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## Question

primary 6 | Maths
| Data Analysis

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Let y be the sum of the 3rd and 4th number.

X+y=48 (sum is 48)

X/2 (ave of the 1st and 2nd number) = 2(2 times) * y/2 (ave of the 3rd and 4th number)

Thus,

X/2=y

X=2y (equation 1)

X+y=48 (equation 2)

Substitute eqn 1 into 2,

2y+y=48

3y=48

Y=16

X=2(16)=32

Finally,

X= 1st number + 2nd number

X= 18 + 2nd number =32

2nd number =14

3rd and 4th no. = 1u

Total sum of 4 no. = 48

3u = 48

1u = 48/3

= 16

2u = 16*2

= 32

2nd no. = 32-18

= 14

Ans: 14

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Students should be able to infer that this means that the sum of the 1st and 2nd numbers is also 3 times that of the sum of the 3rd and 4th numbers. Since both sides will be multiplied by the same amount to find the sum (×2).