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primary 6 | Maths
| Ratio
2 Answers Below
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Please help. Thank you very much
Then,
2 : (B + C) = 1 : 2
==> B + C = 4 units
==> C = 4 units - B
and
C : (B + 3) = 1 : 3
==> (4 - B) : (B + 3) = 1 : 3
==> (B + 3) is three times of (4 - B)
The next steps are done algebraically (are you confident of doing calculations on algebra?), but the level of algebra involved is Sec 1.
B + 3 = 3 times (4 - B)
B + 3 = 4 - B + 4 - B + 4 - B
B + 3 = 12 - 3B
(something like "1 part - 3 = 12 - 3 parts")
B + 3B = 12 - 3
4B = 9
B = 2.25
Recall that C = 4 - B, so C = 1.75
So, A : B : C : D is 2 : 2.25 : 1.75 : 3, which becomes 8 : 9 : 7 : 12 after multiplication by 4,
Choo, for your reference :
① A : (B + C)
= 1 : 2
② C : (B + D)
= 1 : 3
③ A : D
= 2 : 3
One of the concept tested is 'Repeated Identity'
A is repeated. Make the units the same for A in ratio ① and ratio ③
A : (B + C)
= 1 : 2
= 2 : 4 (multiply by 2)
So now we can combine them.
A : (B + C) : D
= 2 : 4 : 3
Next concept : Total unchanged
From this new ratio we know that B + C + D = 4u + 3u = 7u
So A : (B + C) : D : (B + C + D)
= 2 : 4 : 3 : 7
But from ratio ② we see that C : (B + D) = 1 : 3
So this means that C : (B + D) : (B + C + D)
= 1 : 3 : 4
(B + C + D) is also a repeated identity now and should be the same for both ratios.
Make the units the same
4 × 7 = 28, 7 × 4 = 28
Change them to 28 (this is called the lowest common multiple (LCM) , which is already used in P6 but only formally taught in Sec 1.
C : (B + D) : (B + C + D)
= 1 : 3 : 4
= 7 : 21 : 28 (multiply by 7)
A : (B + C) : D : (B + C + D)
= 2 : 4 : 3 : 7
= 8 : 16 : 12 : 28 (multiply by 4)
So now all the units are comparable.
A = 8u
C = 7u
D = 12u
You can find B in a few ways :
B = (B + C) - C = 16u - 7u = 9u
Or
B = (B + C + D) - C - D = 28u - 7u - 12u = 9u
A : B : C : D
8 : 9 : 7 : 12
Plus, the better primary schools already teach their students the solving of primary school questions using basic secondary school algebra, so I cannot rule out the fact that Choo (and other students) could have already learnt some algebraic manipulation.
And that is why Choo would have even encountered such a question in the first place.
Students are expected to employ the covered methods with a bit of lateral thinking to solve them.
This one here isn't your basic ratio question.
Eg. if a = 5, what is the cost of a bag . You are given that the cost is $(3a + 6)
What you're perhaps referring to is the technique of units and parts, but that is usually restricted to 2 variables and word problems.
Ratio is never mixed with algebra or units and parts so far, and there is no need to do multiple substitutions.
And furthermore, they don't deal with ratios that are in decimals or mixed numbers at P6, so reaching such a result is likely to confuse them further.
i.e it wouldn't be intuitive for them to change those mixed numbers or funny-looking decimals (imagine encountering a 1.142857142857....)
So,
given a choice between concepts they are familiar with such as unchanged total and repeated identity, vs algebra which used at a higher and more unfamiliar level, The former is a better choice.
Most students don't really get algebra at this stage.
i.e it is a fast but one-dimensional method since a student would just perform the substitution as a form of drill and practice and going through the motions.
Doesn't help much in honing lateral thinking here.
What is great about Primary school math is that it does focus a lot on logical thinking and problem solving, rationality etc (those true/false/not possible to tell questions esp.)
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