\\\ t &=& \dfrac{9}{2} \\\\ \mathbf{t} &=& \mathbf{4.5~ hours} \\ \hline \end{array}\)

Kevin will start with the integers 1, 2, 3 and 4 each used exactly once
and written in a row in any order.
Then he will find the sum of the adjacent pairs of integers in each row
to make a new row, until one integer is left.
For example, if he starts with 3, 2, 1, 4, then he takes sums to get 5, 3, 5,
followed by 8, 8, and he ends with the final sum 16.
Including all of Kevin's possible starting arrangements of the integers 1, 2, 3 and 4,
how many possible final sums are there?

I assume  5 possible final sums \(\{16,~ 18,~ 20,~ 22,~ 24 \}\)

\(\begin{array}{|cccc|} \hline \text{Start} & a & & b & & c & & d \\ \text{First sum} & & a+b & & b+c & & c+d \\ \text{Second sum} & & & a+2b+c & & b+2c+d & & \\ \text{Final sum} & & & &a+b+c+d+2(b+c)& & & \\\hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{Final sum} &=& a+b+c+d+2(b+c) \quad | \quad a+b+c+d=1+2+3+4= 10\\ \mathbf{\text{Final sum}} &=& \mathbf{10+2*(b+c)} \\ \hline \end{array}\)

\(\mathbf{b+c=\ ?}\)

\(\begin{array}{|c|c|c|c|c|c|} \hline \mathbf{b+c} & c= & 1 & 2 & 3 & 4 \\ \hline b=& & & & & \\ \hline 1 && - &3 & 4 & 5 \\ \hline 2 && 3 &- & 5 & 6 \\ \hline 3 && 4 &5 & - & 7 \\ \hline 4 && 5 &6 & 7 & - \\ \hline \end{array} \)

\(\begin{array}{|c|c|c|c|} \hline & b+c & 2x & +10 \\ \hline 1 & 3 & 6 & 16 \\ 2 & 4 & 8 & 18 \\ 3 & 5 & 10 & 20 \\ 4 & 6 & 12 & 22 \\ 5 & 7 & 14 & 24 \\ \hline \end{array}\)

How many 10 digit numbers are there, such that the sum of the digits is divisible by 5?

The first digit cannot be zero, so 9 ways, the next 8 digits can be anything, so there are 10 ways for each of them.
We can then only choose two digits as last digit to satisfy the condition.

There are \(\mathbf{9\times 10^8 \times 2 = 1~ 800~ 000~ 000}\) numbers.

\(\begin{array}{|c|c|} \hline \text{the sum of the digits 1 until 9} & \text{last digit to satisfy the condition }\\ \hline \ldots 0 & 0~ \text{ or } ~5 \\ \ldots 1 & 4~ \text{ or } ~9 \\ \ldots 2 & 3~ \text{ or } ~8 \\ \ldots 3 & 2~ \text{ or } ~7 \\ \ldots 4 & 1~ \text{ or } ~6 \\ \ldots 5 & 0~ \text{ or } ~5 \\ \ldots 6 & 4~ \text{ or } ~9 \\ \ldots 7 & 3~ \text{ or } ~8 \\ \ldots 8 & 2~ \text{ or } ~7 \\ \ldots 9 & 1~ \text{ or } ~6 \\ \hline \end{array}\)

Example: 1 000 000 08(1) or 1 000 000 08(6)

               4 567 123 98(0) or 4 567 123 98(5)

Three runners Adam, Ben, and Charles all start
at the same time for a 24 km race,
and each of them runs at a constant speed.
When Adam finishes the race, Ben is 8 km behind, and Charles is 12 km behind.
When Ben finishes the race, how far behind is Charles, in km?

Formula \(s=v*t\)

\(\begin{array}{|lrcll|} \hline (1) & 24~\text{km} &=& v_{Adam}*t_{Adam} \\ (2) & 24~\text{km}-8~\text{km} &=& v_{Ben}*t_{Adam} \\ &\mathbf{ v_{Ben}} &=& \mathbf{\dfrac{16}{t_{Adam}} } \\\\ (3) & 24~\text{km}-12~\text{km} &=& v_{Charles}*t_{Adam} \\ & \mathbf{v_{Charles}} &=& \mathbf{ \dfrac{12}{t_{Adam}} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s_{Ben}=24~\text{km} &=& v_{Ben}*t_{Ben} \quad |\quad \mathbf{ v_{Ben}=\dfrac{16}{t_{Adam}} } \\ \\ 24 &=& \dfrac{16}{t_{Adam}}*t_{Ben} \\\\ \dfrac{t_{Ben}}{t_{Adam}} &=& \dfrac{24}{16} \\ \\ \mathbf{ \dfrac{t_{Ben}}{t_{Adam}} } &=& \mathbf{1.5 } \\\\ \hline s_{Charles} &=& v_{Charles}*t_{Ben} \quad |\quad \mathbf{v_{Charles}=\dfrac{12}{t_{Adam}} } \\ \\ s_{Charles} &=& \dfrac{12}{t_{Adam}}*t_{Ben} \\ \\ s_{Charles} &=& 12*\dfrac{t_{Ben} }{t_{Adam}} \quad | \quad \mathbf{ \dfrac{t_{Ben}}{t_{Adam}} = 1.5 } \\\\ s_{Charles} &=& 12* 1.5 \\\\ s_{Charles} &=& 18~\text{km} \\ \hline s_{Ben}-s_{Charles} &=& 24~\text{km}-18~\text{km}\\ \mathbf{s_{Ben}-s_{Charles}} &=& \mathbf{6~\text{km}} \\ \hline \end{array}\)

Charles is \(\mathbf{6~\text{km}}\) behind

At a party. Andy and April shared a cherry pie. Andy ate 2/5 of the pie and April ate 1/3 of the pie.

