Please help my daughter on these challenging questions

**Natasha1** · February 24th 2010, 04:38 AM

My daughter age 11 was given these problems, which she was unable to do. We would be grateful to have answers by tomorrow

Could someone be please kind enough to explain her the answers, in the simplest way possible (she's only 11!).

Here are the problems:

1) Inflation is running at 5% per year. This means that, for example, if a certain amount of any commodity costs £1 this year, the same amount will cost £1.05 in a year's time.

A merchant decides to sell a quantity of gold, then invest the money at an interest rate of a% per year, then after 1 year spends this money (with its interest) on buying gold again. He finds that he can buy 10% more gold than he had originally sold! What was the value of a?

2) A train travels for 10 miles at 30mph, then the next 10 miles at 60mph, and the average speed for the whole journey is 42 6/17mph (this is made of a whole number 42 and a fraction 6/17). What fraction of the distance of the whole journey is the first (30mph) part?

3) In this question x * y means (x+2y)/3 if x < y, and (2x + y)/3 if x is larger or equal to y.

In terms of p and q, find the possible values of (p * q) *p, explaining carefully how you choose which calculations to make.

4) Draw an X and Y plan, with the coordinates of O being your origin (0,0). Plot A with coordinates (2,0), B (3,0) and C (2,1).

Now in this question journeys are made of steps of whole-numbers length in any combination of directions left, right, up or down. Also one may retrace steps. So, for example, it is possible to travel from O to the point (2,1) by journey OAC which has length 3, it is possible to get to (2,1) by a different journey, OBAC which has length 5.

a) Starting from O, how many possible destinations (including O) are there with journeys of length

1) 2?
2) 3?

b) How many different journeys of length 3 are there starting at O and finishing at (1,0)?

c) You are given that the sum 1+2+3+....+100 = 5050
How many journeys of length 102 are there from O to coordinate (100,0)?

**drewbear** · February 24th 2010, 10:27 PM

1) say that the merchant originally purchased $10 in gold. therefore, due to inflation, the same amount of gold will cost $10.50 the next year. plus the extra 10% gets you a cost of $11.55 for the gold he buys the hext year. this means that after putting his $10 in the bank for a year he ended up with $11.55, so...

$11.55=10(1+a)$

$ \frac{11.55}{10} $ $=(1+a)$

$ 1.155=1+a $

$ a=.155$ or 15.5%

I apologize but i am confused on the wording in problem 2, does the train travel more than 20 miles?

I am not sure what you mean when using "*" to the power to or multiply by?

**Natasha1** · February 25th 2010, 01:31 AM

Originally Posted by drewbear

1) say that the merchant originally purchased $10 in gold. therefore, due to inflation, the same amount of gold will cost $10.50 the next year. plus the extra 10% gets you a cost of $11.55 for the gold he buys the hext year. this means that after putting his $10 in the bank for a year he ended up with $11.55, so...

$11.55=10(1+a)$

$ \frac{11.55}{10} $ $=(1+a)$

$ 1.155=1+a $

$ a=.155$ or 15.5%

I apologize but i am confused on the wording in problem 2, does the train travel more than 20 miles?

I am not sure what you mean when using "*" to the power to or multiply by?

The "*" is a sign invented. It is neither a multiplication sign nor a ti the power.

and as for the train here's the question, ignore the original one

A train does the first part of a journey at 30mph, then the second part at 60 mph, and the average speed for the whole journey is 42 6/17mph (this is made of a whole number 42 and a fraction 6/17). What fraction of the distance of the whole journey is the first (30mph) part?

**Soroban** · February 25th 2010, 06:39 AM

Hello, Natasha1!

A train does the first part of a journey at 30 mph, then the second part at 60 mph.
The average speed for the whole journey is $42 \frac{6}{17}$ mph.

What fraction of the distance of the whole journey is the first (30 mph) part?

$\text{Average speed} \;=\;\frac{\text{Total distance}}{\text{Total time}}$

Let $a$ = number of miles in the first part.
Let $b$ = number of miles in the second part.
. . $\text{Total distance} \:=\:a+b\text{ miles}$

The train went $a$ miles at 30 mph.
. . This took: . $\frac{a}{30}\text{ hours}$

The train went $b$ miles at 60 mph.
. . This took: . $\frac{b}{60}\text{ hours}$

Total time: . $\frac{a}{30} + \frac{b}{60} \:=\:\frac{2a+b}{60}\text{ hours}$

The average speed is: . $\frac{720}{17}\text{ mph.}$

We have: . $\frac{a+b}{\frac{2a+b}{60}} \:=\:\frac{720}{17}\quad\Rightarrow\quad \frac{60(a+b)}{2a+b} \;=\;\frac{720}{17} \quad\Rightarrow\quad 1020(a+b) \:=\:720(2a+b)$

. . . . . . . $1020a + 1020b \:=\:1440a + 720b \quad\Rightarrow\quad 420a \:=\:300b \quad\Rightarrow\quad 7a \:=\:5b$

$\text{The ratio of }a\text{ to }b\text{ is: }\;a:b \:=\:5:7$

$\text{Therefore, }a\text{ comprised }\frac{5}{12}\text{ of the journey.}$

**Soroban** · February 25th 2010, 07:10 AM

Hello again, Natasha1!

3) In this question: . $x\star y \;=\;\begin{Bmatrix} \dfrac{x+2y}{3} && \text{if }x < y \\ \\[-3mm] \dfrac{2x + y}{3} && \text{if }x \geq y \end{Bmatrix}$

$\text{In terms of }p\text{ and }q\text{, find the possible values of }\,(p \star q) \star p$

We must consider two cases: . $\begin{Bmatrix} (1) & p \,<\,q \\ (2) & p \,\geq\,q \end{Bmatrix}$

$(1)\;\;p \,<\,q$

Then: . $p \star q \;=\;\frac{p+2q}{3}$

Now we want: . $(p \star q) \star p \;=\;\left(\frac{p+2q}{3}\right) \star p$

Since $q > p,\:\text{ then: }\:\frac{p+2q}{3} \,>\,\frac{p+2p}{3} \:=\:\frac{3p}{3} \:=\:p$

. . That is: . $\frac{p+2q}{3} \;{\color{red}>}\;p$ . . . We must use the second definition.

Therefore: . $\left(\frac{p+2q}{3}\right) \star p \;=\;\frac{2\left(\frac{p+2q}{3}\right) + p}{3} \;=\;\frac{5p + 4q}{9}\quad\text{ for } p < q$

$(2)\;\;p\,\geq\,q$

Then: . $p\star q \;=\;\frac{2p+q}{3}$

Now we want: . $(p \star q) \star p \;=\;\left(\frac{2p+q}{3}\right)\star p$

Since $q < p,\,\text{ then: }\:\frac{2p+q}{3} \:<\:\frac{2p+p}{3} \:<\:\frac{3p}{3} \:=\:p$

. . That is: . $\frac{2p+q}{3} \:{\color{red}<}\:p$ . . . We must use the first definition.

Therefore: . $\left(\frac{2p+q}{3}\right)\star p \;=\;\frac{\frac{2p+q}{3} + 2p}{3} \;=\;\frac{8p + q}{9}\quad \text{ for } p \geq q$

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