勁易F1 Math

15.
Let $y be the amount that David has.
Then, the amount that Carl has = $(y + 9)
and the amount that Hilbert has = $(1/3)(y + 9)

y + (y + 9) + (1/3)(y + 9) = 61

Multiply all terms on the both sides by 3:
3y + 3(y + 9) + (y + 9) = 183
3y + 3y + 27 + y + 9 = 183
7y = 147
y = 21
y + 9 = 30
(1/3)(30) = 10

Ans: David has $21, Carl has $30, and Hilbert has $10.

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16.
Let g be the number of girls.
Then, the number of boys = (4/3)g

g + (4/3)g = 42

Multiply all terms on the both sides by 3:
3g + 4g = 126
7g = 126
g = 18
(4/3)g = 24

Ans: There are 24 boys and 18 girls.
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15. C=Carl's money, D=David's, H=Hilert
C=D+9 (1)
C=3H (2)
C+D+H=81 (3)
Sub (1) and (2) into (3)


16. B=Boy's nubmer, G=Girl's
B+G=42 (1)
B=(4/3)G (2)
Sub (2) into (1)