J's answer to CJ's Primary 6 Maths Data Analysis Singapore question.
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a)
a ★ b = 3a + 2b = 3 × a + 2 × b
So what this operation does is : it multiplies the left term by 3, multiplies the right term by 2, then adds them together.
It is easier to do the operation for the terms in the bracket first. Then perform the operation for the term outside and the result in the brackets.
Likewise to above,
5p ★ (4q ★ 3q)
= 5p ★ (3 × 4q + 2 × 3q)
= 5p ★ (12q + 6q)
= 5p ★ 18q
= 3 × 5p + 2 × 18q
= 15p + 36q (shown)
b)
If p = 1/5 and q = 3,
Then 5p ★ (4q ★ 3q) = 15p + 36q
= 15 × 1/5 + 36 × 3
= 3 + 108
= 111
Note : when you see a number directly on the left of a bracket, it is the same as doing multiplication on the bracket's terms.
Eg. 2(3) = 2 × 3
Eg. 6(3 + a) = 6 × (3 + a) = 6 × 3 + 6 × a = 18 + 6a
So you could also answer part a) like this :
5p ★ (4q ★ 3q)
= 5p ★ (3(4q) + 2(3q))
= 5p ★ (12q + 6q)
= 5p ★ 18q
= 3(5p) + 2(18q)
= 15p + 36q
a ★ b = 3a + 2b = 3 × a + 2 × b
So what this operation does is : it multiplies the left term by 3, multiplies the right term by 2, then adds them together.
It is easier to do the operation for the terms in the bracket first. Then perform the operation for the term outside and the result in the brackets.
Likewise to above,
5p ★ (4q ★ 3q)
= 5p ★ (3 × 4q + 2 × 3q)
= 5p ★ (12q + 6q)
= 5p ★ 18q
= 3 × 5p + 2 × 18q
= 15p + 36q (shown)
b)
If p = 1/5 and q = 3,
Then 5p ★ (4q ★ 3q) = 15p + 36q
= 15 × 1/5 + 36 × 3
= 3 + 108
= 111
Note : when you see a number directly on the left of a bracket, it is the same as doing multiplication on the bracket's terms.
Eg. 2(3) = 2 × 3
Eg. 6(3 + a) = 6 × (3 + a) = 6 × 3 + 6 × a = 18 + 6a
So you could also answer part a) like this :
5p ★ (4q ★ 3q)
= 5p ★ (3(4q) + 2(3q))
= 5p ★ (12q + 6q)
= 5p ★ 18q
= 3(5p) + 2(18q)
= 15p + 36q
Date Posted:
3 years ago
Alternative working to a) ,if you start the operation from the outside first :
5p ★ (4q ★ 3q)
= 3 × 5p + 2 × (4q ★ 3q)
= 15p + 2 × (3 × 4q + 2 × 3q)
= 15p + 2 × (12q + 6q)
= 15p + 2 × 18q
= 15p + 36q (shown)
Alternative working for b) without using the result in a) :
If p = 1/5 and q = 3,
Then 5p ★ (4q ★ 3q)
= 5 × 1/5 ★ ( (4 × 3) ★ (3 × 3) )
= 1 ★ (12 ★ 9)
= 1 ★ (3 × 12 + 2 × 9)
= 1 ★ (36 + 18)
= 1 ★ 54
= 3 × 1 + 2 × 54
= 3 + 108
= 111
5p ★ (4q ★ 3q)
= 3 × 5p + 2 × (4q ★ 3q)
= 15p + 2 × (3 × 4q + 2 × 3q)
= 15p + 2 × (12q + 6q)
= 15p + 2 × 18q
= 15p + 36q (shown)
Alternative working for b) without using the result in a) :
If p = 1/5 and q = 3,
Then 5p ★ (4q ★ 3q)
= 5 × 1/5 ★ ( (4 × 3) ★ (3 × 3) )
= 1 ★ (12 ★ 9)
= 1 ★ (3 × 12 + 2 × 9)
= 1 ★ (36 + 18)
= 1 ★ 54
= 3 × 1 + 2 × 54
= 3 + 108
= 111
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