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Secondary 1 | Maths
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Secondary 1 chevron_right Maths chevron_right Singapore

I keep getting a decimal ://

Date Posted: 3 years ago
Views: 269
J
J
3 years ago
If the total number of flowers can be divided by 42, 63, 108, this number is their common multiple.

42 = 2 × 3 × 7
63 = 3² × 7
108 = 2² × 3³


Fraction of flowers that are yellow
= 1 - ⅛ - ⅔ - 1/12
= 1 - 3/24 - 16/24 - 2/24
= 3/24
= ⅛

So this means that :

The number of flowers can be divided by 12 and 8 such that we can get whole numbers for the different types of flowers.

This number is also a common multiple of 12 and 8.

(We don't need to consider division by 3, since divisible by 12 already implies that)


Now,

12 = 2² × 3
8 = 2³


So all we have left to do is to find the LCM of 8,12,42,63 and 108.


LCM = 2³ × 3³ × 7
= 8 × 27 × 7
= 1512


So minimum number of flowers altogether
= 1512


Minimum number of yellow flowers

=1512 × ⅛
= 8 × 27 × 7 × ⅛
= 27 × 7
= 189

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Lim
Lim's answer
63 answers (A Helpful Person)
Explanation in main comments
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J
J's answer
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3 years ago
but why must we find the lcm of 8,12,42,63 and 108 ?
J
J
3 years ago
Because from the explanation, the number of flowers is divisible by 8,12,42,63,108, to give you a whole number result.

This means that it is a common multiple of all 5 of those numbers.

Since the minimum number of yellow flowers implies the minimum of total number of flowers possible,

Finding the LCM of these 5 numbers would give you your required minimum total number of flowers.


Recall that LCM is Lowest Common Multiple.
J
J
3 years ago
If you look at Lim's working, he found the LCM of 42,63 and 108 first.

He then found the LCM of 3,8,12.

He then took these two results and found their LCM.

The final result 1512 is a number , which is the lowest possible one that is divisible by all 6 numbers.


He did it in 3 separate steps while mine was a combination of all 3. I basically found the LCM of the 5 numbers straight away.

I didn't have to add 3 inside when I do the LCM since 4 out of the 5 numbers contained prime factor 3. This implies my LCM would have at least one factor 3 inside.