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There are two bags of stones labelled A and B. In Bag A, there are 26 black stones and 20 white stones. In Bag B, there are 39 black stones and 30 white stones. How many black and white stones should be transferred from Bag B to Bag A such that 60% of the stones in Bag A and 50% of those in Bag B are black?

$\begin{array}{r|c|l} & black & white \\ \hline A & 26 & 20 \\ \hline B & 39 & 30 \\ \hline sum & 65 & 50 \end{array}$

$\begin{array}{r|c|l} & black & white \\ \hline A & 26 +x & 20 + y \\ \hline B & 39-x & 30-y \\ \hline sum & 65 & 50 \end{array}$

$\begin{array}{rcl} [(39-x)+(30-y)]\cdot 50\% &=& 39-x \\ (39-x)\cdot 0.5 +(30-y)\cdot 0.5 &=& 39-x \\ (39-x)\cdot(1- 0.5) &=& (30-y)\cdot 0.5 \\ (39-x)\cdot 0.5 &=& (30-y)\cdot 0.5 \\ 39-x &=& 30-y \\ 30-y &=& 39-x\\ x &=& 39-30+y \\ \mathbf{x} & \mathbf{=} & \mathbf{9 + y} \end{array}$

$\begin{array}{rcl} [(26+x)+(20+y)]\cdot 60\% &=& 26+x \\ (26+x)\cdot 0.6 +(20+y)\cdot 0.6 &=& 26+x \\ (26+x)\cdot(1- 0.6) &=& (20+y)\cdot 0.6 \\ (26+x)\cdot 0.4 &=& (20+y)\cdot 0.6 \\ 26+x &=& (20+y )\cdot \frac32 \qquad | \qquad x = 9+y\\ 26+9+y &=& (20+y )\cdot \frac32 \\ 35+y &=& (20+y )\cdot \frac32 \\ 35+y &=& 30 +y \cdot \frac32 \\ y \cdot \frac32-y &=&35-30\\ y\cdot \frac12 &=& 5\\ \mathbf{y} & \mathbf{=} & \mathbf{10}\\ \\ \hline \\ x &=& 9 + y\\ x &=& 9 +10 \\ \mathbf{x} & \mathbf{=} & \mathbf{19}\\ \end{array}$

$\begin{array}{r|c|l} & black & white \\ \hline A & 26 +19 = \color{red}{45}& 20 + 10 = \color{red}{30}\\ \hline B & 39-19 = \color{red}{20} & 30-10 = \color{red}{20}\\ \hline sum & 65 & 50 \end{array}$

60 % of (45+30=75) is 45     okay!

50 % of (20+20=40) is 20     okay!

3.5 kg of Grade A coffee powder is mixed with 5.25 kg of Grade B coffee powder. The mixture is then packed into packets of 30g each. (a)how many 30-g packets are there? (b)how many grams of mixture are left over?

$A=3.5\ kg \qquad B=5.25\ kg\\ A+B= 3.5\ kg + 5.25\ kg \\ A+B = 8.75\ kg$

(a) $\frac{8.75\ kg}{0.030\ kg} \qquad |\qquad 30\ g = 0.030\ kg\\ = 291.6\overline{6}~ \text{ packets}$

There are 291 30-g packets

(b)  $0.6\overline{6}*0.030\ kg = 0.02\ kg = 20\ g$

20 g of mixture are left over.

In a month, the average time taken for four weekly club meetings was 1h 15 min. The total time taken for the first two meetings was 2h 5 min. What was the average time taken, in minutes , for the last two meetings?

$\begin{array}{rcl} \frac{m_1+m_2+m_3+m_4}{4} &=& 1\ h ~15\min.\\ \\ \hline \\ m_1+m_2 &=& 2\ h ~5\min.\\ \frac{m_1+m_2}{2 } &=& 1\ h ~2.5\min.\\ \\ \hline \\ \frac{m_3+m_4}{2 } &=& (\frac{m_1+m_2+m_3+m_4}{4})\cdot 2 - \frac{m_1+m_2}{2 }\\ &=& \frac{m_1+m_2}{2 }+\frac{m_3+m_4}{2 }- \frac{m_1+m_2}{2 }\\ \frac{m_3+m_4}{2 } &=& ( 1\ h ~15\min.)\cdot 2 -1\ h ~2.5\min.\\ \frac{m_3+m_4}{2 } &=& 2\ h ~30\min. -1\ h ~2.5\min.\\ \frac{m_3+m_4}{2 } &=& 1\ h ~27.5\min.\\ \frac{m_3+m_4}{2 } &=& 87.5\min.\\ \end{array}$

The average time taken, in minutes , for the last two meetings was 87.5 min.

