heureka

the amount of weight that can be lifted by a championship weightlifter is a 3-digit odd number that does not contain a 9 or a 3 all the digits are different and their sum is 17 the greatest digit is in the middle what is the weightlifters weight ?

$$\big{2~8~7} \quad \text{ or } \quad \big{4~8~5}$$

June had twice as many apples as oranges. Ken had 1/4 as many oranges as apples. The number of oranges Ken had was 2/9 as many as the number of apples June had. Ken had 120 apples and oranges.

(a) express the number of oranges June had as a fraction of the number of apples Ken had. Give you answers in its SIMPLEST FORM.

$$\small{ \begin{array}{lrcl} & O_1 &=& \text{ oranges June }\\ & A_1 &=& \text{ apples June }\\\\ & O_2 &=& \text{ oranges Ken }\\ & A_2 &=& \text{ apples Ken }\\\\ (1) & A_1 &=& 2\cdot O_1\\ (2) & O_2 &=& \frac14 \cdot A_2 \\\\ (3) & O_2 &=& \frac29\cdot A_1 \\\\ (4) & 120 &=& O_2 + A_2\\\\\\ (1), (2), (3) & \frac14 \cdot A_2 &=& \frac29 \cdot A_1 \quad |\quad A_1 = 2\cdot O_1 \\\\ & \frac14 \cdot A_2 &=& \frac29 \cdot 2 \cdot O_1 \\\\ & \frac14 \cdot A_2 &=& \frac49 \cdot O_1 \\\\ & O_1 &=& \frac14 \cdot \frac94 \cdot A_2 \\\\ & \mathbf{O_1} & \mathbf{=} & \mathbf{ \frac9{16} \cdot A_2 }\\\\ \end{array} }$$

The number of oranges June had is 9/16 of the number of apples Ken had.

$$\small{ \begin{array}{lrcl} (A_2?) & 120 &=& O_2 + A_2 \qquad | \qquad O_2 = \frac14 \cdot A_2\\\\ & 120 &=& \frac14 \cdot A_2 + A_2\\\\ &120 &=& \frac54 \cdot A_2\\\\ & A_2 &=& 120 \cdot \frac45 \\\\ & \mathbf{A_2} & \mathbf{=} & \mathbf{96 }\\\\\\ (O_1?) &O_1 &=& \frac9{16}\cdot A_2 \\\\ & O_1 &=& \frac9{16} \cdot 96 \\\\ & \mathbf{O_1} & \mathbf{=} & \mathbf{54 }\\\\\\ (A_1?) &A_1 &=& 2\cdot O_1\\\\ & A_1 &=&2 \cdot 54\\\\ & \mathbf{A_1} & \mathbf{=} & \mathbf{ 108 }\\\\\\ (O_2?) &O_2 &=& \frac14\cdot A_2 \\\\ & O_2 &=& \frac14 \cdot 96\\\\ & \mathbf{O_2} & \mathbf{=} & \mathbf{ 24 }\\\\\\ \end{array} }$$

$$\small{ \begin{array}{|r|r|r|} \hline \text {oranges} & \text{apples} & \\ \hline 54 & 108 & \text{June} \\ \hline 24 & 96 &\text{Ken}\\ \hline \end{array} }$$

wie genau ist dies bei diesem beispiel gemeint: am+bm-cm+zm  ?

$$a\cdot \textcolor[rgb]{1,0,0}{m} +b\cdot \textcolor[rgb]{1,0,0}{m} -c\cdot \textcolor[rgb]{1,0,0}{m} +z\cdot \textcolor[rgb]{1,0,0}{m} \\ =(a+b-c+z)\cdot\textcolor[rgb]{1,0,0}{ m}$$

Der gemeinsame Faktor ist m und kann als gemeinsamer Faktor vorangestellt ( ausgeklammert ) werden.

3/16 of the fruits at a stall are pears. 1/4 of the remaining fruits are oranges. The rest are apples. If there are 585 apples, how many fruits are at the stall??

f = fruits    p = pears   o = oranges   a = apples

(1) $$p=f\cdot \frac3{16}$$

(2) $$o=(f-p)\frac14 \quad \text { or }\quad o = (f-f\frac3{16})\cdot \frac14 = f \frac{13}{16}\cdot \frac 14 = f\cdot\frac{13}{64}$$

(3) $$f = o+p+a$$

$$\small{\text{$ \begin{array}{rcl} a &=& 585 \\ f &=& o+p+585 \\ f &=& o+p+585 \qquad | \qquad o = f\cdot\frac{13}{64} \qquad p=f\cdot \frac3{16}\\\\ f &=& f\cdot\frac{13}{64}+f\cdot \frac3{16}+585 \\\\ f - f\cdot\frac{13}{64}-f\cdot \frac3{16} &=& 585\\\\ f (1- \frac{13}{64}- \frac3{16}) &=& 585\\\\ f (\frac{64-13-3\cdot 4}{64}) &=& 585\\\\ f \cdot \frac{39}{64} &=& 585\\\\ f &=& 585 \cdot \frac{64}{39} \\\\ \mathbf{f} &\mathbf{=}& \mathbf{960}\\\\ \end{array} $}} \\\\$$