What fraction of the pie did they eat altogether

\(\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{2}{5} + \dfrac{1}{3} } \\\\ &=& \dfrac{2*3+1*5}{5*3}\\\\ &=& \mathbf{ \dfrac{11}{15}} \\ \hline \end{array}\)

A coffee shop currently sells 490 lattes a day at $3.00 each.

They recently tried raising the by price by $0.25 a latte, and found that they sold 30 less lattes a day.

a) Assume that the number of lattes they sell in a day, N, is linearly related to the sale price, p (in dollars). Find an equation for N as a function of p.

\(\begin{array}{|rcll|} \hline \dfrac{N-460}{p-3.25} &=& \dfrac{490-460}{3-3.25} \\\\ N-460 &=& (p-3.25)*\dfrac{490-460}{3-3.25} \\\\ N &=& 460+(p-3.25)*\dfrac{490-460}{3-3.25} \\\\ N &=& 460+(p-3.25)*\dfrac{30}{-0.25} \\\\ N &=& 460-(p-3.25)*\dfrac{30}{0.25} \\\\ N &=& 460-p*\dfrac{30}{0.25}+3.25*\dfrac{30}{0.25} \\\\ N &=& 460-p*\dfrac{30}{0.25}+390 \\\\ N &=& 850-p*\dfrac{30}{0.25} \\\\ N &=& 850-p*120 \\\\ \mathbf{N} &=& \mathbf{850 - 120p} \\ \hline \end{array}\)

If the farm has 30 chickens and cows,
and there are 82 chicken and cow legs all together,
then how many chickens and how many cows could the farm have

\(\begin{array}{|rcll|} \hline ch + co &=& 30 \\ ch &=& 30-co \\\\ 2*ch + 4*co &=& 82 \\ 2*(30-co) + 4*co &=& 82 \\ 60-2*co + 4*co &=& 82 \\ 60 + 2*co &=& 82 \\ 2*co &=& 82-60 \\ 2*co &=& 22 \quad | \quad :2 \\ \mathbf{\text{cows}} &=& \mathbf{11} \\\\ ch &=& 30-co \\ ch &=& 30-11 \\ \mathbf{\text{chickens}} &=& \mathbf{19} \\ \hline \end{array}\)

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.
The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.
Find RS

General solution:

\(\text{Let $Center$ is the center of the circle }\\ \text{Let $P-Center =T-Center=v$ }\\ \text{Let $S-Center =w$ }\\ \text{Let $T-Center =w+k=v\qquad k=v-w$ }\\ \text{Let $ST = k =v-w$ }\\ \text{Let $PS = v+w$ }\\ \text{Let $RS = x$ }\\ \text{Let $QS = 25-x$ }\\ \text{Let $Q-Center = u $ }\)

Intersecting chords theorem:

\(\begin{array}{|rcll|} \hline RS*QS &=& ST*PS \\ x*(25-x) &=& (v-w)*(v+w) \\ \mathbf{x*(25-x)} &=& \mathbf{v^2-w^2} & (1) \\ \hline \end{array}\)

Pythagoras:

\(\begin{array}{|lrcll|} \hline & v^2+u^2 &=& 15^2 & (2) \\ & w^2+u^2 &=& (25-x)^2 & (3) \\ \hline (2)-(3): & v^2 -w^2 &=& 15^2-(25-x)^2 \quad | \quad \mathbf{x*(25-x)=v^2-w^2} \\ & x*(25-x) &=& 15^2-(25-x)^2 \\ &25x-x^2 &=& 15^2 -(25^2-50x+x^2) \\ &25x-x^2 &=& 15^2 -25^2+50x-x^2 \\ &25x &=& 15^2 -25^2+50x \\ &25^2-15^2 &=& 25x \\ & 25x &=& 25^2-15^2 \\ & 25x &=& 400 \quad |\quad :25 \\ & \mathbf{x} &=& \mathbf{16} \\ \hline \end{array}\)

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.

The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.

Find RS.

\(\text{Let $C$ is the center of the circle } \\ \text{Let $PS=TS=h$ } \\ \text{Let $RS=x$ } \\ \text{Let $QS=25-x$ } \)

\(\begin{array}{|lrcll|} \hline 1. & \left(25-x\right)^2+h^2 &=& 15^2 \qquad \text{or}\qquad \mathbf{ h^2 = 15^2-\left(25-x\right)^2} \\\\ 2. & x(25-x) &=& h * h \qquad \text{Intersecting chords theorem} \\ & x(25-x) &=& h^2 \\ & x(25-x) &=& 15^2-\left(25-x\right)^2 \\ & 25x-x^2 &=& 15^2-( 25^2-50x+x^2) \\ & 25x-x^2 &=& 15^2-25^2+50x-x^2 \\ & 25x &=& 15^2-25^2+50x \\ & 25^2-15^2 &=& 25x \\ & 25x &=& 25^2-15^2 \\ & 25x &=& 400 \\ & \mathbf{x} &=& \mathbf{16} \\ \hline \end{array}\)