Peter and John had some money. If Peter gave John $360, both would have an equal amount of money. If John gave Peter $180, John would have 1/3 as much money as Peter. How much money did Peter have at first?

$\begin{array}{lrcl} (1) & P-360 &=& J +360 \\ & J &=& P - 720 \\ \hline \\ (2) & \dfrac13 (P+180) &=& J-180 \qquad | \qquad J = P - 720 \\ & \dfrac13 (P+180) &=& P - 720 -180\\ & \dfrac13 (P+180) &=& P - 900\\ & \dfrac{P}{3} +60 &=& P - 900\\ & P-\dfrac{P}{3} &=& 960\\ & \dfrac{2}{3}P &=& 960\\ &P &=& \dfrac32 \cdot 960\\ &P &=& 1440 \end{array}$

5 cents in 1964. It was 49 cents in 2014.  Using your linear model, find the year in which postage was 37 cents. Round to a whole year.

$\begin{array}{rcl} \dfrac{49-5}{2014-1964} &=& \dfrac{37-5}{y-1964} \\ \dfrac{44}{50} &=& \dfrac{32}{y-1964} \\ y-1964 &=& \dfrac{32\cdot 50}{44} \\ y &=& 1964 + \dfrac{32\cdot 50}{44} \\ y &=& 1964 + 36.\overline{36}\\ y &=& 2000.\overline{36}\\ \mathbf{y} & \mathbf{=} & \mathbf{2000}\\ \end{array}$

The year in whitch postage was 37 cents was 2000

What is the half-life (in days) if the decay each day is 2% ?

Also how can I do this

$\begin{array}{rcl} y &=& a \cdot b^t \qquad | \qquad y = \frac{a}{2} \\ \frac{a}{2}&=& a \cdot b^{\frac{h}{1~\text{day}}} \qquad | \qquad h = \text{half-life} \\ \frac{1}{2}&=& b^h \qquad | \qquad \log{} \\ \log{ (\frac{1}{2}) } &=& h\cdot \log{(b)} \\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(b)} } \qquad | \qquad 2~\% \rightarrow 1-\frac{p}{100}=1-\frac{2}{100} = 0.98= b\\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(0.98)} } \\ h &=& \frac{ -0.30102999566 } {-0.00877392431} \\ h &=& 34.3096184915 ~\text{days} \\ \end{array}$

The half-life is 34.3 days

(1)  4x+8=20

(2)  -4x+2y=-30

$\begin{array}{lrcl} (1) & 4x+8 &=& 20 \qquad | \qquad -8 \\ & 4x &=& 20 -8 \\ & 4x &=& 12 \qquad | \qquad :4 \\ &\mathbf{x} & \mathbf{=} & \mathbf{3} \end{array}$

$\begin{array}{lrcl} (2) & -4x+2y &=& -30 \qquad | \qquad x = 3 \\ & -4\cdot 3 + 2y &=& -30 \\ & -12 + 2y &=& -30 \qquad | \qquad +12 \\ & 2y &=& -30 + 12 \\ & 2y &=& -18 \qquad | \qquad :2 \\ &\mathbf{y} & \mathbf{=} & \mathbf{-9} \end{array}$

Fläche Kreis Formeln:

$\begin{array}{lcl} A = \pi \cdot r^2 \\\\ A= \dfrac{ \pi\cdot d^2 } {4 } \end{array}$

"A" ist die Fläche des Kreises

"r" ist der Radius des Kreises

"d" ist der Durchmesser des Kreises. Der Durchmesser ist doppelt so groß wie der Radius.

"π" ist die Kreiszahl 3,14159...