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primary 6 | Maths | Data Analysis
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Lim Zhen Ru
Lim Zhen Ru

primary 6 chevron_right Maths chevron_right Data analysis chevron_right Singapore

help me pls

Date Posted: 4 years ago
Views: 576
Claudine Lim
Claudine Lim
4 years ago
let cost of 1 apple be $a
let cost of 1 pear be $b
let cost of 1 orange be $c

5c + 3a = 3.30 —(1)
4c + 6b = 3.60 —(2)

“1 apple costs 10 cents less than 1 orange” means 1 orange costs 10 cents more than 1 apple so cost of 1 orange is the sum of 10 cents and cost of 1 apple:

c = a + 0.1 —(3)

sub (3) in (2):
5(a+0.1) + 3a = 3.30
5a + 0.5 + 3a = 3.30
8a = 2.8
a=0.35

sub a=0.35 in (3):
c=0.35+0.1=0.45

sub c=0.45 in (2):
4(0.45)+6b=3.60
1.8+6b=3.6
6b=3.6-1.8
6b=1.8
b=0.3

cost of 1 pear = $b = $0.30

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