Question 1160234
.


The answer to (1) is more than easy:


<pre>
    if one raisin bar is 12p more than one nut bar, then 9 raisin bars cost 12p times 9 more than 9 nut bars.


    12p 9 times is 108p.      <U>ANSWER to part (1)</U>
</pre>

Isn't it self-evident ? What is the sense to ask about obvious things ?



====================


To answer question (2), put all 12 + 9 = 21 bars in a row.


Close to each nut bar place 12p (mentally).


Then you will have 21 groups:  9 groups contain raising bars only, each group;

                              12 groups contain nut bar and 12p, each group.


So, you have 21 groups, and each of them "costs" the same amount.


To keep everything in equilibrium, you must add 12*12p = 144p  to  £5.28,  giving you 528 + 144 = 672p.


Last step is to divide 672p by 21 to get the price of one raisin bar  {{{672/21}}} = 32p.


It is your <U>ANSWER</U>: one raisin bar costs 32p.
</pre>

Solved.


Is it clear to you and to your 9-years brother ?



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This problem is close to coin problems.

For coin problems and their detailed solutions see the lessons in this site:

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/Coin-problems.lesson>Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/More-Coin-problems.lesson>More Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/Solving-coin-problem-without-equations.lesson>Solving coin problems without using equations</A> (*)

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Kevin-and-Randy-Muise-have-a-jar.lesson>Kevin and Randy Muise have a jar containing coins</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Typical-coin-problems-from-the-archive.lesson>Typical coin problems from the archive</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Three-methods-for-solving-standard-typical-coin-problem.lesson>Three methods for solving standard (typical) coin word problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/More-complicated-coin-problems.lesson>More complicated coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Solving-coin-problems-mentally-by-grouping-without-using-equations.lesson>Solving coin problems mentally by grouping without using equations</A> (*)

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Non-typical-coin-problems.lesson>Non-typical coin problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Santa-Claus-helps-solving-coin-problem.lesson>Santa Claus helps solving coin problem</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/OVERVIEW-of-lesson-on-coin-word-problems.lesson>OVERVIEW of lessons on coin word problems</A>


You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations. 


For methods without using equations, see the lessons marked (*) in the list.


A convenient place to quickly observe these lessons from a &nbsp;"bird flight height" &nbsp;(a top view) &nbsp;is the last lesson in the list.


Read them attentively and become an expert in this field.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Coin problems</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.