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primary 6 | Maths
| Data Analysis
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Date Posted: 4 years ago
Views: 313
Let x be the sum of the 1st and 2nd number.
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
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
1st and 2nd no. = 2u
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
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
Kayden, the first number is 18. You'll have to edit your working
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We know that the average of 1st and 2nd numbers is 3 times the average of the 3rd and 4th numbers.
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).
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).
Date Posted:
4 years ago
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