At the stall are 960 fruits.

pears = 960 * (3/16) = 180

oranges = (960-180)/4 = 195

apples = 960 - 180 - 195 = 585 ( okay )

Cheryl is 3/5 as old as her mum. In 8 years time, the ratio of her age to her mom's age will be 2:3. How old is Cheryl now??

m = mum      c = cheryl

(1) $$c = \frac 35 m \qquad \text { or } \qquad m = \frac 53 c$$

(2) $$\frac{c+8}{m+8} = \frac 23 \qquad \text{ or } \qquad c = \frac 23 \cdot (m+8) - 8$$

$$\small{\text{$ \begin{array}{lcl} c &=& \frac 23 \cdot (m+8) - 8 \qquad | \qquad m=\frac 53 c\\\\ c &=& \frac 23 \cdot (\frac 53 c+8) - 8 \\\\ c &=& \frac {10}{9} c + 8\cdot \frac 23 - 8 \\\\ c &=& \frac {10}{9} c + \frac {16}{3} - 8 \qquad | \qquad \cdot 9\\\\ 9c &=& 10 c + 3\cdot 16 - 72\\\\ 9c &=& 10 c -24\\\\ 10c - 9 c &=& 24\\\\ \mathbf{c} & \mathbf{=} & \mathbf{24}\\ \end{array} $}} \\\\$$

Cheryl is now 24 years old

There are 2 hotels, Hotel Duxton and Hotel Supreme. There were 219 more guests in Hotel Duxton than in Hotel Supreme. After 57 guests had left Hotel Supreme, the number of guests in Hotel Duxton became 4 times the number of guests in Hotel Supreme. How many guests are left at Hotel Supreme?

d = Hotel Duxton       s = Hotel Supreme

(1) $$d-s = 219$$

(2) $$s_1 = s - 57$$

(3) $$d - s_1 = 219 + 57 = 276$$

(4) $$d = 4\cdot s_1$$

put (4) into (3):

$$\begin{array}{rcl} 4\cdot s_1 -s_1 &=& 276\\ 3\cdot s_1 &=& 276 \\ s_1 &=& 92 \end{array}$$

$$\text{\small{ \begin{array}{lrclr} & Hotel Duxton & Hotel Supreme \\ bevor & 368 = 149 + 219 & 92 + 57 = 149 \\ after & 368 = 92\cdot 4 & 92\\ \end{array} }}$$

$$\small{\text{$\mathbf{n^{th}}$ derivative of $\mathbf{10^{-x}}$}}$$

$$\text{ \small{Formula} $ \boxed{ \begin{array}{lcl} y &=& e^{k\cdot x} \\ y' &=& k \cdot e^{k\cdot x} \qquad y'' = k^2 \cdot e^{k\cdot x} \qquad y''' = k^3 \cdot e^{k\cdot x} \qquad \cdots \\ y^{(n)} & = & k^n \cdot e^{k\cdot x} \end{array} } $} \\\\\\ \small{\text{$ \begin{array}{lcl} y &=& 10^{-x} \qquad \Rightarrow \qquad y = e^{-x\cdot \ln{(10)} } = e^{ \overbrace{\left[ -\ln{10} \right] }^{=\big{k}} \cdot x }\\ \end{array} $}} \\\\ \small{\text{$ \underline{ y^{n} = k^n \cdot e^{k\cdot x} } \qquad \underline{ k = -\ln{(10)} } $}} \\\\ \small{\text{$ \begin{array}{lcl} y^{(n)} &=& [-\ln{(10)} ]^n \cdot \underbrace{ e^{ [-\ln{(10)} ]\cdot x} }_{=\big{10^{-x}}}\\\\ \mathbf{y^{(n)}} & \mathbf{=} & \mathbf{[-\ln{(10)} ]^n \cdot 10^{-x} }\\\\ \end{array} $}} \\\\$$

If x^2+(3a+1)x+2a^2-b^2 is a perfect square, what is a,b?

for example:

$$a=-\dfrac 13 \text{ and } b = \dfrac {\sqrt{2}}{3} \quad \Rightarrow x^2$$

September=1999

combin(31,7)=?

$$\small{ \begin{array}{rcl} \dbinom{31}{7} &=& \dfrac{31}{7} \cdot \left( \dfrac{30}{6} \right) \cdot \left( \dfrac{29}{5} \right) \cdot \left( \dfrac{28}{4} \right) \cdot \left( \dfrac{27}{3} \right) \cdot \left( \dfrac{26}{2} \right) \cdot \left( \dfrac{25}{1} \right)\\\\ \dbinom{31}{7} &=& \dfrac{31}{7} \cdot 5 \cdot \left( \dfrac{29}{5} \right) \cdot 7 \cdot 9 \cdot 13 \cdot 25 \\\\ \dbinom{31}{7} &=& 31 \cdot 29 \cdot 9 \cdot 13 \cdot 25 \\\\ \mathbf{\dbinom{31}{7}} & \mathbf{=} & \mathbf{2~629~575} \end{array} }$$