**Problem**

**Introduction**

Here
we have a numerically challenging problem that involves ratios, and it ultimately
reduces to an algebraic problem with two unknowns. Nevertheless, we are spoilt for choice as
regards to methods of solution:-

(1) Bar Diagram Modelling

(2) explicit letter-symbolic Algebra

(3) “p” and “u”
(parts and units)

(4) Distinguished Ratio Units

Despite the
fact that Bar Diagram Modelling made “Singapore mathematics” famous, let us
remember that it is only one of the ways of solving problem by diagramming,
which is just one of the eleven Primary School heuristics
recommended by the Singapore Ministry of Education.

The methods have a lot in common, and they differ mainly in the form of presentation. However, standard Bar modelling is impractical under high-stakes high-stress examination conditions for this problem, not least because one would have to cut the bars into many pieces. One should not cut off one’s feet just so as to fit the shoes (削足适履), as one Chinese saying goes. We need to be flexible and open-minded. I present a solution using my own Distinguished Ratio Units.

The methods have a lot in common, and they differ mainly in the form of presentation. However, standard Bar modelling is impractical under high-stakes high-stress examination conditions for this problem, not least because one would have to cut the bars into many pieces. One should not cut off one’s feet just so as to fit the shoes (削足适履), as one Chinese saying goes. We need to be flexible and open-minded. I present a solution using my own Distinguished Ratio Units.

**Solution**

**Ans:**735 books

**Commentary**

First off, we
need to equalise the numerators of

^{2}/_{5}and 1^{1}/_{4}=^{5}/_{4}and put them ratio form. This is because the “2” in the^{2}/_{5}represents the same quantity as the “5” in^{5}/_{4}.
We do this adjustment by multiplying the former through
by 5
and the latter through by 2. Thus we deduce that the original number of books in A
and in B are 25 and
8 “heart” units respectively.

Next, we add on the 2 and 3 “triangle” units. By doing a comparison, we can figure out that 1 “triangle” unit must be 45 more than 17 “heart” units. So 2 “triangle” units must be equal to 34 “heart” units plus 90. Replacing the 2 “triangle” units (shown in yellow) with their equivalent, we now know that 59 “heart” units plus 90 gives 444. This allows us to figure out that 1 “heart” is actually 6. Thus, we can work out what 1 “triangle” unit, and then what 5 “triangle” units are worth.

Next, we add on the 2 and 3 “triangle” units. By doing a comparison, we can figure out that 1 “triangle” unit must be 45 more than 17 “heart” units. So 2 “triangle” units must be equal to 34 “heart” units plus 90. Replacing the 2 “triangle” units (shown in yellow) with their equivalent, we now know that 59 “heart” units plus 90 gives 444. This allows us to figure out that 1 “heart” is actually 6. Thus, we can work out what 1 “triangle” unit, and then what 5 “triangle” units are worth.

**Final Remarks**

Due to the
difficulty of the numbers, the solution presented above is about as streamlined
as I can make it to be.

There is
another variation that can be used – equalising the “triangle” units (akin to
the technique of elimination in standard algebra). What we do is we multiply the group with
total 444 by 3 and
to multiply the group with total
489 by 2. This
would give 6 triangle units on each side. Then we can compare the “heart” units and continue
from there. This way of proceeding is
not for those who fear 4-digit numbers.

If there
are nicer or more elegant ways to tackle this question, I would definitely love
to hear from you.

H01. Act it out

H04. Look for
pattern(s)

H05. Work
backwards

H06. Use
before-after concept

H09. Restate
the problem in another way

H11. Solve part
of the problem

**Suitable Levels**

*****Primary School Mathematics

* other syllabuses that involve whole
numbers and ratios

* any problem solver who loves a challenge

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