math(easy`easy(have 10 marks~~!!要例式!!thanks!!

1. Let y be the age of the girl, then (62 - y) be the age of her mother

y -2 = 1/2 (62 - y)
2(y - 2) = 62 - y
2y - 4 = 62 - y
3y = 66
y = 22

Hence, the age of the girl is 22 & the age of her mother is (62 - 22) = 40.

2. Let y be the total amount of money,

y/10 + 30 = y/8
4y + 1200 = 5y
y = 1200

Hence, the total amount of money is $1200.

3. Let y be Peter has, then ( y - 9 ) be Paul has & y/3 be Philip has,

y + ( y - 9 ) + y/3 = 61
2y - 9 + y/3 = 61
2y + y/3 = 70
6y + y = 210
7y = 210
y = 30

Hence, Peter has $30, Paul has (30 - 9) = $21, Philip has (30/3) = $10

1)Let x be the girl's age and y be her mother's age
x-y/2=2...(1)
x+y=62...(2)
From (2) x=62-y...(3)
Put (3) into (1),
62-y-y/2=2
124-2y-y=4
-3y=-120
y=40
Put y=40 into (2)
x+40=62
x=22
Therefore, the girl is 22 years old and her mother is 40 years old.

2)Because each of the 8 attendees gets $30 more after two absent,
so two absent have 8*30=$240
and each of them have $120
so the total amount of money is $120*10=$1200

3)Let x be the money that Peter has,
y be the money that Paul has
and z be the money that Philip has
x-y=9...(1)
x/3=z...(2)
x+y+z=61...(3)
From (1), y=x-9...(4)
Put (4) and (2) into (3)
x+(x-9)+x/3=61
3x+3x-27+x=183
7x-27=183
7x=210
x=30
Put x=30 into (1),
30-y=9
-y=-21
y=21
Put x=30 into (2),
30/3=z
z=10
Therefore, Peter has $30, Paul has $21 and Philip has $10.